*Hi everyone, and welcome to educator.com. Here we are starting our first lesson in algebra based AP physics. I'm Dan Fullerton and I'd like to welcome you to the course.*0000

*To begin with, let's talk about what physics is.*0011

*We are going to talk about how we recognize the questions of physics, we are going to list several disciplines within the study of physics, and finally we are at least going to start to define matter, mass, work and energy.*0014

*These are some of the key concepts that are going to play out throughout this entire course.*0026

*So, what is physics? I like to think of physics as the answer to all of the questions a two year old might ask.*0031

*What does a two year old say constantly? "Why? why? why?"*0039

*The dictionary says physics relates to matter and energy and their interactions.*0044

*Some questions that might come up are: "What is matter?", "What is energy?", "How do they interact?", and most importantly, "Why do we care?"*0048

*The "why" questions are really what start to get more interesting when you take it beyond just a dictionary definition.*0057

*Why is the sky blue? Why does the wind blow? Why does my teacher smell funny? Why do objects fall down instead of up? Why do airplanes fly and why can't I? Why do the stars shine? Or why do I have to eat my vegetables?*0065

*To do this, we have to start talking about what the world, the universe is made up of.*0082

*We are going to start with matter. Matter is anything that has mass and takes up space, where mass is the amount of "stuff" making up an object.*0088

*We can get into a little bit more detailed definition than that, but for now, we think of it basically as anything you can touch. Stars, electrons, Neal Diamond. They are all mass, they are all matter.*0096

*There are actually two types of mass. We could talk about inertial mass, and inertial mass is really how hard it is to accelerate an object.*0108

*Likewise, we could talk about an objects gravitational mass, which refers to how large a gravitational force an object experiences.*0130

*So, inertial mass and gravitational mass. Two types of mass. But what is really slick in physics, anytime we have ever measured anything, the inertial mass and the gravitational mass have always been the same.*0154

*There is no theoretical reason we really understand, yet that says why, but it always works out and, man, is that slick for us as we start our study of mechanics coming up here shortly.*0166

*So let's do a problem. On the surface of the earth, a spacecraft has a mass of 2 × 10*^{4} kilograms.0177

*What is the mass of the spacecraft to the distance of one earth radius above earth's surface?*0185

*We know the mass is 2 × 10*^{4} kilograms on the surface of the earth. We want to know what its mass is up in space.0190

*Mass, the amount of "stuff" an object is made up of, doesn't change. It does not matter where you are, you still have the same mass in the same object. Therefore, our answer must be number two. The same mass.*0201

*We talked about matter, let's talk about energy. Energy is the ability or capacity to do work.But work in physics has a specific definition. Work is the process of moving an object.*0216

*We are simplifying these a little bit, we will get into more depth later. But if we wanted to put those together, we could say that energy, really, is the ability or capacity to move an object.*0227

*A baseball coming at your nose has kinetic energy; it has energy of motion. When it hits your nose, it has the ability to move your nose. That's how you know it has energy.*0251

*On the same token, if we had a bowling ball suspended up above my head, it would have gravitational potential energy. Why? Because it has the ability or capacity to move an object.*0260

*If it were released, it would start to speed up and that potential energy, would become kinetic energy and move faster and faster until it collided with me, in which case, it would move parts of me, in what would probably be a very unpleasant experience.*0271

*In both cases, energy is the ability or capacity to move an object.*0284

*In the early twentieth century, a famous physicist with wild hair, Albert Einstein, formalized a relationship between mass and energy and it's become one of the most famous formulas in physics.*0291

*His relationship says E=mc*^{2}, where what he is saying is that the mass of an object, a key characteristic of matter, is really a measure of its energy.0301

*Energy equals mass times the square of the speed of light. That is just a constant, that is just a number, a fudge factor to make the units work out.*0312

*What we know is that the source of all energy here on earth is the conversion of mass into energy. They are really two different sides of the same coin. Or you could think of it as, mass is a measure of an object's energy and energy is a measure of mass. There is a very, very close relationship there that is going to play out through the world of physics as well.*0320

*To come back to "what is physics?", physics is the study of matter and energy and how they interact, which turns out to be everything.*0341

*Try and think of something that is not related to matter and/or energy.*0350

*Baseball? It's all about physics. Matter, energy, even the roar of the crowd, even the crackerjacks to eat in the stands. You eat matter, you swallow it, you digest it, as you do that, chemical reactions occur. Those chemical reactions are transfers of energy, then that energy allows you to do work later on. Everything is physics.*0354

*That is an awfully big bullet list of things to do for an introductory course in physics, so we have to limit ourselves.*0379

*What we are going to focus on are some of the fundamentals.*0387

*We are going to start with mechanics; talking about how objects move, what makes them move, how they move in circles, how things like gravity work, and work, energy and power, momentum, collisions, explosions.*0390

*Then we will talk about fluids, fluid dynamics, getting to thermo physics, thermodynamics, heat.*0403

*We will talk about electricity and magnetism, circuits. We will talk about waves, sound, optics, light.*0411

*And finally, we will even touch a little bit on a topic known as modern physics. Things like nuclear physics and a couple other small topics that are much more modern. Modern, meaning in the last one hundred years or so.*0418

*That should get us going in algebra based AP physics.*0430

*What I would like you to do before we move on is take just a minute or two and write down three things you would like to learn about in physics. Then, if you can, try and think of ways in which matter and energy relate to those topics. Just a couple of minutes to start to see how all of these things play into our study of physics and the universe.*0446

*Thanks for watching educator.com. Make it a great day.*0465

*Hi and welcome back to Educator.com.*0000

*This lesson is on Newton's Third Law of Motion.*0003

*Our objectives are going to be to explain the meaning of Newton's Third Law of Motion.*0006

*To recognize and identify force pairs, and finally to utilize Newton's Third Law to solve dynamics problems.*0010

*Newton's Third Law of Motion. All forces come in pairs. You can't have a single force.*0021

*If object 1 exerts a force on the object 2, then object 2 must exert a force back on object 1. *0029

*That force that exerted back is exactly equal in magnitude and opposite in direction.*0037

*The force of object 1 on 2 is equal, but opposite in direction to the force of 2 on 1.*0043

*Oftentimes you might have heard this phrased as the Law of Action-Reaction. *0049

*For every action there is an equal and opposite reaction.*0054

*Have to be careful with that terminology. *0058

*What we are really talking about are forces with that law.*0060

*For example, if somebody comes and they punch you in the nose with a force of 100N with their fist, your nose exerts a force of 100N back on their fist. *0064

*The same force exerted on your nose is opposite in direction exerted back on to them.*0073

*Let us take a look at some examples. How does a cat run forward?*0080

*Well, if you want to run forward, don't you push back on the ground to move forward?*0085

*So it pushes backward on ground and the ground is what actually causes the cat to run forward.*0090

*Or if you want to swim forward, which way do you push the water?*0098

*Don't you push it behind you? You push back so that the water pushes you forward?*0102

*Or how do you jump in the air? When you want to jump, you push down on the ground and the ground pushes you back up.*0108

*Newton's Third Law, we use all the time, so much so that it is almost silly to talk about. *0119

*Let us talk about action-reaction pairs or force pairs.*0127

*If a girl is kicking a soccer ball, she has the force of the girl's foot on the ball, then there must also be a reactionary force.*0131

*You have the foot on the ball, the reactionary force must be the ball applying a force on the foot.*0139

*Or a rocketship in space. Hot gases are pushed out by the rocketship.*0149

*Then what is the force pair? The ship is pushed by the hot gases.*0161

*How about the force of gravity on you?*0173

*Earth, the force of gravity on you, is pulling you toward its center.*0175

*Guess what? With the exact same force, you are pulling Earth toward your center.*0189

*Granted you do not see the effect nearly as much because the Earth is so massive, you cause such a tiny, little acceleration.*0194

*Let us take a look at a couple of examples.*0214

*Earth's mass is approximately 81 times the mass of the moon.*0216

*If the Earth exerts a gravitational force of magnitude F from the moon, the magnitude of the gravitational force of the moon on the Earth is, well, Newton's Third Law. *0220

*If we exert a force of one on another, an exact same force but opposite in direction is exerted back, has to be 1F.*0229

*The sailboat example. A 400N girl standing on a dock exerts a force of 100N on a 10,000N sailboat as she pushes it away from the dock.*0239

*How much force does the sailboat exert on the girl?*0249

*Well, the 400N girl -- that is talking about the weight of the girl, that is describing the girl's weight -- exerts a force of 100N, that is the force that she applies to the boat.*0253

*On a 10,000N sailboat, that is the weight of the sailboat. *0265

*So the force that the girl exerts is the 100N, that must be the force that the boat exerts back on her, 100N. *0274

*Do not let these other details screw you up.*0283

*The hammer and nail. A carpenter hits a nail with the hammer.*0289

*Compared to the magnitude of the force the hammer exerts on the nail, the magnitude of the force the nail exerts on the hammer during contact has got to be the same, Newton's Third Law.*0292

*Let us look at one more here.*0303

*If forces only come in pairs, that are equal and opposite, why does all forces not cancel each other out?*0306

*Now, we have to think a little bit.*0314

*Remember the force of object 1 on 2 is equal in magnitude and opposite in direction of the force of object 2 on 1. *0317

*Why don't they cancel each other out?*0330

*They are acting on different objects.*0333

*If they were acting on the same object, of course they would cancel each other out, but since they are acting on different objects, no cancellation.*0338

*Newton's Third Law, we use it all the time every day.*0347

*Very simple concept but so easy to overlook.*0351

*Thanks for watching Educator.com. Make it a great day.*0355

*Hi everyone and welcome back to Educator.com*0000

*This lesson is going to be about friction. *0004

*Now our objectives are going to be to define and identify frictional forces. Yeah friction! *0006

*We will explain the factors that determine the amount of friction between two surfaces and determine the frictional force and coefficient of friction between two surfaces.*0013

*So let us dive in. Friction is a force that opposes motion.*0021

*Kinetic friction is a type of friction that opposes motion for an object that is sliding along another surface.*0027

*Kinetic friction is sliding friction.*0034

*Static friction acts on an object that is not sliding.*0036

*Now the magnitude of the frictional force is determined by two things: the nature of the surfaces in contact -- and we characterize that with μ -- a variable that refers to the coefficient of friction.*0041

*Bigger coefficients of friction, bigger μ -- so you are going to have more friction between the two surfaces.*0059

*Imagine something like -- let us say really flat dress shoes on ice -- very slippery -- compared to two pieces of sandpaper.*0065

*The sandpaper is going to have a much higher coefficient of friction.*0075

*The normal force acting on the object is the other item that determines the magnitude of the frictional force.*0079

*Now as we talk about these types of friction and the magnitudes of these frictional forces, it is important to note that typically, kinetic friction is less than static friction.*0086

*And you have probably observed that before. *0105

*Have you ever tried to push something heavy along the floor -- maybe pushing a sofa or a refrigerator or something heavy -- it takes a lot of work to get it started because you have to overcome static friction.*0108

*Once you have it moving however, now you are into the regime of kinetic friction that usually takes a little bit less force.*0120

*Kinetic friction is usually smaller than static friction. *0125

*Now as we talk about these coefficients of friction, we are going to have a different coefficient depending on whether it's sliding or static then.*0130

*So the coefficient of friction μ -- we are going to talk about the coefficient of kinetic friction μ*_{k} or coefficient of static friction μ_{s}.0137

*So this coefficient of friction is really the ratio of the frictional force and the normal force.*0147

*Coefficient of friction given by the force of friction divided by the normal force.*0152

*It depends only on the nature of the surfaces that are in contact.*0158

*You can look up in many different places approximate coefficients of friction and you can see as we have on the slide here that there are different values for kinetic or static.*0163

*Rubber on dry concrete has a kinetic coefficient of 0.68.*0174

*But on static, when it is not sliding, it is 0.9.*0179

*That means that if you lock your wheels as you are driving down the road on dry concrete -- if there is sliding -- if there is skidding then you have less friction then if they are not sliding.*0182

*This is the reason for antilock brakes in car.*0192

*If you are sliding then you are not getting as much stopping force as you would if you were not sliding, so they try and keep cars from sliding with these antilock brakes -- not allowing them to slide.*0196

*You could look up the coefficient of friction for many different materials.*0207

*So let us take a look at some examples and try and determine which regime of friction they are in -- kinetic or static.*0215

*If we have a sled sliding down a snowy hill -- sliding -- there is our key word -- that must be kinetic friction -- we would use the kinetic coefficient.*0218

*A refrigerator at rest that you want to move -- at rest implies not sliding -- that one is static.*0233

*A car with the tires rolling freely -- well, we just talked about that -- not skidding, therefore static.*0241

*If you are skidding across pavement though, you are going to use kinetic coefficient of friction.*0249

*Let us take an example here. A car's performance is tested on various horizontal road surfaces.*0259

*The brakes are applied causing the rubber tires of the car to slide along the road without rolling.*0264

*They are sliding. *0271

*They encountered the greatest force of friction to stop the car on which of these surfaces -- Dry concrete? Dry asphalt? Wet concrete or wet asphalt?*0273

*Well, first thing we need to realize is, is if we are sliding, we are looking for the kinetic coefficient.*0282

*Which one of these is the biggest -- Rubber on concrete, dry and wet? Rubber on asphalt dry and wet? -- 0.68 is our biggest coefficient, so that would have the greatest force of friction -- dry concrete.*0288

*Another example is we have a block on an incline. *0304

*The diagram shows the block sliding down a plane, inclined at angle θ -- there is θ.*0307

*If angle θ is increased -- as that gets steeper -- What happens to the coefficient of kinetic friction between the bottom surface of the block and the surface of the incline?*0312

*Well, here you have to remember that the coefficient of friction depends on the nature of the surfaces.*0321

*In this case the surfaces have not changed. *0332

*Yes, you are going to have some other different effects, but as far as the coefficient of friction goes, the nature of the surfaces has not changed, therefore the coefficient of friction will remain the same.*0335

*To calculate the force of friction -- again it depends only upon the nature of the surfaces in contact, that coefficient of friction and the magnitude of the normal force. *0348

*We have a nice direct relationship. *0355

*Force of friction equals the coefficient of friction μ times the normal force.*0358

*We can combine this with Newton's Second Law and free-body diagram to solve even more involved problems that we did in our Newton's Second Law discussion.*0364

*While we are here and talking about friction, let us come back to terminal velocity.*0374

*Objects following through Earth's atmosphere experience a force of friction that we call air resistance.*0379

*That is a drag force and as the object goes faster, there is even more of that.*0382

*Eventually, an object gets going fast enough that the force of friction balances the force of gravity on the object. *0388

*When that happens you reach what is known as terminal velocity.*0394

*The net force is zero -- you do not gain any more speed -- the longer you fall.*0397

*So a graph of velocity versus time for an object that we are now taking into account air resistance -- it is going to start -- say we throw somebody out of an airplane -- their vertical velocity starts at zero and it increases, increases, increases that force of friction. *0401

*That force of air resistance -- the faster they go gets greater and greater until eventually they hit this asymptote which we know as the terminal velocity.*0423

*When they do that, at that point where they hit terminal velocity, -- FBD -- the weight of the object and the force of air resistance exactly balance. *0435

*No net force. No acceleration. Constant velocity.*0450

*Let us take a look at another example -- finding the frictional force.*0456

*In the diagram, we have a 4 kg object accelerating at 10 m/s*^{2} on a rough horizontal surface. 0459

*Find the magnitude of the frictional force, (Ff), acting on the object? *0467

*Let us start with our FBD.*0472

*We have the normal force, the object's weight, we have this applied force to the right of 50N, and we have a frictional force to the left.*0475

*All of my forces line up with the axis, so I do not need to draw a P-FBD.*0494

*Since we are looking for the magnitude of the frictional force, I am going to start by writing Newton's Second Law for the x-direction.*0501

*I am going to replace now, net force in the x-direction with all the forces acting in the x-direction.*0510

*I look at my FBD, I have 50N to the right, the applied force, minus force of friction and that must equal my mass- 4 kg times my acceleration 10 m/s*^{2}. 0515

*Or 50N minus force of friction equals 40 kg m/s*^{2} which is a Newton. 0533

*Therefore, force of friction must be equal to 10N.*0543

*Let us take a look at another example. Here we have a box on a wood surface.*0551

*A horizontal force of 8N is used to pull a 20N wooden box moving toward the right along a horizontal wood surface, where we know that the coefficient of kinetic friction there is 0.3.*0559

*We are asked to find the frictional force acting on the box, the net force acting on the box, the mass of the box and the acceleration of the box.*0569

*Well, we will start with our FBD. We have normal force. We have its weight, mg, which it tells us it is a 20N wooden box, so that must be 20. *0580

*We have a force to the right, an applied force of 8N and we must have some frictional force to the left.*0592

*If we want to find the frictional force acting on the box, what I am going to write is the force of friction equals μ times the normal force and by the way, look -- friction is F-U-N -- friction's fun. *0600

*μ is 0.3 and our normal force in this case -- if you look in the y-direction, that must be equal to mg. *0616

*There is no net force in the y-direction, otherwise that box would spontaneously take up off the table or go through it, and we know that does not happen, they have to be balanced.*0625

*The normal force here must be 20N, so 0.3 x 20 or 6N.*0633

*It also asks for the net force acting on the box. *0640

*Net force in the x-direction is just going to be 8N to the right minus 6N to the left, our frictional force or 2N. *0645

*How about the mass of the box? *0657

*Well, we know its weight, mg, is 20N, so if we just divide both sides by g, we should get the mass, which is going to be 20N divided by g(10), -- it is going to be 2 kg.*0660

*And finally, the acceleration of the box.*0675

*Well, acceleration is net force divided by mass. We just determined the net force here was 2N. *0678

*We determined the mass was 2 kg, so the acceleration must be 1 m/s*^{2}. 0689

*Let us take a look at an example where we explore the difference between static and kinetic friction?*0705

*Compared the force needed to start sliding a crate across a rough level floor, the force needed to keep it sliding once it is moving is -- well if needed to start you need to overcome static friction -- once its sliding, it is kinetic.*0710

*Kinetic is less than static, therefore it is going to be less.*0727

*Let us take a look at a drag force. *0734

*An airplane is moving with a constant velocity in level flight. We have an airplane moving with constant velocity in level flight. *0737

*Compare the magnitude of the forward force provided by the engines -- we typically call that thrust -- to the backward frictional drag force. *0749

*Well, let us draw our FBD. *0758

*There is our airplane. We have some thrust forward. *0761

*We have a drag force backwards -- force that is pulling it up, we call lift and we have its weight.*0770

*Now if it is moving at constant velocity in level flight, everything must balance out -- they must be equal. *0781

*So the force of the thrust or the force of the engines must balance the force of drag, therefore they must be equal. *0789

*Another example, have Suzie over here pulling a sled.*0800

*She pulls the handle of a 20 kg sled across the yard with a force of 100N and that is at an angle of 30 degrees. *0805

*The yard exerts a force of 25N on the sled due to friction. *0812

*We are asked to find the coefficient of friction between the sled and the yard and determine the distance the sled travels if it starts from rest and Suzie maintains her 100N force for 5 s.*0816

*Well, let us start off with our FBD -- y, x.*0828

*There is our sled. *0839

*We have weight down -- force of friction to the left 25N. *0840

*We have the normal force from the ground up and we have the applied force of Suzie, which is 100N at an angle of 30 degrees.*0846

*So my P-FBD -- I will draw that down here.*0858

*Right away, let us put in our red vectors -- the ones that are already lined up with the axis.*0865

*We have mg, force of friction, and normal force.*0870

*Now we have to break that up into components, so the x-component of Suzie's applied force is going to be 100N cosine 30 and the y-component 100N sine 30 degrees.*0878

*Now we can go start to solve our questions.*0900

*Find the coefficient of friction between the sled in the yard. *0904

*I am going to start by writing Newton's Second Law and I am going to look in the y-direction-- equals ma*_{y}.0906

*I am going to replace the net force in the y-direction with all the different things I see here. *0915

*I have 100 sine 30 and that is going to be 50, plus the normal force, F*_{n} minus mg. 0919

*And I know -- common sense tells me -- that sled is not going to go spontaneously accelerating up off the ground so that must all equal 0 -- acceleration in the y is 0. *0931

*So I can solve for the normal force -- the normal force then must be mg - 50, which is going to be mass(20)kg x 10 - 50 or 150N.*0941

*μ then, the coefficient of friction, is the force of friction divided by the normal force which is 25 over 150 or 0.167 -- there is our coefficient of friction.*0961

*Now then, it also tells us to determine the distance the sled travels if it starts from rest and Suzie maintains her 100N force for 5 s.*0978

*Well to do that, it would be nice to know the acceleration of the sled in the x-direction. *0987

*Let us go to Newton's Second Law in the x-direction. *0992

*F*_{net}x is going to be 86.6N 100 cosine 30 minus the force of friction, 25N or 61.6N.0996

*Therefore, acceleration which is the net force divided by the mass is 61.6N/20 Kg -- it is going to accelerate at about 3.1 m/s*^{2}.1007

*So, if we want to find out how far it goes in that 5 s-interval, we can go back to our kinematics.*1024

*Δx = V initial T + 1/2AT*^{2} and V initial here -- if it starts from rest is 0. 1032

*So that is just going to be 1/2 x a(3.1) m/s*^{2} x the square of our time, 5 s^{2} or 38.8 m.1041

*Hopefully that gives you a good start with friction, the coefficient of friction.*1062

*Thanks for watching Educator.com. Make it a great day.*1065

*Hi everyone and welcome back to Educator.com.*0000

*We are going to take a look at dynamics applications and problem solving in this lesson.*0003

*Our objectives are going to be to draw and label a free-body diagram (FBD), showing all the forces acting on an object on a ramp.*0008

*We will also draw a pseudo free-body diagram (P-FBD)showing all components of forces acting on the object -- some overlap with what we have done previously in reinforcement.*0016

*We will utilize Newton's Laws of Motion to solve problems of objects on ramps.*0025

*Gain an understanding that tension is constant in a light string passing over a massless, or ideal pulley. *0030

*We will analyze systems of two objects connected by a light string over a massless pulley, and finally, we will determine the reading on a scale in an accelerating elevator.*0036

*So, with that, let us go back to FBDs again -- a quick review. *0047

*FBDs are tools used to analyze physical situations and they show all the forces acting on a single object. *0052

*Then, we draw all the forces on that object and we draw the object as either a box or as a dot. *0060

*When we are drawing FBDs -- what we are going to do is we are going to choose the object of interest and draw it. *0069

*Then label all the external forces and draw them. *0075

*And then sketch the coordinate system choosing the direction of the objects motion as one of the positive axis.*0078

*When we do this for the case of an object on a ramp, that is going to be up or down the ramp, which means typically we are going to have an off-set or a tilted set of axis. *0084

*Quick review -- we have a block sitting on a ramp -- What do we do about the forces acting on it?*0100

*We already said we have the normal force, we have the weight, and the force of friction and we draw them just like they are on the ramp so the answer here would be 4. *0105

*Once we have that down, we are going to complicate matters a little bit. *0116

*With the P-FBDs -- when the forces do not line up with the axis, we draw a new separate FBD and break up those forces into their components that do line up on the axis. *0120

*So here is our box on a ramp. Let us draw the forces -- the FBD, and the P-FBD -- for it sitting on the ramp. *0132

*Then we are going to write Newton's Second Law equations for the x and the y directions. *0141

*So for this box we have its weight down, normal force, and the force of friction, since it wants to slide down the ramp. *0146

*Our FBD -- we will draw our axis -- We have mg down. We have the force of friction and the normal force. *0157

*And as we said -- this weight does not line up with an axis. *0172

*So P-FBD (y,x) -- we have mg perpendicular to the ramp, mg parallel to the ramp, and of course normal force and frictional force do not need to be adjusted.*0178

*A couple of formulas that went with this -- mg parallel -- the component of weight down the ramp or parallel to the objects motion is mg sin θ, mg perpendicular -- the component of weight into the ramp, was mg cos θ. *0199

*With that we could write Newton's Second Law equations. *0216

*In the x direction, the net force in the x direction just means look at the x axis and draw all the forces acting in that direction. *0220

*In this case if I call to the right up the ramp positive, that is going to be the force of friction minus mg parallel or mg sin θ and that is equal to ma. *0228

*In this case since it is just sitting there, there is no acceleration -- that is equal to 0.*0242

*Or in the y direction -- net force in the y direction is the normal force minus mg perpendicular or mg cos θ and in the y direction it is not accelerating either. *0247

*So that is all equal to 0. There is our setup. *0261

*For the box at rest here we have three forces acting on our box on an inclined plane, as shown in the diagram and the vectors are not drawn to scale. *0268

*If the box is at rest, the net force acting on it is equal to...*0277

*Well before you get too involved in a problem like this -- it is at rest. *0281

*At rest means acceleration, 0. It is going to stay at rest. No net force, therefore, answer 4 must be correct. *0287

*Now we have our box held by a force. 5 kg mass is held at rest on a frictionless 30 degree incline by force F. *0300

*What is the magnitude of F? *0309

*Well let us start with our FBD. We have F acting up the ramp; we have the normal force perpendicular to our surface, and we have mg. *0312

*So now I am going to do my P-FBD over here. *0329

*We still have F up the ramp, and we still have our normal force, but now we have mg sin θ or mg parallel down the ramp and mg cos θ. *0334

*So now I can go write my Newton's Second Law equation for the x direction. *0351

*Net force in the x direction is going to be equal to -- if I call this direction positive, that is going to be F minus mg sin θ -- that has to be equal to 0 because it is held at rest. *0359

*Therefore, F must be equal to mg sin θ, which is 5 kg × g to approximate 10 m/s*^{2} × the sin of the angle θ sin 30 degrees. 0376

*We know that sin 30 degrees is half, so that is 50 × 0.5 or 25N.*0392

*Great. Let us take a look at what we call Atwood machines.*0404

*Two objects masses m1 and m2 are connected by a light string over a massless pulley. *0409

*M1, m2 -- pulley of sum radius r and a string -- all connected. *0416

*That is a basic Atwood machine, an experimental or theoretical device designed to help students understand how forces interact, especially when we are talking about Newton's Laws of Motion.*0420

*So, properties of Atwood machines -- they have ideal pulleys. *0433

*If the ideal pulleys are frictionless and massless -- meaning they do not add any inertia to the system -- then you can say that the tension on either side has to be the same. *0437

*That only works because this is a massless pulley but it is constant in the light string since it is an ideal pulley -- it has no mass. *0448

*So tension 1 here must equal tension 2. *0455

*Now as we set these up -- first we are going to adopt the sin convention for positive and negative motion because as one goes up and one goes down it could be a little confusing which way is positive. *0461

*So I like to go draw a direction around the pulley and call that the positive direction.*0471

*Then what we are going to do is analyze each mass separately using Newton's Second Law. *0476

*Here we have our system m1 and m2 -- we have called this way around the pulley, positive y and now we want to know what its acceleration is.*0483

*So the first thing I am going to do is I am going to come in here and I am going to label this tension 1 and that tension 2, just so I do not mix these up later.*0493

*And as I look at mass 1 to draw its FBD -- there is mass 1 and going down we have m1 g, its weight, and we have t1 tension -- a rope can only pull, so that must be up -- there is t1.*0503

*And for this mass, because of our axis over here -- down is the positive y direction. *0521

*Lets do the same thing for the second mass over here for mass 2, we have m2 g down and we have t2 up. *0529

*In this case though, up is going to be the positive direction because of our arrow, the direction that we indicated here. *0540

*So over here positive y is that direction. *0546

*Now what I am going to do is start writing Newton's Second Law equations to see if I cannot solve for the acceleration of the system. *0551

*If I start with mass 1, the net force in the y direction, well m1g in the positive direction minus t1 in the negative y direction must equal m1a.*0557

*Let us write a Newton's Second Law equation for m2.*0577

*We have t2 in the positive direction minus m2g and since m2g is in the negative direction over here, then that must equal m2a. *0581

*Finally, we know because it is an ideal pulley, that t1 must equal t2. *0593

*So what I am going to do now is I am going to see if I cannot combine these equations because I have a couple of unknowns. *0601

*I do not know t1, I do not know (a), and I do not know t2. *0606

*So with three equations and three unknowns I should be able to solve this. *0610

*I will start with m1g - t1 = m1a. Then I am going to add to it our second equation t2 - m2g = m2a. *0615

*Now if the left and right sides are equal and the left and right sides are equal, if I add both left sides and both right sides I should still be equal.*0628

*A little math trick we can pull. *0637

*So if I add the left hand sides here I end up with m1g - t1 + t2 - m2g all equal to... *0640

*And the right hand sides if I add them up m1a + m2a, but I also know that t1 = t2. *0652

*So I am going to replace t1 with t2 in the equation minus t1 + t2, and if those are equal those add up to 0.*0666

*So my new equation m1g - m2g = m1a + m2a. *0675

*And I am trying to solve for a, so I am going to write this as gm1 - m2 on the left hand side equals am1 + m2.*0687

*And if I divide both sides be m1 + m2 I get that (a) is equal to g × m1- m2/m1 + m2.*0700

*I solve for the acceleration of the system by using two separate FBDs. *0715

*As an alternate solution we could look at this as an entire single system. *0727

*I am going to define my system now as m1, the rope, and m2. *0732

*So everything inside that dotted line is part of my single system, my single more complex object.*0741

*And I am defining this direction to be positive y again, so if I re-draw this a little bit I could re-draw this as m1 over here attached to a string, m2 and I am just taking those pieces of the Atwood machine and flattening them out for the purposes of looking at this as a system.*0750

*On m2 I am going to have a force of m2g that passes through that barrier. So we have m2g this way. *0774

*Over here I have m1g passing in that direction. *0783

*So over here I have m1g, again because positive y is pointing this way, that is my definition of the positive y direction.*0790

*Well, if I write Newton's Second Law now for this system, where again I have defined the system as basically what is inside that dotted container, what I get is to the left in the positive direction I have the force m1g, to the right I have the force m2g... *0801

*...so minus m2g because it is in the opposite direction of the positive y.*0823

*And that must equal the mass of my system. The mass of my system is m1 + m2 times the acceleration of the system. *0828

*Now look how slick this is. All I have to do now is divide both sides by m1 + m2 and I end up then with a = gm1 - m2/m1 + m2.*0837

*The same thing I had before, but just an alternate approach -- analyzing the solution as a whole.*0855

*Lets talk a little bit about elevators. *0865

*For some reason physicists seem to love the concept of putting a scale that measures an objects weight in an elevator. *0868

*I do not really know why, they just seem to love it.*0875

*So let us talk about it because you may see a problem or two come up like this.*0878

*To begin with, we need to talk about scales. *0882

*Scales do not really tell you the weight of an object and you should know that because you can go jump on your scale and for a minute it gets a really, really big reading and then it is a light reading and it levels out a little bit.*0884

*So it is not reading your weight the entire time, it is reading something else. *0895

*What it is really reading is the normal force that exerts on you.*0899

*If you put a scale down, you stand on it and once it comes to an equilibrium position as you are standing on the scale, what the scale actually reads on the reading is the normal force that it is exerting, the force it is exerting back on you.*0904

*Scales read the normal force and if we put scales in things like accelerating elevators, we can get some interesting results. *0922

*But we can analyze all of them with the stuff we already know using Newton's Second Law and FBDs. *0929

*So let us take a look here. *0935

*Buddy the dog with the mass of 25 kg is standing on a scale in an elevator when the elevator accelerates upward at 3 m/s *^{2}. 0938

*Probably scared Buddy the dog -- there might have been a little barking there.*0947

*What does the scale read while it is accelerating and what does it read once the elevator has come to a complete stop?*0950

*Well lets draw a FBD of our situation.*0955

*Here we go -- There is Buddy. We have Buddy's weight down mg and the force of the scale, the normal force back on Buddy.*0960

*And let us call up the positive y direction. *0973

*Now Newton's Second Law in the y direction -- F*^{net}y = MAy. 0977

*So in this case net force in the y direction is just going to be the normal force minus mg and that must be equal to MAy.*0986

*And if we want to know what does the scale read -- what we are really looking for is the normal force.*0996

*Therefore the normal force, the scale reading is going to be equal to mg plus (m) times(a) acceleration in the y direction. *1004

*Therefore, the scale reading Fn is equal to Buddy's mass, 25 kg times the acceleration due to gravity g (10) plus Buddy's mass again, still 25 kg times the acceleration of the elevator, 3 m/s *^{2} up, so that is positive.1017

*So we have 25 × 10 = 250 + 25 × 3 = 75, the scale is going to read 320N. *1040

*Considerably more than Buddy's 250N weight while it is accelerating upward. *1051

*What happens when it comes to rest, when it is stopped? *1059

*Well, when it is at rest we can use the same equation -- his Fn = mg + MAy, but as we do that now, when it is at rest Ay = 0. *1062

*Therefore the normal force, the reading on the scale is just mg or 25 kg, Buddy's mass, times his acceleration due to gravity, 10 m/s *^{2}, therefore the scale reads 250N.1083

*Scales in elevators, very popular problems. So let us try and put all of this together for a few minutes. *1106

*We have a truck on a hill here showing a 1 × 10*^{5} Newton truck, at rest, on a hill that makes an angle of 8 degrees with the horizontal. 1112

*What is the component of the truck's weight parallel to the hill?*1121

*Oh, we can go through and do all the FBDs and P-FBDs, or you could recognize that the weight parallel to the hill, it is just asking for mg parallel. *1126

*That is going to be mg sin θ.*1136

*In this case it tells us mg, the trucks weight, is 1 × 10*^{5}N × the sin of 8 degrees or about 1.4 × 10^{4}N. 1140

*The answer is number 3. *1161

*How about a force upper ramp? *1167

*We have a block here weighing 10N on a ramp inclined at 30 degrees to the horizontal. *1170

*A 3N force of friction (ff) acts on the block as it is pulled up the ramp at constant velocity -- that is important, with force f, which is parallel to the ramp as shown. *1175

*What is the magnitude of force f?*1186

*Right away, I start thinking FBDs, let us get ourselves some help here. *1190

*So there we have our axis, x and y and our forces -- we have the normal force perpendicular to the surface, we have the force f up the ramp, we have a force of friction down the ramp, and we have the 10N force, the weight.*1198

*And that does not line up with our axis, so we have to do something about that -- P-FBD time.*1221

*All right, x y -- F up the ramp again, normal force perpendicular to the ramp, force of friction down the ramp -- now 10N, we have a parallel and a perpendicular component. *1231

*The parallel component is going to be 10 sin θ, mg sin θ, which is going to be 10 sin 30 degrees and the perpendicular component 10 cos 30 degrees. *1249

*So to find the magnitude of force f, all I am going to do is write my F*^{net} equation F^{net}x =... 1263

*Well, what I have is f, up the ramp minus force of friction minus 10 sin 30 degrees and that is equal to mass times acceleration.*1276

*But we are at constant velocity a = 0, so that is all equal to 0.*1288

*If I want the force f then, f equals the force of friction plus 10 sin 30 degrees. *1292

*Force of friction is 3N, so that is 3 + 10 sin 30 = 5 for a grand total of 8N. *1303

*How about acceleration down a ramp?*1316

*100 kg block sides down a frictionless 30 degree incline as shown. Find the acceleration of the block.*1319

*Lets start with our FBD. *1326

*We know it is going to go down, so I will call that the x direction. There is our y. *1335

*Now we have a normal force perpendicular to the ramp and we have the block's weight (mg).*1341

*Mg does not line up with an axis, so just like we have been doing -- time to come back to the P-FBD. *1349

*Our axis again, (x, y), normal force. Now we have our components of mg -- we have mg parallel, mg sin θ, down the ramp, and mg perpendicular, mg cos θ end of the ramp.*1358

*So if we want to acceleration of the block, I am going to start with the net force in the x direction. 1387 So net force in the x direction is equal to -- we have mg sin θ, the only thing acting in the x direction, and that must equal MA in the x direction.*1380

*Therefore the acceleration in the x direction must be mg sin θ divided by m, or g sin θ.*1403

*How cool, we did not even need mass to solve this problem. *1416

*All we need to know is acceleration due to gravity, a constant here on the surface of the earth, and the angle. *1419

*The mass does not make a difference. *1424

*So that the acceleration in the x direction is just going to be 10 sin 30 degrees or 5m/s*^{2}. Very slick. 1427

*Let us take a look at another Atwood machine problem. *1443

*Find the acceleration of the 20 kg mass, given that the masses are connected by a light string over an ideal massless pulley.*1446

*And the moment you see "ideal massless pulley" right away you can go and make the assumption that the tensions we have on these are going to be equal. *1454

*Let us call that t, let us call that (t) right there and we will set them as equal now. *1463

*And since that is pretty easy to see the 20 kg mass is going to win here, I am going to define that direction as my positive y. *1467

*Let us call this m1 and we will call this m2. *1475

*So FBD for m1 -- we have m1g, down, and we have tension, up, and for m1, up is the positive y direction. *1480

*Lets do the same for m2 again. For m2 we have m2g, down, we have (t), up, and we will call down the positive y direction.*1496

*So when I write my Newton's Second Law equations for m1, I end up with t - m1g must equal m1a. *1509

*For mass 2, m2g - t must equal m2a.*1521

*We do this the same way we did before. *1533

*We can solve these lots of different ways, but this seems to be working for us right now. *1535

*So when I add these up, I am going to have t and -t that will make 0, so I end up with m2g - m1g = m1a + m2a.*1540

*Or solving for (a), we have (g) on the left hand side, m2 - m1 = a, m1 + m2 or a = g times the quantity m2 - m1/m1 + m2.*1556

*Now I just substitute in my values, a = g (10) × m2 - m1, 20 - 15/m1 + m2, 20 + 15 -- so that is 10 × 5/35 or 1.43 m/s *^{2}.1578

*All right, what happens if we switch up our system a little bit? *1605

*Now we have two masses, m1 and m2, connected by a light string over a massless pulley. *1610

*So again, the tensions can be equal, but now one of them is on a table on a frictionless surface. *1615

*Find the acceleration of m2. *1621

*Let us see what we can do here. *1624

*Right away I can tell that this thing is going to accelerate in that direction. *1626

*So I am going to call that the positive y direction. *1631

*And if we start by our FBD for m1 -- I have down m1g -- I have the normal force on m1, and let us call that t in both places -- can call it the same thing since it is equal -- T to the right. *1634

*And we also have m2 here, where we have m2g down, and t up. *1656

*And here that is the positive y direction and for m1 that is our positive y direction. *1665

*So let us write our equations -- Newton's Second Law over here for m1. *1673

*I am going to look in the x direction and just say that F*^{net} is t, which equals m1a. 1676

*For m2, same idea -- Net force is going to be m2g - t = m2a.*1685

*Let us add those together like we did before.*1697

*Our first equation, t = m1a, and our second equation, m2g - t = m2a. *1700

*When I put them all together and I end up with m2g on the left-hand side equals m1a + m2a or m2g = a × the quantity, m1 + m2.*1713

*Or if I want the acceleration of the system, which will be the acceleration of m2, a = g × m2/m1 + m2.*1730

*Slightly different problem, but we solved it the same way, using those same skills, those same tools.*1745

*What happens if we put our masses and pulleys on a ramp?*1754

*We are getting a little bit more involved every time.*1758

*Well, in this case, it is kind of tough to tell exactly which one is going to win but I am just going to pick a direction to begin with and I am going to call that direction around the pulley my positive y direction.*1762

*So once I have done that, I notice it is a massless pulley again so we can call both of those tensions, the tension is going to be equal, and we are looking for the acceleration of mass2 which is the same as the acceleration of mass1, and it is the same as the acceleration of the system.*1774

*So let us start by drawing our free body diagram for m1.*1789

*It is on a ramp, so let us call that our positive direction.*1794

*We also have the y axis and for our object, we have m1g, always down, we have the normal force on it, Fn, and we have force of tension up the ramp.*1799

*Right away again, we should be thinking P-FBD because m1g does not line up with the axis.*1819

*So let us do that right here. There we go.*1826

*We have tension up the ramp. We have normal force, -- now, m1g, we have got to break up into components -- the component parallel to the ramp is going to be m1g sin 30 degrees and perpendicular to the ramp, m1g, cos sin 30 degrees.*1833

*If we go and we also draw the FBD now for mass2 -- let us do that over here -- we have tension up, m2g down, and we are defining down as the positive y direction.*1857

*So this, Newton's Second Law equation is easy. F*^{net}y is going to be equal to m2g - t which is m2a.1871

*Over here, we have a little bit more work to do.*1885

*If we wanted to write the equation here, let us look in the x direction since that is the direction it is going to be moving -- I have the t - m1g sin 30 degrees = m1a.*1888

*If I rearrange this a little bit, t = m1g sin 30 degrees + m1a.*1904

*All right. Well, I am going to do this one a little bit differently.*1916

*I am going to replace t in this equation with all of that so when I do that, I get the m2g - m1g sin 30 - m1a = m2a.*1919

*We are solving for a again, so let us get all the a's on the same side, m2g - m1g sin 30 degrees = a × m1 + m2. *1939

*Or a = g × the quantity m2 - m1 sin 30 degrees/m1 + m2.*1954

*We are just extending what we have been doing to slightly more complicated situations.*1970

*Let us try one last more to round all this out.*1977

*Let us go back to our elevators problem.*1980

*Darryl the Duck, who has a weight of 230N is standing on a scale in an elevator when the elevator accelerates downwards at 3 m/s*^{2}. What does the scale read?1983

*Remember what we are really looking for here is the normal force at scale.*1994

*Well, FBD for Darryl the Duck, we have mg down, which is 230N -- we know his weight.*1999

*We have the normal force or the force of the scale up on him.*2008

*Let us call down our positive y direction.*2012

*Net force in the y direction then is going to be mg - the normal force and that must equal ma.*2018

*Therefore, the normal force must equal mg - ma or normal force = mg (230N) - ma in this case -- well, we do not know his mass.*2029

*But we know mg = 230N, o if mg = 230, then m must be 230/g or 230/10 which is 23 kg.*2050

*So mass, 23 kg × the acceleration and since it is down and we call down positive -- that is a positive 3 m/s*^{2}.2063

*So the normal force then, 230 - 23 × 3 or about 161N.*2073

*So his typical weight is 230N, but as the elevator accelerates down underneath him, he feels lighter for a second, the scale reads less.*2082

*That goofy feeling you have when the elevator drops out from underneath you and you feel like you are lighter for a second, well you are not lighter, the normal force is actually less on you.*2090

*Imagine you are on the bottom floor and the elevator jolts up with you in it. *2100

*Don't you feel heavy for a second, like you are being compressed into the bottom of the elevator?*2103

*That is when the scale reads more than your typical weight.*2107

*Hopefully, that gets you a good start on some applications of Newton's Second Law and all these different dynamics problems ranging from boxes on ramps to Atwood machines to elevator problems.*2111

*Hope you have gotten something great out of it.*2122

*Thank you for watching Educator.com and make it a great day!*2124

*Hi, everybody and welcome back to Educator.com.*0000

*This lesson is about impulse and momentum. *0003

*Our objectives are going to be to define and calculate the momentum of an object, to define and calculate the impulse applied to an object, and use impulse to solve a variety of problems.*0006

*We will also talk about how we interpret and use force versus time graphs and finish up by talking about how you find the center of mass of a system of point particles.*0018

*So with that, let us start by talking about momentum.*0029

*You probably use this term all the time.*0032

*A car speeds toward you out of control at a velocity of 60 miles per hour (mph), 27 meters per second (m/s).*0035

*Can you stop it by standing in front of it with your hand out? Probably not a good idea.*0040

*Momentum measures how hard it is to stop a moving object.*0046

*That car speeding toward you has too much momentum.*0050

*You are going to wind up in little leaky pieces somewhere on the pavement.*0054

*Momentum is a vector quantity and its symbol is p. *0058

*Don't ask me why -- Momentum, p, one of those physics things.*0061

*The units are kilogram meters per second (kg-m/s) which is equivalent to a Newton-second (N-s).*0067

*The formula for momentum, p = mass × an object's velocity.*0072

*So, since it is a vector, the momentum vector is in the same direction as the object's velocity.*0079

*That is probably obvious, but cannot really hurt to say it.*0084

*By the way, as we look at these units, a N-s -- remember a Newton is a kg-m/s*^{2}, so a N-s times a second is just a kg-m/s.0088

*A little bit of dimensional analysis to reinforce that.*0103

*All right. Let us take a look at a quick example.*0107

*Two trains -- Big Red and Little Blue, have the same velocity.*0110

*Big Red, however, has twice the mass of Little Blue -- Compare their momentum.*0114

*To begin with, we are going to have to use our momentum formula, p = mass × velocity.*0120

*Let us let M equal the mass of Little Blue.*0127

*Having said that then, if we wanted to find out the momentum of Big Red, that is just going to be -- since it has twice the mass of Little Blue, it is going to be two times Little Blue's mass times whatever the velocity is.*0137

*They have the same velocity.*0156

*Compare that to the momentum of Little Blue, which is just equal to its mass times velocity.*0158

*The difference then, of course, is two times larger.*0168

*So, we would say that the momentum of Big Red is equal to twice the momentum of Little Blue.*0172

*Pretty straightforward. Let us take a look at another one.*0183

*The magnitude of the momentum of an object is 64 kg-m/s.*0189

*If the velocity is doubled, what happens to the magnitude of the momentum of the object?*0193

*If our initial momentum is 64 kg-m/s, we are going to multiply that by 2 because of the velocity and that is going to be equal to m × 2V.*0199

*The momentum then is going to be 128 kg-m/s -- double the velocity, double the momentum -- Answer 3.*0216

*Another example where we have a changing momentum.*0230

*We have this D3A bomber with a mass of 3600 kg and it departs from its aircraft carrier with a velocity of 85 m/s due east. What is its momentum?*0233

*Momentum is m × v which is 3600 kg × 85 m/s, so it is going to be 306,000 kg-m/s, and it is a vector -- needs a direction -- East.*0245

*Once the bomber drops its payload, though, its new mass is 3,000 kg and it attains a cruising speed of 120 m/s.*0269

*What is its new momentum?*0277

*Now, it's mass times velocity or 3,000 × 120 m/s or 360,000 kg-m/s, in again, still, East -- it is a direction.*0280

*As you observed in the previous problem, momentum can change.*0302

*A change in momentum is known as an impulse. It is a vector and it gets the symbol J.*0306

*So, J (impulse) is equal to a change in momentum, which is the final momentum minus the initial momentum.*0312

*So, let us take that bomber again which had a momentum of 360,000 kg-m/s East and it comes to a halt on the ground.*0325

*What impulse had to be applied?*0333

*Well, its initial momentum was 360,000 kg-m/s and its final momentum was 0.*0336

*The impulse then, which is change in momentum, which is final momentum minus initial momentum, will be 0 - 360,000 kg-m/s East.*0349

*Therefore, we could say that the impulse is -360,000 kg-m/s East which is equivalent to saying that the impulse was 360,000 kg-m/s West.*0361

*If the bomber was going East, the impulse to stop it has to be applied in the opposite direction -- akes sense.*0390

*Let us take a look at combining impulse and momentum.*0401

*We have a 6 kg block sliding to the east across a horizontal frictionless surface with a momentum of 30 kg-m/s and it strikes an obstacle.*0404

*The obstacle exerts an impulse of 10N-s to the west on the block.*0414

*Find the speed of the block after the collision. Let us start with what we are given here.*0418

*We know the mass of our block is 6 kg and we know that it has a momentum of 30 kg-m/s*0425

*We know the impulse applied, J -- since it is in the opposite direction, it must be -10N-s and N-s and kg-m/s are the same units.*0435

*We are asked to find the final velocity of the object.*0447

*Let us start with the definition of impulse.*0454

*Impulse is change in momentum, which is the final value minus the initial value, but momentum is mass times velocity.*0457

*So, final momentum is mass times final velocity - initial momentum -- mass times initial velocity or V0.*0467

*This implies that impulse plus mV-initial must equal mass times velocity.*0476

*So if I want final velocity, that is just going to be impulse plus mV-initial divided by m, where if I substitute in my variables, velocity is equal to impulse (-10) + mass (6)...*0486

*It does not give us initial velocity. It has the initial momentum -- that is 30, so, 10 + 30 divided by our total mass (6)-- -10 + 30 = 20/6. *0514

*Or 3.33 m/s and in the positive direction which, throughout the problem, has been East.*0531

*That is combining impulse and momentum.*0541

*Now, let us talk about the concept of the impulse-momentum theorem.*0545

*That is a great tool that we can use for problem-solving here in physics.*0549

*We are going to derive it ourselves. *0554

*We are going to start with impulse being equal to change in momentum, but momentum is mass times velocity, so, impulse is change in mass times velocity.*0555

*But hopefully the mass of an object is not going to change.*0567

*If mass is constant, then we can write this as impulse is equal to mass times change in velocity.*0570

*Now we are going to pull one of those math tricks again.*0579

*Remember we can multiply anything by 1 and get the same value.*0581

*We are going to multiply this by 1, but we are going to write 1 a little bit differently again.*0585

*We are going to write 1 as δt/δt.*0590

*That red term there is equal to 1 -- something over something is 1, but if I multiply that, that allows me to make some transformations.*0595

*So then, I am just going to rewrite this and rearrange it a little bit to say that impulse is going to be equal to mass times δV over δt times δt.*0604

*All I did was I slid that δt over. *0620

*What that allows me to do now is, if I look at this -- this piece here -- δv over δt, the rate of change of velocity with respect to time -- that is the definition of acceleration (a).*0623

*So then I can write this as impulse is equal to mass times acceleration multiplied by δt. *0645

*But there is another transformation we can make here.*0655

*Now we have this ma -- Newton's Second Law, net force equals mass times acceleration, so I can now rewrite this again as J (impulse) equals mass times acceleration of force times your time interval. *0660

*So putting it altogether, impulse is a change in momentum, which is equal to a force applied over some time interval.*0680

*How do you apply an impulse? You apply a force for some amount of time.*0693

*How do you change an object's momentum? You apply a force for an amount of time.*0697

*It has some momentum, so if you want to change it, you have to apply a force.*0702

*The longer you apply the force, the greater the change in momentum.*0706

*The larger the force you apply, the larger the change in momentum.*0710

*That is what is known as the impulse-momentum theorem.*0713

*What it is saying again, when an unbalanced force acts on an object for a period of time, a change in momentum is produced and that is known as an impulse.*0722

*Here is our example problem -- A tow truck applies a force of 2,000N on a 2,000 kg car for a period of 3 s.*0735

*First off, what is the magnitude of the change in the car's momentum?*0744

*Well, let us start with number 1 here first.*0748

*Change in momentum, we know, is force times time.*0751

*We applied a force of 2,000N for a period of 3 s, therefore, the change in momentum is going to be 6,000N-s or 6,000 kg-m/s.*0757

*Now for 2 -- it says if the car starts at rest, what will be its speed after 3 s?*0772

*Well, we know δp is mass times change in velocity, which is mass times final velocity minus mass times initial velocity. *0780

*Therefore if we want the final velocity -- that is just going to be δp + mv initial/m -- just a little bit of algebra to rearrange to get v by itself.*0792

*Δp -- we just determined up here, was 6,000N-s plus mass times initial velocity.*0806

*Well, if it started at rest, initial velocity is 0 divided by its mass (2,000 kg), therefore it is going to have a final velocity of 3 m/s.*0816

*But then, what do you do if there is a non-constant force?*0832

*In that case, we can draw a graph of the force versus time.*0836

*The area under the force versus time curve is equivalent to the impulse or the change in momentum.*0840

*The area under a force versus time curve is the impulse or change in momentum.*0846

*Let us determine the impulse applied by calculating the area of the triangle under the curve here for a force that ramps up for 5 s and then back down for 5 s.*0853

*Well, the impulse is going to be the area.*0862

*It is the area of a triangle, so that is one-half base times height.*0865

*One-half the base here is 10 s and our height gets up to 5N.*0872

*So, 1/2 × 10 × 5, I come up with an impulse of 25N-s.*0881

*Another way we can use graphs to help us solve problems.*0890

*Let us talk for a minute about center of mass while we are here.*0895

*Typically, real objects are more complex than these particles, these ideal objects we have been dealing with so far.*0899

*However, what is really nice in physics is we can treat the entire object as if it had all its mass concentrated at a specific point that we know is the object's center of mass.*0906

*To calculate the center of mass, for the x coordinate, all you do is you add up all the different individual masses times their x position and divide by the total mass of the object or the system.*0917

*The y center of mass works the same way -- add up all the individual masses times their y coordinates and divide by the total mass of that system.*0930

*That will give you the x and y coordinates or if you want to go extending to three dimensions, you can go to z as well for the center of mass of an object.*0940

*Let us finish up with a couple of more example problems.*0949

*We have a 2 kg body initially traveling at a velocity of 40 m/s East.*0953

*If a constant force of 10N due East is applied to the body for 5 s, the final speed of the body is what?*0958

*Let us go to our impulse-momentum theorem.*0966

*Impulse is change in momentum or force times time, which implies then that force times time equals mass times the change in velocity, the change in momentum. *0969

*Or it implies that change in velocity then will be force times time divided by the mass.*0983

*Well, δV -- that is V - V-initial, so, V - V-initial equals force times time divided by the mass.*0994

*Or, to get the final velocity by itself, that is just going to be equal to the initial velocity plus force times time divided by mass.*1004

*We are going to come back to that, so I am going to put a little dot there for a second.*1015

*Let us keep going to solve our problem -- V-initial (40) plus our force (10N) times our time (5 s) over the mass of our object (2 kg) and I come up with 65 m/s -- Answer 3.*1019

*But while we are looking at this, let us take a look at that formula for a second -- Force divided by mass -- Newton's Second Law, remember?*1042

*Force divided by mass is acceleration.*1051

*V = V-initial plus force divided by mass(at).*1054

*That is one of our kinematic equations.This all inter-relates.*1062

*Let us take a look as we analyze a motorcycle accident.*1070

*A motorcycle being driven on a dirt path hits a rock. How sad.*1073

*Its 60 kg cyclist is projected over the handlebars at 20 m/s into a haystack.*1077

*Don't worry, it is a nice, soft happy little haystack.*1083

*If the cyclist is brought to rest in half a second, find the magnitude of the average force exerted on the cyclist by the haystack.*1086

*Again, impulse is change in momentum which is force times time.*1094

*If we are looking for force then, force times time equals mδV -- or change in momentum just rewritten -- or that's M x V - V-initial.*1102

*Therefore, our force is equal to the mass times final velocity minus initial velocity all over our time interval.*1116

*So, 60 kg (cyclist), V-final (0) - initial (20 m/s) in the time interval of that 1/2 second means that there must have been a force on our cyclist of -2400N.*1126

*We have a negative sign again. What does that mean?*1145

*The force applied was in the opposite direction of what we called positive -- initially the cyclist's velocity.*1148

*So that is the opposite direction of the cyclist's velocity, which is what you would expect if you are going to bring the cyclist to rest.*1156

*Great. How about an automobile collision?*1169

*In an automobile collision, a 44 kg passenger moving at 15 m/s is brought to rest -- V is 0 -- by an airbag during a 0.1 s time interval.*1175

*That is kind of the point of airbags -- not only do they spread out where the force is applied to decrease the pressure at any given point, but they increase the time interval in which that force is applied.*1187

*If you have an impulse, you are coming to rest at some point -- no matter what, that impulse is going to be applied, so you would rather have it applied over a longer period of time so that you have a smaller force.*1199

*Before the days of airbags, you would hit the windshield -- smack -- really short time, really big force. Game over.*1210

*I am trying to avoid those game overs, those really big forces over short time intervals.*1218

*So increase the time interval, lower the force -- greater survival rates.*1222

*So we are going to find the average force exerted on the passenger during that time.*1227

*Impulse is change in momentum, which is force times time.*1231

*Therefore, force equals change in momentum divided by time which is the final momentum minus the initial momentum divided by time.*1239

*Final momentum (0) came to rest minus the initial momentum (44) mass times velocity (15 m/s) in 0.1 s means that the force exerted on the passenger during that time is -6600 N.*1253

*If there was not an airbag, imagine that was 0.01 s.*1274

*Now we are talking 66,000N exerted on the passenger. Yay, airbags!*1278

*All right. Let us take a look at a couple of center of mass problems.*1286

*Find the center of mass of an object modeled as two separate masses on the x axis.*1291

*The first mass is 2 kg in an x coordinate of 2 and the second mass is 6 kg in an x coordinate of 8.*1295

*Just by looking at this, we should be able to guesstimate where this is going to be.*1303

*First off, it is going to be somewhere between those two objects.*1307

*Since the object over here on the right, the 6 kg mass, is bigger, I would anticipate that we are going to be somewhere closer to the 6 kg mass than they are the 2 kg mass when we find the center of mass.*1311

*So let us go back to our formula for center of mass.*1324

*In the x direction, x center of mass is m1x1 plus m2x2 plus for however many masses we have divided by all of the masses (m1 + m2 + ....).*1328

*In this case, mass 1 is 2 kg, so that is 2 and its x position is 2 -- m2 is 6 kg and its x position is 8.*1344

*We are going to divide that by the sum of the masses, 2 kg + 6 kg.*1354

*So I get 4 + 48/8 -- that is 52/8 which is going to be -- 6.5 should be the position on the scale of our center of mass.*1361

*1, 2, 3, 4, 5, 6 -- So we could draw it in here, right about there.*1383

*We could replace these two masses and treat it as if it is one 8 kg mass centered at a point that has an x value of 6.5.*1391

*The center of mass in the x direction.*1403

*Let us finish off by doing one that is a two-dimensional center of mass.*1406

*Find the coordinates of the center of mass for this system where we have three masses.*1411

*So we are going to have an x and a y coordinate now.*1415

*And right away, let us look at it again and go, "You know, chances are, in terms of the x position, we're going to be between this mass and this mass and in the y position, we're going to be between that one and that one."*1419

*So we are probably looking for a center of mass somewhere in that area.*1429

*That will help us see if we get this right or not.*1433

*The formula for the x center of mass again is m1x1 plus m2x2 and so on divided by the sum of all the individual masses.*1436

*In the y direction, the y center of mass works the same way -- m1y1 plus m2y2 plus however many more divided by all of the masses.*1449

*So let us figure out exactly where the center of mass should lie.*1465

*Start with the x coordinate. That is going to be...*1469

*Well, we have got 3 kg in x coordinate of 1 + 4 kg in x coordinate of 5 + 1 kg in x coordinate of 7 divided by the total mass, 3 + 4 + 1 = 8.*1472

*So that is going to be 3 + 20 + 7 -- that is going to be 30/8 or 3.75 for the x coordinate.*1490

*For the y coordinate -- now we are going to take our 3 kg × the y value (2) + 4 kg × the y coordinate (3) + 1 kg × the y coordinate (1) divided by the the total mass is 8.*1500

*So we have 6 + 12 = 18 -- 19/8 or about 2.38.*1520

*So our center of mass is going to be located at an x coordinate of 3.75 and our y coordinate of 2.38.*1526

*Let us see where that is -- see how that lines up -- 3.75, 2.38 -- centered somewhere in there.*1538

*We could treat this entire system as if we had an 8 kg mass centered at that point -- 3.75 in the x, 2.38 in the y.*1545

*All right. Hopefully, that gets you a good introduction to impulse-momentum and center of mass.*1557

*Thank you so much for your time and for watching Educator.com.*1561

*Make it a great day, everyone.*1565

*Hi everyone and welcome back to Educator.com.*0000

*This topic is collisions and conservation of momentum.*0003

*Our objectives are going to be to use conservation of momentum to solve a variety of problems and also explain the difference between an elastic and an inelastic collision.*0008

*Conservation of momentum -- linear momentum P is conserved in an isolated system.*0019

*Total momentum of a system is constant and this is very useful for analyzing collisions and explosions -- where collision is an event in which two or more objects approach and interact strongly for a brief period of time. *0026

*Or an explosion, which results when an object is broken up into several smaller fragments.*0039

*And the key here -- conservation of momentum says in any of these events, the initial momentum and the final momentum have to be the same, and these are vector sums so we have to add up those momenta in a vector fashion.*0043

*Now the easiest way I know to solve these -- to help keep organized -- is to create a momentum table.*0059

*What we are going to do is we are going to identify all the objects in the system and list them down the left-hand side of our column.*0064

*We will then determine the momenta of the objects before and after the collision. *0070

*We will add them all up using variables for anything we do not know in calculating their momenta, and then we will set the total momentum before equal to the total momentum after.*0074

*Sounds a lot more complicated than it is.*0085

*Let us dive in and see how we would do this.*0088

*As we talk about these, we have to remember that there are two types of collisions.*0092

*In an elastic collision, also known as a "bouncy collision" -- kinetic energy is conserved.*0096

*The total kinetic energy before is equal to the total kinetic energy after -- just like momentum is always conserved before and after.*0103

*In an inelastic collision, kinetic energy is not conserved.*0110

*And in a completely inelastic collision, or a "sticky collision", the objects actually stick together after they collide.*0115

*Let us take a look at a 2,000 kg car traveling at 20 m/s and it collides with a 1,000 kg car at rest at a stop sign.*0124

*If the 2,000 kg car has a velocity of 6.67 m/s after the collision, find the velocity of the 1,000 kg car after the collision.*0133

*Here let us try out that momentum table.*0143

*I am first going to list my objects over here.*0145

*And the objects that I have -- let us call our first car, car A, the second car, car B and we are going to want to look at these in terms of the momentum before the collision in kg-m/s.*0150

*We will put the units down here for this one.*0167

*The momentum after the collision -- momentum after in kg-m/s.*0169

*We will make our table -- and total.*0177

*Now before the collision, car A is traveling at 20 m/s so it is 2,000 as its mass times its velocity (20) for a total momentum of 40,000 kg-m/s.*0185

*Car B is at rest, so the total then -- 40,000 + 0 = 40,000 kg-m/s -- is the total momentum before the collision.*0197

*Now after the collision, car A still has a mass of 2,000, but it has a velocity of 6.67 m/s, so its new momentum after the collision is 13,340 kg-m/s.*0208

*Car B on the other hand, has a mass of 1,000 but we do not know its velocity, so I will put a variable in there.*0227

*Let us call that the velocity of B.*0234

*So when I add these up, I get 13,340 + 1,000 VB.*0236

*Now here is the slick part -- now that we have made the table, conservation of momentum says that these momenta before and after must be equal because there were no external forces.*0246

*Here is the equation that I have to solve in order to figure out what happened.*0256

*If I subtract 13,340 from both sides, I find out that 1,000 VB is going to be equal to 4-3-9-9-9 -- 10 - oops 0, 0 -- 9 -... 10 - 4 = 6, 9 - 3 = 6, 9 - 3 = 6, 3 - 1 = 2, equals 1,000 VB.*0260

*Now, divide both sides by 1,000 and VB -- the velocity of car B after the collision -- is equal to about 26.7 m/s -- Using a momentum table to help organize everything we need to know in these collisions.*0287

*Let us take a look at inelastic collision.*0310

*On a snow covered road, a car with a mass of 1.1 x 10*^{3} kg collides head on with a van having a mass of 2.5 x 10^{3} kg traveling at 8 m/s.0313

*Note that if they colliding head on they must be going in opposite directions.*0324

*As a result of the collision, the vehicles lock together and immediately come to rest.*0328

*Let us calculate the speed of the car immediately before the collision, and again we can neglect friction.*0332

*Let us start off by listing our objects.*0338

*We have the car, we have the van, and the momentum before and the momentum after.*0342

*Now before the collision, the car has a momentum of -- its mass is 1,100 times some velocity we do not know.*0355

*Vcar, that is what we are trying to find.*0366

*The van has a mass of 2,500 and its velocity is -8 because it is going in the opposite direction of the car.*0369

*Now after the collision, they stick together and essentially they become one object.*0377

*However, they are at rest so their momentum afterwards must be 0.*0383

*As we make our table, let us fill out our row for total.*0390

*Momentum before must be 1,100 Vcar - 2,500 × 8 and all of that must be equal to the momentum after -- 0.*0396

*If I solve this equation then, 1,100 Vcar = 20,000.*0411

*If I divide both sides by 1,100, then the velocity of our car must be 18.2 m/s.*0418

*A nice easy way to organize our thoughts here with those momentum tables.*0428

*Let us take a look at one involving recoil velocity.*0434

*If you shoot a gun, initially before you do anything with it, it has a momentum of 0 -- you are holding it still.*0437

*Then you shoot a bullet out one end of it -- very, very quickly -- it has a momentum.*0444

*Conservation of momentum says the total momentum of the system must remain constant.*0450

*So if a bullet comes out one way with some momentum, the rest of the gun must have a momentum back.*0454

*That is the kick on a gun or the recoil.*0460

*We call the velocity it kicks back with the recoil velocity.*0462

*So a 4 kg rifle fires a 20 g shell with a velocity of 300 m/s. Find the recoil velocity of the rifle.*0464

*Our objects -- here we have a rifle and we have a bullet.*0476

*We have a momentum before and a momentum after.*0488

*Now before this explosion that fires the bullet -- in essence, the rifle and bullet are really one object and they are at rest.*0493

*So their momentum before is 0.*0504

*After the incident, after they are fired, the rifle has a mass of 4 kg and it has some recoil velocity -- Vrecoil -- we do not know what it is, we are trying to find it.*0507

*The bullet has a mass of 20 g -- 0.02 grams -- and it has a velocity of 300 m/s.*0518

*That is going to be equal to 6.*0531

*So as I make my table here -- total on the left is 0 and on the right I have 4 Vrecoil + 6.*0534

*Momentum before must equal momentum after, therefore if I subract 6 from both sides, -6 = 4, Vrecoil, or Vrecoil, the recoil velocity of the rifle is -1.5 m/s.*0545

*What does the negative mean? It is in the opposite direction of the bullet's velocity.*0563

*So conservation of momentum in two dimensions -- in this problem we have a cue ball that Bert strikes with a mass of 0.17 kg giving it a velocity of 3 m/s to the right.*0569

*When the cue ball strikes the 8-ball of mass 0.16 kg, the 8-ball gets deflected in this direction with an angle of 45 degrees and the cue ball comes to this direction with an angle of 40 degrees.*0580

*We need to find the velocity of the cue ball and the 8-ball after the collision.*0591

*A great problem for momentum tables but it is all about staying organized -- taking our time and doing it right.*0596

*Here we go -- First off, let us say that the velocity of the cue ball is Vc -- that is after the collision, that is what we are trying to find.*0604

*We will call V8 the velocity of the 8-ball after the collision.*0614

*So when we make our momentum table in the x direction for momentum, our objects are -- we have a cue ball, we have an 8-ball, and of course we have our line for total.*0619

*We will take a look at the momentum before the collision in the x direction and the momentum after the collision in the x direction.*0638

*--Long pause--*0645

*The cue ball before the collision has a mass of 0.17 and a velocity of 3, so its total momentum before is 0.51.*0662

*The 8-ball is at rest so its total momentum before in the x direction is 0.*0674

*Our total then is just 0.51 before the collision and that is going to equal our momentum after in the x direction.*0679

*Now the cue ball has the same mass after the collision.*0686

*We do not know its velocity but we are going to call that Vc and we need to take the x component of that, so that is going to be the cos 40 degrees.*0690

*The 8-ball has a mass of 0.16 kg -- its velocity after the collision is V8 and we need its x component so that is going to be the cos 45 degrees.*0701

*When I add these up I get 0.17 cos 40 degrees × Vc -- that is going to be 0.13 Vc + 0.16 cos 45 × V8, which is 0.113 V8.*0712

*Now let us make our momentum table for the y direction.*0733

*Again we have the same objects -- the momentum before in the y direction and the momentum after in the y direction.*0737

*And our objects are our cue ball, our 8-ball, and a row for the total -- so objects. *0748

*All right, we have a nice little momentum table here.*0768

*The cue ball -- its momentum in the y direction before the collision is 0 because it has no velocity in the y direction and the 8-ball is at rest so the total must be 0.*0774

*Now after the collision in the y -- P after in the y direction -- the cue ball still has a mass of 0.17, its velocity is Vc but now it is going down in that direction, so that is going to be times the sin -40 degrees.*0784

*The y component of the 8-ball's momentum -- well its mass is 0.16 times its velocity, V8, times the sin 45 degrees to get its y component.*0809

*So 0 is going to be equal to 0.17 sin -40 × Vc is going to be -0.109 Vc + 0.16 × sin 45 × V8 = 0.113 V8.*0823

*If you look here I have two equations now and two unknowns that I can then solve to find my unknown unknown values.*0842

*What I am going to do to begin with is let us solve this and see if we can find what Vc is equal to.*0851

*If I add 0.109 Vc to both sides, the left-hand side becomes 0.109 Vc = 0.113 V8.*0858

*If I divide both sides by 0.109, I find that Vc equals about 0.104 V8.*0868

*Now what I am going to do is I am going to take that value up here to my Equation 1.*0878

*So up here in Equation 1 -- let us give ourselves a little room here.*0883

*We can now write that 0.51 = 0.13, and I am going to replace Vc with 0.104 V8 + 0.113 V8.*0887

*A little bit of algebra -- 0.51 equals -- well that is going to give me 0.248 V8.*0904

*If I divide both sides by 0.248, thenI find out that the velocity of the 8-ball is 2.06 m/s.*0914

*Great start. Now that we have that, we can take that value and we can put it back in here.*0925

*Velocity of our cue ball then is 1.04 × velocity of our 8-ball, 2.06 m/s, so the velocity of our cue ball then comes out to be 2.14 m/s.*0933

*Same basic strategies -- now we are just using momentum tables for the x components and the y components of momentum.*0953

*Let us take a look at one more -- an atomic collision -- but in this case we are dealing with an elastic collision.*0962

*A proton with some mass (m) and a lithium nucleus with some mass (7m) undergo an elastic collision.*0968

*Elastic collision means that kinetic energy is conserved -- the kinetic energy before the collision is equal to the kinetic energy after the collision.*0975

*Find the velocity of the lithium nucleus following the collision.*0984

*Well since this is coming straight on, we really only have to deal with one dimension here.*0988

*So let us take a look.*0993

*Our objects are our proton, our lithium nucleus, and a row for total, and the momentum before and momentum after.*0995

*For our proton before the collision, it has a velocity of 1000 and a mass of m, so its momentum must be 1000 m.*1019

*Lithium nucleus is just sitting there nice and happy at rest, 0, so the total before must be 1000 m.*1029

*Now after the collision, the momentum after -- well the mass of the proton does not change but it has some velocity, Vp, the velocity of our proton.*1038

*The lithium nucleus has some mass (7m) times the velocity of L, the velocity of our lithium nucleus.*1048

*So 1000 m is going to be equal to mVp + 7 mVL and I can right away divide the mass out of all of that to say that 1000 = Vp + 7 VL. *1056

*There is one equation.*1073

*Now we need to bring in that this is an elastic collision.*1075

*That means that the total kinetic energy before the collision equals the total kinetic energy after the collision.*1079

*Kinetic energy is 1/2 mass times velocity*^{2}, so kinetic energy before -- we have 1/2 times (m) times the square of its velocity.1088

*Let us call that V-initial*^{2}.1102

*That must equal the kinetic energy after.*1104

*Now both of these are in motion -- so that is going to be 1/2m times Vp*^{2} plus 1/2 our 7m, our lithium nucleus, times its speed squared.1106

*And once again I can make a nice simplification and divide out 1/2m out of all of those to say that V0*^{2} = Vp^{2} + 7 VL^{2}.1120

*So we have two equations, two unknowns because really we know this V0 is 1,000, so V0*^{2} is 10^{6}.1136

*So, 10*^{6} = Vp^{2} + 7 VL^{2}.1143

*All right to take that a little bit further then, we can then say that 10*^{6} -- let us come over here first and solve for Vp.1152

*If we do this, we could say Vp = 1000 - 7 VL, so over here, 10*^{6} now equals -- I am going to replace Vp with 1000 - 7 VL^{2} + 7 VL^{2}.1165

*All right, so 10*^{6} = 1000 - 7 VL^{2}, which is going to be 10^{6} - 7000 VL - 7,000 VL, so that is minus 14000 VL + 49 VL^{2} + 7 VL^{2}.1189

*That implies then, that 56 VL*^{2} - 14,000 VL = 0.1210

*We can factor out a VL to say that VL × 56 VL - 14000 = 0, so either VL = 0 or 56 VL - 14,000 = 0.*1222

*So, 56 VL - 14,000 = 0, add 14,000 to both sides, divide by 56 and I can find that VL = 250 m/s.*1238

*And if VL = 250 m/s, I can take that back over here to say that Vp = 1,000 - 7 × 250... *1255

*... or Vp = 1000 - 7 × 250 = -750 m/s.*1271

*So what do we have here? Velocity of our lithium nucleus is 250 m/s -- velocity of our proton is -750 m/s.*1281

*Why negative? It is going in the opposite direction that it first started.*1291

*It comes this way to the right, bounces off, and comes back to the left with a speed of 750 m/s to the left.*1295

*This brings in the fact that it is an elastic collision to give us one more equation -- this kinetic energy before equals kinetic energy after.*1303

*Hopefully this gets you started with analyzing collisions and looking at conservation of momentum.*1311

*I appreciate your time. Thanks for watching Educator.com*1316

*Hi and welcome back to Educator.com*0000

*I'm Dan Fullerton and I would love to talk to you now about describing circular motion.*0003

*Our objectives are going to be to calculate the speed of an object traveling in a circular path or a portion of a circular path.*0008

*And also to calculate the period and frequency for objects moving in circles at constant speed.*0014

*So uniform circular motion has to do with an object traveling in a circular path at a constant speed.*0020

*First thing we have to know. The distance around the circle is its circumference.*0030

*That is equal to 2π times the radius of the circle.*0034

*Or it is also equal to π times the diameter or the diameter is the distance through the circle.*0038

*The average speed formula we learned from kinematics still applies.*0046

*If average speed is the distance travelled divided by time, for something moving in a circle you have to take into account it is 2π times the radius, the distance around the circle, divided by the time.*0050

*Frequency is a number which describes the number of revolutions or cycles that you can get in in 1 second. *0063

*So if an object travels 3 times in a circle in 1 second, it would have a frequency of 3 cycles per second, or 3 hertz.*0069

*H-E-R-T-Z, abbreviated Hz, where a Hertz is the same as 1 over a second.*0081

*The symbol for frequency is **f*, so frequency is the number of cycles per second, or the number of revolutions per second.0087

*In similar fashion, period is the time it takes for an object to take one complete trip around the circle, or do one complete revolution or cycle.*0095

*Its symbol is capital T and its units are seconds. The amount of time it takes to go once around the circle.*0105

*Now what is really nice here is that frequency and period of course are closely related.*0112

*Frequency is 1 divided by the period and period is 1 divided by the frequency.*0117

*Once you know one, you automatically know how to figure out the other.*0123

*Let us take an example by looking at a car on a track.*0129

*Miranda drives her car clockwise around a circular track of radius 30 meters.*0131

*She completes 10 laps around the track in 2 minutes.*0139

*Let us find Miranda's total distance travelled and average speed and centripetal acceleration.*0142

*Well to find the distance travelled, her distance is 2π times the radius and she does 10 laps.*0150

*So circumference times 10 laps is going to be 2π times 30 meters, her radius times 10 laps. 300 times 2π is about 1885 meters.*0160

*Average speed then is the distance she travels divided by the time, or 1885 meters divided by 120 seconds, 2 minutes or 15.7 meters per second.*0174

*Now centripetal acceleration. That is going to describe how quickly she is accelerating.*0190

*Since she is moving in a circle, her velocity is changing, there must be an acceleration.*0195

*Centripetal acceleration, we will talk more about it later, is given by the formula v*^{2} over r. Square the speed divided by the radius.0199

*So in this case, the speed is 15.7 meters per second squared, divided by the radius of the circle, 30 meters.*0209

*That is going to be 8.22 meters per second squared.*0219

*And the direction of centripetal acceleration is always going to be toward the center of the circle.*0226

*All right, let us take a look at a race car problem.*0234

*The combined mass of a race car and its driver is 600 kilograms.*0237

*Travelling at constant speed, the car completes one lap around the circular track of radius 160 meters in 36 seconds. *0241

*Calculate the speed of the car.*0253

*All right, well, average speed is just distance travelled divided by the time, completes 1 lap or 1 circumference, 2πr in 36 seconds.*0255

*So that is going to be 2π times 160 meters over 36 seconds or about 27.9 meters per second.*0266

*Take a look at the toy train.*0286

*A half-kilogram toy train completes 10 laps of its circular track in 1 minute and 40 seconds.*0288

*If the diameter of the track is 1 meter, let us find the train's period and frequency.*0293

*Let us start with period. Period is how long it takes for 1 lap.*0299

*It takes it 100 seconds to go 10 laps, so the period must be 10 seconds per lap or revolution, and once we know period, frequency is easy.*0304

*Frequency is 1 divided by the period or 1 divided by 10 seconds which is 0.1, 1 over seconds, which is also equal to 0.1 hertz.*0318

*Named of course after the famous rental car company. I'm kidding.*0333

*Another example here, a roundabout on a playground.*0339

*Allan makes 38 complete revolutions on the playground roundabout in 30 seconds.*0342

*If the radius of the roundabout is 1 meter, let us determine the period of motion, the frequency of the motion and the speed at which Allan revolves.*0347

*Well, here is our diagram. Roundabout, the radius of 1 meter.*0357

*The period of the motion, how long it takes to go once around.*0364

*He makes 38 revolutions in 30 seconds, so that is 30 seconds for 38 revolutions or 0.789 seconds.*0367

*Frequency then 1 over period, or 1 over 0.789 seconds which is 1.27 hertz.*0379

*And the speed at which Allan revolves.*0393

*Speed is distance travelled divided by the time, that is going to be 2π times the radius 1.*0396

*He does 38 laps, 38 times around, and the total time to do all that is 30 seconds.*0403

*So 2π times 1 times 38 divided by 30, I get a speed of about 7.96 meters per second.*0411

*Hopefully that gets you going with describing circular motion.*0423

*We are worried about things like speed, period, frequency and we'll tackle a little bit more about centripetal acceleration coming up quickly.*0427

*Thanks for watching Educator.com. Make it a great day.*0435

*Hi, folks, and welcome back to Educator.com.*0000

*I am Dan Fullerton. In this lesson, we are going to talk about centripetal acceleration and force.*0003

*Our objectives are going to be to explain the acceleration of an object moving in a circle at a constant speed, to solve problems involving calculations of centripetal acceleration...*0008

*...to define centripetal force and recognize that it is not a special kind of force, but that it is provided by forces such as tension, gravity and friction -- something always causes the centripetal force.*0018

*Finally, of course, we are going to solve problems involving calculations of centripetal force.*0030

*So, let us get back to uniform circular motion.*0037

*The big question "Is an object that is undergoing uniform circular motion accelerating?"*0040

*If it is moving in a circle at constant speed, is it accelerating?*0045

*To answer that, we really need to understand very well what acceleration is.*0056

*Acceleration is a change in velocity and for an object going around the circle -- at some point there -- its velocity at that instant in time is tangent to the circle.*0062

*There its velocity is tangent to the circle and there it has a velocity tangent to the surface. *0072

*Notice that the direction of the velocity keeps changing.*0078

*Since the direction of velocity is changing -- yes, it is accelerating. Absolutely!*0082

*An object undergoing circular motion, even though it is moving at constant speed, is accelerating.*0095

*What about the direction of its acceleration then?*0102

*To do that, I am going to take you through sort of a kind of quasi-proof.*0105

*If acceleration is the change in velocity divided by time, that is the final velocity (vF) minus the initial velocity (vI) divided by time.*0112

*Or, vF - vI -- final velocity - initial velocity for an object going around a circle -- is just going to be the same as vF plus the opposite of vI.*0125

*So if we have an object going around a circle here, its initial velocity at a point in time -- tangent to the circle that way and an instant or two later that way.*0140

*Let us see if we cannot add up those vectors to see what we get.*0150

*vF is easy. We have already got that in here. It is pretty much in that direction.*0155

*vI, we have pointing up, so the opposite of vI would be pointing down.*0160

*To do that, I have to line them up tip to tail and draw it there, so that would be -vI.*0165

*The sum of those are vector addition and again, we go from the starting point of the first to the ending point of the last.*0173

*That must be the direction of vF + -vI which is the direction of the acceleration.*0181

*So where am I going to draw that?*0187

*Since I did that for these two points, I am going to draw that vector when it is right in between those two and we see if I draw it, the (a) points towards the center of the circle.*0189

*That is what we mean by centripetal acceleration.*0199

*'Centripetal' actually means center-seeking. It is always toward the center of the circle- Centripetal, center seeking. *0202

*For an object moving in a circle, it is accelerating even if it is moving at constant speed in the direction of that object's acceleration toward the center of the circle.*0218

*Now, the magnitude of centripetal acceleration -- we talked about very briefly in our previous lesson, but finding the magnitude is straightforward.*0230

*It is the square of the speed divided by the radius; Ac = V*^{2}/r.0240

*Centripetal force is something that causes a centripetal acceleration.*0248

*If the object is traveling in a circle, we know it is accelerating toward the center of the circle, and for it to accelerate, there must be a net force.*0253

*Remember, net force equals mass times acceleration -- Newton's Second Law.*0259

*We call this force a centripetal force because it too is pointed toward the center of the circle. It is center-seeking.*0265

*So a net force toward the center of the circle causing a centripetal acceleration is a centripetal force, which we label Fc.*0274

*So we had F*_{net} = MA and we talked about breaking that down into the x direction, F_{net}x = MAx.0282

*Then we did that in the y direction, F*_{net}y = MAy.0290

*We could even do that for the centripetal direction, where the centripetal direction is just pointing toward the center of the circle -- F*_{net}c = MAc.0295

*It is important to note here -- a centripetal force is not some new magic force.*0306

*An object moving in a circle does not automatically have some magic force that causes it to move in a circle called the centripetal force.*0310

*Something else must be causing it to move in a circle, something like gravity or attention or somebody pushing on it.*0317

*You have to have something causing it to move in a circle.*0325

*Think of a car going around a track really fast in a circle.*0328

*As it is moving in a circle, something has to be applying a force toward the center of the circle. 0337 What is it that is doing that?*0332

*Try and imagine what the car would do if some of those forces went away.*0339

*If it is going around in a circle and it is a car -- if all of a sudden, a friction between the tires and the road goes away, the car goes careening off in a straight line; it cannot go in a circle.*0344

*For a car moving in a circle, what is causing the centripetal force is the force of friction.*0353

*You have to understand, you do not label something like centripetal force on a free body diagram (FBD).*0359

*A centripetal force is just a net force pointed toward the center of the circle. *0363

*It is a label we put on another force because it is pointed toward the center.*0368

*Calculating centripetal force -- We already said that the net force is equal to mass times acceleration, and since it is in the centripetal or toward the center of the circle direction, F*_{net}c = MAc.0374

*But we also just learned that the magnitude of centripetal acceleration is the speed*^{2} divided by the radius.0388

*Therefore, F*_{net} c = M × V^{2}/r.0397

*If we want to check the units -- their dimensional analysis -- mass is in kg, velocity is in m/s that is squared divided by (r), which is in meters.*0408

*So I am going to have in the top kg-m*^{2} per m/s^{2}.0420

*One of the meters, we will make a ratio of 1 and I will be left with kg-m/s*^{2} which is a Newton (N).0428

*That makes sense. It is a force that we are after.*0436

*So F*_{net} c = MV^{2}/r -- just combining Newton's Second Law with the magnitude of the centripetal acceleration.0440

*If a car is accelerating, is its speed increasing?*0451

*That is another one of those thought questions.*0455

*Well, if we have a car over here and it is traveling with a velocity to the right and it is accelerating to the right, its speed is going to be increasing.*0459

*Acceleration and velocity in the same direction -- it is going to be speeding up. Absolutely, yes!*0475

*On the other hand, if we have a car and it has a velocity to the right but an acceleration vector pointing to the left, is it accelerating?*0482

*No, what is going to happen in the next instant in time -- that velocity is getting smaller.*0497

*For the accelerations pointing to the left, our next velocity is going to be that way.*0502

*Then it is eventually going to stop and then it is eventually going to be going the other direction. *0506

*So in this case, no, or we could also look at the situation of the car traveling in a circle.*0510

*Remember, the entire time it is traveling in a circle, it is accelerating toward the center of the circle even if it is doing so at constant speed.*0519

*Another sample question -- In the diagram below, a cart travels clockwise at constant speed in a horizontal circle. *0534

*At the position shown right here, which arrow indicates the direction of the centripetal acceleration of the cart?*0541

*Again, centripetal acceleration, a vector always points toward the center of the circle. The correct answer must be A.*0548

*A ball attached to a string is moved at constant speed in a horizontal circular path. *0560

*A target is located near the path of the ball, as shown in the diagram here. *0565

*At which point, along the ball's path, should the string be released, thereby removing the centripetal force, if the ball is to hit the target?*0569

*So the string's tension is what is causing the force to keep it moving in a circle. *0577

*Where do we let go of that so it hits the target?*0580

*Well, if the ball is moving this way, the moment that centripetal force goes away, there is no longer any net force on the ball. *0584

*It is going to travel in a straight line -- Newton's First Law. *0591

*There is no net force, no acceleration -- constant velocity, straight line.*0596

*I would let go at B where that line is now tangent to the target.*0600

*Another example -- A 1,000 kg car travels a constant speed of 20 m/s around a horizontal circular track. *0609

*Which diagram correctly represents the direction of the car's velocity and centripetal force at a particular moment?*0615

*Again, velocity at a particular moment is going to be tangent to the circle -- centripetal force -- centripetal must point toward the center of the circle.*0623

*It must be answer 1. *0633

*Again, since it is a car, what provides the centripetal force? *0635

*Well, if all of a sudden, the tires give way and you hit an oil slick, that car right here is going to keep going in a straight line. *0638

*It cannot turn anymore. So it must be the frictional force that is providing the centripetal force.*0644

*Here, we have an example with the Demon Drop, a popular amusement park ride. *0656

*Maybe you have been on it.*0659

*When you get on the ride, they have you stand up against the wall and they start to spin you. *0660

*They spin you faster and faster and faster and then the bottom -- the floor -- drops out. *0665

*You stay up there! You do not fall with it.*0670

*It is called different names at different amusement parks, but the diagram here shows a top view of this with a 65 kg student spinning.*0673

*Right now they are at point A. It is a radius of 2.5 m and a constant speed of 8.6 m/s.*0683

*The floor is lowered and the student remains against the wall without falling to the floor.*0688

*Draw the direction of the centripetal acceleration of the student on the diagram.*0694

*Centripetal acceleration -- by now, drawing these are probably getting pretty easy.*0699

*There we go -- centripetal acceleration -- toward the center of the circle.*0703

*Now, however, we want to determine the centripetal acceleration of the student and the centripetal force acting on the student.*0709

*The centripetal acceleration -- AC = V*^{2}/r; that is 8.6 m/s^{2} divided by the radius (2.5 m) or about 29.6 m/s^{2}.0716

*How about the centripetal force acting on the student?*0737

*To find that -- F*_{net}c = MAc. Mass is 65 kg. We just determined Ac was 29.6. 0740

*If we multiply those together and I come up with the centripetal force of about 1923N.*0755

*What causes that centripetal force?*0764

*Again, imagine we are looking down overhead. *0766

*Well, what is pushing in on the student must be the normal force, the perpendicular force, from the wall.*0769

*What force keeps the student from sliding to the floor?*0779

*Now we have to think a little bit more.*0782

*Let us draw a FBD for the student.*0785

*There is the wall -- while they are going around in this circle -- FBD -- there is our student.*0788

*Of course, we have their weight -- the force of gravity down and we have a force from the wall, the normal force -- perpendicular -- toward the center of the circle.*0797

*And if the student wants to slide down, he must have a force of friction pointing up.*0809

*It moves so quickly that we have a large normal force, therefore, we have a large frictional force.*0816

*So what force keeps the student from sliding to the floor?*0825

*That frictional force balances the weight to keep the student on the wall. *0827

*How do they do that?*0833

*They spin so fast you have a big normal force and since the force of friction is μ times the normal force -- if you have a very big normal force, you get a big force of friction.*0834

*Here, we have an example where we are looking at graphs to best represent the relationship between the magnitude of the centripetal acceleration and the speed of an object moving in a circle of constant radius.*0851

*The first thing I do when I see these sorts of problems is I try and find the relationship we are looking at, so I want a formula that has centripetal acceleration and speed in it -- relationship between the two.*0863

*I know that Ac = V*^{2}/r, where the variables I am interested in are speed on the x and centripetal acceleration.0873

*What happens as V gets larger? Well, Ac gets larger.*0885

*Right away, we can get rid of those two.*0888

*Now is this a linear relationship? No, it is a squared relationship. Therefore, the correct answer must be 2.*0891

*Another one -- We have a half kg object moving in a horizontal circular path with a radius of one-quarter meter at a constant speed of 4 m/s. *0904

*What is the magnitude or size of the object's acceleration? *0913

*Ac = V*^{2}/r, so that is going to be 4 m/s^{2}/0.25 m -- 16/0.25 will be 64 m/s^{2}.0916

*For something to be in uniform circular motion, it does not even have to go all the way around a circle, it just has to be traveling in a circular path.*0936

*So let us take a look at the example of a running back. *0945

*An 800N running back turns a corner in a circular path of radius 1 m at a velocity of 8 m/s. Find the running back's mass, centripetal acceleration and centripetal force.*0947

*To find the mass, let us start off -- we are given the weight of the running back -- mg = 800N. 0968 Therefore, the running back's mass is 800N/g -- 10 m/s*^{2} -- or about 80 kg.0960

*To find the running back's centripetal acceleration -- Ac = v*^{2}/r. 0984

*The Velocity is 8 m/s*^{2}/radius (1 m) -- 64 m/s^{2}.0991

*The centripetal force -- F*_{net}c = MAc. 1006

*So, the mass of our running back (80 kg), the centripetal acceleration (64 m/s*^{2}) -- I come up with a net force, a centripetal force of 5,120N.1014

*Now we have a car at an intersection. *1034

*A 1200 kg car traveling at a constant speed of 9 m/s turns at an intersection. *1038

*The car follows a horizontal circular path with a radius of 25 m to point P. *1045

*At point P, the car hits an area of ice and loses all frictional force on its tires. *1051

*What is the frictional force on the car before it reaches point P?*1058

*To answer that, we first have to realise that the frictional force is what is causing the centripetal force.*1062

*So the frictional force, F*_{net}c, is the frictional force, which is MV^{2}/r.1069

*Our mass is 1200 kg, our speed is 9 m/s*^{2} and our radius is 25 m, so I get a frictional force of 3,888N.1078

*What path does the car follow on the ice?*1099

*Once it hits the ice, no longer does it have a centripetal force -- anything pushing into the circle is going to travel in a straight line. *1102

*It is going to keep going straight ahead, the same direction it was and it is going to follow path B.*1109

*One of my favorite demonstrations -- In the diagram, we have a 5 kg bucket of water that swung in a horizontal circle of radius 0.7 m at a constant speed of 2 m/s. *1120

*What is the magnitude of the centripetal force on the bucket of water?*1132

*F*_{net}c = MAc, which is MV^{2}/r.1138

*Our mass is 5 kg, our speed is 2 m/s*^{2}/radius (0.7 m) or 28.6N.1145

*Now we have the 5 kg bucket of water swung in a vertical circle of radius 0.7 m.*1161

*Now it is being swung vertically with a speed of 3 m/s. *1167

*What is the magnitude of the tension on the string at the top of the circle and at the bottom of the circle?*1171

*Let us start at the top and what I am going to do is I am going to draw the FBD for the bucket when it is at the top of the circle.*1176

*There is my bucket and forces acting on it -- we have the tension in the string and of course we have the weight of the bucket, mg.*1184

*So Newton's Second Law in the centripetal direction, F*_{net}c, which is all the forces now pointing toward the center of the circle is going to be t + mg and F_{net}c is always equal to MV^{2}/r.1192

*So if I want to solve for the tension, tension is going to be mv*^{2}/r - mg. 1209

*The mass is 5, speed is 3*^{2} divided by the radius (0.7) minus the mass (5) times g (10).1217

*So at the top of the circle, I come up with a tension of about 14.3N on the string.*1229

*Let us try the same thing at the bottom of the circle.*1237

*At the bottom of the circle, the FBD is a little bit different.*1243

*When the bucket is down here, now we have gravity still pulling down but the tension is pulling up.*1246

*So now the net force in the centripetal direction, we have (t) toward the center of the circle and mg away from the center of the circle, so that is -mg.*1255

*Again, always equal to MAc or MV*^{2}/r.1265

*Therefore, tension = mg + mv*^{2}/r or tension = 5 kg × 10 m/s^{2} + 5 × speed (3^{2})/radius (0.7)...1270

*...so our tension now is 114.3N, which is probably what you expect.*1289

*Try this one sometime -- take a bucket of water, tie a string to it or even hold it in your hand -- use your hand instead of the string -- and swing it up and down.*1297

*You will feel a lot more force on your arm when it is at the bottom of the vertical circle than when it is at the top.*1305

*One last problem here -- It is actually possible for a car to turn on a banked curve without friction if the speed of the car and the angle of the bank are just right.*1314

*This is a terrific physics application for folks who live in icy areas, especially when the roads freeze over, they try and design the banks at angles so that the cars do not need a whole lot of friction.*1325

*Let us determine the required speed of the car for a given bank angle.*1336

*I am going to start off by drawing my banked angle and on here, we will put our car and I am going to look at it from behind again.*1340

*There is the license plate and we have some angle θ to our incline.*1352

*So we are going to start off with our FBD.*1358

*Even though the car is on a banked angle, because it is turning in a circle, that means that the motion of the car, the acceleration, is toward the center of the circle.*1361

*That way we are not going to tilt our axis. We are going to keep the acceleration going straight toward the center of the circle.*1372

*So when we draw our FBD in this case, I am going to...*1379

*... put my y axis, my x axis -- here's the car and we of course have mg down -- its weight -- and the normal force, but the normal force is here at an angle and that angle there is θ.*1384

*So when we draw our psuedo free body diagram (P-FBD), now we have to break up the normal force in the components.*1401

*There is my x, there is my y, and we still have mg down. *1410

*Now, if we want the x component of the normal force, that is going to be the opposite side of this triangle. *1418

*So the x component here is going to be Fn -- opposite side -- sin θ and the y component again is going to be the adjacent side, Fn cos θ.*1425

*Once I have that -- my P-FBD -- I can come back here and I can write my Newton's Second law equation, F*^{net}c equals...1445

*Well, I am going to replace F*^{net}c with all of the different forces I have pointing toward the center of the circle. 1455

*In that case, it is Fn sin θ.*1460

*So Fn sin θ points toward the center of the circle and I know that must be equal to MV*^{2}/r.1463

*Let us do my F*^{net}y equation over here.1478

*Net force in the y direction is going to be Fn cos θ minus mg and all of that must equal 0because the car is not going to spontaneously go flying up off the bank or down into it, not in the situation here where we are saying that it is just right that it is going to stay on the banked curve.*1481

*So I can say that the normal force, cos sin θ, must equal mg or the normal force equal mg/cos sin θ.*1500

*Now let us go back over here. *1513

*With that information, knowing that Fn = mg/cos θ, I am going to replace normal force here with mg/cos θ.*1514

*So I have mg/cos θ and I still have the sin θ here, times the sin θ and all of that equals MV*^{2}/r.1527

*We can do a simplification here and divide both sides by m.*1542

*By the way, sin θ/cos sin θ -- that is the tangent function.*1546

*Therefore, g × the tangent of θ = V*^{2}/r or to solve for the velocity, V^{2} = g(r) tangent of θ, or to get V by itself, take the square root, V = the square root of g(r) tangent of θ.1552

*I do not need any friction whatsoever when I have the situation where the velocity is equal to the acceleration due to gravity times the radius of the curve times the tangent of the bank angle of the curve -- square root.*1576

*Hope that gets you a good start on centripetal force and centripetal acceleration.*1589

*Thanks for watching Educator.com and make it a great day!*1593

*Hi, folks. I am Dan Fullerton and I would like to welcome you back to Educator.com.*0000

*Today's lesson -- gravity and gravitation.*0004

*Our objectives are going to be to utilize Newton's Law of Universal Gravitation to determine the gravitational force of attraction between two objects.*0007

*We are going to determine the acceleration due to gravity near the surface of the earth, calculate gravitational field strength and explain apparent weightlessness for objects in orbit.*0016

*So with that, why not dive right in?*0027

*Universal gravitation -- All objects that have mass attract each other with a gravitational force.*0029

*For example, right now you are attracted to me. *0036

*Yes, I know, that is kind of creepy, but any two objects that have mass, no matter how far apart they are, all have some level of attraction.*0039

*The bigger the masses, the more the attraction and the closer the masses are to each other, the closer the attraction, which is why we have a very, very, very, very tiny amount of attraction between us at the moment... *0047

*...or probably a long way away, our masses are relatively small and there is not much gravitational force there.*0059

*Between you and the earth, for example, the earth has a very big mass and you are relatively close to it, so you have a very measurable gravitational force of attraction there.*0065

*If we wanted to look at this in terms of our math, the force of gravity is gm1m2/r*^{2} in the direction of our hat and the negative just says that it is an attractive force.0076

*If we have one object over here -- let us call this mass 1 -- over here, we have some object, mass 2, and the distance between their centers of mass, we are going to call (r).*0090

*In this case, (r) is not specifically a radius; it is a distance between the two centers of mass.*0094

*Then you are going to have a force of Object 2 on Object 1 and you are going to have a force of Object 1 on Object 2 and they will be equal in magnitude and opposite in direction.*0109

*We know that because of Newton's Third Law.*0124

*If we wanted to get just the magnitude of the force, which is typically how this relationship is used, we say that the force of gravity is equal to mass 1 × mass 2/r*^{2}.0128

*If you do that and your masses are in kilograms and your distance is in meters, the units do not work out to anything overly useful, so we put in this fudge factor, this universal gravitational constant (G). 0152 That is equal to 6.67 × 10*^{-11} Nm^{2}/kg^{2}.0139

*It is there to make the units work out.*0158

*So, how do we calculate g, the acceleration due to gravity? *0162

*Let us see if we cannot use what we know to find out.*0168

*The mass of the earth is approximately 6 × 10*^{24} kg and its radius is about 6.38 million meters.0171

*The force of gravity -- we typically write -- is mg in a constant gravitational field.*0179

*Universal Gravitational Law says that G times the first mass times the second mass divided by the square of the distance between them.*0186

*Over here, we are assuming the second mass is already the mass of the earth.*0194

*Let us rearrange this a little bit.*0200

*What we can do is realize that we have the mass of the object here. *0202

*That is a mass of the object, so we are left with the mass of the earth, therefore g equals G, that constant, times the mass of Object 2, the earth, divided by the square of the distance between the objects between their centers of mass.*0207

*Therefore, g = 6.67 × 10*^{-11}Nm^{2}/kg^{2} × the mass of the earth -- 6 × 10^{24} kg divided by the square of the distance between them -- 6.38 × 10^{6} m -- roughly the radius of the earth.0223

*Do not forget to square that. That is a big mistake that students make.*0244

*Go through that and you should get an answer right around 9.8 m/s*^{2} or 9.8N/kg; the units are equivalent.0247

*Of course, that is what we expect. That is the acceleration due to gravity we have been using here on earth.*0259

*For the AP test, we typically round that to 10 to make the math a little simpler but you can see that it works out.*0264

*The force of gravity decreases with the square of the distance between the centers of the masses. 0277 This is called an inverse square law.*0273

*The force of gravity is gm1m2/r*^{2}.0279

*We are going to see lots of relationships in physics that have this inverse square relationship based on the distance between them.*0281

*As the distance gets bigger, the force gets smaller and it gets smaller by the square of that distance between the objects.*0289

*Graph of force versus distance is distance gets bigger -- the force tails off very, very quickly. 0304 That distance is an important factor because it is squared.*0298

*So this graph would be proportional to 1/r*^{2}.0308

*So then the question then, what happens to the force of gravity if you double the distance from the centers of mass?*0313

*Let us take a look at how we could answer that.*0340

*If the initial force of gravity Fg*_{i}initial is gm1m2/r^{2}, the final is going to be gm1m2 over... 0344

*We are going to double that distance, so this becomes 2 times whatever our initial was squared and that becomes gm1m2/initial r*^{2}, but the 2 is squared there too over 4.0361

*So if you rewrite this a little bit, you could write this as 1/4 × gm1m2/r*^{2}, but notice this is the force of gravity initial.0383

*So the final gravitational force is 1/4th the initial gravitational force.*0402

*If you double the distance, you get 1/4th the gravitational force.*0412

*If you halve the distance, you get 4 times the gravitational force.*0417

*If you triple the distance between objects -- 1/9th the gravitational force.*0420

*If you cut the distance between them into 1/3rd -- 9 times the gravitational force.*0426

*Whatever that factor is that you change the distance by, you square it in order to find out what happens with that new force.*0430

*Here are some problem-solving hints as we go through a lot of these gravity problems.*0442

*Try and substitute values in for variables at the end of the problem only.*0446

*Because you oftentimes have some pretty unwieldy numbers, the longer you can keep the formula in terms of variables, the fewer opportunities there are to make mistakes.*0450

*Secondly, before using your calculator to find an answer, it is oftentimes valuable to try and estimate the order of magnitude of the answer.*0458

*We will have to go through and calculate the whole thing but try and get a guess as to roughly where your answer is going to be and that way, if you make a goofy calculator error, it is pretty easy to pick up.*0470

*Finally, once your calculations are complete, take a second to make sure your answer makes sense by comparing your answer to some sort of known or similar quantity where you can. *0475

*If your answer does not make sense, stop, take just a second and see if you made a goofy calculator error or math mistake because lots of the problems I see are not with the physics here, it is with making goofy mistakes on calculators and calculations.*0483

*Example 1 -- What is the gravitational force of attraction between two asteroids in space if each has a mass of 50,000 kg and they are separated by a distance of 3800 m?*0500

*The force of gravity -- we are going to worry about the magnitude -- is equal to gm1m2/r*^{2}, where g is 6.67 × 10^{-11}Nm^{2}/kg^{2}... 0513

*...that is given to you for the exam -- × the first mass, m1 (50,000 kg) × the second mass, also 50,000 kg divided by the square of the distance between their centers of mass, 3800 m*^{2}.0527

*When I go through and do this, I get an answer of around 1.15 × 10*^{-8}N. 0544

*Why so small a force? You need a very, very, very big mass in order to have an appreciable gravitational force.*0555

*If we wanted to take this problem and do a quick order of magnitude estimation -- just to show you how you have done that -- what I do, is I would look at this expression here and try and estimate it quickly.*0564

*We have 10*^{-11}. We have -- that is something times 10^{4}, so I would say times 10^{4}, that is times 10^{4} divided by...0576

*...Well, those are 10*^{3}, so 10^{3}^{2}. Okay, 10^{8}. Then, 10^{-11}, 10^{-3}. 0586

*And you have 10*^{6} down here, so I would say you are roughly talking in the order of magnitude of something in the 10^{-9} and look you are only off by a factor of 10. 0597

*You are in the ballpark. You probably did not make a really goofy calculator error.*0607

*So that is how I would do an order of magnitude estimation here.*0611

*Example 2 -- Meteor and earth -- As a meteor moves from a distance of 16 earth radii to a distance of 2 earth radii from the center of earth, the magnitude of the gravitational force between the meteor and the earth becomes...*0618

*We have a couple of different solutions to choose from.*0630

*The biggest problem I see students having with questions like this has to do with reading the question and understanding what it is talking about.*0634

*Let us draw Earth here.*0642

*The meteor starts at a distance of 16 earth radii away, so it is going to be way over there. *0643

*There is its initial position -- 16 r's away.*0650

*Now if this is one (r) right there, then when it is 2 earth radii away from the center of the earth, there is 1 (r), there is the second (r), so it is moving from 16r to 2r.*0656

*The distance (r) is going from 16r to 2r -- the distance is 1/8th -- that's great.*0668

*The first thing I do here, say, is the force going to get bigger or smaller? As it gets closer together, you expect a bigger force. Right away, we can make answer 1 go away.*0677

*Because we have got that inverse square law with distance, our factor is not going to be 1/8th, it is going to be 1/8th squared, which is 164 and we said this is going to be bigger.*0690

*Because the distance is in the denominator, it is going to be 64 × that's great.*0701

*Another way you could do this is you could say the initial gravitational force is gm1m2/16 r*^{2}, which will be gm1m2/256r^{2}.0709

*What I am going to do is I am just going to take this gm1m2/r*^{2} and I am going to call that x. 0728

*So my initial force is going to be 1/256x.*0734

*Now the final gravitational force is gm1m2/2r*^{2} which is gm1m2/4r^{2}. 0741

*I am going to pull the same trick again and call that x. So that is 1/4th x.*0757

*If we want to know the ratio then -- what happens -- we will take the final gravitational force over the initial gravitational force, which is 1/4th x/256x or 256/4 which is a factor of 64 times larger.*0764

*Which diagram best represents the gravitational forces (Fg) between a satellite (s) in the earth?*0780

*First thing -- gravity only attracts, it never repels. *0801

*So over here in number 1, the satellite is being attracted, but earth is being repelled. *0804

*Nope, that does not work.*0809

*Number 2 -- they are both being repelled.*0811

*Number 3 -- they are both being attracted -- that is looking promising and they are both being attracted with the same force. *0813

*Even more promising, Newton's Third Law says that the force on one must be equal in magnitude to the force on the other just opposite in direction, so 3 must be our answer.*0816

*Let us talk for a minute about gravitational fields.*0831

*Gravity is what is known as a non-contact or a field force. *0834

*We cannot see it. We cannot go touch it. We cannot detect it with a special scope.*0838

*We just know it is there by putting an object there and then seeing what happens to it -- observing the force on some test particle that we would put out in space to see if there is a field there.*0844

*The closer objects are to large masses, the more gravitational force they experience and the denser the force vectors, as shown here, the force that you would see on a test object, the stronger the gravitational force.*0854

*So we could say that the gravitational field is weaker the further away you are if the lines are less dense and stronger as you get closer, where the lines are closer together.*0867

*Now, you can use that the gravitational force or the weight of an object is mg when you are close to earth -- where the change in the radius is negligible or really what we are talking about is a constant gravitational field strength.*0879

*Universally, this one always works -- gm1m2/r*^{2}, which is why it is called Newton's Law of Universal Gravitation.0899

*Going a little bit further into this gravitational field strength concept, if the magnitude of the gravitational force is gm1m2/r*^{2} and that is equal to m1g, assuming that we do not have a big change in that distance -- that we are in a constant gravitational field... 0909

*...then in that instance, we could take a look and say that g therefore must equal gm2/r*^{2} and the units of that are going to be N/kg or m/s^{2}.0927

*This is what we call gravitational field strength.*0945

*Wait -- you might say -- We have been calling g the acceleration due to gravity.*0952

*Yes, they are the same thing.*0957

*N/kg, m/s*^{2}, gravitational field strength, acceleration due to gravity -- they are the same thing, just different ways, different approaches of looking at the same phenomenon.0959

*So those are equivalent -- the acceleration due to gravity and gravitational field strength.*0971

*Let us take a look at an example.*0979

*Suppose we have 100 kg astronaut feeling a gravitational force of 700N when placed in the gravitational field of a planet. What is the gravitational field strength at the location of the astronaut?*0981

*The force of gravity is mg, therefore, we could find gravitational field strength -- the force of gravity divided by the mass or 700N/100 kg, should be 7N/kg or 7 m/s*^{2}.0996

*What is the mass of the planet if the astronaut is 2 × 10*^{6} m from its center?1018

*To do that, let us go to the Universal Law of Gravitation -- Fg = gm1m2/r*^{2}. 1023

*If we want the mass of the planet, that is going to be the force of gravity times the square of the distance between their centers of mass divided by G times the mass of our astronaut.*1032

*Our force is 700N. *1048

*Our distance is going to be 2 x 10*^{6} -- do not forget to square that-- divided by G, 6.67 × 10^{-11}Nm^{2}/kg^{2} × the mass of our astronaut, 100 kg... 1051

*...therefore, I come out with a mass of the planet of about 4.2 × 10*^{23} kg.1067

*Now, what happens if we talk about gravitational potential energy?*1085

*Two masses separated by some distance exhibit an attractive force on each other. *1091

*They want to move closer together because that gives them gravitational potential energy.*1095

*In a uniform gravitational field, the gravitational potential energy can be found by mg -- the weight of the object times the height, and we will talk about that more when we get to energy and work and a couple of other topics.*1101

*If the height is varying significantly to where we are not looking at a uniform gravitational field, we need something more general, a Universal Law for Gravitational Potential Energy.*1112

*That is -gm1m2/r. What does that minus mean?*1123

*Typically, we assume that potential energy equals 0 when you are infinitely far away from all other objects -- a long, long, ways away, you do not have any other influences.*1130

*Practically, you cannot get there; theoretically, you can.*1139

*If you were to take -- and we have a planet here and we have an object infinitely far away and we bring it closer and closer and closer and closer and closer, it wants to get sucked in -- gravity attracts.*1143

*If it had 0 potential energy way out there -- well, to get it back to the point where it is completely free of this planet's influence, you would have to add energy to free it.*1153

*It is almost like it is in energy debt before it is free, while it is trapped in the gravitational field here. *1164

*That is where the negative sign comes from.*1168

*Let us take a look at how orbits work.*1176

*This is a very interesting discussion problem because lots of folks have seen videos of astronauts and the space shuttle and they are floating around and the question often comes up, "Why are they floating around? They must be weightless."*1180

*No, they are not weightless and to understand that, you really have to know how orbits work.*1197

*We are going to go back to a thought experiment that Isaac Newton proposed many years ago.*1201

*He said, "Let us imagine that we have this hypothetical mountain, huge mountain, so high that at the very top of it, you are above the atmosphere of the earth."*1207

*You do not have any friction because there is no air to slow anything down.*1216

*At the top of this mountain, we are going to place a cannon. *1219

*I know the cannon is not going to work without an atmosphere, but just hang with me for the purposes of the thought experiment.*1222

*While we are up there, if we were to shoot a cannon ball, it is going to follow some projectile path down to the earth. *1228

*But if we shot it a little bit faster, it is going to travel a little bit further as it follows that parabolic trajectory.*1237

*Give it a little bit more velocity, it is going to travel even further, but eventually you are going to come to a point where you shoot it fast enough that at the rate it is falling, it is also falling around the earth because the earth is a circular path.*1247

*Yes, it is constantly falling. It is falling all the time, but it is moving so fast horizontally that by the time it falls, the earth has moved underneath it and it stays at the same altitude above the earth.*1262

*That is what happens in orbit.*1275

*They are not weightless. They are falling. *1278

*They are just moving so fast horizontally that by the time they fall and the earth has moved around underneath them and because the earth is a sphere, they maintain the same altitude.*1280

*Let us take a look and see if we cannot prove that a little bit.*1295

*If the space shuttle orbits the earth at an altitude of 380 km above the surface of the earth, what is the gravitational field strength due to earth at that altitude? *1298

*At what speed does the shuttle have to travel to maintain that orbit?*1306

*Let us start with the gravitational field strength.*1311

*The force of gravity is mg, which equals gm1m2/r*^{2}. 1315

*Therefore, the gravitational field strength (g) must be g times mass, which is going to be the mass of the earth divided by r*^{2}, where g, we know is that constant 6.67 times10^{-11}Nm^{2}/kg^{2}. 1323

*The mass of the earth is 6 × 10*^{24} kg over the distance between their centers.1339

*To find the distance between their centers -- if this is 380 km above the surface of the earth, we also have to account for the radius of the earth.*1347

*The radius of the earth is 6.37 × 10*^{6} m roughly + 380,000 m^{2} or about 8.78 m/s^{2} or 8.78N/kg.1356

*Compare that to 9.8, what we have here on the surface of the earth. *1377

*That is not a huge reduction. There is still an awful lot of gravitational field out there where they are orbiting.*1381

*What speed does the shuttle travel to maintain that orbit?*1389

*To do that one, let us take a look at the force of gravity, which is gm1m2/r*^{2} = mv2, mv^{2}/r because it is moving in a circular path -- centripetal force.1393

*Therefore, the square of our velocity if we rearrange these is going to be -- we have rgm1m2/mr*^{2} and I can do a little bit of simplifying here.1410

*We have rn and r*^{2}, we have a mass and a mass, so that will leave me with g times the mass of the earth divided by r.1428

*If that is v*^{2}, then v itself must be g times the mass of the earth over (r) square root.1440

*When I substitute in my values, that is 6.67 × 10*^{-11}Nm^{2}/kg^{2}, -- mass of the earth is about 6 × 10^{24} kg and the distance between their centers, 6.37 × 10^{6} radius of the earth + 380,000 m above the surface of the earth. 1449

*The square root of all that and I come up with a velocity of about 7700 m/s or that is greater than 17,000 miles per hour (mph).*1473

*To put that in perspective, that is more than 23 times the speed of sound at sea level. *1494

*That is fast!*1505

*Let us take a look at another example.*1513

*Calculate the magnitude of the centripetal force acting on earth as it orbits the sun, assuming a circular orbit of radius 1.5 × 10*^{11} m in an orbital speed of 3 × 10^{4} m/s. 1515

*Use that to determine the mass of the sun.*1528

*Let us start out with the magnitude of the centripetal force.*1532

*Centripetal force is mv*^{2}/r or 6 × 10^{24} × our velocity, 3 × 10^{4})^{2}/1.5 × 10^{11}... 1536

*... which gives me a value of about 3.6 × 10*^{22}N.1552

*Let us use that to determine the mass of the sun. *1562

*If that is the force, we know gravitational force is gm1m2/r*^{2}, where one of those is mass of the sun -- one is mass of the earth and that is equal to 3.6 × 10^{22}N.1565

*Therefore, we could say the mass of the sun is equal to 3.6 × 10*^{22}N × r^{2}/G × the mass of the earth. 1580

*Or 3.6 × 10*^{22} or 1.5 × 10^{11}^{2}/G, 6.67 × 10^{-11}Nm^{2}/kg^{2} × the mass of the earth, about 6 × 10^{24} kg.1593

*If I plug that all into my calculator very carefully and I find that the mass of the sun is right around 2 × 10*^{30} kg.1616

*So you can see we are using the same equations and relationships over and over again. *1628

*The tricky part is keeping all of your values well taken care of, being careful with the calculator -- very fastidious in your calculations.*1632

*Example 7 -- The diagram shows two bowling balls, A and B. *1645

*Each has a mass of 7 kg and they are 2 m apart. *1649

*Find the magnitude of the gravitational force exerted by ball A on ball B.*1653

*The gravitational force is gm1m2/r*^{2} where 6.67 × 10^{-11} × mass 1 (7) mass 2 (7)/the square of the distance between them -- 2 m^{2}) or about 8.2 × 10^{-10}N.1660

*Example 8 -- A 2 kg object is falling freely near earth's surface. *1680

*What is the magnitude of the gravitational force that earth exerts on the object?*1693

*If it is near earth's surface, we can do this one a simple way.*1698

*Force of gravity or the object's weight is mg, which is going to be 2 kg; g is 9.8 or let us round that to 10 to make it easy -- about 20 N. *1702

*Nice, simple, straightforward because it is near the earth's surface.*1714

*Let us do an example finding g. *1719

*What is the acceleration due to gravity at a location where a 15 kg mass weighs 45N?*1722

*Weight, mg = 45N, therefore, g must equal 45N/mass (15 kg) or 3 m/s*^{2}. 1728

*Just some very simple interpretation problems.*1744

*Let us take a look at a space vehicle on Mars. *1749

*A 1200 kg space vehicle travels at 4.8 m/s along the level surface of Mars. *1753

*If the magnitude of the gravitational field strength on the surface of Mars is 3.7 N/kg -- that is g -- find the magnitude of the normal force acting on the vehicle.*1759

*When I see normal force, right away I start thinking FBD.*1771

*We have the weight down (mg) -- normal force which we will call Fn -- pointing up -- and they must be balanced -- we call that +y direction.*1774

*It is not accelerating spontaneously up off the surface of the planet or going down through it. 1790 Therefore, the net force in the y direction must be 0 and the normal force and mg must be matched, therefore net force in the y direction is the normal force minus mg must equal 0.*1784

*Therefore, the normal force equals the object's weight (mg) or its mass (1200 kg) × g (3.7 N/kg) for a force of around 4,440N.*1803

*Let us take a look at a graphical analysis problem.*1825

*This graph represents the relationship between gravitational force and mass for objects near the surface of the earth. *1828

*What does the slope represent? The slope is rise/run.*1834

*Rise is going to be change in gravitational force and our run is going to be change in mass.*1844

*Change in gravitational force, as long as we are near the surface of the earth is δmg/δm and that is just going to give us g.*1851

*Then the slope is the acceleration due to gravity.*1861

*All right. Let us go back to Mars. A 2 kg object weighs 19.6N on Earth. *1874

*If the acceleration due to gravity on Mars is 3.71 m/s*^{2}, what is the object's mass on Mars?1878

*I love these questions! They are so simple but meant to trick you and it is so easy to fall into the trap.*1886

*It asks you what is the object's **mass* on Mars. The mass has not changed.1891

*The weight may have changed, but its mass is still 2 kg. Do not get suckered into those tricks!*1897

*Your find is the same as your given.*1905

*One more -- Here we have two satellites. *1910

*The diagram shows the two satellites, both of equal mass, A and B, in circular orbits around a planet here. *1913

*Compare the magnitude of the gravitational force of attraction between A and the planet. *1919

*Find the magnitude of the gravitational force of attraction between B and the planet.*1925

*First thing -- since B is further away, it should be pretty obvious that it is going to have a smaller force. Okay?*1929

*Right away -- twice as great, four times as great we can eliminate.*1938

*Because of that Inverse Square Law, we are going from radius (r) to 2r as we are doubling the distance and we must have 1/4th the force.*1942

*The answer is number 3: Inverse Square Law.*1950

*There are lots of different ways you can go through and solve that. *1953

*You can go through and do it analytically or you can make up numbers for them, but the easiest way is if you understand the Inverse Square Law, you can realize right away if the distance doubles, the force becomes 1/4th.*1956

*Hopefully, that gets you a great start on gravity and Newton's Law of Universal Gravitation.*1968

*Thank you so much for your time and make it a great day!*1973

*Hi everyone and welcome back to Educator.com.*0000

*I am Dan Fullerton and today we are talking about rotational kinematics.*0003

*Our objectives are going to be to understand the analogy between translational and rotational kinematics, to use the right-hand rule to associate angular velocity with a rotating object, and to apply equations of translational and rotational motion to solve a variety of problems.*0008

*Let us start by talking about radians and degrees.*0027

*In degrees, one time around a circle is 360 degrees.*0029

*In radians though, once around a circle is 2π, where a radian measures the distance around an arc equal to the length of an arc's radius.*0034

*So distance around a circle -- oftentimes written δS -- is the circumference, which is 2π radians or it would be 360 degrees if you are looking at an angular measurement.*0043

*Let us convert 90 degrees to radians.*0058

*If we start off with 90 degrees -- if we want to convert that to radians, we are going to multiply this by -- well we want degrees to go away, so 360 degrees = 2π radians.*0060

*So the degrees will cancel out and I will be left with 90/360, that is 1/4 and that is going to be π/2 radians or 1.57 radians.*0075

*Let us convert 6 radians to degrees, going the other way.*0091

*We have 6 radians, and we are going to multiply that -- we want radians to go away -- I know there are 2π radians in a complete circle and 360 degrees in a complete circle.*0096

*So we will be left with 6/2π × 360 or 344 degrees.*0109

*As we do this, let us talk for a few minutes about linear versus angular displacement.*0120

*Linear position displacement, is given by δR δS.*0125

*If we talk about angular position or displacement though, we can talk about how much this angle changes. *0129

*That is given by δ θ, and there is a conversion between these.*0135

*The linear distance is equal to R × θ, or δS = R × δθ.*0139

*Multiply the angular displacement, δθ by the radius to get a linear displacement.*0145

*We can also look at this for linear versus angular velocity.*0155

*Linear speed or velocity is given by the symbol V.*0159

*Angular speed or velocity is given by ω, kind of a curly W.*0162

*Now whereas velocity was the change in displacement over time, angular velocity is the change in angular displacement over time.*0168

*Dθ, DT, or δθ with respect to T.*0176

*If we want to take a look at the direction of angular velocity, we use the right-hand rule.*0186

*And the way we do that is you wrap the fingers of your right hand in the direction of the angular velocity -- your thumb will point in the direction of that vector.*0191

*Having a typical vector is not going to work because angular velocity, the direction linearly, is constantly changing, so you have to define it with something perpendicular.*0200

*Wrap the fingers of your right hand around the circle -- your thumb will point in the direction of the angular velocity vector.*0210

*In this case, as we have here on the screen, angular velocity is around this way, so as I wrap the fingers of my right hand around that direction, my thumb points out toward me.*0217

*I show that by showing a dot coming toward me, almost as if there is an arrow being pointed toward me -- that is what I would see.*0228

*So that is out of the plane of the board.*0235

*If it was into the plane of the board, the way I would draw it would be an x like I am looking at the fletchings of an arrow as it is moving away from me, so that would be into the plane.*0239

*In this case though, angular velocity points out of the plane.*0249

*Angular velocity is the cause of counterclockwise rotations, typically referred to as positive, and those that cause clockwise, negative.*0251

*How do we convert linear to angular velocity?*0263

*Well, linear velocity is just equal to angular velocity times the radius or angular velocity equals linear velocity divided by the radius.*0268

*Let us take an example.*0282

*Let us find the magnitude of Earth's angular velocity in terms of radians per second (rad/s).*0283

*Angular velocity is going to be a change in angular displacement divided by the time.*0290

*The Earth goes once around on its axis or 2π radians every 24 hours, once a day.*0296

*And let us multiply that to get radians per second -- let us convert hours into seconds.*0305

*One hour is 3,600 s, so my hours make a ratio of 1 and I am left with ω = 2π/24/3,600 or 7.27 × 10*^{-5} rad/s.0312

*As we look at linear versus angular acceleration, linear acceleration is given by A, and angular acceleration is given by the symbol α and it too is a vector.*0336

*Just like linear acceleration is change in velocity over time, angular acceleration is change in ω over time.*0346

*The rate of change of the angular velocity with respect to time, or we can write that as δ ω/δT.*0353

*The conversions between them are pretty straightforward as well -- A = Rα or α = A/R.*0362

*Another example -- angular acceleration.*0376

*A clown rides a unicycle. If the unicycle wheel begins at rest and accelerates uniformly in a counterclockwise direction to an angular velocity of 15 rpms in a time of 6 s, find the angular acceleration of the unicycle wheel.*0378

*Let us start by converting this 15 rpms to radiants per second. We have 15 rpms or revolutions per minute. *0393

*We need minutes to go away, so I will put minutes on the top and I want seconds here, so I know 1 minute is 60 seconds and now I have revolutions per second (rps).*0403

*So, I also need to multiply to make the revolutions go away.*0413

*One revolution is 2πradians.*0417

*Unit conversions then -- minutes make a ratio of 1, revolutions make a ratio of 1 and I am left with 15 × 2π/60, or 1.57 rad/s.*0422

*Similarly, the angular acceleration -- now I can find as change in angular velocity divided by time.*0440

*That is going to be final angular velocity minus initial angular velocity over time or 1.57 - 0/6s, which is 0.26 rad/s*^{2}.0448

*So let us put this all together to talk about kinematic variable parallels.*0467

*We talked about displacement in the translational or linear world.*0472

*Displacement -- we are writing as δS or D, or δX or we would even have it as R.*0478

*In the angular world, it is δθ.*0485

*Velocity is V. Angular velocity is ω, acceleration is A, angular acceleration is α, and time is the same translationally and in the angular world.*0490

*And there are more parallels we can draw, such as kinematic variables -- we can convert them.*0509

*Displacement is S = Rθ or if we want the angular version we have θ = S/R. *0514

*Velocity or V = Rω, angular velocity is ω = V/R, acceleration linear is A = Rα, angular is α = A/R, and same as before, time equals time.*0522

*So it is very easy to translate back and forth to these variables.*0543

*We even have parallels with the kinematic equations -- translational kinematic equations, V final = V initial + AT.*0548

*In the rotational world, we have kinematic equations too -- we just replace the variables with their angular equivalents.*0559

*So ω = ωinitial + αt and for translational, δX = V initial T + 1/2 AT*^{2}.0565

*For rotational, we have δθ = ωinitial T + 1/2αT*^{2} and final velocity^{2} = initial velocity^{2} + 2 × acceleration × δX.0580

*The rotational equivalent -- final angular velocity*^{2} = initial angular velocity^{2} + 2 α δθ.0600

*So really there is not a whole lot new to learn here. It is just using different variables to cover the rotational kinematics as opposed to just the linear kinematics.*0614

*Let us take an example of a medieval flail. *0625

*A knights swings a flail of radius 1 m in 2 complete revolutions. What is the translational displacement of the flail?*0627

*Well S = R θ, so R is going to be 1 m, θ is 4π radians, twice around the circle.*0636

*So that is just going to be 4π × 1 or 12.6 m.*0647

*Or let us look at a CD player.*0657

*A compact disc player is designed to vary the disc's rotational velocity so that the point being read by the laser moves at a constant linear velocity of 1.25 m/s.*0659

*What is the CD's rotational velocity in revolutions per second when the laser is reading information on an inner portion of the disc when the radius is 0.03 m?*0669

*Angular velocity is linear velocity divided by radius, so that is going to be 1.25 m/s over the radius of 0.03 m which is 41.7 rad/s.*0680

*We want that in revolutions per second so let us convert it.*0699

*41.7 rad/s times -- there are 2π radians in one revolution, so radians make a ratio of 1, 41.7 × 1/2π -- I get 6.63 rps.*0701

*We can even look at a carousel.*0733

*A carousel accelerates from rest to an angular velocity of 0.3 rps in 10 s. What is its angular acceleration?*0735

*Well, just like we did in kinematics, we can make a table -- ω initial = 0, ω final is 0.3 rad/s -- δθ, α and we know t is 10 s.*0743

*What is its angular acceleration? We can use our kinematic equations.*0759

*α = ω - ω initial/t or 0.3 rad/s - its initial 0/10 s for an angular acceleration of 0.03 rad/s*^{2}.0763

*What is the linear acceleration for a point at the outer edge of the carousel 2.5 m from the axis of rotation?*0783

*Well to do that we just need to find the linear acceleration from the angular acceleration.*0790

*A = Rα, where R = 2.5 m and our α -- we just determined, 0.03 rad/s*^{2} gives me a linear acceleration of 0.075 m/s^{2}.0795

*Or a circular saw example -- a carpenter cuts a piece of wood with a high powered circular saw.*0814

*The saw blade accelerates from rest with an angular acceleration of 14 rad/s*^{2} to a maximum speed of 15,000 rpms.0821

*What is the maximum speed of the saw in rad/s?*0830

*Well 15,000 rpms or revolutions per minute -- let us convert those minutes to seconds -- 1 min is 60 seconds, and instead of revolutions, we need this in radians.*0835

*So we have 2π radians per revolution -- revolutions make a ratio of 1, minutes make a ratio of 1 and I come up with 1,570 rad/s.*0852

*How long does it take the saw to reach its maximum speed? Well, that is a kinematics problem.*0871

*ω initial = 0 -- final, its maximum speed is 1,570 ras/s, δθ, α -- which we said was 14 rad/s*^{2} and time.0878

*If we are looking for how long -- we are looking for time -- I will use the formula ω = ω0 + αT and rearrange this for the time.*0896

*Time = ω - ω0/α or 1,570 - 0/14 rad/s*^{2}, which gives me a time of 112s.0907

*Hopefully that gets you started with rotational kinematics.*0928

*It is really similar to what we did with linear kinematics, it is just we have some slightly different variables dealing with objects going around the circle.*0932

*Once you know the change -- the parallels with the variables -- and you know the equations already, it is just a matter of being careful with your variables.*0939

*Thanks so much for your time and thanks for watching Educator.com.*0949

*Make it a great day.*0952

*Hi folks and welcome back to Educator.com. *0000

*This lesson is on torque. *0003

*Our objectives are to calculate the torque on a rigid object and apply conditions of equilibrium to analyze a rigid object under the influence of a variety of forces. *0006

*As we do this, let us start by defining torque. *0014

*Torque is a vector -- τ is a force that causes an object to turn. *0018

*Now in order for it to cause a rotation, torque must be perpendicular to the displacement. *0024

*So if we look at this diagram of a wrench, we are trying to turn it around this point here.*0029

*What we need to do is we need to have a force that is perpendicular to this line of action. *0037

*The stronger the force, the more torque and the further away you are from that point, the more torque.*0044

*Because of that, the further away you are, you obtain more leverage, so that line, that distance is called the lever arm.*0052

*Now, officially, torque is a cross-product; it is a vector product -- a vector multiplication of the r-vector, which is the vector from the point of rotation to where the force is applied, and the force vector. *0060

*Now for our purposes, instead of getting into detail around cross-products, let us focus on the magnitude of the force (F).*0073

*The magnitude is going to be rF sin θ and the reason is since this force -- only the portion that is perpendicular to this line of action counts. . . . *0080

*...if we were to draw the component of the force here that is perpendicular to the line of action -- if that is angle θ, that then must be the opposite side, so that is where we get the sin of(θ). *0089

*So the magnitude of the torque vector is the distance (r) × the force (F) × the sin of the angle between the line of action and the force.*0102

*The direction of the torque vector again is a little tricky -- kind of like the angular velocity angular acceleration vectors.*0113

*The direction of the torque vector is perpendicular to both the position vector and the force vector and again we figure it out using the Right Hand Rule. *0119

*For example, if we have a position vector (r) from the point of rotation to where the force is applied, let us call that (r) and then we have a force (F).*0129

*The way we find the direction of the torque vector is we take the fingers of our right hand, point those in the direction of (r), bend the fingers toward (F) and your thumb will point in the direction of the positive torque vector.*0140

*It is always perpendicular to both (r) and (F).*0152

*Now, positive torques cause counterclockwise rotations while negative torques typically cause clockwise rotations.*0156

*What is really nice here is -- once again -- just like when we were talking about rotational kinematics, we have the same sort of parallels as we talk about torque.*0167

*The net force on an object in a linear sense was the mass times linear acceleration.*0175

*The net torque on an object is equal to (I), the moment of inertia, or also known as the rotational inertia times the angular acceleration α. *0180

*So we start to see all these parallels again.*0192

*In the linear world, we have force (F); in the rotational world, we have torque.*0195

*In the linear world, we have mass -- a measure of inertia -- in the rotational world, we have moment of inertia or rotational inertia.*0201

*In the linear world, we have acceleration; in the rotational world, we have angular rotation.*0208

*In the linear world, we have velocity (V); in the angular world, angular velocity.*0216

*We also have displacement, linear displacement, δ x; -- in the angular world, angular displacement δ θ.*0222

*All of these parallels just keep coming up. *0233

*Let us talk for a minute about types of equilibrium.*0237

*Static equilibrium -- that implies that the net force and the net torque on an object are 0 and the system is at rest. *0240

*Dynamic equilibrium implies that the net force and net torque again are 0, but the system is moving at constant translational and rotational velocity.*0248

*No linear acceleration. No angular acceleration.*0258

*Rotational equilibrium implies that the net torque on an object is 0, therefore, no angular acceleration.*0262

*Let us take a look at a couple of examples.*0271

*A pirate captain takes the helm and turns the wheel of his ship by applying a force of 20N to a wheel spoke.*0273

*If he applies the force at a radius of 0.2 m from the access of rotation and at an angle of 80 degrees to the line of action, what torque does he apply to the wheel?*0280

*Well, that is a straightforward calculation where the magnitude of the torque vector is rF sin θ, where our (r) -- radius 0.2 m -- the distance from the point of rotation to where the force is applied -- × the force itself (20N) × the sin of our angle (80 degrees).*0290

*So he applies a torque of 3.94Nm -- units of torque -- N x m.*0312

*Let us take a look at another one -- our auto mechanic.*0322

*A mechanic tightens the lugs on a tire by applying a torque of 100Nm at an angle of 90 degrees to the line of action. *0326

*What force is applied if the wrench is 0.4 m long?*0334

*Well again, magnitude of the torque vector is rF sin θ, therefore, if we want the force -- that is just going to be the torque/r sin θ.*0339

*If our torque is 100Nm, our (r) is 0.4 m × the sin of our angle (90 degrees) -- sin 90 = 1, so 100/0.4 = 250N.*0355

*How long must the wrench be if the mechanic is only capable of applying a force of 200N?*0371

*Well if we want the length there, the torque = Fr sin θ again, therefore, r = the torque/F sin θ, which is 100Nm/F (200N) × sin 90 degrees, which is 0.5 m.*0376

*Example 3 -- We have a 3 kg cafe sign hung from a 1 kg horizontal pole, as shown in the diagram.*0404

*We have attached a guy-wire to prevent the sign from rotating. Find the tension in the wire.*0414

*Well to start off with -- let us draw a diagram of our situation.*0420

*There is our pole. It is attached over here to the pivot.*0424

*We have the weight of the pole down; its mass is 1 kg.*0429

*So the force on it -- down is 1 mg, so 1 × g.*0435

*We also have the 3 kg sign which is over here at a distance of 3 m from the pole, so that is 3g.*0442

*We also have the tension in our guy-wire here at this angle of 30 degrees.*0450

*How could I find the tension in the wire? I am going to use Newton's second law for rotation.*0457

*The net torque is going to be equal to -- well, counterclockwise, we will call positive.*0463

*We have t sin 30 degrees × the distance from the point -- our reference -- 4.*0471

*Now going the other direction -- going in the clockwise direction or the negative direction, we also have -3g F × its distance from our point -- 3.*0480

*We also have this 1g F - 1g, at a position of 2.*0493

*All of that must equal 0 since it is in rotational equilibrium.*0499

*So I could solve for (t) to say that (t) must be -- we have 9g + 2g -- 11g/4 sin 30 degrees, which is going to be 11 × 10 m/s*^{2}/4 sin 30 degrees or 110/sin 30 1/2, 2, which is going to give us 55N as the tension in that wire.0504

*Let us take a look at one more -- the seesaw problem.*0537

*A 10 kg tortoise sits on a seesaw 1 m from the fulcrum. *0542

*Where must a 2 kg hare sit in order to maintain static equilibrium and what is the force on the fulcrum?*0546

*First off, we are going to assume it is a massless seesaw. *0553

*It does not tell us anything about the mass, so let us just assume the mass of the seesaw does not come into play.*0557

*Let us draw what we have here. We have a seesaw. We have a fulcrum.*0563

*Over here on one side, we have a 10 kg tortoise, so its weight -- the force on it -- is 10g and that is 1 m from the fulcrum -- 1 m. *0568

*On the other side, we have our hare -- 2 kg, so its force is 2g and it is some unknown distance from our fulcrum.*0582

*If it is in static equilibrium though, we know that the net torque must equal 0.*0595

*Looking at our torques, the 10g over here -- that force times the distance (1 m) -- 10g × 1.*0602

*Now, the negative direction, -2g times whatever that distance happens to be from the fulcrum (x) must equal 0, so 10g - 2g(x) = 0, 10g = 2g(x) or x = 10g/2g, which must be 5 m.*0610

*The hare must sit 5 m from the fulcrum.*0630

*What is the force on the fulcrum?*0636

*To find the force on the fulcrum, all we have to do now is look at all of the different forces that we have here.*0638

*Newton's Second law says that we must have some force from the fulcrum pointing back up.*0646

*If we write Newton's Second Law for that object, F*^{net} in the y direction, we have Fp - 10g - 2g = 0.0651

*Therefore, the force of that pivot point, the fulcrum, must be 12 g or 120N.*0665

*Hopefully, that gets you a good start on torque.*0673

*Thanks for watching Educator.com.*0676

*Looking forward to seeing you next time. Make it a great day!*0677

*Hi folks, and welcome back to educator.com. What I'd like to do now, is take a few minutes to go through a review of some of the math skills we are going to need to be successful in this course.*0000

*In our outline, we are going to talk about the metric system and the system international, or SI units, which is the unit system that we use in physics.*0010

*We will talk about significant figures, scientific notation, and finally, the difference between accuracy and precision, and why they are so important.*0019

*The objectives are: convert and estimate SI units, recognize fundamental and derived units, express numeric quantities with correct significant figures so we understand how accurate and how precise our measurements are going to be.*0028

*We will use scientific notations to express physical values efficiently, and finally, differentiate between accuracy and precision.*0043

*So, why do we need units? Well physics involves the study of prediction and analysis of real world events and real world events have quantifiable numbers.*0052

*In order to communicate these to other people accurately, we need to have some sort of standards. Whether it be a sound was this loud, or this quiet. We need to put a number on that so we can communicate to people. The light was this bright, or this dim.*0062

*How do we put numbers around that? We have to decide on a set of standards and physicists have agreed to use what is known as the system international, which is a subset of the metric system.*0080

*You will also sometimes see it referred to as the MKS system because the basic units include meters, kilograms, and for time, seconds*0090

*Let's talk about it. The system international is comprised of seven fundamental units. It is based on powers of 10 because it is a subset of the metric system and all other units are derived from these basic seven.*0109

*The fundamental units are the meters, the kilograms, the second, hence the MKS system, the ampere, the candela, kelvin and the mole, which you may be familiar with from chemistry.*0122

*So let's start with the meter. The meter is a measure of length similar to the yard in the English system. For measurements smaller than the meter, use a centimeter which is about the width of your pinky finger perhaps. A millimeter is 1/10 of that, micrometer which is often times written μm, and nanometer, nm.*0137

*For measurements larger than a meter, typically we use kilometers, kilometers, 1000 meters.*0158

*The kilogram on the other hand, is roughly equivalent to 2.2 English pounds. For measurements smaller than a kilogram, we often times use grams or milligrams. A gram is about a paperclip.*0166

*For measurements larger than a kilogram, we could use things like a megagram, also known as a metric ton. That is 1000 kilograms.*0179

*In time, and everyone is probably familiar with this one, the base unit of time is a second. And unlike the rest of the metric system, time is a little funny. It is not based on units of 10. We have, instead, things like minutes, which is 60 seconds. Hours, which is 60 minutes. Days, which is 24 hours, and years, 365¼ days. But most of us are so familiar with this, it is not really a big deal.*0189

*For shorter times, we go back to base 10. For example, things like milliseconds, microseconds, and nanoseconds and so on...*0212

*We can take and we can make other units from these fundamental units. A unit of velocity or speed, for example is a meters per second, or if we take that further, could be a kilometer per hour. In the English system it might be a mile per hour.*0219

*Acceleration is a meter/second*^{2} which is really just a meter per second every second.0237

*Force is measured in newtons. But a newton is really just a kilogram times a meter divided by a second, divided by a second. That is kg×m/s*^{2}.0244

*These are derived units. They are comprised of combinations of those seven fundamental units.*0253

*As we talk about the metric system and these powers of ten, we need to look at the prefixes. *0260

*If we talk about something like a *__kilo__gram, a kilogram gets the symbol k in front of the g, for gram, kilogram would be 10^{3} grams.0267

*A gigagram would be 10*^{9} grams. Micrometer would be 10^{-6} meters, and this table is awfully helpful for converting units.0276

*Let's talk about how we convert fundamental units. If we have something like 2,480 meters and we want to convert it to kilometers, here is a nice and easy way to convert these.*0282

*Even if you can do it in your head, it is probably pretty good to learn this method because later on, when the units get more complicated, it will still work out for you.*0302

*Let's start off with what we have right now. 2,480m, and I am going to write that as a fraction so it is 2480/1.*0310

*I want the meters to go away, so I am going to multiply by something where I have meters in the denominator on the right hand side.*0320

*The units that I want are kilometers. To fill in the rest of this, what I have to realize, is that I can multiply anything by 1 and I get the same value.*0328

*If I multiply 3,280 by 1, I get 3,280. If I multiply 6 pigs by one, I get 6 pigs. The trick is, I can write 1 in a bunch of different ways.*0341

*I could write 1 as 0.5/0.5, that is equal to 1. I could write 1 as 3 apples/3 apples, that is still equal to 1.*0353

*So, I am going to use this math trick and I am going to multiply this by 1, but I am going to pick how I write 1 very carefully.*0363

*To do this, what I am going to do, is, I am trying to convert to kilometers, k. So I go over to my table of prefixes and I find k for kilo.*0370

*I see that it means 10*^{3}, so I am going to write 10^{3} over here on the bottom because on the bottom, there is no prefix in front of the unit.0380

*If I put 10*^{3} here, I am going to put 1 on the other side. What I have now made is a ratio 1km/10^{3}m and 1km is 1000× - 10^{3}m. 0392

*What I have really written here is 1 but I've written it in a special way so when I multiply this through, my meters make a ratio of 1. 2,480×1km/10*^{3} is going to leave me with 2.48 and my units that are left are kilometers.0405

*2,480m is 2.48km. It is a nice, simple way of converting units. Let's try another one.*0428

*5.357kg. Let's convert that to grams. I start by writing what I have. 5.357kg, and I write it as a ratio over 1, 5.357kg/1 ×,I want kg to go away so I will write kg in the denominator and I want grams in the numerator.*0439

*Now, I go to my prefix table and look up kilo, k, again is 10*^{3}. I am going to write that on this side that does not have a prefix. So that goes on the top this time and put a 1 on the bottom.0460

*Now kg and kg make a ratio of 1, or cancel out. What I'm left with is 5.357×10*^{3}g/1. So 5.357×10^{3};is just going to be 5,357 grams.0476

*There we go, converting fundamental units. Let's take a look at a 2 step conversion. Sometimes you have to do this in a couple of different steps.*0499

*We want to convert 6.4×10*^{-6}milliseconds to nanoseconds. I start by writing what we have. 6.4×10^{-6}ms/1. I want ms to go away.0509

*I put ms on the bottom and I will convert to my base unit, seconds on the top. I look up what milli, m, means and it means 10*^{-3}. Again, I write that on the side that does not have a prefix.0526

*So 10*^{-3} up there and 1 on the other side. Milliseconds would make a ratio of 1 and we are left with seconds but I do not want just seconds. I want nanoseconds, so I need to do another step.0541

*Multiply by, I want seconds to go away, so I will put that in the denominator and I want units of nanoseconds.*0556

*Now I go look up nano, n, 10*^{-9}. That again goes on the side without a prefix. I put a 1 on the other side and when I go look back here, seconds are going to cancel out.0562

*When I multiply this through, 6.4×10*^{-6}×10^{-3}/10^{-9} and the units I'm left with should be nanoseconds. I come up with 6.4 nanoseconds.0578

*So 6.4×10*^{-6}ms is 6.4 nanoseconds. A two step conversion.0595

*Let's go back the other way just to verify we have got this down. We already know what the answer should be here because we just did this problem, just in the other direction. Let's verify that it works.*0604

*6.4ns/1 and we are going to multiply. We want nanoseconds to go away so we are going to put that on the bottom and we'll go to seconds.*0616

*I look up nano, which means 10*^{-9} so I write 10^{-9} over here on the side that does not have a prefix. I put 1 on the other side.0628

*Now I'm left with seconds, but I want milliseconds so I do it again. If I want seconds to go away, I want milliseconds so I go to my table and look up milli which is 10*^{-3}.0640

*It goes on the side without a prefix, I put a 1 on the other side, and as I look here, nanoseconds cancel out, seconds will cancel out, and I should be left with milliseconds.*0654

*So I multiply through. 6.4×10*^{-9}/10^{-3} gives me 6.4×10^{-6} and the units I'm left with are milliseconds.0666

*There is my answer. It is exactly as we expected. Let's do one with some derived units.*0682

*We have 32m/s and we want to convert that to something like kilometers per hour. We are going to follow the same basic path again. We are going to write 32m/s as a fraction and if I want to convert to kilometers per hour, I can convert either the meters or seconds first, it does not really matter.*0690

*Let's start by converting the meters into kilometers. I want meters to go away, so that goes into the denominator and I want kilometers in the numerator.*0710

*I go to my handy dandy table over here and find that kilo means 10*^{3}. That goes on the side without a prefix and 1 goes on the other side.0719

*Now I'm going to be left with kilometers per second but I want kilometers per hour. So I have another step. The seconds here in the denominator, I need those to go away so I put seconds up here and it would be nice to put hours down here but I do not really know how many seconds are in an hour, but I know how many seconds are in a minute.*0730

*So I will do this first. I will say that there are 60 seconds in 1 minute. Now when I look at my units, my seconds will cancel out and I'm down to kilometers per minute.*0750

*I had best do another step here. So if I want minutes to go away, I will put that in the numerator. I want hours and I know that there are 60 minutes in 1 hour. I check my units again and minutes make a ratio of 1 and what I should be left with for units is going to be kilometers in the numerator per hour.*0761

*I am all set to go do my math. 32×60×60/10*^{3}should give me about 115.2 kilometers per hour. A derived unit conversion problem.0782

*Let's take a look at a multi-step conversion. One last unit conversion problem. Let's see how many seconds are in one year. I have no idea but it is kind of a fun problem to take a look at.*0804

*Let's start with 1 year, we will make that as a ratio. I do not know how many seconds are in a year but what I do know is that there are 365¼ days in 1 year. Years make a ratio of 1 and I am left with units of days.*0816

*We are still not to seconds but what I happen to know is if the days go away, there are 24 hours in 1 day. Days will make a ratio of one and I am down to hours. We are still not to seconds. *0836

*So in another step, I want hours to go away so I will convert to minutes. I know there are 60 minutes in 1 hour. Hours will make a ratio of 1 and I am down to minutes. We are getting closer.*0853

*I want minutes to go away so I will put minutes in the denominator. I want seconds and there are 60 seconds in 1 minute. Minutes will make a ratio of 1 and I am left with my units of seconds.*0867

*When I go through and I do all of this math, 1×365¼×24×60×60, I come out with about 3.16×10*^{7}seconds. That is a lot of seconds in 1 year.0880

*Another useful tool or skill is being able to estimate some of these units. For example, estimate the length of a football field. Well that is pretty big but just a rough ballpark figure is maybe about 100 meters.*0901

*If you are familiar with the English system, 100 yards and 100 meters are roughly the same thing. Or the mass of a student is maybe 60-70kg for a typical student. *0919

*The length of a marathon is somewhere in the ballpark of about 40, 42km or the mass of a paperclip, I think we mentioned this one previously is somewhere in the ballpark of about one gram.*0931

*So as you walk around and see different objects see if you can take an estimate of what their mass, their length, their time is in various units. It is a useful skill.*0946

*Let's talk about significant figures. Significant figures represent the manner of showing which digits in a number you know with some level of certainty. *0959

*For example, If you are walking along and see a garden gnome in someone's yard, significant figures can help you understand to what exactness you know the height of that garden gnome.*0970

*14cm, 14.3827482cm, or 14.0cm? These three numbers are all telling you slightly different things. What do they mean? Well, the key to significant figures is following these rules: Write down as many digits as you can with absolute certainty.*0982

*Once you have done that, go to one more decimal place, one more level of accuracy and try to take your best guess. The resulting value is your quantity in significant figures.*1003

*Now reading the significant figures, you start with the value in scientific notation and we will talk about that here very shortly. All non zero digits are significant. All digits that are in-between non zero digits are significant.*1015

*Zeros to the left of significant digits are not significant but zeros to the right of significant digits are significant.*1031

*As an example, how many significant digits are in the value 43.74km? Well we have 1,2,3,4 non zero digits so we must have 4 significant figures. We know at for certainty to 43.7 and that 4 is our best guess on the next level of accuracy.*1040

*How many significant figures are in the value of 4,302.5 grams? Well we have 4 non-zero digits and zeros between significant figures are significant so we have a total of 5 significant figures.*1062

*How many significant figures are in the value of .0083s? Well those are significant but zeros to the left of significant figures are not significant so here we have 2 significant figures.*1081

*How many significant figures are in the value 1.200×10*^{3}kg? Zeros to the right of significant figures are significant so we have 1,2,3,4 significant figures.1094

*Having gone through this, let's talk now about scientific notation. The need for scientific notation has to do with the tremendous variation in units, in magnitudes of these units, and their sizes.*1111

*For example, when we talk about length, we could talk about something like the width of a country, like the United States, which is probably a pretty big number, but we also have to talk about the thickness of human hair, all with the same base measurement of meters.*1126

*Even smaller, how about the transistor on the integrated circuit. Those are getting so small, it is smaller than a wavelength of light. So small that there is no optical microscope in the world that can ever see some of those features.*1142

*Huge ranges in orders of magnitude for these different measurements. Scientific notation can helps us express these efficiently and make it much easier to read.*1154

*For example, which of these numbers is easier to read. 4000000000000 or 4×10*^{12}. That is obvious, that is a lot easier to read and there is much less chance of making a mistake.1167

*Or, which is easier here .0000000001m or 1×10*^{-9}m? I think it's easy to see that those are a lot more accurate and less error prone. It is almost tough to read these numbers with all of the zeros because it's so easy to lose your place in them.1182

*So, using scientific notation. First off, show your value using the correct number of significant figures. Then, move the decimal point so that one significant figure is to the left of the decimal point.*1200

*Finally, show your number being multiplied by 10 to the appropriate power so that you get the same quantity, the same numerical value.*1214

*And finally let's talk about accuracy and precision. There is a difference between these two and in everyday speech, we often times use them interchangeably but in the world of physics, the world of science, There is an important distinction.*1223

*Accuracy is how close a measurement is to the target value. Precision, on the other hand, is how repeatable your measurements are. I like to look at these from the metaphor of target practice with a bow and arrow.*1236

*If we are aiming over here towards our first target and we are kind of all over the place here with our arrows and, by the way, they are nowhere close to the target and nowhere near each other, we have low accuracy and low precision which is typically not what you are after.*1248

*Over here, however, we have pretty high accuracy, we are starting to get close to the target but we are still not repeatable. We are accurate, close to the target but not repeatable therefore we have high accuracy and low precision.*1262

*Over here we are nowhere close to the target but we can hit that same spot nowhere close to the target every time. We are extremely precise, but our accuracy is off. High precision and low accuracy.*1277

*Finally, the nirvana of measurement, we have high accuracy, we are very near the target and we are repeatable, we have high precision. We can get near the target and we can get near the target every time.*1291

*With that, let's take a look at a couple more examples. Let's show this number 300,000,000 in terms of scientific notation assuming we know 3 significant figures.*1303

*We will find that 3 significant figures and I want to show this in scientific notation, I have one digit, one number to the left of the decimal place and I know 2 more significant figures so I write that as 3.00 to give me my 3 significant figures and I multiply it by 10 to the appropriate power which would be 1,2,3,4,5,6,7,8. 3.00×10*^{8}.1316

*How about showing this number, .000000... There is no way I can read this whole thing... 282 in scientific notation. Well, we have 3 significant figures so this must be 2.82×10 to some power. What power is that going to be? Well we have to move the decimal place 15 places to the right. So it would be 10*^{-15}. Isn't that a lot more efficient and easier to read?1343

*How about here? Express the number .000470 in scientific notation. We have 3 significant figures, so 4.70×, and the power is going to be, 1,2,3,4 to the right, so 10*^{-4}.1387

*And one last one, let's see if we can expand 1.11×10*^{7}. We have 1.11 and we need to move the decimal place 7, so 1,2,3,4,5,6,7. So I would write that as 11,100,000. 11 million, 100 thousand.1408

*Hopefully this gets you a good start on some of the basic math skills we are going to need here in physics especially around scientific notation, significant figures, units, converting units, and accuracy and precision. Thanks for watching educator.com, we will see you next time and make it a great day!*1434

*Hi, folks. I am Dan Fullerton and I would like to welcome you back to Educator.com.*0000

*Our next topic -- Rotational Dynamics. *0004

*Our objectives are going to be to understand the moment of inertia or rotational inertia of an object or system -- depends upon the distribution of mass within the object or system, to determine the angular acceleration of an object when an external torque or force is applied.*0006

*We will calculate the angular momentum for a point particle, utilize the Law of Conservation of angular momentum and analyzing the behavior of rotating rigid bodies, and finally calculate the kinetic energy of a rotating body.*0025

*With that, let us talk about types of inertia. *0039

*So far, we have talked about inertial mass or translational inertia, which is an object's ability to resist the linear acceleration.*0041

*Well, in the rotational world, we have an analog of that as well. It's called moment of inertia or rotational inertia.*0050

*That is an object resistance to a rotational acceleration or an angular acceleration.*0057

*Now, objects of that most of their mass near their center of rotation tend to have smaller rotational inertias than objects with more mass farther from their axis of rotation.*0063

*Think of a figure skater spinning on the ice. While their arms are out, they tend to go slower.*0071

*To go faster they pull their arms in; they are shrinking their moment of inertia as they do that.*0077

*Smaller moment of inertia means easier to accelerate.*0082

*The formula for moment of inertia is the sum of mass times the square of the radius.*0086

*Now, if you have an object that is more complex than a simple particle, you have to add up all of the little bitty pieces of mass times the square of their distance from that axis of rotation.*0093

*Add them all up and you get the moment of inertia.*0104

*Let us talk about the moment of inertia for a couple of common objects. *0108

*For any object, if you take the sum of the all masses times the square of the distance from the axis of rotation that formula will work for any object.*0111

*But that is not always easy to apply, so for some common objects -- things like a disc, the moment of inertia is 1/2 times the mass of the disc times the square of its radius, assuming it is a uniform mass density distribution.*0120

*A hoop on the other hand is mr*^{2}. A solid sphere is 2/5 mr^{2}.0134

*A hollow sphere on the other hand, where all of the mass is on the outside almost like a spherical shell, is 2/3 mr*^{2}.0143

*A rod rotated about its center is going to be about 1/12 mL*^{2} where L is the length of the rod, but if you rotate it about its end, then it becomes 1/3 mL^{2}.0149

*The moment of inertia goes up because more of the mass is situated away from that axis further away from that axis of rotation.*0165

*Let us take a look of how could we calculate moment of inertia.*0174

*We have two 5 kg bowling balls joined by a meter long rod and we are going to say that rod is of negligible mass.*0177

*If we rotate it about the center of the rod we can find its moment of inertia this way.*0184

*The moment of inertia is going to be the sum for all the different particles of mr*^{2}, which in this case -- let us call this m1 and we will call this m2.0189

*We will call this distance r1 and this distance r2.*0200

*That is going to be m1(r1)*^{2} + m2(r2)^{2}.0208

*In this case, m1 here is going to be 5 and if this whole distance -- we call 1 m then r1 must be 1/2, so that is .5*^{2} + m2(5) × r(.5)^{2}.0217

*This gives me a moment of inertia equal to 2.5 kg × m*^{2}.0237

*Now, let us take the same object and rotate it now by the end under one of the bowling balls -- so putting more of the mass further away.*0247

*That to me, just theoretically, I would think you know that is going to be harder to spin.*0256

*I am thinking we are going to have a larger moment of inertia. Let us find out.*0261

*Once again, moment of inertia, capital I, is the sum of mr*^{2}, which will be m1r1^{2} + m2r2^{2}.0266

*Once again, m1, m2, but now r1 is this entire distance, 1 m and r2 is going to be 0. *0277

*So I end up with mass1(5) × r1(1)*^{2} + mass2(5) × distance from the axis of rotation, 0^{2}0287

*That is just going to be 5 kg-m*^{2}.0297

*So the moment of inertia here doubled compared to when we spun it about its center of mass.*0302

*We can take a look at this in terms of Newton's Second Law as well.*0310

*Newton's Second Law said that the net force on an object was equal to its mass -- its linear inertia times the acceleration.*0314

*The angular acceleration of an object, on the other hand, was the net torque applied divided by the object's moment of inertia. *0324

*Again we have the same parallels -- force, torque. Linear inertia -- rotational inertia. Linear acceleration -- rotational acceleration*0331

*It all works the same way. Let us take another example here. Let us talk about a rotating top.*0345

*A top with moment of inertia .001 kg-m*^{2} is spun on a table by applying a torque of .01N-m for 2 seconds.0352

*If the top starts from rest find the final angular velocity of the top.*0362

*Well, let us figure out what information we know to begin with.*0367

*The initial angular velocity is 0. It starts at rest.*0371

*We are trying to find the final angular velocity. *0374

*We do not know the angular displacement; we do not know α, and we don't know time. *0379

*Pardon me. We do know time, it is 2 s. *0388

*Well, it is sure be helpful to know that angular acceleration.*0390

*Let us take a look and say that the net torque is equal to I(α) -- Newton's Second Law for Rotation.*0394

*That means then that alpha is going to be the net torque divided by the moment of inertia and our net torque was .01N-m and our moment of inertia .001 kg-m*^{2}.0403

*That tells me then that my angular acceleration must be 10 rad/s*^{2}.0419

*I can plug that in over here for my alpha as 10 rad/s*^{2}.0427

*Now, I can use my kinematics to find what final angular velocity is.*0435

*Final angular velocity is initial angular velocity plus alpha times time*0443

*That is going to be -- well this is 0, so 10 rad/s*^{2} × 2s is going to give us a final angular velocity of 20 rad/s.0449

*That is Newton's Second Law in kinematics, all put together -- This time though for rotation.*0468

*Let us take another example. *0475

*What is the angular acceleration experience by a uniform solid disc of mass 2 kg and radius .1 m when the net torque of 10N-m is applied?*0476

*Assume the disc spins about its center, which we can see from the diagram there as well.*0486

*Well, net torque is moment of inertia or rotational inertia times angular acceleration.*0492

*Now, because this is a disk we can look up its moment of inertia which is going to be 1/2mr*^{2}, where there is our (r) and it has some total mass (m).0499

*The net torque equals -- well I, we have 1/2mr*^{2} × α.0516

*Therefore, alpha must be equal to 2 times our net torque divided by mr*^{2}.0529

*Now, we can substitute in our values to find that alpha is equal to 2 times our net torque (10N-m) divided by the mass (2 kg) times the square of the radius .1 m*^{2}.0540

*20/2 × .1*^{2} = .01, which should give us 1000 rad/s^{2}. 0556

*The same basic sort of problem -- now, we are just solving for angular acceleration and we had to go look up the formula for the moment of inertia, which you saw a couple of slides ago for some common objects.*0569

*Linear momentum -- the product of an object's inertial mass and its velocity -- is conserved in a closed system. *0583

*That is the conservation of linear momentum. We have talked about that already.*0588

*Linear momentum describes how difficult it is to stop a moving object.*0594

*There is an analogy in the rotational world, too.*0597

*Angular momentum -- a vector (capital L), which is the product of an object's moment of inertia or rotational inertia and its angular velocity about the center of mass -- is also conserved in a closed system when there are not any external torques.*0600

*That describes how difficult it is to stop a rotating object.*0606

*We have angular momentum equals moment of inertia times angular velocity.*0620

*That fits right along with our analogy, linear momentum equals linear inertia (mass) times linear velocity. *0627

*Here are the analogs -- angular momentum, linear momentum; rotational inertia, linear inertia; angular velocity, linear velocity -- same sort of parallels again. *0636

*How do we calculate angular momentum?*0652

*Well, what we are going to do is, we are going to talk about a mass moving along with some velocity (v) at some position(r) about point (Q).*0655

*Angular momentum depends on their point of reference.*0663

*We are going to start by setting a reference point (Q).*0669

*In that case, the object has some angular momentum (L) about (Q) and we could find that by multiplying the vectors (r) and (p) with the vector cross product -- the vector product, which will give us another vector, which is a lot like we did the talking about torque.*0671

*The angular momentum vector (r) cross (p) -- we determine its direction by the right-hand rule.*0687

*Point the fingers of your right hand in the direction of (r) where (r) is the vector from your reference point to the object.*0693

*Now, bend your fingers in the direction of the velocity. Your thumb then will point in the direction of the positive angular momentum.*0701

*It is another right-hand rule, so that would be into or out of the plane of the page.*0709

*In this case, if I point the fingers of my right hand in the direction of (r), bend them in the direction of (v), my thumb is going to point into the plane of the page or the screen here.*0710

*The direction of the angular momentum vector would be into the plane of the page.*0723

*Its magnitude is given by (mvr) sin(θ) -- mass times velocity times its distance (r) times the sine of the angle between this continued line and velocity -- very, very similar to torque.*0730

*We have two ways to find out our angular momentum.*0745

*Now, total angular momentum -- if we have a bunch of particles -- is just the sum of all the individual angular momenta.*0749

*Let us take a look quickly at a special case here -- what about for an object traveling in a circle?*0755

*Now, if we have some mass traveling in a circle with some velocity at a given point (V) and it is located some radius (r) from the center of the circle -- and let us call that point (C), our reference point...*0762

*...Then the angular momentum about point (C) is going to be (mvr) sin(θ).*0775

*But notice because (v) is always going to be tangent to the circle and (r) is always 90 degrees from that -- sin(θ) is always going to be 90 degrees -- sin 90 degrees is 1. *0787

*So that is just going to be (mvr), but, also remember when we do our translation between linear and angular variables that (v) is equal to = omega(r).*0796

*I can replace (v) with omega(r) so that is (m) omega(r) times another (r) or (r)*^{2}.0811

*If I rewrite that, I could rewrite that as omega (mr)*^{2}, but if you recall for a point particle mr^{2} is the moment of inertia.0819

*(L) about point (C) is equal to omega times (I), or as we wrote it earlier that is I(ω). *0832

*That is where that comes from.*0844

*Angular momentum is equal to rotational inertia or moment of inertia times angular velocity.*0847

*Let us take a look at how we could calculate angular momentum for a couple of particles.*0855

*We are trying to find the angular momentum for a 5 kg point particle located at 2-2 with a velocity of 2 m/s to the East.*0860

*We want to find it about three different points though, so first, let us find it about this point (O).*0868

*The angular momentum about point (O) -- and let us just stick with this magnitude now to make life nice and simple. *0874

*The magnitude of the angular momentum about point (O) is going to be equal to (mvr) sin(θ), where our mass is 5 and our velocity is 2 m/s. *0881

*Our distance from our point -- well if this is 2 and this is 2, the Pythagorean Theorem says right here that our hypotenuse must be 2 square roots of 2.*0895

*The sin of θ -- well that is going to be an angle here of 45 degrees and that is equal to square root of 2 over 2.*0910

*When I do all of this -- 5 × 2 = 10 × 2 = 20 and square root of 2 × and the square root of 2/2 = 1, so I end up with 20 kg-m*^{2}/s. 0918

*Now, let us find it about point (P).*0934

*Angular momentum about point (P) -- same formula (mvr) sin(θ). *0937

*Our mass is still same 5 and our velocity is still 2. *0944

*Now, about point (P) though -- our (r) distance is just 2 units (2) and the sine here is going to be sin 90 degrees which is 1, so 5 × 2 = 10 × 2 = 20 × 1 = 20 -- 20 kg-m*^{2}/s. 0949

*So for the moment of inertia about (O) and about (P), you get the same thing.*0971

*Now, let us do it about point (Q) -- Moment of inertia about point (Q) is going to be (mvr) sin(θ), but in this case, about point (Q), notice our (r) vector and (v) vector are in the same direction -- the angle between them then is 0. *0976

*Since the sine of 0 degrees equals 0, the angular momentum about point (Q) is going to be 0.*0995

*Angular momentum depends on your point of reference. *1005

*Let us take a look at an example with a rotating pedestal.*1012

*Angelina spins on a rotating pedestal with an angular velocity of 8 rad/s. *1016

*Bob throws her an exercise ball which increases her moment of inertia from 2kg-m*^{2} to 2 1/2 kg-m^{2}.1021

*What is Angelina's angular velocity after she catches exercise ball?*1029

*We are going to neglect any external torque from the ball just to keep the problem simple.*1033

*Well, what I do here is realize by conservation of angular momentum -- since she is spinning about her center, her axis of rotation -- we can say that the total angular momentum before she catches the ball must be equal to the total angular momentum after she catches the ball.*1038

*So, (L)initial equals (L)final, but angular momentum is moment of inertia times angular velocity initial, so that must equal moment of inertia final times angular velocity final.*1055

*Well, initial moment of inertia, we know is 2 and final is going to be 2 1/2, so omega must change.*1075

*In this case, (I)initial is 2, (ω)initial is 8, so that must equal (I)final (2.5) times whatever her final angular velocity is.*1082

*16 divided by 2 1/2 -- I am going to come up with an angular velocity of 6.4 rad/s.*1097

*By increasing her rotational inertia, her angular velocity decreases.*1100

*Let us take some example of some rotating discs. *1120

*We have a disc with moment of inertia 1 kg-m*^{2} spinning about an axle through its center.1122

*It has an angular velocity of 10 rad/s. *1130

*An identical disc which is not rotating is slid along the axle until it makes contact with the first disc.*1132

*If the 2 discs then stick together, what is their combined angular velocity?*1139

*Well, I will go back to conservation of angular momentum, which will work because they are rotating about their centers of mass.*1144

*Initial angular momentum equals final angular momentum or initial moment of inertia and initial angular velocity must equal final moment of inertia, final angular velocity.*1150

*I want to know what the final angular velocity is. *1166

*That is going to be equal to I(0), omega(0) over I-final.*1168

*I-initial was 1, omega-initial was 10, and I-final -- well if we double that, it is going to go from 1 kg-m*^{2} to 2 kgm^{2}, so 10/2 = 5 rad/s. 1177

*This should make some amount of intuitive sense -- one objects spinning at 10 rad/s, the other is still, but identical object, and you put them together -- What happens?*1195

*Again you get the twice the mass, twice the rotational inertia, and half the angular velocity.*1204

*Let us talk about angular momentum with respect to heavenly bodies.*1214

*Really what we are talking about here is orbits. *1216

*We want to develop a relationship for the velocity and radius of a planet in an elliptical orbit about any point in that orbit.*1219

*Now, right away when we look at this, we know angular momentum must be conserved because there is no external torque in the system.*1227

*We will go put something like a planet over here, and call up the mass.*1240

*It has some velocity right at that point. *1245

*At this point, it has some (r) vector r(1), we will call that v(1) and there is our mass.*1251

*Another point in time -- say it is down over here -- now it has velocity (2) and it has a different position vector r(2).*1259

*Since the net torque is 0 though, the total angular momentum must be the same.*1273

*The angular momentum about point (S) is going to be -- well when it is at point (1) that it is going to be (m1v1r1) sin(θ)(1) where that angle there is θ(1).*1278

*But that also must be equal to the angular momentum over here at point (2) -- (m2v2r2) sin(θ)(2).*1295

*But the mass is the same. That has not changed.*1307

*We can divide out the mass and then state that (v1r1) sin(θ)(1) must equal (v2r2) sin(θ)(2).*1311

*Our relationship between the velocity is the distance from the Sun and the angle at any point in that orbit.*1328

*Now for the special case, when the planet is at this point, which is known as the apogee point -- so let us call that point (A) or when it is over here at perigee, you can call that point (P).*1335

*Well, at those points we have a special situation, because if you look here -- the velocity and the (r) vectors -- we are going to have an angle of 90 degrees and the same thing over here.*1352

*We have (r) versus velocity and our angle here again is (θ) 90 degrees, so at apogee and perigee, we can simplify this even further.*1365

*The velocity at apogee times the radius of the apogee times the sine of theta at apogee must equal the velocity at perigee times the radius with a position vector at perigee times the sine of θ(P).*1380

*But since these are both 90 degrees and the sine of 90 degrees is 1, we can simplify this to say that the velocity at (A) times the length of the position vector at (A) must equal the velocity at perigee times the position vector at perigee.*1394

*That works when we are at these special points where we have got that 90 degree angle.*1409

*It is a nice relationship between velocity and the position vector and the angle.*1414

*All right, let us talk now about types of kinetic energy.*1422

*We briefly talked about the kinetic energy of an object as the energy an object has due to its state of motion.*1425

*The translational kinetic energy we talked about was 1/2 mv*^{2} -- mass times the square of speed.1432

*Objects traveling with a translational energy must have a translational kinetic energy.*1439

*Similarly again another parallel to rotational motion, objects that are spinning must have a rotational kinetic energy.*1443

*Again as we look here, rotational kinetic energy is 1/2 instead of mass or linear inertia -- we have rotational inertia or moment of inertia.*1452

*Instead of linear velocity squared, we have angular velocity squared.*1462

*Same parallels again just swapping the linear variables for the rotational variables.*1467

*If we wanted to put this all together into a nice table -- displacement in the translational world, we called δ(s) or δ(x) depending on what we were talking about.*1475

*In the angular world, δ(θ) - angular displacement; velocity (v) - angular velocity ω; linear acceleration A - angular acceleration α; and time, the same in both worlds...*1483

*...Force (F) linear, the angular equivalent torque and mass or moment of inertia -- (m) in the translational world is (I) -- rotational inertia in the angular world.*1498

*In our equations, we can expand too -- (S) = r(θ), θ equals (s) over (r).*1509

*We have done this translations between linear and angular quantities before.*1515

*Time is the same, but now Newton's Second Law -- F = ma and torque = I(α).*1521

*For momentum -- linear momentum (P) equals mass times velocity and angular momentum equals moment of inertia times angular velocity.*1527

*And kinetic energy -- kinetic energy is 1/2 mv*^{2} and rotational kinetic energy is 1/2I(ω)^{2}.1535

*Let us put this together to talk about the kinetic energy of a basketball.*1545

*A .62 kg basketball flies through the air with a velocity of 8 m/s. *1549

*Find its translational kinetic energy.*1554

*Well, kinetic energy to the translation linear kinetic energy is 1/2 mv*^{2}, which is going to be 1/2 times our mass (.62 kg) times our velocity (8 m/s^{2}) or 19.84 and the units of energy are joules (J). 1557

*The same basketball -- knowing that its radius is .38 m -- also spins about its axis as it is traveling with an angular velocity of 5 rad/s.*1582

*Let us determine its moment of inertia and its rotational kinetic energy. *1591

*Well, we can model it as a hollow sphere and going back to our table of formulas for moments of inertia, the moment of inertia of a hollow sphere is 2/3 mr*^{2}. 1596

*That is going to be 2/3 times its mass (.62) times its radius (.38*^{2}) or .0597 kg-m^{2}.1609

*Determine its rotational kinetic energy. *1629

*Well, kinetic energy for rotational motion is 1/2 I(ω*^{2}).1630

*For our moment of inertia, we just determined as .0597 and its angular velocity is 5 rad/s, so 5*^{2} -- multiply that out and I come up with 0.75 J.1638

*What is the total kinetic energy of the basketball?*1657

*Well, to get its total kinetic energy, all we are going to do is we are going to combine its translational and its rotational.*1661

*Total kinetic energy is the translational kinetic energy plus the rotational kinetic energy, so that is going to be 19.84 J + 0.75 J or about 20.6 J in total.*1668

*That is kinetic energy of a rotating object that is also moving translationally.*1691

*Let us take a look at a playground roundabout again. *1698

*A roundabout on the playground with a moment of inertia of 100 kg-m*^{2} -- (I) = 100 kg-m^{2} -- starts at rest and is accelerated by a force of 150N at a radius of 1 m from its center.1700

*If the force is applied at an angle of 90 degrees from the line of action for a time of.5 s that equals 90 degrees times half of a second, what is the final rotational velocity of a roundabout? *1723

*Well, as I look here, any time I start seeing final angular velocity in initial, I am starting to think about 'You know probably looking at a kinematics equation.'*1740

*But it would sure be nice to have the angular acceleration. *1749

*Well to do that I probably need to go to Newton's Second Law for Rotation.*1752

*Net torque equals I(alpha), therefore, alpha is going to be equal to our net torque over our moment of inertia. *1757

*I do not have net torque, but I do have force, radius and the angle. *1768

*Our net torque is going to be F(r) sin(θ) over our moment of inertia.*1772

*Now, I can substitute in to find angular acceleration equal to our force -- 150 times our radius (1) times the sine of 90 degrees and that is going to be 1 all over our moment of inertia -- 100.*1779

*So I get 150/100 or 1.5 rad/s*^{2}. 1796

*We want the final rotational velocity of the roundabout though.*1808

*I am going to go back now to my kinematics for rotation and say that final angular velocity is initial angular velocity + alpha angular acceleration times time.*1811

*That is going to be 0 + α -- we just determined is 1.5 rad/s*^{2} times our time of 0.5 s. 1822

*Therefore, our final angular velocity is going to be 1.5 × 1/2 or 0.75 rad/s.*1834

*Let us take a look at another one. *1852

*The ice skater is a famous problem in physics around rotational dynamics and moment of inertia.*1853

*Here without getting into the numbers, we have an ice skater that spins with a specific angular velocity.*1860

*She brings her arms and legs closer to her body reducing her moment of inertia to half of its original value.*1865

*What happens to her angular velocity?*1871

*Well, as the skater pulls her arms and legs in, moment of inertia is going to decrease to the point that it is half of its original value, but angular momentum remains constant.*1873

*As she spins around, her center of mass remains constant. *1887

*Why? There is no external torque in this problem. *1892

*Therefore, angular momentum about the center of mass -- the axis of rotation to the center of mass -- is conserved.*1899

*If (L) equals I(ω) and we are going to cut (I) in half, (L) must remain the same -- it is conserved.*1905

*In that case, omega must double, so if we cut that in half omega doubles.*1917

*All right, that explains what happens to our angular velocity, but what about a rotational kinetic energy?*1925

*Well, for that, let us go to our formula for rotational kinetic energy. It is 1/2 I (ω)*^{2}.1931

*In this case again, we have 1/2 -- (I) became a lot smaller. It got cut in half, but omega doubled.*1941

*Do not forget omega is squared, so if that cut in half and that got doubled -- 1/2 × 2 × 2 -- we are going to double the rotational kinetic energy of the entire system. *1951

*Kinetic energy rotational doubles while angular velocity gets cut in half.*1966

*Wait. Where did that energy come from?*1975

*This rotational energy doubled as she pulled her arms in. *1978

*Well, the skater must have done work to pull her arms in.*1981

*That must have required a force applied for some distance in order to do that. *1984

*That is where we got this extra rotational kinetic energy.*1988

*Let us take a look at the example of a bowler. *1995

*Gina rolls a bowling ball of mass 7 kg -- m = 7 kg and radius (10.9 cm), which is .109 m, down a lane with a velocity of 6 m/s.*1998

*Find the rotational kinetic energy of the bowling ball assuming it does not slip. *2012

*What is its total kinetic energy?*2016

*Well, the first thing I am going to do -- it is a solid bowling bowl, and we will assume the mass is uniformly distributed.*2020

*I am going to find the moment of inertia of the bowling ball by modeling it as a solid sphere.*2025

*Moment of inertia first for a solid sphere is 2/5 mr*^{2}. 2031

*That will be 2/5 times its mass (7) times the square of its radius (.109*^{2}) or about 0.033 kg-m^{2}.2037

*Now, it would be helpful to find its angular velocity and we can do that by recognizing angular velocity as its linear velocity divided by the radius, assuming it is not slipping and we can make that assumption. It does not slip.*2054

*That is going to be our 6 m/s divided by its radius (.109 m) or about 55 rad/s.*2066

*Well, from here let us find its rotational kinetic energy.*2080

*Rotational kinetic energy is 1/2 I(ω*^{2}) or 1/2 × our (I) .033 kg-m^{2}× our angular velocity (55 rad/ms^{2}) or about 50 J.2086

*What about its total kinetic energy?*2111

*Well, kinetic energy -- total is going to be 1/2 mv*^{2} + the rotational, 1/2 I (ω^{2}), which is going to be 1/2 × our mass (7) × the velocity (6^{2}) +... 2114

*Well, 1/2 I(ω*^{2}) -- we all ready said was 50 J, so 1/2 × 7 × 6^{2}, 36 + 50, I come out with about 176 J for its total kinetic energy -- rotational + translational.2135

*Thanks for watching Educator.com. *2154

*Hopefully this gets you started with rotational motion and conservation of angular momentum and putting that all together with rotational dynamics as well.*2156

*Make it a great day. We will see you again.*2164

*Hi everyone. I am Dan Fullerton and I am thrilled to welcome you back to Educator.com.*0000

*Today we are going to talk about work and power.*0004

*Now, our objectives are going to be, first, to define work, to calculate the work done by a force, to utilize Hooke's Law to describe the force that you get from compressing or stretching a spring, recognizing power as the rate at which work is done...*0007

*...and finally, calculating the power supplied for a variety of situations.*0023

*So with that, why don't we dive in and talk about what is work.*0027

*Well, you do work on an object when you move it and the rate at which you do work is your power output.*0031

*When you do work on an object you transfer energy from one object to another.*0038

*That is a key point here -- work transfers energy.*0043

*So work is the process of moving an object by applying a force.*0046

*An object must be moving when you apply a force, therefore you do work.*0051

*Now to give some examples of work -- A stunt man in a jet pack blasts through the atmosphere accelerating to higher and higher speeds.*0056

*We have a force causing an object to move.*0063

*The jet pack is applying a force causing it to move, the hot expanding gases are pushed backwards out of the jet pack, and the reactionary force -- Newton's Third Law of the Gas -- is pushing the jet pack forward causing a displacement.*0066

*You need to have that displacement for work.*0078

*And the expanding exhaust gas, therefore is doing work on the jet pack.*0081

*Let us take another example -- A girl struggles to push her stalled car, but cannot make it move.*0088

*She expends a lot of effort; she is sweating; she is feeling like she is doing a lot of work, but from a physics' perspective, no work is being done since the car is not moving.*0092

*Very different definition -- every day work, compared to the physics' definition of work.*0104

*Another example -- We have a child in a ghost costume at Halloween, carrying a bag of candy across the yard.*0109

*If the child applies a force horizontally upward on the bag, but the bag is moving horizontally, the forces of the child's arms on the bag are not what causes the displacement.*0115

*The force of the child's hands on the bag is up -- the displacement is horizontal, therefore, no work is being done by the child's arms due to the force that is upwards.*0124

*When we want to calculate work quantitatively, we will use the formula -- work is equal to the force times the object's displacement.*0136

*And the units of work are going to be Newton-meters (N-m), force times distance, or joules (J).*0144

*Only the force in the direction of the displacement counts for calculating quantitatively the work.*0154

*When the force and displacement are not in the same direction, you must take the component of the force that is in the direction of the displacement.*0161

*So you could write work as F cos(θ), where θ is the angle between the object's displacement vector and the force vector times that displacement vector, so F cos(θ) times δr or F(δr)cos(θ).*0167

*And of course, if the force and the displacement are in the same direction, θ is 0, cos(θ) is 1, and that term is just going to cancel out -- you will just have F(δr).*0183

*So let us take a look at an example of moving a refrigerator.*0196

*An appliance salesman pushes a refrigerator 2 m across the floor by applying a force of 200N.*0200

*Let us find the work done.*0206

*Well let us start off with our formula -- work equals force times displacement (δr), which is going to be 200N and our displacement is 2 m or 400N-m.*0208

*And as we just discussed 400N-m is also known as a 400 J, so our answer there would be 400 J.*0225

*How about liberating a car? A friend's car is stuck on the ice.*0239

*You push down on the car to provide more friction for the tires by way of increasing the normal force -- remember back from dynamics, the frictional force is μ times the normal force.*0243

*If you push down, you will get more normal force up, which means you are going to get more frictional force.*0254

*However, that allows the car's tires to propel it forward 5 m on the less slippery ground.*0258

*How much work did you do?*0263

*Well, this is kind of a tricky question because the force you are applying is in the downward direction and the car's displacement is horizontal.*0265

*The force is not in the direction of the displacement, therefore no work is done.*0274

*Or if you wanted to do that mathematically -- if there is our force vector -- here we have our displacement vector, δr and the angle between them is 90 degrees.*0285

*So if work is F(δr) cos(θ) -- well since θ equals 90 degrees, and cos(90 degrees) is 0... *0294

*...then you could say that work equals 0 in this instance.*0311

*Let us take a look at another example.*0320

*Let us say that we push a crate up a ramp with a force of 10N.*0322

*Despite our pushing however, the crate slides down the ramp at a distance of 4 m.*0326

*How much work did you do?*0331

*And here is where we are going to have to re-define or maybe clarify that definition of work a little bit.*0332

*Let us draw a ramp to begin with. There is our ramp on here.*0339

*Let us put our crate -- and what is going to happen is despite all of our efforts, it is going to move some displacement of δr = 4 m.*0345

*And as we do this, we are going to apply a force on the box of 10N up the ramp.*0360

*How much work did we do?*0369

*Well let us go back to our mathematical definition -- Work = F(δr) cos(θ).*0371

*Our force is 10N, δr was 4 m and the angle between them -- well if the force is going up the ramp and the displacement is going down the ramp, our angle is going to be 180 degrees.*0383

*Cos(180 degrees) is -1, so we are going to get an answer of 10 × 4 = 40 × -1 or -40 J.*0404

*We have done negative work on the box. What does that mean?*0413

*Well that means that the force was in the opposite direction of the displacement.*0419

*So we are kind of re-defining that initial definition of work or clarifying that definition.*0422

*All right, let us take a look at lifting a box.*0432

*Now we want to find out how much work is done in lifting an 8 kg box from the floor to a height of 2 m above the floor.*0434

*Let us start with our box -- there it is -- We are going to apply some force (F) in order to make it move, a displacement of about 2 m.*0444

*Well what force do we have to apply to lift that box off the ground?*0455

*We have to overcome the force of gravity.*0459

*So the force we apply has to be equal to mass times the acceleration due to gravity here on the surface of the earth (mg).*0461

*The work then is going to be F(δr) cos(θ) or in this case, (F) is mg, so we have mg(δr).*0470

*Force and our displacement are in the same direction, so cos(0 degrees), therefore is 1.*0483

*The cosine term goes away and we just have mg(δr), or this implies then that work is equal to our mass 8 kg times the acceleration due to gravity.*0489

*Let us round that off and say that that is roughly 10 m/s*^{2} times the displacement of 2 m -- 8 × 10 = 80 × 2 = 160 J of work.0500

*Let us take a look at an example now where we are applying a force that is not specifically in the direction of the displacement.*0519

*Barry and Sidney pull a 30 kg wagon with a force of 500N, a distance of 20 m.*0525

*The force acts at an angle of 30 degrees here above the horizontal. Calculate the work done.*0531

*We will go back to our definition again.*0539

*Work equals F(δr) times the cosine of the angle between those vectors (θ), so that is going to be 500N, our applied force, times our displacement (20 meters) cos(30 degrees).*0541

*So we have 500 × 20 × cos(30) is about 8660 J.*0561

*Let us take a look at force vs. displacement graphs.*0574

*The area under a force vs. displacement graph is the work done by the force.*0577

*So if you have a force vs. displacement graph -- if you want to know the work done, just take the area underneath it.*0582

*Let us consider the situation of a block being pulled across a table with a constant force of 5N for a displacement of 5 m -- so that part of the graph -- and then, over the next 5 m, that force tapers off to 0 in a linear fashion.*0589

*Find the work done.*0604

*To do that, all we have to do is take the area of these two sections of our graph.*0605

*Over here in this section, we have a rectangle -- so the area is going to be the base times the height -- 5 m × 5N = 25 J.*0610

*Over here we have a triangle.*0624

*The area of a triangle is one-half base height, or 1/2 × 5 × 5 -- 1/2 of 25 will be 12.5 J for our area here.*0626

*So the total work done is going to be the area of the first part of our graph, 25 J plus the area of the second part of our graph, 12.5 J, therefore, our work must be 37.5 J.*0639

*Let us take a look at work from a varying force with an example.*0660

*A box is wheeled to the right with a varying horizontal force.*0664

*The graph below represents the relationship between the applied force and the distance the box moves.*0667

*What is the total work done in moving the box that displacement of 10 m?*0673

*We have to find the area under the graph in order to find the total work done.*0678

*And there are a lot of different ways we could break this up, but I like to find nice, simple shapes myself.*0682

*So what I would probably do is look at something like -- looks like we have a triangle over here and it should be easy to find the area of that purple triangle.*0687

*Looks like we have another triangle over here. *0697

*If we find the area of that green triangle, then that will just leave us down here a rather long red rectangle and that will give us the entire area under the graph.*0701

*The area over here of this triangle -- 1/2 base height.*0714

*So we have 3 m × 5 = 15 and 1/2 of that will be 7.5 J there.*0718

*Over here in our green triangle, we have a base of 5 m and we have a height of 3N, so 5 × 3 = 15 and we have 7.5J here again.*0725

*And our red rectangle has a height of 1 and a length of 10, so that is going to be 10 J there, so our total work is going to be -- our total area or 10 + 7.5 + 7.5 or 25 J.*0739

*Let us talk a little bit about springs.*0763

*The more you stretch or compress a spring, the greater the force of the spring.*0765

*The more you push on it -- the more you compress it, the harder it pushes back or the more you stretch it, the more it wants to return to its equilibrium or its happy position.*0771

*The spring's force then, is opposite the direction of its displacement from its equilibrium.*0779

*And we can model this as a linear relationship where the force applied by the spring is equal to some constant, which we will call the spring constant -- kind of how strong the spring is...*0783

*...multiplied by the spring's displacement from its equilibrium, or rest or happy position -- whatever you want to call it.*0794

*This is known as Hooke's Law.*0800

*The force on the spring is equal to the opposite of the spring constant (K) -- how strong it is times its displacement and that negative sign just means it is restoring force.*0802

*If you pull it this way, the force wants to go back.*0813

*If you compress it this way, the force wants to push it back to where it started.*0815

*That displacement is always from its equilibrium position.*0819

*The negative tells you it is a restoring force.*0823

*So as an example, if we have a nice spring here -- there it is -- and we will start with an axis and call that distance its happy or equilibrium position -- we will call that x = 0.*0826

*If we then go and we try and stretch our spring out -- now at this point we have an (x) that is significantly greater than 0, so the force of the spring is going to be in the opposite direction.*0841

*There is the negative sign again -- why that is a restoring force.*0855

*So how do you find the spring constant of a spring?*0862

*Well the easiest way is probably to look it up on the box when you go buy a spring. *0865

*But assuming you do not have the box anymore, make a graph of the force required to stretch the spring against its displacement from its equilibrium position.*0871

*This is not the length of the spring here, this is how far you have stretched it from its happy position.*0881

*So force vs. displacement graph -- the slope is going to give you the spring constant (K) in newtons per meter (N/m).*0886

*So slope, which is rise over run -- for something like this, let us pick a couple of points.*0892

*These are easy points to pick, so let us pick that point right there, and that point right there.*0902

*So the rise is going to be going from 0 to 20N and that will be 20N and the run, we will go from 0 to 0.1 meters, so over 0.1 meters will give us a spring constant of 200N/m.*0905

*The bigger that (K) value, the stronger the spring.*0921

*So let us take a look at the work done in compressing a spring.*0928

*Here we have a force vs. displacement graph.*0931

*If we want the work done in compressing the spring -- well notice a force vs. displacement graph -- we have here an area.*0934

*The area under the force vs. displacement graph, still works; it still gives us the total work done.*0942

*So in this case, our work is going to be the area of our triangle -- nice big triangle there.*0947

*Work done will be 1/2 base times height or 1/2 × 0.1 m × the height (20N), or 2 × 1/2 = 1 J.*0957

*So let us do an example where we are finding the spring constant.*0978

*A spring is subjected to a varying force and its elongation is measured.*0985

*Determine the spring constant of the spring.*0990

*We have a bunch of points here to plot so let us start with that.*0993

*We have (0,0), we have an elongation of 0.3 with the force of 1, so somewhere right around there.*0997

*We have 0.67 with a force of 3 so that will be somewhere right around here -- make another point.*1006

*At 1 m, we have a force of 4N, at 1.3 m we have a force of about 5N and finally at about 1.5 m we have a force of 6N.*1015

*So the first thing I will do is use a straight edge to draw a best fit line here -- something like that -- use a straight edge yourself.*1031

*And when we do that now we have to find the slope of that line.*1040

*What we will do is pick a couple of points that are on that line and let us say we have a point right there -- is an exact point on the line and (0,0) is there, so that will make it pretty easy.*1044

*Our slope is our rise over our run or 6N/1.5 m or 4N/m.*1056

*Pretty easy to find the spring constant, just by taking our graph.*1074

*Let us take another example where we are calculating the spring constant again.*1078

*We have a 10N force -- F = 10N -- compressing a spring 0.25 m from its equilibrium position, So x = 0.25m.*1082

*Find the spring constant (K).*1092

*We will start off by writing Hooke's Law, and let us just worry about the magnitude for now.*1095

*We will worry about direction later.*1100

*So F = Kx, therefore the spring constant (K) is F/x or 10N/0.25 m for a spring constant of 40N/m.*1102

*Let us talk about power for a couple of minutes.*1122

*If work is the process of moving an object by applying a force, power is the rate at which that force does work.*1126

*Power is the rate at which work is done.*1132

*The units of power are joules per second (J/s), which we also call a watt (W).*1134

*Now you have to be careful using watts as your units because 'work' is capital W, and now we have the unit watts as capital W.*1140

*So you have to be careful and understand what you are doing when we write these.*1148

*Our formula for power is going to be work over time, the rate at which work is done.*1152

*And since power is the rate at which work is done, it is possible to have the same amount of work done but with a different supplied power if it has two different time intervals.*1157

*For example, Robin Pete move a sofa 3 m across the floor by applying a combined force of 200N horizontally.*1167

*If it takes them 6 s to move the sofa, what amount of power did they supply?*1174

*Well the power supplied is going to be the work done divided by the time it took, which is going to be F(δr) -- the displacement and force are in the same direction, so we do not have to worry about that cos(θ) term -- divided by t.*1180

*So we have 200N as our force, they moved it 3 meters, our displacement in a time of 6 s, so 200 × 3/6 is just going to be 100 J/s or 100 W.*1193

*At the same time though Kevin pushes another sofa 3 m across the floor by applying a force of 200N.*1209

*Kevin, however, takes 12 s to push the sofa.*1215

*What amount of power did Kevin supply?*1218

*Well the same formula -- Power will be F(δr)/T, which is going to 200 × 3/12 s this time or 50 W.*1221

*So same amount of work done, took him twice as much time so he had half the power output.*1234

*When we are calculating power, there are a couple of different ways we can do this.*1242

*We already talked about power as being the work done divided by the time, but that is also F(δr) cos(θ) divided by time.*1246

*But take a look, we have δr over (t) here, displacement over time.*1260

*That looks like velocity.*1266

*Velocity is δr/t, so we could rewrite this as power is equal to force times velocity times the cosine of the angle between those.*1269

*Another version of that same formula, another way to write it, another way to calculate it.*1282

*Let us take a look at that with an example.*1286

*Motor A lifts a 5,000N steel crossbar upward at a constant velocity of 2 m/s.*1291

*Motor B lifts a 4,000N steel support upward at a constant 3 m/s.*1297

*Which motor supplies more power? Let us figure out the power from each one.*1301

*The power from motor A is going to be the force times velocity, no cosine-theta term needed because they are in the same direction again.*1307

*That is 5,000N, our force, times our velocity of 2 m/s or 10,000 W, which we could write as 10 kilowatts (kW).*1315

*The power for motor B on the other hand, we calculate the same way, but now we have a force of 4,000N and we are doing this at a velocity of 3 m/s for 12,000 W or 12 kW.*1329

*Which motor supplies more power? Well obviously it must be motor B.*1350

*Let us take a look at an example with a cyclist.*1361

*A 70-kg cyclist develops 210 W of power while pedalling at a constant velocity of 7 m/s East.*1363

*What average force is exerted eastward on the bicycle to maintain this constant speed?*1371

*Let us start with our givens.*1376

*We know the mass is 70 kg; we know that the power is 210 W; our velocity is 7 m/s in eastward direction and we are trying to find an average force.*1378

*Power is force times velocity, therefore if we want just the force we will rearrange this as power over velocity or 210 W divided by 7 m/s.*1403

*It should give us a force of 30N, and of course, that is going to be in the eastward direction as well if we make that a vector and we are going to track our direction -- 30N East.*1419

*Let us take a look at work on a spinning mass.*1433

*A 5 kg ball is spun by a chain in a horizontal circle of radius 2 m at a speed of 3 m/s.*1436

*So a horizontal circle being spun pretty quickly. What is the work done on the ball by the chain?*1443

*First thing, let us draw a graph of this, let us draw a diagram.*1449

*If we look at it from the top, our horizontal circle, that is my best attempt at a circle.*1452

*Any point in time -- there is our object -- it has some velocity tangent to the circle.*1458

*The force is always toward the center of the circle because it is a centripetal force.*1463

*The force is always perpendicular to the displacement in the velocity.*1469

*Because of that, no work is done on the ball by the chain.*1475

*You cannot do any work because the force is toward the center of the circle.*1479

*The velocity, the displacement, at any instantaneous point in time is always 90 degrees from that, it is always perpendicular.*1482

*So you cannot do any work on that spinning mass, not by that force.*1488

*That force is changing its direction, keeping it moving in a circle, but it is not doing any work on the object; it is not causing that displacement.*1494

*Let us take a look at one where we are talking about work done by friction now as we again explore that definition of work in the force having to cause that displacement and how we are just going to massage that a little bit.*1505

*We have an 80 kg wooden box pulled 10 m horizontally across a wood floor at a constant velocity by a 250N force at an angle of 37 degrees above the horizontal.*1519

*If the coefficient of friction between the floor and the box is 0.3, find the work done by friction.*1531

*Wow! There is a lot there. Let us start by exploring this problem a little bit more.*1537

*First of all we know it is being pulled at a constant velocity.*1542

*The moment I see that, right away I think, 'You know we must have 0 acceleration.'*1546

*The net force must be 0 by Newton's Second Law.*1553

*Let us draw our box here -- it is an 80 kg box with a 250N force that is being applied in an angle of 37 degrees above the horizontal.*1560

*It is going to be pulled at a displacement of 10 m, and we know the coefficient of friction between the box and the floor is 0.3.*1578

*Let us start off with a free body diagram (FBD) here because we have a lot going on.*1594

*There is my box. I must have its weight (mg) down, a normal force opposing that.*1597

*Our applied force of 250N at an angle of 37 degrees, and we must have our frictional force opposing that motion, Ff.*1606

*Now I am going to make my pseudo free body diagram (P-FBD) and get all my forces to line up with an axis.*1618

*So I am going to break this 250N up into components and when I do that, we will have (mg) down still.*1624

*I still have my normal force up -- 250 times the sine of 37 to give me the vertical component is going to give me 150N up.*1633

*Its horizontal component 250(cos37) is going to be 200N and I have the frictional force opposing that.*1645

*Let us write Newton's Second Law in the (x) direction.*1655

*Net force in the (x) direction and I look at my P-FBD.*1660

*I have 200N to the right, minus the frictional force to the left and that must all be equal to 0 because the acceleration is 0; it is moving at a constant velocity, therefore, the frictional force must be 200N.*1665

*The work done by friction then must be that frictional force times the displacement.*1680

*Frictional force is going to be opposite in direction to the displacement, so I could write that as 200N, displacement (10 m), but I have to bring in my cosine-theta term -- Cos(180 degrees) which will be -1 or -2000 J of work done by friction.*1686

*Why a negative? Because the box's displacement is in one direction, the force of friction is in the opposite direction.*1709

*Let us explore the units of power a little bit. *1718

*Determine the unit of power in terms of fundamental units -- Kilograms (kg), meters (m), and seconds (s).*1720

*Let us start by using our definition of power.*1725

*Power is work over time, which is going to be force times displacement divided by time.*1728

*And force, by the way, Newton's Second Law, is mass times acceleration. *1736

*So we have broken this down into some more detailed -- a different definition based on fundamental units, let us find the units of these.*1743

*Units of mass are kilograms, acceleration is meters per second squared (m/s*^{2}), displacement will be meters -- we will have a squared there -- and time, well, we have seconds down here again.1752

*So our total unit must be kg × m *^{2}/s^{3} and that all must be equal to a watt.1766

*What are the units for power? Watt.*1778

*One last example problem -- the frictional force on a sled.*1782

*Bob supplies 2000 W of power, P = 2000 W, in pushing a heavy sled across a frozen lake at a constant speed of 2 m/s.*1787

*So, constant speed right away, I think, acceleration = 0, it must be at equilibrium and that speed is 2 m/s.*1796

*Find the frictional force acting on the sled.*1803

*Let us take a look at the FBD for this case.*1806

*We have the force of Bob acting in one direction.*1810

*We have the frictional force acting in the opposite direction.*1815

*We know they must balance out because it is a constant speed of 2 m/s and of course we must have also here the normal force and the gravitational force, the object's weight.*1819

*Because it is moving at a constant speed, the force of Bob must be equal to the frictional force, since the acceleration up here is 0.*1832

*And if power is force times velocity, then we could say that the force of Bob must be equal to the power over the velocity, or 2000 W over 2 m/s, which is going to give us 1000N as the force of Bob.*1841

*And since the force of Bob equals the frictional force, we could then say that the frictional force is also equal to 1000N -- it is just in the opposite direction.*1861

*Hopefully that gets you a good start on work and power.*1872

*Thanks for watching Educator.com. We will talk to you soon. Make it a great day.*1876

*Hi folks! I am Dan Fullerton and I am thrilled to welcome you back to Educator.com.*0000

*Today's lesson is on energy.*0004

*Our objectives or goals are going to be to calculate the kinetic energy of a moving object, to calculate the gravitational potential energy of a system, and to analyze the relationship between the work done on or by a system and the energy gained or lost by that system.*0007

*So with that let us dive right in.*0021

*What is energy? From a physics' perspective, energy is the ability or capacity to do work, but if you remember, work is the process of moving an object.*0024

*So energy is really the ability or capacity to move an object.*0034

*Now there are lots of types of energy and a bunch of different ways we can break these up, but just as a starting point and a very, very, very strong oversimplification, we are going to say that all energy is broken up into either potential or kinetic...*0039

*...where potential energy is energy due to condition or position -- things like gravitational energy, where the amount of energy you have depends on how far you are from another mass.*0054

*Chemical energy is having to do with the energy involved in your bonds.*0065

*Elastic energy is energy due to how much a spring or something like a spring is compressed or stretched.*0069

*Electrical energy is where we have to deal with electric potential difference and we will get to that in a little bit, that is a lot of fun.*0076

*Nuclear potential energy... *0082

*Kinetic energy on the other hand, is energy of motion, having to do with the movement or velocity of an object.*0084

*Electrical energy -- you will notice -- is in both places because over here we have moving electrical charges.*0092

*Light -- and this is a big oversimplification -- moving photons; wind -- moving air molecules; thermal...*0098

*Thermal energy is the energy caused by the vibration of the molecules or atoms making up an object, or sound, again vibrating air molecules.*0106

*So, types of energy -- we are going to break down into these two main types, potential or kinetic.*0115

*Energy of condition or position over here on the potential side, and energy of motion over here on the kinetic side.*0120

*Energy can be transformed from one type to another, and you can transfer energy from one object to another.*0128

*The way you transfer energy from one object to another is by doing work.*0134

*So the Work-Energy Theorem is going to be a big part of the course.*0139

*The work done on a system by an external force changes the energy of a system.*0143

*If I take the pen -- if I apply a force on it, if I do work on the pen, I am going to give it energy.*0149

*If I do work sideways on the pen, I am going to give it kinetic energy; it is going to go flying that way.*0155

*If I do work on the pen in this direction, I am changing its gravitational potential energy, but when I do work on an object, I transfer energy.*0160

*That is the Work-Energy Theorem in a nutshell.*0168

*The units of energy are the same as the units of work.*0172

*They must be because if I do work on an object, I am giving it energy.*0174

*They are two sides of the same coin.*0179

*The units are going to be joules (J), therefore, where a joule is a newton-meter (N-m), or 1 kg-m*^{2}/s^{2}.0181

*Let us start with kinetic energy as we dive into these in more detail.*0189

*Kinetic energy is energy of motion.*0193

*If energy is the ability or capacity to move an object, kinetic energy is the ability or capacity of a moving object to move another object.*0196

*Say a baseball is coming toward your nose at a 100 mph. It has a lot of kinetic energy.*0207

*It has a lot of kinetic energy because when that baseball hits your nose, it has a lot of ability to move another object, namely your nose.*0212

*It might not be pleasant, but it is going to transfer energy; it is going to do work on your nose and cause motion there.*0220

*The ability or capacity of a moving object to move another object is kinetic energy.*0226

*Translational kinetic energy -- we can quantify as 1/2 times the mass of the object times the square of the velocity.*0231

*Larger objects have more kinetic energy and faster moving objects have more kinetic energy.*0239

*If you are standing out on the road, do you want to get hit by a mosquito coming at you at 1 mph or 1 m/s, or do you want to get hit by a Mack Truck coming at you at 100 km/h, 60 mph.*0245

*Well, unless you have a death wish, you are probably after the mosquito, a lot less unpleasant.*0258

*It has a smaller mass, a smaller velocity and therefore a smaller kinetic energy.*0263

*A smaller ability to do work on you, a smaller ability to move you.*0267

*In the rotational world -- and we talked a bit about this briefly already -- the kinetic energy is 1/2 times the moment of inertia, times the square of the angular velocity.*0272

*And we have talked already about how (m) and the (i) were equivalents from the translational to rotational world and velocity and angular velocity are equivalents there as well.*0281

*Now a single object all by itself isolated and lonely can only have kinetic energy.*0292

*Potential energy requires an interaction between objects.*0298

*You need at least two objects to have any type of potential energy.*0302

*Let us take a look at the kinetic energy of a motorcycle with a simple example.*0309

*A frog speeds along on its motorcycle -- a frog-sized motorcycle of course -- at a constant speed of 30 m/s.*0312

*If the mass of the frog and motorcycle is 5 kg, find the kinetic energy of the frog-motorcycle system.*0320

*Kinetic energy, KE or K, is 1/2mv*^{2}, so that is 1/2 times the mass (5 kg), times the square of the speed (30^{2}) -- 900 × 5 = 4500 × 1/2 = 2,250 and again our units of energy and work, joules (J).0326

*Now potential energy is often times written as (PE) or sometimes you will see it abbreviated with a capital (U) and the AP tends to prefer the capital (U).*0360

*That is an energy an object possesses due to its position or condition.*0370

*Potential energy exists within a system if the objects in that system interact with conservative forces.*0374

*Gravitational potential energy (Ug) is the energy an object possesses because of its position in a gravitational field.*0381

*The pen right here, has some amount of potential energy because if I let go of it due to its position, it is going to accelerate downward, and as it does that, that potential energy is going to become kinetic energy as it goes faster and faster and faster until it will eventually hits you...*0390

*...it will probably make a little of noise, create just a little bit of heat and we are going to convert that energy into other types.*0405

*Elastic potential energy, on the other hand, (US), typically for spring, is the energy an object possesses due to its condition of being compressed or stretched.*0411

*We take a spring, we do work on it to compress it, we compress it more and more and more and more and more, and now it has a bunch of energy.*0421

*I know that because if I let go of it, it is going to have a tremendous ability to move another object, to do work on something else.*0427

*If I let go of it, it goes flinging off to the side.*0434

*It has a lot of elastic potential energy.*0436

*As we talk about gravitational potential energy -- In a constant gravitational field we have to worry mostly about relative changes in gravitational potential energy.*0442

*What we are going to call 0 energy is really just an arbitrary point.*0453

*If I were to drop the pen onto the table here, I would worry about this distance.*0457

*I would call the table an energy level of 0 and this some other energy level.*0463

*On the other hand, if I were to drop it off the edge of the table, I could call the top of the table some energy level and I could call the ground 0.*0468

*It is all arbitrary where you set 0 so we are going to worry about differences in gravitational potential energy, δUg, where that is going to be mass times the acceleration due to gravity times the height difference from your high point to the point you are calling 0.*0476

*Or you could write that if you would prefer as mg(δh) and that works if you are in a constant gravitational field.*0492

*More universally, however, things get a little bit more complicated where the potential energy due to gravity is minus the gravitational constant times the first mass times the second mass divided by (r).*0500

*What is that negative sign about?*0513

*When we are talking about the universal calculation of gravitational potential energy, we really need some reference point to call 0, to measure all things against.*0516

*And what we are going to do to try and find a 0-point that makes sense is we are going to say imagine you have an object infinitely far away from all other objects, so far away that there are no other forces that interact with it -- infinitely far away.*0526

*That is what we have to call 0.*0540

*Now if we bring that object, say it is way, way, way out there in space -- here is Earth.*0542

*We bring that object closer and closer and closer and closer and closer to Earth. *0547

*As we do that, its gravitational potential energy must be changing because now it wants to go toward the earth.*0551

*If we add 0 a long, long, ways away, now it is kind of captured by Earth that wants to go toward it.*0558

*So the gravitational potential energy that we have talked about while it is here on Earth -- the universal gravitational potential energy is negative because it is captured by Earth's gravitational field.*0564

*It has a negative energy of -1,000,000 J, that means if we did 1,000,000 J of work on it, we could completely free it from Earth's gravitational field and get it back out to infinity.*0576

*That negative sign just has to do with that reference point that we set way out there at an infinite distance away where no other objects interact with it.*0589

*If we want to calculate elastic potential energy -- for a linear spring -- something that obeys Hooke's Law, the potential energy in a spring is 1/2 times that spring constant times the square of the displacement from the equilibrium position or the displacement from its happy position.*0598

*Internal energy of a system includes the kinetic energy of the objects that make up the system and the potential energy of the configuration of the objects that make up the system.*0617

*If we were looking here at just the pen, and saying just looking at the pen, right here as it sits right now, its internal energy... *0628

*...We could characterize by taking and looking at the average kinetic energy of all of the molecules making up this pen as they vibrate if we looked at it with a really, really, good, good, good, good, tremendously amazing microscope.*0634

*Now a change in a system's internal structure can result in changes in internal energy.*0650

*And we will see how that works out as we go through a couple of examples.*0655

*If we want to calculate gravitational potential energy in a constant gravitational field, let us set a 10 kg box on the floor -- Floor, box (10 kg).*0659

*And what we are going to do here is we are going to set its current position to ground level as a reference point as 0.*0672

*So we are going to call that while it is on the ground, its gravitational potential energy right there is 0.*0679

*We have set an arbitrary 0 that is going to make sense to us.*0686

*Now if we want to come over and we want to do something with our box -- if we want to do something like take our 10 kg box and we want to bring it up there somewhere, some height difference (h) from the ground, well to do that, we have to do work on it to lift it up there, right?*0690

*The work that we do on it has to be the force times that displacement or in this case, the force that we have to overcome to lift it is its weight, the force of gravity on it.*0709

*So that is the gravitational force and its displacement is going to be (h).*0722

*Well the force of gravity -- if we are in a constant gravitational field -- the weight we can write as (mg), so that is going to be (mgh).*0728

*Therefore, the potential energy that we have given this must be (mgh) -- gravitational potential energy in lifting that up -- its mass times the acceleration due to gravity here on the surface of the Earth, we can round that to 10 m/s*^{2} times the height which we have raised it.0739

*Now, an important point -- the source of all energy on Earth is the conversion of mass into energy.*0762

*Ultimately that is where it all comes from.*0768

*Think of where you get your energy. *0771

*The gas in your car. Where did the gas come from? Well, refineries from oil in the ground.*0772

*Where did the oil come from? Critters, plants, long, dead compressed under lots of pressure for a long, long, long time. *0779

*Where do they get their energy? Well eating other things.*0786

*The sun -- Where did the sun get its energy?*0789

*The sun is a giant nuclear reaction; it is a conversion of mass into energy.*0791

*So a lot of energy from the sun, anything from the sun is conversion of mass into energy.*0797

*What is our other source of energy?*0802

*Well straight up nuclear energy which is a conversion of mass into energy.*0803

*So the source of all energy on Earth is the conversion of mass into energy. *0808

*Mass and energy are intimately related.*0812

*You are going to explore that in more detail later on toward the end of the course as well.*0815

*So let us take a look now at another example where we look at potential energy.*0822

*The diagram here represents a 155N box on a ramp.*0825

*An applied force (F) causes the box to slide from point (A) to point (B).*0830

*What is the total amount of gravitational potential energy gained by the box?*0835

*Well right away this could be an intimidating problem until we think about what we really need to do in order to find its change in potential energy.*0840

*The change in potential energy here is just going to be (mgh), or δh if you prefer.*0848

*Mass and mg, its weight, is just a 155N, and the height it is raised to 1.8 m -- 155 × 1.8 is about 279 J.*0857

*Energy of a system -- Which situation describes a system with decreasing gravitational potential energy?*0882

*A girl stretching a horizontal spring -- as you stretch that spring you are going to be giving the spring more energy because when you let go it is going flinging back that way.*0889

*You are giving it the ability to cause motion. It cannot be that.*0898

*Two -- a bicyclist riding up a steep hill.*0902

*A bicyclist is doing a lot of work going up the hill, up the hill, gaining the gravitational potential energy. It cannot be that one. *0906

*A rocket rising vertically from Earth goes up and up and up and up -- changes its height and the height gets bigger and bigger and bigger; it is gaining gravitational potential energy.*0914

*But a boy jumping down from a tree limb is converting gravitational potential energy into kinetic.*0922

*His gravitational potential energy is being used up, so our correct answer must be 4.*0927

*A hippopotamus is thrown vertically upward. Do not ask me why.*0937

*Which pair of graphs best represents the hippo's kinetic energy and gravitational potential energy as functions of its displacement while it rises? Key -- while it rises.*0942

*Well let us think about that.*0954

*We have our hippo -- something, maybe a superhero, throws the hippo up, hippo goes upward.*0955

*As the hippo is going up, it starts off with a lot of velocity so it must have a lot of kinetic energy initially, and as it goes higher and higher and higher, it slows down, slows down, slows down, slows down -- stops.*0966

*Had a lot of kinetic energy here, now it has gravitational potential energy.*0976

*So kinetic energy was high here, very low, 0 here.*0982

*Gravitational potential energy was low here on the ground, but it was high up here.*0986

*Which graph shows us that? Must be number 1.*0991

*Start off with a lot of kinetic energy and as you have more and more and more displacement, you get less kinetic energy.*0995

*Start off with very little potential energy and as you add more and more displacement, you convert to potential energy.*1001

*And the key here being this is only looking at while the hippo is rising.*1007

*Let us take a look at one last example.*1014

*A pendulum of mass (m) swings on a light string of length (L).*1017

*If the swing hanging directly down is set as the 0-point of gravitational potential energy or I should say if the pendulum hanging directly down, is the 0-point of gravitational potential energy, find the gravitational potential energy of the pendulum as a function of θ and (L).*1021

*This is going to require a little bit more thinking here I believe. *1039

*Well, let us think about what is going on here. *1043

*At the highest point here, it has gravitational potential energy. *1045

*As it swings down, it all becomes kinetic energy. *1051

*We are trying to find the gravitational potential energy of the pendulum in terms of θ and (L).*1056

*Well, we know the change in gravitational potential energy is going to be (mgh). *1061

*The trick then is going to be finding out what this (h) is as it moves from its lowest point to its highest point. *1069

*To do that we are going to have to analyze this with a little bit of Geometry and Trig. *1077

*The first thing I do here is I take a look and notice -- this is (L), then this length, also must be (L). *1081

*If I draw a straight line over to the center of the ball on my pendulum though, this length now is no longer (L), it has gotten a little bit shorter, so (h) is the difference between (L) and this red line. *1089

*If we could find the length of that red line, which is the adjacent side of this right triangle, we would be golden. *1108

*So let us see if we cannot do that. *1115

*Theta is there and we know the co-sine -- we want to know the adjacent side -- we know the hypotenuse.*1118

*Cos(θ) is the adjacent side divided by the hypotenuse (L), therefore, the adjacent side must be equal to (L)cos(θ), so this is (L) cos(θ). *1125

*What we really want to know is this distance here, that (h). *1146

*(H) then must be the total length of our pendulum (L) minus this (L) cos(θ) or I could factor out the (L)'s and say that that is (L) times 1 minus the cos(θ)*1151

*Now when I go to put that back into my formula here for potential gravitational potential energy, UG = mg, and instead of (h) here I am going to put (L) times 1 minus the cos(θ). *1166

*So there is the gravitational potential energy of our pendulum when it is over here at this point. *1186

*Its weight times the length of the pendulum times 1 minus the cos(θ), which really is just giving you the height difference from its lowest point to its highest point. *1195

*Hopefully, that gets you a good start on energy. *1205

*We will talk more about that in our next presentation on conservation of energy. *1209

*Thanks so much for your time and make it a great day!*1212

*Hi everyone! I am Dan Fullerton and I would like to welcome you back to Educator.com*0000

*Today we are going to talk about conservation of energy.*0003

*Our goals and objectives for this unit are to recognize situations in which total energy and mechanical energy are conserved and to apply conservation of energy -- to analyze energy transitions and transformations in a system.*0009

*So the Law of Conservation of Energy -- one of the big ideas in Physics.*0021

*Energy cannot be created or destroyed; it can only be changed to different forms.*0026

*So mechanical energy is kinetic plus gravitational potential energy, plus spring potential energy.*0032

*And we also have conservation laws for total energy and if there is no friction, conservation of mechanical energy.*0039

*Let us see how these work.*0046

*Let us assume as we talk about conservation of energy that we have a jet fighter with a mass of 20,000 kg, and it is coasting through the sky at an altitude of 10,000 m with a velocity of 250 m/s.*0050

*Let us try and figure out what its total energy is.*0061

*Well its total mechanical energy (Etotal) is going to be its gravitational potential energy because it is so high up, plus the kinetic energy it has due to its velocity, so that is going to be mgh + 1/2 mv*^{2}.0064

*Now its mass is 20,000 kg -- (g) we are going to estimate at about 10 m/s*^{2} with an altitude or height of 10,000 m plus 1/2 times its mass (20,000 kg) times the square of its velocity (250^{2}).0083

*Or that implies then that its total mechanical energy will be about 2.63 times 10*^{9} J.0102

*Now that jet is going to dive to an altitude of 2,000 m -- it is going to go from 10,000 to 2,000 m. *0117

*Find the new velocity of the jet -- and we are going to assume that we are not losing any energy to friction.*0124

*We are not running the engines at the moment so we are not gaining any energy or converting any other types of energy -- just a simple first pass calculation.*0133

*So now for our total energy, we are going to follow the same formula -- gravitational potential energy plus kinetic, which is still (mgh) + 1/2 mv*^{2}.0141

*That is going to be -- then as we solve for this -- let us see if we can find out its new velocity; we will get velocity all by itself.*0153

*I could say that mv*^{2} then -- if I multiply both sides by two and rearrange this a little bit -- mv^{2} is going to be 2 × total energy - (mgh).0161

*Multiply all of this by 2 and then subtract the (mgh) from one side.*0174

*Now then if I want just the velocity -- velocity is going to be 2 × total energy - (mgh), all divided by that mass, and I need to take the square root of that.*0180

*When I substitute in my values -- that is 2 times our total energy, which was 2.63 × 10*^{9} J...0197

*...We have a little bit more to put in there -- × 10*^{9} - mgh (20,000) × g (10) × our new height (2000) all divided by the mass (20,000 kg) and we need the square root of all of that.0207

*Therefore, our velocity must be -- when I plug all of that into a calculator -- not doing that one in my head -- it comes out to be about 472 m/s.*0230

*So it was going 250 m/s -- it dove -- it converted some of that gravitational potential energy into kinetic energy and therefore increased its speed.*0243

*And that is why oftentimes when you are talking about aerial combat, a saying among pilots is that altitude is life; altitude is energy. It really is.*0252

*That is what allows them to convert very quickly that altitude into velocity, which is so important for winning dog fight scenario battles.*0262

*Let us take a look at how we can analyze the motion of an object from an energy approach and then from a kinematics approach -- two different ways of solving the same sort of problem -- things we have been doing.*0272

*Let us drop an object -- any object from a height of 10 m and see if we can find its velocity right before it hits the ground, that split second before it contacts the ground.*0284

*If we start with an energy approach, we know that the energy at the top of its path -- when it is at 10 m -- must equal the energy at its bottom by conservation of energy.*0296

*When it is at the top -- that is gravitational potential energy, and at the bottom it is all converted into kinetic energy.*0307

*Therefore, we could write that (mg) times the height -- when it is at its highest point -- must equal 1/2 mv*^{2} at its lowest point.0315

*If I multiply both sides by 2 here and -- this is nice -- I can divide out the (m)'s on both sides and then I will get 2gh = V*^{2} or V = square root of 2gh.0323

*And if I substitute in my values -- V = 2 × -- let us assume (g) is roughly 10, our height (10), and the square root of 200 is going to be about 14.1 m/s.*0343

*Let us do that from a kinematics approach, back from some of our earlier lessons.*0363

*If we drop an object from a height of 10 m -- well if we do that, let us look at the vertical analysis.*0367

*Our initial velocity -- V0 must be 0 m/s and V final is what we are looking for, so δy is going to be 10 m, which we will call down our positive direction.*0373

*Our acceleration is going to be about 10 m/s*^{2} down and the time -- we do not know either.0387

*For solving for final velocity, we will use one of our kinematic equations, and the one that is probably most helpful right now will be to write that Vf*^{2} or V^{2} = V0^{2} + 2A(δy).0395

*Or again, this nice little trick -- our initial velocity is 0 so that term goes away -- V*^{2} = 2A(δy) or V = the square root of 2A(δy).0412

*But let us look at this for a second -- (a) is our acceleration which is the acceleration due to gravity.*0428

*So we could write then that (a) = (g), and δy is the change in height -- so δy = h.*0436

*I can then rewrite this equation as V = the square root of 2, and instead of (a) here, I am going to write (g), and instead of δy, I am going to write (h).*0445

*That should look mighty familiar -- V = the square root of 2(gh).*0456

*And of course when I plug in my values again -- V = square root of 2 × 10 × 10 -- I will get the same numeric answer as well, 14.1 m/s.*0465

*Two different ways of solving the same problem -- one with the conservation of energy approach, one with your traditional kinematics approach.*0477

*Let us take a look at a problem of a pendulum.*0488

*A pendulum comprised of a light string -- meaning we are going to ignore its mass, it is an ideal pendulum -- of length (L) swings mass (m) back and forth.*0492

*So this must be our mass (m), it is going to go back and forth and it has some length (L) -- same length here.*0500

*And as it does this -- at its highest point here, it has gravitational potential energy.*0508

*At its lowest point, it must have kinetic energy -- that is where it is going the fastest.*0514

*Then it is going to convert that kinetic energy back into potential energy and for a split second it is going to stop here with no kinetic energy -- all potential -- and back and forth, and back and forth.*0520

*So if I were to make a graph of what this would look like from an energy perspective -- if we put energy on this axis vs. (x) displacement here -- well when it is in its middle position all of its energy is kinetic.*0530

*So right there let us put some amount of kinetic energy, and kinetic energy we will make it green.*0545

*At its farthest (x) displacement, its kinetic energy must be 0 -- over here and over here -- corresponding to these points where it is no longer moving.*0550

*So our graph of kinetic energy is probably going to look something like this upside down U shape.*0560

*It should be pretty symmetric -- should be perfectly symmetric -- my art skills are a little off.*0569

*On the other hand, if we wanted to take a look at potential energy, we know that at these points at its maximum displacement, all of its energy is potential.*0573

*So then what I am going to do is at that same point I am going to put potential energy (PE) or (U) over here, which at its lowest point here, all of its energy is kinetic -- none of it is potential.*0585

*So we have 0, and back over here, when it is at its highest point on the left, again, all of the energy is potential and kinetic is 0 again.*0600

*Over here we get an alternate curve that looks kind of like this.*0608

*And what we are really doing is as the pendulum swings back and forth is trading off gravitational potential energy and kinetic energy.*0613

*The entire time though, assuming we do not have any non-conservative forces, we do not have any friction -- we are going to have a constant total energy.*0621

*It is going to remain the same the entire time, just different amounts, and different divisions of gravitational potential energy vs. kinetic energy.*0632

*We could look in a little bit more detail over here as well -- so all potential energy here, potential energy here, and kinetic here.*0643

*And we have already done this derivation once, but the potential energy due to that height difference (h) -- well we determine that this adjacent side was L cos(θ) here on our string...*0650

*...If that entire length there is (L), then (h) is equal to (L) - L cos(θ), which is (L) times the quantity -- 1 - cos(θ).*0665

*Its maximum potential energy is going to be (mgh) or (mg) and (h) is (L) × the quantity -- 1 - cos(θ).*0678

*Its kinetic energy -- max on the other hand -- is 1/2 mv*^{2}, which also must be equal to the potential energy maximum because of conservation of energy, so that also has to be equal to (mgL) × 1 - cos(θ).0695

*So we could solve to find the maximum velocity of our ideal pendulum here by solving this equation for velocity.*0716

*And hey! We are here. Why not do it?*0725

*I could write then that if this is the maximum velocity -- what I am going to do is rearrange this a little bit... *0728

*...I multiply both sides by 2, we get this nice little -- divide (m) out of both sides, so I can write that as 1/2 v*^{2} = (gl) × 1 - cos(θ), multiplying both sides by 2, then v^{2} = 2gl × 1 - cos(θ).0735

*Just remembering that this is the maximum velocity, therefore, the maximum velocity is going to be -- take the square root of both sides -- it is going to be the square root of 2gl, 1 - cos(θ).*0757

*So we could use conservation of energy that way to solve for the maximum velocity of our ideal pendulum, and of course that is going to occur when our pendulum is down in that position or its maximum kinetic energy is 0 potential energy.*0773

*All right, let us look at an example of a cart compressing a spring.*0789

*The diagram here shows a toy cart possessing 16 J of kinetic energy traveling on a frictionless horizontal surface toward a horizontal spring.*0792

*If the cart comes to rest after compressing the spring a distance of 1 m, find the spring constant of the spring.*0801

*Well, we can use conservation of energy to solve this.*0808

*The way I would do that is I would look and say that its initial kinetic energy, which is 16 J, must equal its final total energy which is all spring potential energy.*0811

*Therefore, 16 J must equal -- well the formula for the potential energy stored in a compressed spring is 1/2 Kx*^{2}.0823

*We want to rearrange this to solve for (K), the spring constant, so I could write that as (K) is equal to...*0833

*...multiply both sides by 2 and we get 32/x*^{2}, which is going to be 32 over our displacement of (1 m), 1^{2} or 32N-m for my spring constant.0838

*Looking at a little bit more involved example -- here we have a pop-up toy compressing a spring.*0859

*The pop-up toy has a mass of 0.020 kg and a spring constant of 150N-m.*0867

*A force is applied to the toy to compress the spring 0.05 m. *0873

*Calculate the potential energy stored in the compressed spring.*0878

*All right, well our first step there -- the potential energy stored in the compressed spring is 1/2 Kx*^{2} or 1/2 × 150N-m, our spring constant, × 0.050 m^{2} or about 0.1875 J.0881

*Let us take this one a little bit further -- that toy is then activated and all that compressed spring's potential energy is converted to gravitational potential energy.*0906

*The spring unleashes and the toy pops up; it starts going very quickly and it slows down, slows down, slows down until it gets to its highest point then it stops.*0915

*Let us find the maximum vertical height that it was propelled to, and we can do that by conservation of energy.*0923

*In this case, our initial potential energy, which was all stored in a spring is turned into final potential energy, which is gravitational.*0929

*Therefore, 0.1875 J must equal (mgh), so if we solve for the height, (h) is going to be 0.1875 J/mg, or 0.1875 J/mass (0.2) × (g), which we are going to estimate as 10, which gives us a maximum height of about 0.9375 m.*0939

*Let us see if we cannot take this to another type of problem. *0972

*A car initially traveling at 30 m/s slows uniformly as it skids to a stop after the brakes are applied. *0978

*Sketch a graph showing the relationship between the kinetic energy of the car as it is being brought to a stop and the work done by friction in stopping the car.*0985

*Well our car starts out and its initial velocity is 30 m/s; its final velocity -- if it is coming to a stop -- is 0 m/s, and we want a graph of the kinetic energy vs. the work done by the force of friction.*0993

*Now when no work is done by friction, of course it is going to have all or its maximum kinetic energy, so we should expect a nice, high point here.*1017

*When it has come to a stop after a lot of work has been done by friction -- it has no velocity, it must have no kinetic energy.*1025

*Well we have a uniform slow down, therefore that is all we need.*1033

*Let us take a look at accelerating an object -- the work done in accelerating an object along a frictionless horizontal surface is equal to the change in the object's momentum? No. Velocity? No. Potential energy? No. Kinetic energy -- this is the Work-Energy Theorem.*1048

*When you do work on an object, you change its energy. *1070

*If you are doing work on a frictionless horizontal surface to it, what type of energy are you giving it?*1073

*It must be kinetic energy or energy of motion. You are increasing its velocity.*1077

*Let us take a look at a block on a ramp.*1084

*A 2 kg block sliding down a ramp from a height of 3 m above the ground reaches the ground with a kinetic energy of 50 J.*1087

*Find the total work done by friction on the block as it slides down the ramp.*1093

*Well let us start by making a diagram -- there is our ramp, we will put our block on it.*1099

*Now we know that its gravitational potential energy at the top must be equal to its kinetic energy at the bottom plus whatever work was done by friction.*1106

*So if we want the work done by friction then -- that is just going to be the potential energy due to gravity at the top minus its kinetic energy at the bottom, which will be (mgh) at the top minus the kinetic energy (50 J)...*1116

*...therefore the work done by friction will be m (2 kg) × g (10) × our height difference (3 m) - 50 -- 2 × 10 = 20 × 3 = 60 - 50 for a total of 10 J.*1135

*Let us take a look at a little bit more creative, fun example.*1160

*Andy the adventurous adventurer, while running from evil bad guys at an Amazonian rainforest, trips, falls, and slides down a frictionless mudslide of height 20 m as depicted here.*1164

*Once he reaches the bottom of the mudslide he has the misfortune to fly horizontally off of a 15 m cliff.*1175

*He has gravitational potential energy up here, but as he slides down, all of that is going to be converted into kinetic energy and he is going to be moving completely horizontally, so all of the velocity here are corresponding to his kinetic energy.*1182

*And at that point he becomes a projectile. How far from the base of the cliff does Andy land?*1196

*Well let us treat this as two problems -- a conservation of energy problem here to find his horizontal velocity and to review our projectile problems down here to figure out where he lands.*1202

*First step -- let us find what his potential energy due to gravity is up here.*1213

*That is going to be (mgh) and that is going to be equal to his kinetic energy down here -- 1/2 mv*^{2} by conservation of energy.1221

*So the height difference is 20 m, so as we do some of our cancellations -- (m)'s cancel out -- we can say again that V = square root of 2gh, or that is going to be square root of 2 × 10 × its height (20 m) from when he goes over the cliff, or 20 × 10 = 200 × 2 = 400, and the square root of 400 = 20 m/s.*1231

*He is going to fly off the cliff horizontally with a velocity of 20 m/s.*1256

*Now we have a projectile problem. *1261

*Horizontally, he is going to have a constant velocity of 20 m/s. *1265

*If we can find how long he is in the air, we can solve for his displacement horizontally, (d) or δx, which is going to be his velocity times the time.*1271

*To figure out how long he is in the air, then we have to look at vertical motion.*1281

*Vertically, his initial velocity is 0;, his final velocity we do not know;, and δy is going to be 15 m from the time he goes off the mudslide to hitting the ground at the bottom of the cliff.*1286

*So that is 15 m -- we will call down the positive y direction again; (a) our acceleration is going to be 10 m/s*^{2} down and let us see if we can solve for time.1300

*The equation that I would use to do this is I would say that δy = V initial × t + 1/2 at*^{2}.1312

*V initial = 0, so that whole term goes away. *1322

*We could then write that (t) must be 2 δy/a(square root) or 2 × 15 m/10 m/s*^{2}(square root).1327

*So (t) equals -- when I plug all that into my calculator I come up with the time in the air of about 1.73 s, and if he is in the air 1.73 s vertically, he must be in the air 1.73 s horizontally.*1343

*So d or δx is just going to be velocity × time -- δx (d) will be Vt or that is going to be 20 m/s × our time of 1.73 s; it is going to give him a displacement horizontally of about 34.6 m.*1357

*This distance right there must be 34.6 m from the base of the cliff.*1379

*That is conservation of energy problem combined with our knowledge of projectile motion.*1387

*Hopefully that gets you a good start with conservation of energy.*1392

*Thank you for your time and for watching Educator.com.*1395

*Make it a great day.*1399

*Hi everyone. I am Dan Fullerton and I am thrilled to welcome you back to Educator.com. *0000

*Today's topic is simple harmonic motion. *0004

*Our objectives are going to be to sketch and analyze a graph of displacement as a function of time for an object undergoing simple harmonic motion, to write down an appropriate expression for displacement of the form A-cos(ωt) or A-sin(ωt), where ω is going to be that angular frequency. *0008

*We will state the relations between displacement, velocity, and acceleration and determine the points in the motion where these quantities are minimum, 0, and maximum for an object undergoing simple harmonic motion, determine the total energy of an object in simple harmonic motion and sketch graphs of kinetic and potential energies as functions of time or displacement, and determine the period of oscillation for an ideal pendulum as well as a mass on a spring with a horizontal mass and a vertical mass. *0026

*That is what we are going to try and accomplish here. *0052

*All right. What is simple harmonic motion? *0056

*Simple harmonic motion is nature's typical reaction to a disturbance. *0059

*It is all over in this world in many, many places. *0064

*When you disturb something it typically responds with simple harmonic motion. *0068

*That could be something as simple as walking past a tree -- if you brush a branch, the branch gets a force done and starts oscillating back and forth. *0073

*Simple harmonic motion -- it is all over .*0081

*A displacement which results in a linear restoring force results in simple harmonic motion.*0084

*And the simple items we are going to focus on for today are going to be an ideal pendulum, a pendulum where the string has no mass and a mass on a spring, going back and forth due to that spring. *0088

*Let us start with a review of springs. *0101

*When a force is applied to a spring, the spring applies a restoring force and a spring can be compressed or it can be stretched. *0106

*When the spring is in equilibrium, it is unstrained; it is in its happy position; it is just thrilled to be there. *0113

*The factors affecting the force of a spring -- well the spring constant (k) is how tough it is to compress or stretch the spring. *0120

*The bigger the spring constant, the tougher it is to compress or stretch and that is measured in Newton's per meter (N/m) and the displacement is always measured from equilibrium. *0126

*We could make a graph of the force of a spring vs. the displacement and when we do that, for an object obeying Hooke's Law, we should get a straight line where the slope of that line which is rise over run, gives us our spring constant. *0136

*Or we could make the same type of graph -- force of a spring versus displacement get the same basic shape and if we want to get the work done in compressing and stretching that spring, all we have to do is come back and take the area under that curve and the area will be the work. *0155

*Now as we analyze that work, the work that we do on it must be the potential energy stored in the spring, so that is 1/2 base times height or 1/2 our base is going to be (x) and our height is going to be the force. *0178

*But if (F) is (kx), then that is going to be 1/2 x × our force, because it is a spring, (kx), or you could say that the potential energy stored in the spring is 1/2 kx*^{2}. 0195

*That is where that formula comes from -- one way we can derive the energy stored in a compressed or stretched spring. *0211

*All right. Let us talk about oscillations. *0221

*Repeated motions back and forth are called oscillations -- something going back and forth, back and forth is oscillating. *0225

*Now one revolution or one round trip, one complete cycle all mean the same thing and the period of the oscillation is the time it takes for one complete cycle or one complete revolution. *0232

*That is going to be an important vocabulary word, period. *0243

*Frequency, on the other hand, (F), is the number of cycles per second and it is measured in 1/seconds or a unit known as Hertz (Hz). *0247

*Period (T) is the number of seconds divided by the number of cycles to give you the time for each cycle. *0261

*Frequency is the number of cycles divided by number of seconds to give you the number of cycles per second and they are very closely related because Period is 1/frequency and therefore frequency is 1/Period. *0269

*Let us take a look at the spring block oscillator. *0283

*Imagine that we have a mass and we are going to connect it by a spring to some immovable object like a wall.*0286

*If we pull it one way and let it go, it is going to go back and forth and back and forth and oscillate.*0295

*That is a spring-block oscillator. *0297

*Factors affecting the period of its oscillation are the mass of its block (m) and the spring constant (k). *0299

*As we analyze this in a little bit more detail, we will call this the x = 0 position, the happy position, the equilibrium position -- maximum displacement of A or -A. *0306

*Now the period of a spring-block oscillator, period of a spring ST*_{s} is going to be 2π times the square root of the mass divided by the spring constant. 0318

*Now the frequency of that spring-block oscillator which is always 1/period is just going to be 1/2π square root then of k/m. *0332

*Now we could also rearrange this a little bit to say that 2π times the frequency equals the square root of k/m. *0344

*Where there is 2π times the frequency is often times called the angular frequency (ω). *0355

*Angular frequency = 2πF, therefore we could write the angular frequency (ω) is the square root of k/m. *0366

*A couple of definitions to go along with our spring-block oscillator system. *0384

*All right. Let us take a look at an example with this system. *0390

*We have a block of mass 5 kg and it is attached to a spring, whose constant is 2,000 N/m.*0393

*Find the period of oscillation, the frequency, and the angular frequency. *0399

*Well, let us start with the period. *0406

*Period for a spring-block oscillator -- we know is 2π square root m/k or 2π square root -- mass is 5 kg, (k) is 2,000 N/m or when I plug that into my calculator, I get about 0.314 s for my period.*0408

*Let us find the frequency. *0435

*Frequency is 1/Period, so that is going to be 1/0.314 = 3.18, 1/seconds or 3.18 Hz.*0436

*And the angular frequency (ω) is 2π times the frequency, 2πF or 2π × 3.18 Hz, or 20 and the units are rad/s, although radians are not an official unit -- 20 rad/s will be our angular frequency. *0457

*Let us take a look at how we might analyze this in even more detail. *0487

*Here we are going to show our spring-block oscillator, mass (m), and we are going to look at it at three different positions -- at its equilibrium, position (A), at maximum displacement to the right (B), back to (A) to its minimum displacement or maximum displacement on the left (C), and back to (A). *0491

*So it is going to go back and forth, (A) to (B) to (A) to (C) to (A) and back and forth and it is going to displace a distance (x) to the right or (-x) to the left.*0508

*Well, if we want to look and see if we have different displacements, at point (A), we have 0 displacement, that is its equilibrium. *0518

*When it is at (B), its displacement is (x) and when it gets all the way to (C), its displacement is (-x). *0526

*Let us take a look at velocity now .*0536

*When it is at (A), it is going to have maximum velocity because all of its energy is going to be kinetic, so this will be maximum velocity here at (A). *0538

*Over here at (B), it is not going to have any velocity. *0548

*That is where all of its kinetic energy is converted into spring potential energy, so that will be 0 and (C) is the same way just on the other side.*0551

*There is no velocity at the end points.*0558

*It goes back and forth and for a split second it stops, turns around, speeds up, slows down, slows down, slows down, stops, speeds up, speeds up, speeds up, speeds up, slows down, stops, and back and forth as it goes on its oscillating path.*0560

*Let us take a look then at the force.*0573

*Well, while it is at point (A), the force is going to be 0 because there is no force to the spring on it because there is no displacement.*0576

*When it is at (B), it is going to have a maximum force, but it is a restoring force bringing it back the other direction, bringing it back towards its equilibrium position, so we will call that the negative max force.*0585

*And at (C) it is going to have the maximum force back to the right in the positive direction, so that will be maximum.*0597

*Because force = mass × acceleration, or acceleration = force/mass, the acceleration charge should look extremely similar.*0604

*At (A) there is no acceleration, at (B) it has its negative maximum acceleration, and at (C) it has its positive max acceleration.*0612

*Let us look at energy now. Spring potential energy over here is (U).*0623

*At (A), there is no displacement, so the spring potential energy must be 0.*0631

*It will have its maximum spring potential energy here at (B) and (C), so there is a maximum at (B) and (C).*0638

*If we want to look at kinetic energy, well that is going to be basically the inverse, right?*0647

*At (A), it is going to have its maximum velocity so it will have its maximum kinetic energy over there at (A), so maximum kinetic energy at (A), and at (B) and (C), it stopped at those points, it has no velocity for a split second there, so 0 and 0.*0652

*Let us take a look at what would happen if we made a graph of some of these.*0671

*I am going to start by taking a look at the displacement (x) and we are going to look at it as the object goes from (A) to (B) to (A) to (C) to (A) to (B) and back again.*0675

*Let us make this point (A) down all of these graphs, then we will draw another line all the way down here from (A) to (B)...*0687

*...another one here back to (A), and it goes to (C), and back to (A).*0697

*I think you quickly get the idea of what is going on here.*0714

*From (A) of course and back to (B) and so on, on its oscillating journey.*0722

*Well, if we look at displacement -- at (A) its displacement is 0, so anywhere we have (A) we can fill in the dots for 0.*0730

*At (B), its displacement is going to be positive (x), so it must come up like that and come back to (A).*0738

*At (C), its displacement must be negative (x) -- back to (A) and so on.*0747

*A graph of the displacement versus time.*0754

*If we want to do that for velocity, however, let us take a look and see what we know here.*0759

*For velocity, we know =at (B) and (C) we have 0, so anywhere we have (B) or (C) we must have 0 velocity.*0764

*At (A) we have maximum velocity and initially it is going to the right, so we will make that a positive velocity as it comes through.*0773

*When it comes back to (A), it has velocity going in the opposite direction at the same magnitude and back up and you can quickly see the pattern here for velocity of our spring-block oscillator.*0781

*Let us go to force and acceleration, let us put those over here -- a nice purple color, maybe for our force.*0798

*If we want to look at force, we know anywhere we have an (A), the force is 0.*0810

*So 0 here, 0 here, 0 here, and we have a maximum force when we are at (C), so let us fill that in -- maximum force over here at (C) and the negative maximum force when we are at (B).*0815

*So at (B) over here, we have negative maximum force, negative maximum force, and we can quickly plot our force versus time graph looking something like that.*0832

*And acceleration is just going to follow that -- force = mass × acceleration, so almost a mirror image of the graph, just different values but same shape.*0848

*Now, let us take a look and finish off our graph by looking at potential energy and kinetic energy versus time.*0865

*Let us look at potential energy in the spring first over here...*0877

*...there is (T), we have (A) here, goes to (B), back to (A), to (C)...*0878

*...from (C) we go back to (A), and from (A) we go back to (B).*0895

*All right. Potential energy in the spring is 0 at any of the (A)'s.*0906

*So there you go, a 0, 0 , 0; when it is at (B) it is at a maximum, so we will fill in our points there, and when it is at (C), it is also at a maximum.*0913

*Potential energy is scalar; it does not have a direction, so our graphs can look kind of like this.*0925

*When we look at kinetic energy, let us go there right underneath it.*0935

*Kinetic energy is going to have values of 0 at (B) and (C) when it is not moving.*0942

*So for (B), (C) -- (B) is at 0 and it is going to have maximum value when it is at (A), so it is going to look almost like the inverse of the potential energy graph.*0947

*Something kind of like that, and if we were to add the kinetic and potentials, they would add to a constant value because we are neglecting friction in this problem, using all conservative forces so we have conservation of mechanical energy.*0967

*Let us take a look at solving some problems with this and we will start off with a detailed harmonic oscillator analysis.*0982

*A 2 kg block is attached to a spring. A force of 20 N stretches the spring to a displacement of half a meter (0.5 m). 0996 Find the spring constant.*0989

*Well, we know that F = kx, therefore (k) must equal F/x or 20 N/0.5 m, which is 40 N/m.*0998

*The total energy is going to be the spring potential energy when it is at its maximum displacement or 1/2 kx*^{2}...1015

*...which is 1/2 × 40 N/m (k) and its maximum displacement (0.5m*^{2}) for a total energy of 5 J.1026

*About the speed at the equilibrium position, well at that point we have converted all of that spring potential energy into kinetic.*1041

*So we could start solving this one by saying that the spring potential energy is equal to the kinetic energy at the equilibrium position or 1/2 mv*^{2} and that must equal that 5 J.1048

*Therefore, the velocity = 2 × 5 divided by the mass, or 2 × 5/2 square root of 5 for about 2.24 m/s.*1064

*And how about the speed when it is at (x) = 0.3 m? *1083

*Well to do that, we are going to have to look at an energy analysis again.*1088

*The total energy is the spring potential energy plus the kinetic, therefore the total energy = 1/2 Kx*^{2} + 1/2 mv^{2}.1092

*And if I am going to try and find the speed, let us get (V) isolated; I will multiply both sides by 2...*1107

*...2 there and I will subtract the kx*^{2}, therefore mv^{2} = 2 total energy minus kx^{2}.1115

*Divide by (m) and take the square root to find that the velocity will be 2 × the total energy - kx*^{2}, all divided by the mass, square root, and finally I can substitute in my values. 1128

*I have 2 × 5 J (total energy) - 40 (spring constant) × 0.3 (x-value), our displacement*^{2} divided by 2 (mass) and the square root of that entire thing gives me a velocity, a speed at 0.3 m of about 1.79 m/s.1144

*And I will do a quick check to see if that makes sense.*1165

*It should be less than the speed at the equilibrium position, the maximum speed, and it is less than 2.24 m/s.*1168

*So that makes sense. Excellent!*1175

*Let us go a little bit further with this one.*1177

*Let us try and find the speed now at x = -0.4 m.*1181

*We can use the same formula we just had, but plug in a different displacement value.*1185

*Velocity is going to be equal to 2 times total energy minus kx*^{2} divided by (m), square root...1191

*...that is going to be 2 × 5 J, (total energy) - 40 (spring constant) × -0.4*^{2} (new displacement/2 kg (mass); square root of all of that is about 1.34 m/s.1202

*As it is a little closer to its full extension, it is slowing down even more -- less than our velocity when we were at 0.3 m. That, too, makes sense.*1222

*Let us find the acceleration at the equilibrium position.*1231

*Well when we were at x = 0, the force must equal 0 by Hooke's Law, and if the force is 0, then Newton's Second Law (F = ma) tells us that the acceleration must also be 0.*1234

*But if we want the acceleration at 0.5 m -- well to do that I am going to use Hooke's law (F = -kx), where (k) again is 40 N/m and a displacement at 0.5 m or -20 N.*1249

*We can apply Newton's Second Law again -- acceleration is force divided by mass, or -20 N/0.2 kg for an acceleration of -10 m/s*^{2}.1267

*What does the negative tell you? *1283

*When it is at its furthest positive displacement, the acceleration is back towards its equilibrium position, going in the opposite direction, hence the negative.*1284

*Let us take this one even further. Let us find the net force at the equilibrium position.*1298

*Well at the equilibrium position we already determined the acceleration was 0, so the net force there must be 0 N.*1305

*How about the net force at half a meter?*1312

*Well, we could use Hooke's Law again (F = -kx) or -40 N/m × 0.25 m (displacement) or -10 N. *1315

*Where does kinetic energy equal potential energy?*1330

*Well if our total energy is 5 J, that is going to be the spot where the kinetic energy is 2.5 J and the potential energy is 2.5 J.*1334

*So if that is the case, we could figure that out -- the spring potential energy there is 2.5 J -- that is 1/2 kx*^{2}, where (x) is what we are solving for.1344

*Therefore, (x)*^{2} = 2 × 2.5/40 (K), and as I solve that then, that is equal to 0.125 and if I take the square root of both sides...1356

*...(x) then equals the square root of 0.125 or about 0.354 m.*1373

*Pay special attention here -- note at this point where the kinetic energy and potential energy are equal is not midway between the equilibrium and the maximum displacement positions.*1384

*It does not work out that way, you cannot just guess halfway in between the two, you actually have to go through and solve; that is not halfway between the equilibrium and maximum displacement positions.*1396

*So let us take a look at circular motion versus simple harmonic motion and how they are related.*1407

*We have already talked about rotational motion for an object moving in a circle where we have some radius or amplitude of its motion (A) -- the angle θ measured from the horizontal, angular velocity ω and its position vector are, given by its (x) coordinate A-cos(θ), its (y) coordinate A-sinθ.*1411

*Well you could think of that almost as a projection down in one dimension to the spring-block oscillator system.*1431

*As you look at that system, imagine the object moving in a circle.*1438

*If you could shine a light down on it so you were just getting the projection in one dimension, you would see the exact same motion as what you see here in one coordinate.*1443

*Let us take a look at the (x) coordinates here -- (x) = A-cos(θ), but as we learned previously, θ = ω × (T).*1452

*Therefore we could write that as x = A-cos(ωT).*1465

*We also know that ω = 2π/period (T), so we could write that as x = A--cos(2π)/period × time.*1472

*Or going back to this equation, we also know that ω = 2π × the frequency, so we could write that -- replacing ω with 2π × the frequency as x = A-cos(2π), frequency, × time.*1487

*A bunch of different ways of looking at the same simple harmonic motion.*1504

*Instead of having our block start over here at a maximum displacement, what happens if we want to have it start here at x = 0?*1509

*Well we could use the sine function for that.*1518

*If we are going to have the block start at x = 0 at its equilibrium position -- start at x = 0 at time (t) = 0, then we could have the exact same equation -- we are just going to replace cosine with sine.*1520

*So x = A-sin(θ) or A-sin(ωt).*1536

*All it is, is a face shift, a sliding of the graph one way or another, depending on what you are calling your starting point.*1543

*As we look at graphing simple harmonic motion in that system, let us take a look at what happens to our graph of x = A-cos(ωt).*1552

*We will graph the (x) displacement and as we do that, we are going to look at it in terms of radians at 0, at π/2, at π, at 3π/2, 2π and so on.*1567

*I will make those marks on our graph right now -- π/2, π, 3π/2, 2π, and we will copy the same notches down here as well.*1580

*All right, for x = A-cos(ωt), we will do this as a function over here of ωt.*1598

*What are we going to graph?*1604

*Well, when our argument is 0, cosine here at times 0 is going to be 1, so we are going to get (A), so our maximum value here is (A).*1606

*As we get to π/2 right here, well our (x) coordinate is now 0, so we come down here to 0. *1618

*As we get over here to 2 full π, now we are adding negative maximum displacement, -(A).*1628

*Let us mark that on our graph, -(A), and back to 3π/2, 2π again and the cycle repeats.*1636

*We get ourselves -- just like the curve we had expected -- we get our cosine curve.*1645

*Looking at that, if we wanted to use the sine function instead, x = A-sin(ωt), we are going to have the same maximum amplitudes of course -- (A) and -(A).*1655

*But we are going to start -- if we look at the y-coordinate or for the sine function -- we are going to start at 0 for our (y), so 0 at π/2, we are at maximum value here for the (y), so positive (A)...*1670

*...down here to 0 at π, down here to -(A) at 3π/2 and back to 0 at 2π -- so we get our sine curve.*1687

*But as you look at the graphs, notice they are really the same shape; they are just off set by that π/2 amount, so you can use either one depending on which starting point you prefer to work with.*1698

*All right. Let us take a look at an example where we are looking at the position of an oscillator.*1711

*A spring-block oscillator makes 60 complete oscillations in 1 minute or 60 seconds.*1715

*Its maximum displacement is 0.2 m. *1723

*What is its position at time (t = 10 s) and at what time is it at position x = 0.1 m? *1727

*Well let us start up here with question A.*1735

*If x = A-cos(ωt) and we know that ω = 2πF, we could write this as x = A-cos(2πFt).*1738

*If we have 60 cycles in 60 seconds, then we know that our frequency must be 1 Hz.*1759

*So if frequency is 1 Hz and we know that our maximum amplitude (A) is 0.2 m, we can fill in our function a little bit to say that X = 0.2 cos(2π) × 1 (frequency) × 10 s (time) or putting that all together I get about 0.2 m.*1765

*It is at its maximum displacement.*1794

*Moving down here to B, at what time is it at position x = 0.1 m?*1797

*All right, s = A-cos(2πFt), but now we are solving for the time, so this implies then that 2πFt must be equal to the inverse cosine of x/A.*1804

*Or if we want just (t) but itself let us isolate the variable we want to find -- t = the inverse cosine of x/A divided by 2πF.*1827

*Now we can substitute in our variables, so that will be the inverse cosine of 0.1/0.2 or 1/2/2π × 1 Hz (frequency) or about 0.167 s.*1841

*Now it is important to note here for an oscillating system, that is not going to be the only time it is in that position, but it is one answer to that question.*1860

*Let us take a look at a vertical spring-block oscillator system.*1872

*Once the system settles the equilibrium where we are hanging our block from a spring instead of having it on a horizontal surface, we are going to displace the mass by pulling at some amount either +A -- pull it down, let it jump up -- or -A -- lift if up a little bit, drop it and let it oscillate up and down.*1877

*This is a really slick derivation and a neat analysis.*1893

*If we looked at a free body diagram (FBD) when it is at its equilibrium position, we have gravity pulling down and the force of the spring, (ky) pulling it up, where (y) is the equilibrium point. *1898

*So since it is at equilibrium at that point, we could write that the net force in the y-direction is going to be (mg), calling down positive, minus (ky) and since it is at equilibrium, let us say that that is mg - ky (equilibrium point), that must equal 0.*1910

*Therefore we could solve to say that the y-equilibrium point must be mg/k.*1933

*Now when we displace it by some amount (A), the net force in the y-direction, as we pull it down, is going to be mg - K × whatever that (y) would happen to be, which is going to be Y-equilibrium point plus that (A) amount we pulled it down.*1941

*We can distribute that through -- multiply that (k) through -- to find that it is mg - ky-equilibrium position - kA*1964

*Here is the slick part though -- Notice then as we do this that we had up here mg - ky-equilibrium = 0.*1976

*That means this part mg - ky-equilibrium must be equal to 0 and we could rewrite this then as F*_{net}Y = -kA.1989

*There is our big result. What does that mean? *2005

*That is the same analysis you would do for a horizontal spring system with a spring constant (k), displaced horizontally some amount (A) from its equilibrium position.*2008

*We just made this so much simpler.*2018

*In short, to analyze a vertical spring system, all you do is find the new equilibrium position of the system, treat that -- once you have taken into account the effect of gravity -- and treat that as if that is the only force you have to deal with in the system -- just the spring force.*2021

*So once you find its new equilibrium position you could almost pretend that you turned it on your side and it is a horizontal spring oscillator system again.*2037

*You do not have to continue dealing with that force of gravity.*2044

*A really, really slick way to analyze a vertical spring-block oscillator.*2048

*Find that new equilibrium position and then ignore the effects of gravity from there; treat that as your new equilibrium position, just find it using the effect of gravity first.*2052

*Let us take an example to make sure we have this.*2066

*A 5 kg block is attached to a vertical spring, with a spring constant of 500 N/m. *2069

*After the block comes to rest it is pulled down 3 cm and then released.*2075

*What is the period of oscillation?*2079

*Well, period is 2π square root m/k or 2π square root 5/500, or 0.628 s -- that straight forward.*2082

*What is its maximum displacement of the spring from its initial unstrained position?*2101

*Well let us first look at it when it is at rest.*2106

*At that point we have (mg) down for our FBD, and we have (k) times -- let us call that displacement (d) at that point, so that the net force in the y-direction is 0, since it is at equilibrium because it is just sitting there, which implies that KD = mg or D = mg/k, which is 5 kg × 10 m/s*^{2}/spring constant (500) or 0.1 m.2110

*So once you are hanging it there it hangs down 0.1 m.*2144

*Now you are going to go displace it; you are going to pull it down 3 cm from that equilibrium position.*2148

*If you then pull it down 3 cm, your maximum displacement in the y-direction is going to be that 0.1 m that you had from when it came to rest due to gravity...*2155

*...and that extra 3 cm that you added on as you pull it down or 0.13 m or 13 cm. *2170

*A nice straight forward analysis, once you take into account and figure out what its new equilibrium position is with gravity and then ignore the effects of gravity and treat it as a standard horizontal spring-block oscillator system.*2183

*All right. Example 5 -- another fairly involved example.*2199

*We have a 60 kg bungee jumper stepping off a 40 m high platform.*2203

*The bungee cord behaves like a spring, of spring constant 40 N/m.*2209

*Find the speed of the jumper at heights of 15 and 30 m above the ground.*2213

*And as we do this, we are going to have to assume that there is no slack in the system.*2217

*Find the speed of the jumper at heights 15 and 30 m above the ground.*2222

*All right. So first thing, let us draw a diagram here.*2227

*Here is our jumper and the jumper is on a 40 m foot high platform, so down here somewhere is the ground.*2231

*We want the speed of the jumper 15 m above the ground, so we will call that position (A) -- that is 15 m above the ground.*2243

*Position (B) is 30 m above the ground, which means we must have another 10 m here -- 10, 15 m, 15 m.*2255

*All right. The key here is we are going to try and do this through conservation of energy at this point.*2267

*So the gravitational potential energy at the top must be equal to the gravitational potential energy at (A) plus the spring potential energy (A) -- UAG, UAS plus the kinetic energy at (A), 15 m above the ground.*2273

*Or mg × H-initial, the initial gravitational potential energy must equal the gravitational potential energy at (A), mg × 15 + 1/2 k.*2296

*At (A) the displacement is 10 + 5, so the spring is stretched 25 m, 25*^{2} (kx^{2}) + 1/2 mv^{2}.2312

*All right. A little bit of math here.*2325

*We could then say that mass (60), G (10), and height (40)...*2327

*...must equal 60 × 10, = 600 × 15 + 1/2 × 40 (k) × 25*^{2} = 625 + 1/2 × 60 (mass) × v^{2}...2336

*or let us see here, that is going to be 24,000 = 9,000 + 12,500 + 30 v*^{2}.2355

*Or 30 v*^{2} will equal about 2,500, so I get a velocity of about 9.13 m/s at (A), so vA = 9.13 m/s.2373

*We also need the velocity at 30 m above the ground. *2394

*To do that then, we will follow the same basic idea but we are going to have different values for our height.*2399

*So we will have 24,000, our initial total energy, equal to mg × (height) 30 m + 1/2 k -- at (B) the spring is only stretched 10 m, so 10*^{2} + 1/2 mv^{2}.2405

*Therefore 24,000 = 600 (mg) × 30 + 1/2 × 20 (k) × 100 = 2,000 + 1/2 mv*^{2}. 2427

*Therefore 4,000 = 30v*^{2}; and in solving for V, I get about 11.55 m/s.2449

*Makes sense -- it is going a little bit faster at (B), a little bit slower at (A), hopefully slows down before hitting the ground.*2460

*All right, let us go a little bit further with this problem. Let us see what happens next.*2469

*How close does the jumper get to the ground?*2476

*Well, to do that we are going to have to figure out where the kinetic energy becomes 0.*2482

*So as we do this one, we will say potential energy due to gravity total equals the potential energy due to gravity at some point (C), which is where they stop, plus potential energy due to the spring at point (C).*2491

*No kinetic energy -- it is 0; that is where the person stops.*2507

*Following along with our calculations, 24,000 (total energy) = mgH + 1/2 k, and now how far has that person been displaced? *2512

*That is going to be 40 minus whatever height is left, so 40 - H*^{2}.2525

*So that implies then that 24,000 = 600 (mg) × H + 1/2 × 40 (k), so that is going to be 20 × 40 - H*^{2}.2532

*That is starting to look like a quadratic equation, so let us get it into that form.*2553

*24,000 = 600H + 20 × -- well, if we square this, we get 1600- 80H + H*^{2}...2557

*...or 24,000 = 600H + 32,000 - 1600H + 20H*^{2}.2576

*We rearrange this to fit the quadratic formula, H*^{2} - 50H + 400 = 0.2595

*A couple of ways you can solve that, but the quadratic formula is probably my favorite.*2606

*I come up with a height of 10 m above the ground.*2609

*And let us just test that to make sure we did not make any mistakes here.*2615

*H*^{2}, 10^{2}, 100 - 50 × 10, 100 - 500 = -400 + 400 = 0.2618

*So how close does the jumper get to the ground? 10 m.*2627

*Let us take a look at the pendulum again.*2635

*Now from the perspective of oscillations and simple harmonic motion.*2639

*Mass (m) is attached to a light string that swings without friction about a vertical equilibrium position.*2644

*For all of these, we are going to assume that this θ is relatively small.*2650

*Well, we have length (L) here again -- as it comes down here where there is all kinetic energy, its height has changed and we have derived a couple of times now that this height (H) is going to be L - L-cos(θ)...*2655

*...or this is L-cos(θ) compared to our entire length of our string (L).*2673

*Here we have all potential energy; here we have all kinetic and this (H) as we just said is going to be L - L-cos(θ).*2685

*If we take a look and start analyzing this with energy involved too though, again assuming a small θ, here we have all potential energy -- U = mgH, which is mgL × 1 - cos(θ).*2696

*Here it is all kinetic energy, and back here again it is all potential.*2716

*And if we wanted to find the velocity of our pendulum when it is at this lowest point, well we could say that the kinetic energy at the bottom must equal the potential energy at its highest point -- the top.*2723

*Or 1/2 mv*^{2} = mgH where (H) is L - L-cos(θ), and we can divide now our masses.2735

*So v*^{2} = 2gL × 1 - cos(θ) or just velocity itself -- I take the square root and that will be the square root of 2gL, 1 - cos(θ).2748

*Going a little further here, let us look at some other quantities that might be of interest.*2765

*Here at the highest point, you have the maximum force, you have the maximum acceleration, you have the maximum gravitational potential energy, but you have no kinetic energy and no velocity.*2769

*Here on the other hand, you have no force, (F = 0), your acceleration = 0, your gravitational potential energy = 0 -- assuming that is what we are calling 0 in this problem which should make sense.*2786

*Our maximum kinetic energy occurs there and we have maximum velocity there.*2799

*And what is causing our restoring force to put this in simple harmonic motion?*2805

*Well our restoring force is going to be based on the gravitational force pulling this down.*2810

*If we look right here, we have its weight pulling it down, but that is not what is causing the displacement; it is only a portion of that.*2815

*The portion that is going to be perpendicular to our string, mg-sin(θ) here is what is causing our restoring force.*2826

*So another way that we could graph this in terms of energy, is if we looked at energy on the y-axis versus (x) position on the (x), we know of course the total energy must remain the same. *2844

*We are dealing with conservative forces, conservation of energy, and we are not dealing with friction at this point.*2854

*Here at 0 displacement, everything is kinetic energy.*2860

*So as we draw this U-shape, anything above the U-shape between the E (total line) and our parabola is kinetic energy.*2865

*Anything below it is potential.*2873

*The two always sum up to the total energy.*2875

*So if I wanted to look at another point on the graph -- let us say we wanted to look right here.*2878

*In this case anything up here would be kinetic and down here would be our potential.*2884

*Use the graph that way -- what is above the line is kinetic, what is below is potential.*2895

*Another way of representing the same information; it is that important.*2899

*So what happens when we are talking about this pendulum and we have to start dealing with frequency and period?*2905

*Well, the period of an ideal pendulum is 2π square root of L/g.*2910

*The length of the pendulum is your variable. *2915

*The mass on the end does not matter, the length is what matters.*2918

*In your grandfather clocks at home, the length is all set, that is why those are so big.*2922

*Or the frequency is just going to be 1/period or 1/2π square root of g/L.*2929

*Now for all of these, again, we have to assume that θ is small due to a small angle approximation in the mathematics.*2934

*So as we are looking at this, let us take a look and see what would happen if we tried to graph period versus length.*2941

*If you did that you would probably get something that looks kind of like that because period is proportional to the square root of (L).*2951

*So if you wanted to get a nice linear graph that you could do something with, if you wanted to try and determine something like the acceleration due to gravity, for example.*2960

*You could take a pendulum, graph the period versus the square root of the length and you should get a nice linear graph and if you take the slope of that, the slope is going to be rise/run, which is going to be T/square root of L, which turns out to be 2π/square root of (g).*2971

*So if you want to go to the moon, figure out what the acceleration due to gravity is, make a bunch of different pendulums of different lengths, have them go back and forth, measure their periods, come graph the period versus square root of the length, find the slope and you can calculate the acceleration due to gravity that way.*2995

*Let us take a look at an example.*3015

*We have a 1 kg mass suspended from a 30 cm string that creates a simple pendulum.*3017

*The mass is displaced at an angle of 12 degrees from the vertical equilibrium position.*3023

*First thing we have to do, is if we are going to use any of these formulas, is to make sure that θ is small, and 12 degrees is small enough for our purposes.*3029

*Find the frequency and period of the pendulum.*3037

*Well, the period is 2π square root L/g, or 2π square root (L) 0.3 m/10 m/s*^{2} (g)...3040

*...or about 1.09 s and frequency then is just 1/period or 1/1.09 s, which is going to be about 0.92 Hz.*3056

*All right, find the height of the pendulum above equilibrium when at maximum displacement.*3070

*Well we know the height is L × 1 - cos(θ), so that is going to be 0.3 × 1 - cos(12 degrees) or about 0.0066 m.*3076

*And find the speed of the pendulum at the equilibrium position.*3093

*V = square root of 2GH if we want to use conservation of energy, which is square root of 2gL, 1 - cos(θ)...*3097

*...we have derived that a couple of times at this point, or the square root of 2 × 10 (g), (l) 0.3 × 1 - cos(12 degrees) or about 0.632 m/s.*3107

*Carrying this one a little further -- Find the restoring force at maximum displacement.*3130

*All right. The force at maximum displacement -- we just said was mg-sin(θ), the component of the weight pulling it back down -- mg-sin(θ), which is going to be its mass of 1 kg × 10 m/s*^{2} (g)-sin(12 degrees), or about 2.08 N.3136

*And how about the tension in the string at the equilibrium position?*3158

*Well at that point we can make our FBD.*3162

*There is our tension; there is our weight -- if it is in equilibrium, those must match -- or net force in the centripetal direction is mAC which implies that (t), which is in the direction toward the center of the circle... *3166

*...tension - mg = mv*^{2}/r, which implies then that the tension is mg + mv^{2}/r.3181

*Therefore, the tension must be (mass) 1 × 10 (g) + 1 (mass). *3193

*Our velocity we found was about 0.362 m/s*^{2} over our radius (0.3 m) or about 10.44 N of tension in that string.3201

*How long should the pendulum be in order to keep perfect time with a period of 1 second?*3220

*Well let us start there and we are going to assume again that it is an ideal pendulum, no friction and everything is perfect, and no mass in the string.*3227

*Period is 2π square root L/g and we want a time, a period of 1 s, so we are solving for (L).*3235

*Let us take t*^{2} = 4π^{2} (L/g).3247

*Therefore L = gt*^{2}/4π^{2}, which implies then that the length (L) should be G (10 m/s^{2})...3254

*...with a period of 1 s*^{2}/4π^{2} or about 0.253 m, a quarter of a meter.3270

*How long should the pendulum be if the period is to be half a second?*3285

*Well, same thing -- let us just plug in a couple of different values here.*3289

*L = gt*^{2}, so (g) × 0.5^{2}/4π^{2}, where g = 10 m/s^{2}, which gives us about 0.633 m.3293

*A lot shorter. A lot shorter -- one-fourth.*3308

*Let us take a look then at a pendulum on the moon.*3313

*How long must a pendulum of period 1 second be on the moon if the acceleration due to gravity on the moon is about 1.6 m/s*^{2}?3316

*Well we can use our same formula, L = gt*^{2}/4π^{2}, where (g) = 1.6 m/s^{2} on the moon × our period of 1 s^{2})/4π^{2}, which is about 0.405 m.3325

*All right. Doing great. Hang in there. One last sample problem.*3355

*Mass (m) is placed on a horizontal frictionless surface and attached to a spring with spring constant (k).*3360

*The mass is pulled back a distance (x) and released to oscillate horizontally.*3368

*What is the kinetic energy and potential energy of the mass at a displacement halfway between the equilibrium position and maximum displacement?*3373

*Well, let us draw a picture first of what our situation is going to look like.*3382

*Horizontal spring-block oscillator -- let us color that in there.*3386

*We will put our mass over here (m), some spring with spring constant (k), and we will start this at some displacement 0.*3391

*Here is our (x) and it is somewhere in there we are going to have a point (A), and what we know is at the maximum energy is 1/2 kx*^{2}.3405

*The potential energy due to the spring at (A) is going to be 1/2 × (k) -- well (A) is halfway between these two, so that is going to be at x/2*^{2}.3420

*That will be 1/2 k × x*^{2}/4 or 1/8 kx^{2}, which is 1/4 × 1/2 kx^{2}.3438

*Why did I write it that way?*3451

*Well, 1/2 ks*^{2} is our maximum spring potential energy, so this then says that UA must equal 1/4 of UMax.3454

*So at (A) it has 1/4 of its maximum energy, so where is that other 3/4 of the energy? *3471

*That has to be kinetic energy at (A), so that must be 3/4 of the maximum spring potential energy.*3476

*Notice here that halfway between the two, the spring potential energy and the kinetic energy are not the same -- they are not equal.*3485

*You have to go through the steps to go solve for points in between those two.*3495

*Do not take shortcuts.*3499

*Hopefully this is a good start for simple harmonic motion.*3501

*Thank you so much for your time and for watching Educator.com. Make it a great day everyone!*3504

*Hi everyone and welcome back to Educator.com.*0000

*Today we are going to talk about fluids -- starting a new unit, specifically about density and buoyancy. *0003

*Our objectives are going to be to calculate the density of an object, to determine whether an object will float given its average density, and to calculate the forces on a submerged or partially submerged object using Archimedes' principle. *0010

*As we start this new topic of fluids, let us talk about what a fluid is. *0026

*Fluid is matter that flows under pressure -- things like liquids -- a great example might be water -- gases, like air; and even plasmas like what you would get from an arc welder.*0031

*Now fluid mechanics is going to be the study of fluids and how they move -- fluids at rest, fluids in motion, forces applied to fluids and then the forces exerted back by fluids. *0044

*So to begin, let us get into density. *0056

*Density, which gets the symbol, Greek letter ρ, is the ratio of an objects mass to the volume it occupies. *0059

*Less dense fluid flow on top of more dense fluids and less dense solids will float on top of more dense fluids. *0065

*Now if density is the ratio of an objects mass to volume -- density (ρ), is mass over volume and the units are going to be kilograms per meter cubed. *0072

*Let us take a look at an example with a density of water. *0077

*A single kilogram of water fills a cube of length, 0.1 meter. What is the density of the water?*0090

*Well we have our cube of water and the length of each side is 0.1 meter. What is its density? *0096

*Well density is mass divided by volume, so that is going to be 1 kg and the volume of the cube (length × width × height) is going to be 0.1 m × 0.1 meter × 0.1 meter for a total of 1,000 kg/m*^{3}.0107

*That is probably a good one to remember -- density of fresh water is 1,000 kg/m*^{3}.0132

*Let us take a look at the volume of gold. It has a density of 19,320 kg/m*^{3}. It is very dense. 0140

*How much volume does a single kilogram of gold occupy?*0147

*Well, if density is mass over volume, then volume will be mass over density or 1 kg/19,320 kg/m*^{3}. 0169 Or about 5.18 × 10 ^{-5}m^{3}. 0151

*Let us take a look at an example of things that are floating. *0184

*Fresh water has a density of 1,000 kg/m*^{3}, we just calculated that. 0188

*Which of the following materials will float on water? *0193

*Ice has a density of 917 kg/m*^{3} and if you have ever had a glass of ice water, you already know the answer -- ice will float on water. 0196

*It is less dense than the water. *0206

*Magnesium has a density of 1740 kg/m*^{3}. It is more dense than water, so it is going to sink. 0209

*Cork, of course, 250 kg/m*^{3}, is going to float. That is why we make bobbers out of cork when we go fishing. 0215

*Glycerol is 1260 kg/m*^{3} is more dense so it is going to sink. 0224

*So those two will float because they are less dense. *0229

*Let us talk a little bit about buoyancy now. *0234

*Buoyancy is a force exerted by a fluid on an object and it opposes the objects weight when it is in that fluid. *0236

*The buoyant force, typically written as F*_{b}, is determined using what is known as Archimedes' principle. 0242

*The buoyant force is equal to the density of fluid times the volume of the fluid displaced times the acceleration due to gravity. *0248

*Let us spell these out because they are easy to mix up. *0256

*(G) of course is the acceleration due to gravity. *0259

*The volume is going to be the volume of the fluid displaced by your object. *0266

*Typically, you just see this written as density, but I like to put the fluid after it to remind me that it is the density of the fluid that we need in this calculation and not the density of the object, so that is going to be the density of the displaced fluid. *0280

*And of course, F*_{b} is the buoyant force. 0302

*So let us do an example with the buoyant force. *0312

*What is the buoyant force on a 0.3 m*^{3} box, which is fully submerged in freshwater if it has a density of 1,000 kg/m^{3}.0314

*Well the buoyant force, F*_{b}, is equal to the density of the fluid, ρ, times the volume (v) of the fluid displaced times (g). 0323

*So the density of the fluid displaced is 1,000 kg/m*^{3} and we have a volume of 0.3 m^{3}, and (g) -- we will estimate as 10 m/s^{2} for a force of about 3,000N. 0335

*Let us get exciting. Let us talk about a shark tank. *0356

*A steel cable holds a 120 kg shark tank 3 m below the surface of salt water. *0359

*Salt water is slightly more dense than freshwater at 1025 kg/m*^{3}. 0365

*If the volume of water displaced by the shark tank is 0.1 m*^{3}, what is the tension in the cable? 0369

*Well, let us start with our free body diagram (FBD). *0376

*There is our object and we have its weight pulling it down, pulling it down, and we have the buoyant force opposing that and we also have a cable on it that has some tension in it (t). *0379

*We want to find the tension in the cable. *0392

*If it is just sitting there 3 m below the surface of the saltwater, it is not accelerating, so the net force must be 0. *0394

*We could write Newton's Second Law for the y direction -- F*_{net}y equals tension plus the buoyant force minus the weight and all of that must be equal to 0. 0400

*If we want tension, that must be equal to the weight minus the buoyant force. *0414

*This implies then that the tension is equal to the weight minus -- well the buoyant force is the density of our fluid -- (ρ) fluid -- times the volume of the fluid displaced times (g). *0421

*So that is going to be mass (120), g (10 m/s*^{2}) minus density of our fluid (1025 kg/m^{3}) times the volume of the water displaced by the tank (0.1) times g (10 m/s^{2}).0434

*Therefore, our tension must equal about 175N. *0454

*Let us go to a favorite project in physics -- building a concrete boat that floats. *0468

*A rectangular boat made out of concrete with a mass of 3,000 kg floats on a freshwater lake. *0473

*The density of freshwater, again, is 1,000 kg/m*^{3}. 0480

*If the bottom area of the boat is 6 *^{2} meters -- it is a pretty big boat -- how much of the boat is submerged? 0484

*Well, let us start with a FBD. *0490

*We have the weight of the boat down (mg) and we have the buoyant force holding it up. *0494

*Net force in the y direction then, must be the buoyant force minus (mg) and because it is not accelerating up or down, that must be equal to 0, therefore the buoyant force must be equal to (mg). *0500

*But we also know that the buoyant force is equal to the density of the fluid times the volume of the fluid displaced times (g).*0517

*Therefore, we could say then that density of the fluid times the volume times (g) equals (mg). *0527

*I can see right away that there is a simplification that we can make -- we can divide (g) out of both sides. *0537

*We could also say then that the volume of our boat is going to be its area times its depth. *0543

*The volume of the fluid displaced is going to be the area of the boat times how much of it is submerged -- that depth submerged (d). *0549

*Therefore the density of our fluid -- (ρ) fluid -- times our volume (ad) must equal (m). *0559

*We rearrange this to find (d), the depth of the boat submerged. *0568

*(D) is going to be equal to the mass over density of the fluid times the area or 3,000 kg... *0572

*...about 3 tons divided by density of our fluid (1,000) and the area (6 *^{2}m) or about 0.5 m. 0582

*About 1/2 a meter is going to be under the water, submerged. *0597

*All right, let us take a look at a problem of apparent mass. *0604

*A cubic meter of bricks have an apparent mass of about 2400 kg when they are submerged in saltwater with a density of 1025 kg/m*^{3}. 0610

*What is their mass on dry land? *0619

*Well, what does that mean? Let us think about this for a second. *0623

*If we were to go make a scale and on it we are going to put a bunch of bricks -- there we go. *0627

*And our scale has a reading here. *0646

*It has an apparent mass of 2400 kg when it is submerged. *0648

*The scale reads like it is 2400 kg there, so what would its actual mass be on dry land?*0654

*What would its -- well the mass is not actually changing, so what would the scale read on dry land? 0665 What would its weight be on dry land? *0660

*I am going to start off with a FBD. *0668

*We have the weight down and while it is submerged, we have the buoyant force up and we have the normal force from the scale and as you know, scales tell you the normal force. *0672

*We wanted to write Newton's Second Law equation -- normal force plus the buoyant force minus (mg) must equal 0, because acceleration is 0 because the bricks are just sitting there on the scale; they are not accelerating. *0683

*Therefore, we could write that -- well knowing that the apparent mass is 2400 kg, that means the normal force which must be (mg), it must read that that is normal force of 24,000N... *0698

*...so 24,000 plus the buoyant force -- density of our fluid, times the volume of the fluid displaced, times (g) minus (mg) equals 0. *0716

*This implies then that 24,000 plus density of our fluid (1025) times the volume of our fluid displaced... *0731

*...which was 1 times g (10) - 10 m must equal 0. *0742

*Therefore, 24,000 + 1025 × 1 × 10 = 10 m. *0760

*Therefore (m) must equal -- Well divide both of these by 10 -- 2400 + 1025 = 3,425 kg is the actual mass. *0773

*It appears to have a lower mass when it is in water because the buoyant force is helping lift it on the scale, providing some upward force to counteract that weight. *0788

*Let us take a look at the volume of a submerged cube. *0802

*We have a cube of volume 0.002 m*^{3} submerged in a glass of freshwater and attached to the bottom of the glass by a massless string. 0805

*If the force of gravity on the cube is 10N what is the tension in the string? *0814

*Let us see if we cannot draw this out a little bit first -- feeble attempt at drawing a glass. *0819

*There it is and somewhere in the glass we have our cube and it is attached by a string to the bottom there and we have some sort of freshwater in our glass. *0826

*All right it is time for FBD again. *0847

*Here we have our object -- we have its weight (mg) down, we have the tension in the string down, and we have the buoyant force up and again because it is just sitting there, it is not accelerating up or down, it must be an equilibrium -- the net force must be 0. *0850

*So net force in the y direction, which is the buoyant force minus (mg) minus (t) must equal 0. *0867

*Solving for the tension then -- tension equals the buoyant force minus (mg) which implies then that the tension must be the buoyant force...*0876

*...density of our fluid times the volume of the fluid displaced times (g) by Archimedes' principle minus (mg) or that is going to be 1,000 kg/m*^{3} since it is freshwater...0889

*...volume displaced is 0.002 and (g) is 10 minus mg -- well, it gives us the force of gravity on it which is 10N, which is its weight, which is (mg) minus 10. *0904

*Therefore, the tension comes out to be 1,000 × 0.002 × 10 -- 20 - 10 = 10N. *0919

*Let us try one more practice problem here -- determining density. *0935

*This one is a little bit more involved. *0941

*The density of an unknown specimen may be determined by hanging the specimen from a scale in air and in water and then comparing the two measurements. *0943

*If the scale reading in air -- we are going to call (Fa) -- and the scale reading in water is (Fw), let us develop a formula for the density of the specimen in terms of the scale reading in air, in water, and the density of the fluid. *0952

*I am going to start with a FBD. *0967

*When we are in air, we have (mg) down and we have (Fa) on the scale up and those will be balanced because it is sitting on the scale at equilibrium. *0970

*When it is in water, our FBD is going to look similar, but a little bit different. *0983

*We still have the weight down, but now we have the buoyant force up along with the force of the scale (Fw). *0991

*Starting with the water, we have (Fb), the buoyant force, plus the force of the scale when it is in water must be equal to its weight because it is the equilibrium. *1001

*And we also know that the force of air is equal to (mg), so I could rewrite this as (Fb) + (Fw) = (Fa). *1011

*But that buoyant force is equal to ρ fluid (v)(g). *1027

*I could rewrite this then as (Fa) - (Fw), with a little bit of rearranging, must equal density of our fluid times our volume displaced times (g). *1034

*Now we have to take another step that is maybe not quite so obvious. *1048

*Let us take a look and let us say that the density of our object is equal to the mass of the object divided by the volume of the object. *1053

*Therefore the volume of our object is equal to the mass of the object over the density of the object. *1064

*I am going to use that as I rewrite this equation to say that (Fa) - (Fw) = -- we have our density of our fluid, but I am going to replace the volume displaced with my new formula for volume -- mass of the object over the density of the object. *1073

*We have mass of the object over density of the object times (g). *1090

*Now it is just a little bit of Algebra to prove that the density of the object is going to be equal to the density of our fluid times the mass of the object times (g) divided by (Fa) - (Fw)...*1101

*...just a little bit of rearrangement to get the density of the object all by itself and finally one more step, that (Fa) -- we can change that a little bit, we can rearrange things. *1126

*Let us then say then that the mass of the object times (g) -- that is just its weight in air (Fa) -- so mass of the object times (g) right here -- I am going to replace with (Fa) to write the density of our object is equal to... *1137

*Well we have (Fa) times the density of our fluid divided by the scale reading in air minus the scale reading in water. *1160

*Hopefully that gets you a good start on density and buoyancy as we start this new section on fluids. *1180

*I appreciate your time and make it a great day everyone. *1185

*Hi everyone. I am Dan Fullerton and I would like to welcome you back to Educator.com. *0000

*Today we are going to continue our study of fluids as we talk about pressure and Pascal's principle. *0004

*Our objectives are going to be to calculate pressure as the force a system exerts over an area, to explain the difference between gauge pressure and absolute pressure, and explain the operation of a hydraulic system as a function of equal pressure spread throughout a fluid. *0009

*Pressure -- pressure is the effect of a force acting upon a surface. *0025

*It is a scalar. It is a force per unit area and its units are Newtons per meter squared (N-m*^{2}), which are also known as Pascal's, which we typically abbreviate as (Pa). 0030

*If pressure if force per unit area, our formula for pressure is P = F/A. *0040

*Now it is important to know that the force is always perpendicular to the surface it is acting on. *0047

*Exerting pressure -- all states of matter can exert pressure. *0063

*You walk across an ice covered lake, you exert pressure on the ice equal to your weight, divided by the area, which contacts the ice. *0067

*That is why if you do not want to crack the ice, they teach you to spread your hands and feet out to spread out that force over a larger area, so you get less pressure. *0074

*If you walk on snow with snow shoes with large areas of contact, you increase the area; you reduce the pressure and you walk on top of the snow, that is why they are so big -- larger area. *0085

*Now, fluids exert outward pressure in all directions on the sides of any container holding the fluid. *0096

*Even Earth's atmosphere exerts pressure. *0102

*Atmospheric pressure is about 101,325 Pa, which will typically round you about 100,000 Pa or 10N/cm*^{2}. 0105

*And you can even experience this by riding in an airplane as you change altitudes -- go up and down -- you may experience a popping sensation in your ears. *0114

*The pressure inside your ear, balances the pressure outside your ear in a transfer of air through some small tubes connecting your inner ear to your throat. *0123

*When that happens, you hear that or feel that popping sensation. *0131

*Let us take a look at our first example -- pressure on a keyboard. *0138

*Air pressure is approximately 100,000 Pa, what force is exerted on your keyboard when it is sitting flat on a desk if the area of the keyboard is 0.035 m*^{2}. 0141

*Well if pressure is force divided by area, then that means force is pressure times area. *0153

*If our pressure is about 100,000 Pa × 0.035 m*^{2} (area), that will give us a force of about 3500N. 0162

*Let us take a look at an example of a sleepy fisherman. *0183

*A fisherman with a mass of 75 kg falls asleep on his four-legged chair of mass (5 kg).*0186

*If each leg of the chair has a surface area of 2.5 × 10*^{-4}m^{2} in contact with the ground, what is the average pressure exerted by the fisherman and chair on the ground? 0192

*Well, the force applied is going to be the force of gravity, so our pressure is going to be force/area, which is going to be mg/a.*0205

*Now our mass is 75 kg, the fisherman, plus 5 kg for the chair, times (g) about 10 m/s*^{2} all over the area...0217

*...which is going to be the four corners of the chair, the four legs of the chair times the area of each leg, 2.5 × 10*^{-4}m^{2} or about 800,000 Pa. 0226

*Let us take a look at another one. *0250

*A scale which reads 0 in the vacuum of space is placed on the surface of Planet Physica.*0252

*On the planet's surface, the scale indicates a force of 10,000N. *0258

*Calculate the surface area of the scale given the atmospheric pressure on the surface of Physica is 80,000 Pa. *0263

*If pressure is force/area, then area is force/pressure. *0271

*That will be 10,000N/80,000 Pa (pressure), or about 0.125 m*^{2}. 0279

*Let us see if we cannot rank some pressures. *0297

*Rank the following from highest pressure to lowest pressure upon the ground. *0301

*The atmosphere at sea level -- well that we know is right around 100 Pa. *0305

*A 7,000 kg elephant with total area of 0.5 m*^{2} in contact with the ground -- well pressure will be force/area or that will be 7,000 kg × g (10)/area of 0.5 or about 140,000 Pa. 0313

*A 65 kg lady in high heels with a total area of 0.005 m*^{2} in contact with the ground -- if pressure is force/area, that will be 65 kg × 10 m/s^{2}/area (0.005) or about 130,000 Pa. 0336

*And finally a 1600 kg car with a total tire contact area of 0.2 m*^{2} -- the pressure equals force over area, so that will be 1600 kg × 10 m/s^{2}/0.2 m^{2} or about 80,000 Pa. 0363

*So, if I were to rank these from highest pressure to lowest pressure, I would start with the elephant (B), go to the lady in high heels, atmospheric pressure, and finally the car, so (B), (C), (A), (D). *0384

*Let us talk about pressure on a submerged object. *0404

*The pressure a fluid exerts on an object submerged in that fluid is determined by multiplying the density of the fluid by the depth submerged, all multiplied by the acceleration due to gravity. *0407

*We call that the gauge pressure, which is ρ (density) × (g) × (h). *0417

*If there is also atmosphere above the fluid, such as the situation here on Earth, you can determine the absolute or total pressure by adding in the atmospheric pressure which we will write as P0, which is about 100,000 Pa. *0423

*So absolute pressure is atmospheric pressure plus gauge pressure -- P0 + ρ, where ρ is the density of the fluid, (gh). *0435

*Let us take an example of gauge pressure. *0448

*Samantha spots buried treasure while scuba diving on her Caribbean vacation. *0450

*If she must ascend to a depth of 40 m to examine the treasure, what gauge pressure will she read on her scuba equipment? *0454

*The density of sea water is 1025 kg/m*^{3}. 0461

*Well, we want gauge pressure, so that is going to be ρ (fluid), (gh) or 1025 kg/m*^{3} × g (10 m/s^{2} × her depth of 40 m or about 410,000 Pa. 0466

*If we want to look at absolute pressure here, let us find the absolute pressure for Samantha in the same scenario. *0493

*Now we are looking for absolute pressure and that is P0 + gauge pressure and we are going to say atmospheric pressure is about 100,000 Pa, for simplicity...*0500

*...so 100,000 + 410,000 Pa (gauge pressure) = 510,000 Pa of pressure as the absolute pressure. *0513

*Let us talk about Pascal's principle. *0530

*When a force is applied to a contained incompressible fluid, the pressure increases equally in all directions throughout that fluid. *0533

*This is the foundation for hydraulic systems and things like barber shop chairs, construction equipment, and even car brakes. *0540

*In car brakes, in the movies, you will even sometimes see people cut the brake lines so the brakes do not work -- the fluid leaks out and the brakes no longer work because the fluid is no longer contained. *0548

*It must be a contained fluid or incompressible or nearly incompressible fluid. *0558

*Let us talk about force multiplication using Pascal's principle, also known as the basis of hydraulics. *0566

*To begin with, we have a force (F1) that we are applying to a piston of area (A1) and that is going to create a pressure of (P1), so (P1) is caused by force (F1) applied to a piston of area (A1) on a contained, incompressible fluid -- we have a closed container, incompressible fluid here. *0573

*Now over on the right hand side, the pressure on this piston, must be F2/A2. *0594

*Why? By Pascal's principle, you must have the same pressure anywhere throughout the fluid. *0601

*Well when you do that, let us take a look at the ramifications. *0607

*If P1 = P2, by Pascal's principle, then that means F1/A1 = F2/A2 or if I cross-multiply (F1)(A2) = (F2)(A1) or F2 = A2/A1 × F1. *0611

*What does this mean? You have increased the force -- if you apply a force (F1) and you have a different area on your two pistons, you can increase that force by the ratio of the areas. *0637

*If (A2) is five times larger than (A1), and you apply force (F1), you get five times that force on (F2).*0650

*You have effectively increased your applied force. *0658

*Now you do not really get anything for free here. *0661

*What you are going to end up having by conservation of energy is you are also going to have to push this piston five times further or you will get 1/5 the displacement that you would over here for the same displacement over there. *0665

*The total work done on each side has to be the same for conservation of energy, and that is going to be by the same ratio as the area multiplier, but you can multiply a force if the areas are a ratio of 100:1, you have increased your force by a factor of 100 and that is the principle behind hydraulic systems. *0678

*Let us take the example of a barber's chair when we apply this. *0699

*A barber raises his customer's chair by applying a force of 150N to a hydraulic piston of area 0.01 m*^{2}. 0702

*If the chair is attached to a piston of area 0.1 m*^{2}, how massive a customer can the chair raise?0711

*Assume the chair itself has a mass of 5 kg. *0717

*Well, to solve this problem let us first determine the force applied to the larger piston. *0721

*We know (F2) must be equal to the ratio of the area, A2/A1 × F1, therefore, F2 = 0.1(A2)/0.01(A1) = 10 × F1 (150N) = 1500N. *0728

*This is the largest applied force you can have. *0750

*Now then, if we want to know how massive a customer the chair can raise, if our force is 1500N, that must be equal to the weight -- the maximum that we can handle -- therefore, the mass that you can handle is 1500N/g (10 m/s*^{2}) or 150 kg. 0758

*Now that 150 kg -- five of those kg have to be the chair, therefore, you could lift a customer of 145 kg, which is about 300 lbs. *0782

*Let us take a look at a hydraulic auto lift. *0806

*A hydraulic system is used to lift a 2,000 kg vehicle in an auto garage. *0808

*If the vehicle sits on a piston of area 0.5 m*^{2}, and a force is applied to an area of 0.03 m^{2}, what is the minimum force that must be applied to lift the vehicle? 0813

*Well, starting with Pascal's principle, we know that P1 = P2, assuming we have a contained incompressible fluid, therefore F1/A1 = F2/A2. *0825

*We are looking for (F1), so that is going to be A1/A2 × F2, which is 0.03 m*^{2}/0.5 m^{2} × (F2)... 0838

*...which is the weight of our vehicle, 2,000 kg × the acceleration due to gravity (10), therefore F1 must be 1200N. *0855

*We have to be able to apply 1200N in order to lift that 20,000N vehicle. *0871

*Let us take a look at the pressure on a penny. *0881

*A penny with a diameter of 19.05 mm sits on the bottom of the ocean, where we have a saltwater density of 1025 kg/m*^{3} at a depth of 340 m. 0883

*What is the force on the penny?*0894

*Let us figure out the area of the penny first. *0898

*The area for the penny is πr*^{2} and the radius of the penny will be half the diameter or half of 19.05 mm... 0900

*...and that is 0.009525 m*^{2} or about 2.85 × 10^{-4}m^{2}. 0913

*Now, if we are looking for the force on the penny, let us start by finding the pressure. *0928

*The absolute pressure is the atmospheric pressure plus the gauge pressure (ρgh)... *0935

*...so that will be about 100,000 Pa plus our density of our fluid (1025), the acceleration due to gravity (10 m/s*^{2}) and our depth of 340 m or about 3,585,000 Pa. 0943

*Now to find the force if P = F/A, then that means F = (P)(A), or 3,585,000 Pa × 2.85 (area) × 10*^{-4}m^{2} = 1,022N (force). 0965

*That is quite a force on a little penny. *0995

*How about depth in freshwater -- a diver's pressure gauge reads 250,000 Pa in freshwater. *1000

*How deep is the diver? *1007

*Well, gauge pressure is (ρgh), therefore h = P/ρg or 250,000 Pa/1,000 kg/m*^{3}, our density of freshwater × g (10) or 25 m. 1010

*One more -- A pressure gauge reads 350,000 Pa, what is the absolute pressure?*1040

*Well, just to review, absolute pressure is atmospheric pressure plus (ρgh)...*1049

*...or 100,000 Pa (atmospheric pressure) + 350,000 Pa (gauge pressure) = 450,000 Pa (absolute pressure). *1058

*All right. Hopefully that gets you a great start with pressure and Pascal's principle*1078

*Thank you so much for your time and for watching us on Educator.com. *1081

*Looking forward to seeing you again. Make it a great day!*1084

*Hi everyone and welcome back to Educator.com.*0000

*Today we are going to continue our study of fluids as we talk about the continuity equation.*0003

*Our objectives are going to be to apply the continuity equation to fluids and motion.*0008

*To explain the continuity equation in terms of conservation of mass flow rate.*0012

*Conservation of mass for fluid flow. When fluids move through a full pipe, the volume of fluid entering the pipe must be equal to the volume of fluid leaving the pipe.*0017

*The Law of Conservation of Mass for Fluids.*0027

*This holds true even if the diameter of the pipe changes.*0030

*In short, what we call the volume flow rate remains constant throughout the pipe.*0034

*And we will look through a couple of applications of that here.*0055

*Volume flow rate. The volume of fluid moving through the pipe can be quantified in terms of volume flow rate.*0059

*The volume flow rate is the area of the pipe times the velocity of the fluid, and it must be constant throughout the pipe.*0065

*So over here on the left-hand side, if we are looking at a pipe with a changing diameter, we have Area1, where the fluid has Velocity1.*0072

*Over here on the right-hand side, we have Area2 and Velocity2.*0079

*A1V1, the volume flow rate on the left-hand side, must be equal to A2V2, the volume flow rate on the right-hand side.*0083

*What that means practically is that you must have a higher velocity or a faster flow over here and a slower flow over here.*0091

*Let us look at some examples and applications.*0104

*Water runs through a water main of cross-sectional area 0.4 square meters with a velocity of 6 meters per second. *0107

*Calculate the velocity of the water in the pipe when the pipe tapers down to a cross-secitonal area of 0.3 meters squared.*0114

*Well, continuity equation for fluid says A1V1 must equal A2V2.*0122

*Therefore, Velocity2 at the skinnier section of the pipe must be equal to A1 over A2 times V1. *0128

*Or 0.4 divided by 0.3 square meters times that 6 meters per second or 8 meters per second.*0129

*It gets a little narrower, it gets a little faster.*0150

*Let us take a look at the garden hose example.*0156

*A lot of folks have probably done this before.*0158

*As you are watering the garden or playing with the hose, you want the water to come out a little bit faster so you cover up the end of the nozzle with your thumb a little bit.*0160

*You decrease that cross-sectional area so that the water has to come out faster to maintain that volume flow rate.*0167

*In this problem, the water enters a typical garden hose of diameter 1.6 centimeters with the velocity of 3 meters per second.*0174

*Calculate the exit velocity of water from the garden hose when a nozzle of diameter half a centimeter is attached to the end.*0181

*First let us figure out what the cross-sectional areas are.*0188

*When it is entering the pipe, A1 is πr1*^{2}, or π times. If our diameter is 1.6 centimeters, our radius must be 0.8 centimeters. 0192

*So that is 0.008 square meters, or an area of about 2.01 times 10*^{-4} square meters.0204

*Area 2 at the nozzle is πr2*^{2} or π times. Well its diameter is 0.5 centimeters so its radius is half of that, 0.25 centimeters, 0.0025 meters squared.0215

*Which is 1.96 times 10*^{-5} square meters.0231

*Now we can apply our continuity equation for fluids.*0239

*A1V1 equals A2V2. This implies then that V2 equals A1 over A2 times V1.*0245

*Or 2.01 times 10*^{-4} square meters over 1.96 times 10^{-5} square meters.0259

*All times V1 which was 3 meters per second, for a total of about 30.8 meters per second.*0270

*It comes out a lot faster when you decrease that area.*0282

*Let us take a look at an oil pipe line problem.*0287

*Oil flows through a pipe of radius (r) with speed (v).*0291

*Some distance down the pipe line, the pipe narrows to half its original radius.*0294

*What is the speed of the oil in the narrow region of the pipe?*0299

*Well, A1 we will call πR*^{2}. A2 is going to be πR/2^{2}, which is going to be πR^{2}/4 or π/4R^{2}.0303

*Now as we apply the continuity equation for fluids, A1V1 = A2V2, which implies then that V2 = A1/A2(V1).*0325

*A1 = πR*^{2}, A2 = π/4R^{2} times V1 which we will just call V.0341

*We are going to have some simplifications, R*^{2}, R^{2}, π, π.0356

*We have 1/, 1/4 times V, which is going to be equal to 4V.*0361

*That is 4 times faster.*0370

*One last problem here.*0375

*So we look at the roots of the continuity equation, which statement below best describes the continuity equation for fluids?*0377

*Energy is conserved in a closed system? Mass is conserved in a closed system? Linear momentum is conserved in a closed system?*0384

*Angular momentum is conserved in a closed system? Or charge is conserved in a closed system?*0391

*Well, we are really talking about a mass conservation here.*0398

*The volume flow rate is basically saying, the continuity equation is saying that the mass that goes in must come out.*0404

*Therefore, mass is conserved in a closed system. *0409

*Hopefully this gets you a good start on the continuity equation for fluids.*0414

*Thanks for watching and make it a great day.*0418

*Welcome back to Educator.com. *0000

*Today we are going to finish up our study of fluids, by talking about Bernoulli's principle.*0002

*Our objectives are going to be to understand how Bernoulli's principle describes the conservation of energy in fluid flow and apply Bernoulli's principle to problems of fluids in motion. *0008

*Let us start by talking about Bernoulli's principle qualitatively. *0021

*Bernoulli's principle states that fluids moving at higher velocities lead to lower pressure and fluids moving at lower velocities lead to higher pressures. *0024

*This comes into play quite a bit when you talk about the design of an airplane wing. *0035

*If the airplane's moving forward to the right, then as the air goes over the wing, it has a longer path up above the wing. *0040

*And because it has a longer path compared to below the wing, where it has a shorter path above the wing with a longer path, it has a higher relative velocity. *0049

*If you have the higher air velocity over that wing, you get lower pressure and where you have the shorter path, you end up with a lower velocity where the air does not have to travel as far in the same amount of time, so you end up with a higher pressure. *0070

*If you have higher pressure below and lower pressure above that wants to equal out and you get a net force up. *0090

*That force is one of the components of lift. *0099

*Now note that it is not the only component and that there is a lot more to lift and flight than just Bernoulli's principle in the airplane wing shape, but that is one component of it -- one application of Bernoulli's principle. *0102

*Another one is called a venturi pump. The idea here is it that you can make a vacuum pump or another type of pump using fluid flow. *0116

*If you have fluid or an incompressible fluid in a closed system coming through from the left to the right here, and as it goes there, you have a narrow opening, you must have faster flow here -- we know that from our continuity equations for fluids. *0125

*And if we have faster flow, we must have lower pressures here. *0140

*As we have lower pressures there, what we are going to get is a pumping action in which we can start sucking things up this way to join that fluid flow. *0144

*So we force a lot of water or some other fluid through the pipe this way, constricted here and you are going to get a sucking action, you are going to get a pumping action. *0157

*These are used in venturi pumps -- applications of those -- things like carburetors... *0168

*...It is responsible for sailboat propulsion, gas delivery systems, or even for some folks who have sump pumps in case the power ever goes out.*0173

*Often times you will see these as a backup sump pump. *0183

*When the power goes out, if the water level gets too high in the sump, what happens it is lifts a float that turns on a bunch of water pressure and the water pressure comes running through here and this is connected to that cistern and sucks up the water to pull it out with that pumping action. *0186

*This is a way to help keep your basement dry even when you do not have power to keep that sump pump running. *0205

*All right. Bernoulli's equation quantitatively looks a little bit scarier. *0212

*It relates the pressure velocity and height of a liquid in a tube at various points. *0217

*Do not let it scare you, it is all a fairly simple equation, it just looks like a lot all at once. *0221

*Pressure 1 plus 1/2 the fluid density times the square of its velocity plus the fluid density times (g) times its height equals (P2 plus 1/2 ρv2*^{2} plus ρgy2^{2}. 0226

*Really what this is is a statement of conservation of energy. *0241

*Notice how similar this looks -- 1/2 ρv*^{2} to 1/2mv^{2} -- similar to kinetic energy -- ρgy is similar to (mgh), which is similar to gravitational potential energy. 0244

*These are some pretty close parallels here. *0256

*It really is a version of conservation of energy. *0258

*What it says is the pressure at any point in the tube plus 1/2 the density times the square of the velocity added to (ρgy) must be the same anywhere at any point in the tube. *0262

*As we check this out, let us take a look and use it to derive what is known as Torricelli's theorem. *0276

*If we have water sitting in a large jug at a height of 0.2 m above the spigot, what is the pressure on the spigot and at what velocity will the water leave the spigot when the spigot is opened? *0282

*Well, the first thing we need to realize is that (P1) up here and (P2) up here are both open to atmosphere. *0294

*So we are going to say P1 = P2 = Atmosphere, and we are not going to worry about the difference in height compared to the overall atmospheric pressure as that is going to be negligible. *0300

*Now as we start to look at this, we will start by writing down Bernoulli's equation -- P1 + 1/2ρv1*^{2} + ρgy1 = P2 + 1/2ρv2^{2} + ρgy2. 0311

*Now as we look at this, up here at the top of our fluid, because we have so much fluid there, we can assume that (v1) is approximately equal to 0.*0335

*Now, (P1) and (P2) are both atmosphere, so we can subtract both of those right out and that will simplify right there. *0346

*We said (v1) is approximately 0, and that term goes away, so what we are left with now is ρgy1 = 1/2ρv2*^{2} + ρgy2. 0354

*All right, we are getting closer already. *0372

*I can divide out the density to say that gy1 = 1/2v2*^{2} + gy2... 0374

*...or solving for (v2) -- v2*^{2} = 2g × y1 - y2 (the quantity) = 2 × 10 and our height difference from y1 to y2 is just 0.2 m...0387

*...so that implies that v2*^{2} is going to be equal to 4 or take the square root of v2 must equal 2 m/s. 0405

*And what we have really done here is we have derived at what is known as Torricelli's theorem -- this part right here v2*^{2} = 2gy1 - y2 or more commonly written -- take the square root, v2 = 2g × y1 - y2, square root...0418

*...calculating the velocity coming out of a container of liquid like this, Torricelli's theorem. *0439

*We can look at an example with gauge pressure here too. *0448

*Water flows through a large diameter pipe at Point (A) before it is constricted into a smaller diameter pipe at Point (B). *0450

*How does the gauge pressure compare at Points A and B? *0458

*Well, if the water is going through this pipe and it is being constricted, it must be going faster right? *0462

*So the velocity at (B) must be greater than the velocity of (A). *0468

*We know that from out continuity equations for fluids. *0472

*If it is going faster at (B) then (B) must have lower pressure than (A), therefore (A) must have a higher pressure than (B). *0476

*Let us take a look at the shower problem. *0497

*A water main of area 0.003 m*^{2} at ground level flows at 2 m/s into Kate's house. 0501

*At the second floor shower head, 5 m above ground level, the pipe has an area of 0.001 m*^{2}. 0507

*Find the velocity of the water in the pipe as well as the gauge pressure just prior to the shower head if the water main's pressure gauge reads 2 atmosphere. *0513

*Well let us start with a diagram here -- water main of area 0.003 -- so we are going to start over here with a pipe over here at ground level and we know if we put a gauge on it, that it is going to read 2 atmosphere's right there. *0523

*It has a cross-sectional area of 0.003 m*^{2} and it is flowing at 2 m/s into the house -- that is at section one. 0537

*Now, somewhere up here it has a height difference of about 5 m and before it comes out, it is now down to a cross-sectional area of 0.001 m*^{2}. 0548

*We need to figure out, here at section two, area two, what the velocity is of the water in the pipe as well as the gauge pressure just prior to that shower head. *0564

*To start off to find the velocity of the pipe, I am going to use the continuity equation for fluids. *0573

*a1v1 = a2v2, therefore the velocity here at Point 2 is going to be equal to a1/a2 × v1 or 0.003/0.001 × 2 m/s (v1) = 6 m/s. *0579

*All right, now let us see if we can find out the gauge pressure, just prior to that shower head. *0611

*We will use Bernoulli's equation, P1 + 1/2ρv1*^{2} + ρgy1 = P2 + 1/2ρv2^{2} + ρgy2. 0618

*Now, if we set (y1) over here and call this ground level, y1 = 0, that term becomes 0, and that goes away. *0640

*Now let us start substituting in our values to see if we cannot find what (P2) is going to be. *0649

*Over here at (P1), we already know our pressure is 2 atmospheres or 200,000 Pa + 1/2 × 100,000 kg/m*^{3} (density of freshwater) -- v1 is 2 m/s^{2}... 0655

*...equal to P2 + 1/2(1,000) (density), v2*^{2} (6 m/s^{2}) + ρ (1,000) × g(10) and y2 is 5 m higher, so 5. 0677

*So a little bit of math here -- 200,000 + 1,000 × 4 -- 1/2 of that will be 2,000 = P2 + 36,000 × 1/2 (18,000) + 10,000 × 5 = 50,000...*0701

*... so solving for P2 then = 202,000 - 68,000 or 134,000 Pa...*0721

*...or approximately 1.34 atmospheres. *0734

*Great! Let us take a look at a water fountain example. *0747

*Sandy is designing a water fountain for her front yard. *0751

*She would like the fountain to spray to a height of 10 m -- that is a pretty impressive water fountain. *0754

*What gauge pressure must her water pump develop? *0759

*Well, let us start with a diagram again. *0763

*We will start over here at her pump and that is going to go to a point where it is going to release the water up at ground level and once it is there, we want the water to go up to a height of 10 m or so. *0765

*We will start again with Bernoulli's equation -- P1 + 1/2ρv1*^{2} + ρgy1 = P2 + 1/2ρv2^{2} + ρgy2. 0783

*If we call this section over here on the left (1) and over here (2), right away we can make some simplifications. *0802

*At (1), we will assume that we have so much water that the velocity there in the pipe is roughly 0 -- that term goes away. *0813

*We are also doing this at ground level, so y1 = 0. *0823

*On the right hand side at its highest point right here, we want the velocity of the water to be 0 and that is what happens when it gets to its highest point, so velocity (2) will go to 0, therefore, P1 = P2 + ρgy2. *0826

*We are looking for P1 and P2 is open to atmosphere, so we know that is going to be 100,000 Pa + ρ(1000) × g (10 m/s*^{2}) × y2...0848

*...we want that 10 m high, so P1 = 100,000 + 1,000 × 100 for a total of 200,000 Pa. *0862

*So that is the pressure that we need total, so 200,000 Pa is equal to P1, which is equal to atmospheric pressure + ρgh... *0879

*...where P0, here, is our atmospheric pressure (100,000 Pa) and ρgh here is our gauge pressure. *0894

*So if 200,000 = 100,000 + gauge pressure, that means our gauge pressure must be 100,000 Pa in order to make the water fountain shoot the water 10 m high. *0908

*Let us take a look at one more -- an elevated cistern problem. *0925

*We have a water cistern that is elevated 15 m above the ground and it feeds a pipe that terminates horizontally 5 m above the ground as shown. *0930

*With what velocity will the water leave the pipe and how far from the end of the pipe, will the water strike the ground?*0938

*The first thing I am going to do is try to come up with a strategy here. *0945

*I think I can use Bernoulli's principle to find the velocity of the water right here at what we will call Point (2) and then it becomes a projectile problem as to where it is going to land. *0949

*If we call this our Point (1), we have a height of 15 m here and a height of 5 meters here, to find the velocity, why do I not just bring this back and if I call Point (2) ground level for the first part of the problem for figuring out the velocity, then the height here will be 10 m because that is the difference. *0960

*So let us apply Bernoulli's equation and see how this is all going to look and work out. *0980

*Bernoulli's equation -- P1 + 1/2ρv1*^{2} + ρgy1 = P2 + 1/2ρv2^{2} + ρgy2. 0985

*As we look at that, some simplifications we can make -- P1 is open to atmosphere; P2 is open to atmosphere, so they will have the same pressure and we can subtract those out of both sides -- v1 is going to be roughly 0, so we can make that go away. *1003

*On the right hand side, if we are calling this the 0 height level for the first part of our problem, setting that as our 0 and the height here is 10, so we can make that term go away. *1021

*So we have simplified Bernoulli's equation to say that ρgy1 = 1/2ρv2*^{2} or as we substitute in our values -- first off we can get rid of the rho's. 1031

*We have gy1 = 1/2v2*^{2} or v2^{2} = 2gy1 or v2 = the square root of 2gy1. 1050

*Notice how similar that looks -- v = square root of (2gh), the conservation of energy value we found in order to determine how fast something is moving after it has been dropped some distance...*1062

*...V = square root of 2gh by kinematics or by conservation of energy approach. *1077

*It is the same idea, so (v2) will be equal to the square root of 2 × g (10 m/s*^{2}) × y1 (10 m) or the square root of 200, that is about 14.1 m/s.1083

*Our water is going to be leaving the pipe down here with a horizontal velocity of 14.1 m/s. *1098

*Now we have ourselves a projectile problem, where we have a height of 5 m and we need to find the horizontal distance the water travels. *1106

*All right. Well let us first figure out how long that water is going to be in the air. *1116

*That is a vertical kinematics problem, where V-initial vertically is 0, δy will be 5 meters, acceleration will be 10 m/s*^{2} and we need to find the time. 1120

*I would use the equation δy = V-initial(t) + 1/2 at*^{2}, but again, V-initial is 0, so that term goes away. 1135

*(T) then becomes 2δy/a (square root) or 2 ×5/10 (square root) or 1 s. *1148

*Now I can use my horizontal kinematics to figure out how far it goes. *1162

*Horizontally, the velocity is going to be a constant 14.1 m/s and it is going to be in the air for 1 s, so δx is just going to be velocity × time, 14.1 × 1 or 14.1 m. *1168

*Putting a couple of these concepts together to get a big picture solution. *1187

*All right. Hopefully that gets you a great start with Bernoulli's principles and Bernoulli's equations. *1192

*Thanks for watching. Make it a great day everyone!*1197

*Hi everyone and welcome back to Educator.com. *0000

*I am Dan Fullerton and I am thrilled to be opening up our unit today on thermophysics. *0003

*We are going to start with heat temperature and thermal expansion. *0008

*Our objectives are going to be to calculate the temperature of an object given its average kinetic energy, to describe the temperature of a system in terms of a distribution of molecular speeds, and describe thermal equilibrium as a probability process where energy is typically transferred from high to low energy particles. *0011

*We will also explain heat as the process of transferring energy between systems at different temperatures, and finally calculating the linear and volume metric expansion of a solid as a function of its temperature.*0027

*Let us talk about thermophysics. *0042

*Thermophysics explores the internal energy of objects due to the motion of the atoms and molecules comprising the objects. *0045

*It explores the transfer of this energy from object to object, known as heat, a transfer mechanism for energy. *0052

*Let us start with temperature. *0061

*The internal energy of an object, known as its thermal energy is related to the kinetic energy of all the particles comprising the object. *0063

*The more kinetic energy the constituent particles have as they move in their vibrations as part of that object, the greater the objects thermal energy. *0071

*For most systems, the kinetic energy of the constituent particles is not the same, it is a distribution, therefore the system is modeled as a distribution of kinetic energies, typically using Maxwell Boltzmann's statistics. *0080

*As we talk about temperature and phases of matter, in solids the particles comprising the solids are held together very tightly, therefore their motion is limited to just vibrating back and forth in their given positions. *0103

*In liquids, the particles can move back and forth across each other, but the object itself does not have a defined shape. *0116

*In gases, the particles move throughout the entire volume available, but in all cases the total thermal energy is the sum of the kinetic energies of the constituent particles. *0122

*Average kinetic energy and temperature -- actual kinetic energies of individual particles may vary significantly and the average kinetic energy we can find by taking 3/2 times this constant K*_{b}, known as Boltzmann's constant times the temperature and that temperature should be in Kelvins (K), our si unit of temperature. 0136

*If K*_{b} is Boltzmann's constant, that is 1.38 × 10^{-23} J/K -- the temperature is in Kelvins, the si unit of temperature again, not Celsius, not Fahrenheit, but Kelvins. 0157

*Now it is important to note that even though two objects can have the same temperature and therefore the same average kinetic energy, they may have different internal energies, depending on what those particles that are moving are. *0172

*Let us take a look at some temperature scales. *0186

*We have Fahrenheit (F) on the left, Celsius (C) in the center, and Kelvins (K) on the right. *0189

*Now they all work in the same basic way, but they have different values at different key temperature readings. *0194

*The Fahrenheit scale has water freezing at 32 ° F and water boiling at 212 ° F and if you extrapolate that back, you get to what is known as absolute 0 at about -459.7 ° F, where absolute 0 is a theoretical minimum temperature. *0199

*It is the point on a volume vs. temperature graph on a gas where the extended curve would hypothetically reach 0 volume. *0220

*It is not specifically the absolute lack of motion of particles, it is a theoretical minimum, but for our purposes, really cold and you do not get any colder than that. *0230

*For Celsius, water boils at 100 ° C, freezes at 0 ° C, and absolute 0 would be -273.15 ° C. *0240

*Now Kelvins, the scale we are going to use here in physics -- and Kelvins has the same size of its main unit, a Kelvin -- 100 ° between water freezing and boiling, but the only difference is we are going to start 0 at absolute 0. *0250

*That means that water freezes at 273.15 K and it boils at 373.15 K. *0268

*Converting between temperature scales is fairly straight forward. *0277

*If you know Celsius, to get Kelvins, you add 273.15. *0280

*If you know Celsius and you want Fahrenheit, multiply the Celsius temperature by 9/5 and add 32. *0286

*If you know Fahrenheit and you want Celsius, take the Fahrenheit temperature, subtract 32 and multiply by 5/9. *0293

*Now let us talk about heat as the transfer of thermal energy from one object to another object due to their difference in temperature. *0304

*That is typically accomplished through some sort of particle interactions or collisions in which momentum is transferred from one object to another. *0313

*That is conduction. *0319

*Now energy is typically transferred from higher energy to lower energy particles and after many collisions, both systems of particles likely have the same average temperature. *0322

*That does not mean that every collision works this way, but on the average it goes in that general direction. *0330

*Because the particles comprising objects have a distribution of particles, velocities, and energies, on the microscopic scale, this transfer of energy is a probabilistic process. *0335

*So as you look at it more and more closely, you have to get more and more into statistics of distribution. *0345

*So methods of heat transfer -- We can transfer heat from one object to another by three different methods. *0353

*Conduction is the transfer of energy along an object to the particles comprising the object colliding. *0359

*Think of sticking an iron rod in the fire. *0364

*Okay, the fiery end is going to get hot real quick, but if you hold that long enough, the other end that started off cool is going to get pretty hot. *0368

*That is by the transfer of the energy from particle to particle to particle in that object. *0374

*Convection is the transfer of energy as a result of energy or heated particles moving from one place to another, like the convection ovens -- heated air molecules move from one place to another. *0379

*And radiation is the transfer of energy through electromagnetic waves. *0389

*Now as we try and quantify heat transfer in conduction, we can get a look at the rate of heat transfer, (H) in J/s or also watts (W). *0397

*Heat is K × A × δt/L, where δt is your temperature gradient, the difference in temperature; (A) is the cross-sectional area, (L) is the length of your object, and (K) is a thermal conductivity depending on the material, typically something you would look up, a material property. *0407

*Now, I have put in here a table of some thermal conductivities of selected materials and on the left we have materials such as aluminum, concrete, copper, glass, stainless steel, and water and on the right they are thermal conductivities in J/(s-m-K). *0432

*For example, copper has a much higher thermal conductivity than something like water. *0447

*Let us see how we can put this into practice. *0455

*What is the average kinetic energy of the molecules in a steak at a temperature of 345 K?*0458

*Well the average kinetic energy is given by 3/2 times Boltzmann's constant times the temperature. *0464

*So that will be 3/2 × 1.38 × 10*^{-23} (Boltzmann's constant) and a temperature of 345 K is going to give us an average kinetic energy of about 7.14 × 10^{-21} J. 0474

*Let us take a look at another example this time dealing with body temperature. *0498

*Normal canine body temperature is 101.5 F. What is normal canine body temperature in degrees C and K?*0504

*Well let us convert temperature in degrees C -- is 5/9 times the temperature in degrees F minus 32. *0513

*So that will be 5/9 × 101.5 - 32 = 38.6 ° C. *0523

*Now let us convert that to K. *0539

*Temperature in K is the temperature in degrees C plus 273.15, so that will be 38.6 + 273.15 = 311.75 K. *0543

*Let us look at the temperature of space. *0570

*The average temperature of space is estimated as roughly -270 ° C, that is cold. *0574

*What is the average kinetic energy of the particles in space? *0579

*Well, first thing we are going to do is convert to K, so the temperature in K is the temperature in degrees C plus 273.15, so that will be -270 + 273.15 = 3.15 K. *0583

*Average kinetic energy then, is going to be 3/2 times Boltzmann's constant times our temperature, or 3/2 × 1.38 × 10*^{-23} × 3.15 K (temperature) or 6.5 × 10^{-23} J. 0608

*All right, let us look at the temperature of the sun. *0644

*Given the average kinetic energy of the particles comprising our sun is 1.2 × 10*^{19} J. 0648

*Find the temperature of the sun in K. *0654

*Well, if average kinetic energy is 3/2 K*_{b}t, then that means the temperature must be 2 times the average kinetic energy divided by 3 times that Boltzmann's constant, K_{b}. 0658

*Or 2 × 1.2 × 10*^{19} J/3 × 1.38 (Boltzmann's constant) × 10^{-23}, or 5800 K. 0675

*Let us take a look at a heat transfer problem. *0698

*Let us find a rate of heat transfer through a 5 mm thick glass window with a cross-sectional area of 0.4 m*^{2} if the inside temperature is 300 K and the outside temperature is 250 K. 0701

*Well the rate of heat transfer (h) is Kaδt/L, where if we look up (K) for glass, we can find that the thermal conductivity of glass is about 0.9. *0713

*So that is going to be 0.9 times the cross-sectional area (0.4) times δt, the change in temperature, is 50 K (temperature gradient) divided by our length (5 mm or 0. 0.005 m). *0729

*So our heat transfer rate is going to be about 36 J/s or 3600 W. *0748

*Let us look at heat transfer across a rod. *0761

*One end of a 1.5 m stainless steel rod is placed in an 850 K fire. *0764

*The cross-sectional area of the rod is 1 cm and the cool end of the rod is at 300 K. *0770

*Calculate the rate of heat transfer through the rod. *0776

*Well, first let us figure out that cross-sectional area. *0779

*Area is πr*^{2}, so that is going to be π times our radius (1 cm), so that is 0.1 m^{2} or 3.14 × 10^{-4}m^{2}. 0782

*Now we are also going to need the thermal conductivity of steel and there are different conductivities depending on the types of steel, but let us just assume an average thermal conductivity of steel, rough estimate of about 16.5. *0801

*Our rate of heat transfer (h) is Kaδt/L, where (K) for steel is 16.5. *0818

*Our cross-sectional area, we just determined was 3.14 × 10*^{-4}m^{2} and our temperature gradient from 850 K to 300 K is 550 K...0828

*...divided by the length of our rod 1.5 m or 1.9 J/s or 1.9 W. *0842

*All right, so you know when objects are heated, they tend to expand and when they are cool, they tend to contract and at higher temperatures, objects have higher average kinetic energies so their particles vibrate more. *0857

*At those higher levels of vibrations those particles are not bound as tightly to each other, so the object expands -- exact opposite, as it cools down, they do not vibrate as much and they are bound a little bit more tightly, so they contract. *0870

*This is why if you have a stuck jar of pickles or something and you are trying to open it and you cannot quite untwist it, go try and run it under hot water because if you run it under hot water, the lid is going to start expanding. *0885

*It is going to expand at a faster rate than the glass, so if you run it under hot water, you give yourself a little bit more room and you loosen it up so hopefully, now you are strong enough to undo the lid. *0898

*Linear expansion -- the amount of material expands is characterized by the materials coefficient of expansion. *0912

*One-dimensional expansion, we use the linear coefficient of expansion which gets the symbol α.*0918

*So the change in an object's length due to linear expansion is this -- linear coefficient of expansion times its initial length times its change in temperature (δt). *0925

*For volumetric expansions, the amount of material that expands is again characterized by the coefficient of expansion, but if it is three-dimensional expansion, you use the volumetric coefficient of expansion, which gets the symbol β. *0938

*The change in volume is that coefficient of expansion, the volumetric coefficient of expansion, β times the initial volume times the change in temperature. *0951

*In most cases the volumetric coefficient of expansion is roughly 3 times the linear coefficient of expansion and that change in temperature can be provided in either ° C or K because the sign of the individual units are the same and we are looking at a relative change, not an absolute C or K -- it does not really matter for these problems. *0963

*So, some coefficients of thermal expansion again. *0984

*We have aluminum, concrete, diamond, glass, stainless steel, and water and we have the linear coefficient of expansion and the volumetric coefficient of expansion. *0987

*Now water is a little bit tricky here. *0996

*Although I have included it here, it actually expands when it freezes, so calculations near the freezing point of water require a little more detailed analysis than is provided here. *0999

*There is a window of a couple of degrees in water, that make it a little bit more complicated, so just keep that in mind, that this is not the full story for water. *1008

*Let us take a look at a contracting railroad tie. *1020

*A concrete railroad tie has a length of 2.45 m on a hot sunny 35 ° C day. *1023

*What is the length of the railroad tie in the winter when the temperature dips to -25 ° C? *1029

*Well, if it is a concrete railroad tie, let us find the linear coefficient of expansion for concrete and for concrete, that just so happens to be about 12 × 10*^{-6}.1036

*So, δL is equal to α L-initial δT. *1050

*That is 12 × 10*^{-6} × 2.45 m (initial length) × -60 (temperature change) or a total of -0.0018 m. 1059

*So what is the new length of the railroad tie? *1078

*Well, δL is equal to (L) - L-initial -- δ anything is the final value minus the initial.*1080

*Therefore the final value is going to be δL + L-initial...*1087

*...which will be -0.0018 m + 2.45 m, so its new length will be about 2.448 m. *1095

*All right, pretty straightforward. *1115

*Let us take a look at the expansion of an aluminum rod. *1117

*An aluminum rod has a length of exactly 1 m when it is at 300 K. *1121

*How much longer is it when placed in a 400 ° c oven? *1125

*Well, a couple of things I am going to need to know here. *1130

*First I am going to need to convert this temperature to K, because I start at K and then I am crossing over to C, that is kind of tough to tell the temperature difference between the two. *1133

*First, let us convert that -- our temperature in K is our temperature in ° C + 273.15, so that is going to be 400 ° C +273.15 or 673.15 K. *1143

*For dealing with the expansion of aluminum, I am also going to have to know that α, the linear coefficient of expansion for aluminum is about 23 × 10*^{-6}, so now I can find that shift in length. 1164

*Δ L is α L-initial δT or 23 × 10*^{-6} × 1 m (initial length) × 673.15 (change in temperature) to 300 K... 1182

*...is going to be about 373.15 or a total change in length of about 0.0086 m. *1200

*How about looking at some volumetric expansion. *1218

*A glass of water with volume 1 liter is completely filled at 5 ° C. *1221

*How much water will spill out of the glass when the temperature is raised to 85 ° C? *1227

*Well, we have to realize here that both the glass and the water are going to expand, so let us see how much each expands and find the difference between those two. *1234

*If we start with the water, the change in volume is going to be β, the volumetric coefficient of expansion times its initial volume times our difference in temperature and the volumetric coefficient of expansion for water (β) is 207 × 10*^{-6}, so that is 207 × 10^{-6} × 1 L × 80 ° or about 0.0166 L. 1244

*The glass is a slightly different story. *1267

*Change in volume is (β)(V0δT) again, but the volumetric coefficient of expansion for the glass is 27 × 10*^{-6} × 1 L × 80 ° or about 0.0022 L. 1291

*We have considerably more expansion from the water than the glass, so how much is going to spill out? *1312

*We are going to take the difference of these two, 0.0166 L - 0.0022 L to find out that we have 0.0144 L spilling out. *1318

*All right, let us take a look at an example problem where we are looking at some graphs of average kinetic energy vs. temperature. *1338

*Which graph best represents the relationship between the average kinetic energy of the random motion of the molecules of an ideal gas in its absolute temperature. *1344

*Well, first of all, let us write down that relation. *1353

*The average kinetic energy is 3/2 Boltzmann's constant times (t). *1356

*Notice that we have a direct linear relationship between the average kinetic energy and the temperature. *1365

*As temperature goes up, average kinetic energy goes up. *1372

*There it is -- our direct linear relationship. *1376

*Let us take a look at one more. *1382

*Jodie cannot remove her wedding ring. *1387

*If she runs the entire ring under hot water, what is going to happen to the hole in the middle>? *1389

*Will it expand, contract, or stay the same? Well, here is how we are going to treat this. *1394

*We are going to find what happens if we treat this as two rings, an outer ring and an inner ring. *1400

*Let us treat it as a circle, a bigger circle and a more little circle. *1407

*The big circle is going to expand and the inner circle is also going to expand. *1410

*Let us expand them both and then we are going to recombine them and when we do that what we are going to find is if the inner one has expanded and the outer one has expanded, of course this is where the finger goes inside that one and that one is expanded as well, therefore, they both expand. *1418

*In linear expansion, every linear dimension of an object changes by the same fraction when it is heated or cooled. *1434

*That is a good way to get the ring off -- run it under hot water, hopefully it expands -- maybe try a little bit of dish soap or some lubricant there as well. *1441

*Hopefully that gets you a good start on temperature, heat, and thermal expansion. *1450

*Thank you so much for your time and make it a great day! *1454

*We will see you soon.*1457

*Hi everyone, and welcome back to educator.com. This lesson is on vectors and scalars.*0000

*Our objectives are going to be to differentiate between scalar and vector quantities, to use scaled diagrams to represent and manipulate vectors, being able to break up a vector into x and y components, finding the angle of a vector when we are given it's components...*0006

*...and finally, performing basic mathematical operations on vectors such as addition and subtraction.*0021

*When we talk about vectors, what we are really talking about are different types of measurements in physics, different quantities, and there are really two types. Scalars and vectors.*0028

*Scalars are physical quantities that have a magnitude or a size only. They do not need a direction. Things like temperature, mass, and time. I know what you are thinking. Time has a direction, right? forward or backward. Well, not really.*0039

*When we are talking about direction, we are talking about things like north, south, east, west, up, down, left, right, over yonder, over yander.*0055

*That sort of direction. Forward or backward when we are talking about just a positive or negative value is not what we are talking about here.*0061

*On the other hand, vectors are quantities that have a magnitude and a direction. They need a direction to describe them fully, things like a velocity. You have a velocity of 10 m/s in a direction.*0071

*A force is applied in that direction. Or a momentum, you have a momentum in a specific direction.*0083

*Vectors, we typically represent by arrows. The direction of the arrow tells you the direction of the vector, obviously, and the length of the arrow represents the magnitude or size. The longer the arrow, the bigger the vector.*0091

*Let's take a look at a couple of vector representations. Let's call this nice happy blue one A, and this red one down here B.*0105

*Notice they both have the same direction but A is much smaller than B. A has a smaller magnitude than B. B is the longer arrow, with larger magnitude.*0118

*Now the other thing that is nice about vectors is as long as you keep their magnitude, their size and their direction the same, you can slide them around anywhere you want. You can move a vector as long as you do not change it's direction or its magnitude.*0128

*So if we want to, we could take vector A and instead of having it there, we can slide it somewhere over here, for example, give it the same direction and magnitude, make this one go away, and now there is A.*0141

*With the same value, same direction, same magnitude, we are allowed to move them like that.*0152

*Let's talk about how we would add up two vectors. A vector such as A and B. The little line over that means it's a vector. If we want to try and put together, A and add it to vector B, to get sum vector, C. The sum of those two vectors.*0161

*Well, graphically, here is the trick. Take any vectors you want to add, however many there are and if we slide them around so they are lined up tip to tail, we can then find the resultant, the sum of the vectors.*0184

*So here we have A and B but they are not lined up tip to tail. So, what I am going to do is I am going to redraw these so I am going to put A over here and B, I am going to line up so that it is now tip to tail with A. Hopefully something roughly like that.*0199

*So now we have A and B lined up so that they are tip to tail. To find the sum of the two vectors, all we have to do is draw a line from the starting point of the first to the ending point of our last vector, that must be the sum of the vectors, C.*0217

*Alright, does it make a difference what order you add things? Well if you think back to math, B plus A should be the same thing, and it is.*0240

*Lets prove it. We are going to redraw this now but we're going to do B first.*0256

*So what I am going to do is I am going to draw B down here, there's roughly B and now I am going to put A on it but I am going to line them up tip to tail, in this direction this time so B comes first and then A.*0262

*Once again when I go to draw the resultant, I go from the starting point of the first to the ending point of the last. Notice that I have the same thing. Same magnitude, same direction, same vector, same result.*0277

*Alright, how about graphical vector subtraction? Here we have A again and B. Put the line over them to indicate they are vectors. What do we do for A minus B?*0297

*The trick here is realizing that A minus B is the same as A plus the opposite of B.*0313

*What is the opposite of B? Well it is as simple as you might guess.*0319

*If we have B pointing in this direction with this magnitude, all I have to do is switch it's direction and there is negative b.*0324

*So if I want A plus negative B, let's just redraw them again, tip to tail. We will start A down here. There is A. Now negative B goes something like this.*0333

*So A plus negative B, or A - B, we go from the starting point of our first again to the ending point of our last. A plus negative B equals C. Basic vector manipulation.*0349

*Now when we have these vectors and they are lined up at angles, often times we can simplify our lives from a math perspective if we break them up into component vectors or pieces that add up to the sum.*0368

*If we pick those pieces carefully, so they line up with an axis, the math gets a whole lot easier and I am a huge fan of easier math.*0381

*Let's assume that we have some vector, A right here at some angle Θ from the horizontal. We could replace this with a vector along the X axis and a vector along the Y axis.*0389

*Notice that the blue vector plus the green vector, if we add them together, gives us that A vector, the vector we started with.*0405

*So we are going to take this A vector and we are going to replace it with this blue one and this green one. Two vectors that are a little simpler to deal with mathematically.*0412

*Let's call this the X component of A and let's call this the Y component of A. How do we figure out what those are?*0420

*If you notice, here, we have made a right triangle. Here is our hypotenuse, this is the opposite side because it is the opposite the angle and AX must be the adjacent side, it's beside the angle.*0434

*Now I can use trig to figure out AY is. AY, since it's the opposite side is going to be equal to A, the hypotenuse times the sine of that angle.*0449

*On the same note, AX is going to be A times the cosine of Θ because this is the adjacent side. remember SOHCAHTOA?*0461

*Sine of Θ equals the opposite side divided by the hypotenuse, cosine of Θ equals the adjacent side divided by the hypotenuse and tangent of Θ equals the opposite side divided by the adjacent.*0474

*All we are doing is we are finding out what this opposite and this adjacent side happens to be. So we can break up this vector A into components AX and AY that are going to be much easier to deal with mathematically.*0488

*We could also go back to finding the angle of the vector. If we know two of the three sides of these triangles, if we know both of the components, we can find the angle, if we know the hypotenuse and the opposite, we can find the angle.*0501

*How do we do that though? Well we have to go back to our trig functions.*0511

*Tangent of Θ equals the opposite over the adjacent side. Therefore Θ must be the inverse tangent of the opposite side divided by the adjacent side.*0518

*But what if we do not know opposite or adjacent? Well sine Θ is equal to the opposite over the hypotenuse. So if we know opposite over hypotenuse, we can find Θ by taking the inverse sine of the opposite side divided by the hypotenuse.*0533

*So if you know any two sides of this right triangle you are making with components, you can find the angle using basic trigonometry.*0582

*Let's talk for a few minutes about vector notation. You can express vectors in many different ways.*0589

*You can just draw it on a sheet of paper, you can express it mathematically, but want to do this as efficiently as possible. so I am going to show you some examples in 3 dimensions but you can always scale those back to just two dimensions*0597

*Let's start off by making an axis. We've got YX and lets have a ZX coming out towards us. If we have some vector A, we could express it as having an X component, a Y component, and a Z component.*0609

*On the other hand though, we could also look at in terms of what are known as unit vectors.*0628

*If we take a vector of length 1 along the X axis, magnitude of 1 along the X axis, we are going to call that specific vector ihap, length one along the X axis.*0636

*In the Y axis, we will do the same thing. A vector of unit length, of length 1 in the Y direction, we will call jhap.*0647

*In the Z direction, same idea. A vector of length 1 in the Z direction we'll call khap. Specific vector constants. So we could write A now, as some value, X value times ihap plus its Y value times jhap plus its Z value times khap.*0658

*So whatever the X value is, you multiply it by a vector unit length 1 in the X direction.*0687

*Y value times the unit vector of length one in the Y direction and the Z value times the unit vector of length one in the Z direction.*0693

*Another way to express vectors. That can be very useful when we get to the point of doing vector addition. Let's assume we have our axis here again.*0703

*Y, X and Z. And here let's put a vector that is 4 units in the X, 3 in the Y and out toward us, 1.*0713

*So let's call this point P, 4,3,1, which is defined by some vector P which is 4 units in the X, 3 in the Y and 1 in the Z.*0724

*Let's also define another vector Q. Let's go 2 units in the X, we won't go any in the Y, zero in the Y and let's come out toward us in the Z direction 1,2,3,4.*0741

*Let's call that point Q which is 2, 0, 4 and we'll label the vector from that origin to that point vector Q.*0752

*How do we add these vectors in multiple directions? Well, what we could say is that vector R is going to be equal to Vector P plus vector Q.*0762

*Therefore, let's write P as equal to 4,3,1 in this bracket notation for vectors and vector Q is equal to 2,0,4 in vector bracket notation.*0778

*Well if the left hand side is equal to the right here and the left hand side is equal to the right here, then we can add the left hand sides and add the right hand sides, they should still be equal.*0799

*What we can say then, if we add those two, and add those two. Therefore P plus Q which is equal to our R must be equal to, well, in vector bracket notation, we add up the X components 4 plus 2 is 6.*0809

*We add up the Y components, 3 plus 0 is 3, and we add up the z components, 1 plus 4 is 5.*0838

*So the resultant, R, would be 6 units in the X, 3 in the Y and then 5 towards us. Something like that in 3 dimensions.*0845

*Adding up vectors using that vector notation can make things a lot simpler especially when you don't want to go drawing all of the time.*0861

*Let's take a look at a vector component problem. A soccer player kicks a ball with the velocity of 10m/s with an angle of 30 degrees above the horizontal. Find the magnitude of the horizontal component, and vertical component of the ball's velocity.*0869

*I am going to start off with a diagram here. A Y axis, and an X axis and realize that the soccer player is kicking the ball with an initial velocity of 10m/s, so there is our vector, 10m/s at an angle of 30 degrees above the horizontal.*0886

*We want to know the horizontal component and the vertical component. As you recall, if we want the vertical component, if this is our initial velocity, P, then the Y component of that velocity is going to be V 10m/s times the sine of 30 degrees.*0906

*10m/s sine 30 should be 5m/s.*0934

*In similar fashion, the X component of velocity V is going to be V cosine Θ again or 10m/s times the cosine of 30 degrees.*0937

*Cosine 30 is 0.866 so 10 times that is going to be 8.66m/s. We have broken up V into it's X and Y components.*0951

*Alright, another one. An airplane flies with the velocity of 750 kmph 30 degrees south of east. What is the magnitude of the plane's eastward velocity?*0962

*Well let's draw a picture again. North, south, east and west. The airplane flies with a velocity of 750kmph 30 degrees south of east. That means start at east and go 30 degrees south.*0977

*So I am going to draw it's velocity as roughly that. 750kmph at an angle of 30 degrees south of east. If we want it's eastward velocity, the eastward component, that means we want the X component here.*0996

*X component of it's velocity, the X is going to be V cosine Θ or 750kmph times the cosine of 30 degrees 0.866 should give us something right around 650kmph.*1020

*Let's take a look at another one where we have to deal with vector magnitudes. A dog walks a lady 8 meters due north then 6 meters due east, I'm sure you've all seen that before. A big dog, a little person trying to walk it but really the dog is in charge?*1040

*Determine the magnitude of the dog's total displacement. Well if the dog walks the lady 8m due north, we'll have a vector 8m north and then 6m east. Determine the magnitude of the dog's total displacement.*1057

*Well if we start it down here and line these up tip to tail so that the total displacement is a straight line from where you start to where you finish, is going to go from here right to there. That's the displacement.*1075

*How do we find the magnitude of that? Well if we look, that's a right triangle. We can use the pythagorean theorem. A*^{2} plus B^{2} equals C^{2} where A is our 8m, B is our 6m, C is going to be our hypotenuse or the displacement.1092

*Therefore, this is going to have a magnitude of the square root of 8m*^{2} plus 6^{2} or the square root of 64 plus 36, square root of 100 is going to be 10 meters.1110

*Say we wanted to know what this angle is. If we wanted to know that, we could take a look and say, you know, the X component of that green vector is going to be 6m, the Y component must be 8m.*1126

*Therefore if we wanted that angle, Θ is going to be the inverse tangent of the opposite side over the adjacent which is the inverse tangent of 8m over 6m which comes out to be about 53.1 degrees.*1144

*That would be our angle, Θ there too.*1162

*If it had asked us for the angle, it only asked us for the magnitude of the dog's total displacement which we found to be 10 meters.*1165

*Let's take a look at some more vector addition. A frog hops 4m at an angle 30 degrees north of east.*1174

*He then hops 6m at an angle of 60 degrees north of west. What is the frog's total displacement from his starting position?*1184

*This just screams for us to draw a picture here first. So let's draw our axis here. I have a Y axis, and an X and as we look at this, There's our X, here is our Y, The frog starts out 4 meters at an angle of 30 degrees. There is 4 meters at an angle of 30 degrees.*1191

*Then he is going to go and hop 6m at an angle of 60 degrees north of west. So 6m at an angle of 60 degrees north of west is probably something kind of like that.*1202

*That angle is 60 degrees north from west and that is 6m long, thats 4m long. What is the frog's total displacement from the starting position.*1240

*Well, I could find that out graphically, by drawing a line from the starting point of the first to the ending point of the last.*1252

*Or, if I wanted to do this analytically, or a little bit more exactly, I could take a look if our blue vector is A, A is equal to it's X component is going to be 4m cosine 30 degrees, and it's Y component is going to be 4m sine 30 degrees.*1264

*Our B vector, there in red, is going to be, well we have got 6m cosine 60 degrees for it's X component, but it is to the left, so let's make sure that's negative and it's Y component is 6m sine 60 degrees.*1280

*So if I wanted to find the resultant, the sum, vector C. C is just going to be equal to A plus B, so that's going to be 4m cosine 30 degrees. The X component of A, plus the X component B, negative 6m cosine 60 degrees. So that will give us the X component of C.*1299

*For the Y component, we add their Y components together. 4m sine 30 degrees from A plus 6m sine 60 degrees from B.*1324

*When I do the math here, I find out that C equals 4 cosine 30 plus negative 6 cosine 60, that is going to be about 0.46m and the Y component 4m sine 30 degrees plus 6m sine 60 degrees comes out to be 7.2 meters.*1339

*So there is our C vector. 0.46, so not much in the X, 7.2 in the Y. While we are here, let's find out it's magnitude and angle.*1360

*The magnitude of C, I take the C vector and take it's absolute value, I can find out by using the pythagorean theorem again since I know it's components.*1371

*That is going to be the square root of 0.46*^{2} plus 7.2^{2} it comes out to be about 7.21m.1380

*If we wanted it's angle as well, I am expecting a big angle here just by looking at the picture. Θ is going to be equal to the inverse tangent of the opposite side over the adjacent side.*1392

*The opposite side is the Y, 7.2 over the adjacent 0.46 for an angle of 86.3 degrees which is over here 86.3 degrees north of east.*1410

*So we could express the vector with magnitude and a direction or we could express it just by leaving it in the vector bracket notation. If we wanted to we could have even written it as 4 6m ihap plus 7.2m jhap. They are all equivalent.*1421

*Let's take a look at one more sample problem, the angle of a vector. Find the angle Θ depicted by the blue vector below given the X and Y components.*1442

*Since I am given the opposite side, opposite the angle Θ and the adjacent side, the side next to the angle, but not the hypotenuse, I am going to use the tangent function since tangent of Θ equals opposite over adjacent.*1455

*Therefore Θ is going to be the inverse tangent of the opposite side over the adjacent side. Or Θ equals the inverse tangent of the opposite side 10 divided by the adjacent, 5.77 or 60 degrees.*1468

*Hopefully this gets you a good start on vectors and scalars. We will be using them throughout the entire course. They are very important.*1494

*Thanks for watching educator.com. Make it a great day.*1501

*Hi everyone. I am Dan Fullerton and I would like to welcome you back to Educator.com, as we continue our study of thermophysics and thermodynamics, by talking about ideal gases. *0000

*Now our objectives for this lesson are going to be utilizing the ideal gas law to solve for pressure, volume, temperature and quality of an ideal gas, and explaining the relationship between root mean square velocity and the temperature of a gas. *0010

*With that, let us talk about ideal gases. *0023

*Ideal gases are theoretical models of real gases which utilize a number of basic assumptions to simplify their study. *0027

*The first assumption is that the gas is comprised of many particles moving randomly in a container. *0034

*One or two molecules in a container is not a really good model. *0039

*We need to have some substantial amount of gas. *0042

*The particles are on average, far apart from one another. *0045

*They are not combined and almost in a liquid state. *0049

*In the particles, do not exert forces upon one another unless they come in contact in an elastic collision. *0052

*So we can neglect things like the gravitational force of attraction between these tiny particles. *0058

*Now, this works well for most gases at standard temperatures and pressures, but it does not hold up so well for very heavy gases at low temperatures or very high pressures, but for most of the things we would want to use it for, it works just great. *0064

*The ideal gas law relates pressure, volume, number of particles and temperature of an ideal gas in a single equation. *0079

*You can see this written in a number of different forms. *0086

*Pressure times volume equals (NRT) -- PV = NRT or NK*_{b}t depending on how you want to see it written and we will talk about what these values are. 0089

*Now (n), the number of moles of a gas, (n), is (N), which is the number of molecules divided by Avogadro's number or 6.02 × 10*^{23} something's per mole. 0100

*In this equation, pressure (P) is given in Pascal's, the volume (V) is in m*^{3}, and if you use (n), that is the number of moles of a gas and we use that over here. 0114

*(R), then is the universal gas constant or 8.31 J/mol K. *0126

*Use that if we are using the number of moles in a gas version and (T) is the temperature in Kelvins. *0132

*On the other hand, if you want to use this version, PV = NK*_{b}T, where (N) is the number of molecules that you have and K_{b} is Boltzmann's constant -- we talked about that previously -- 1.38 × 10^{-23} J/K, and T, again is your temperature in Kelvins. 0137

*Now before we move on, it is probably important to note that one mole of a gas at standard temperature and pressure has a volume of just about 24 L. *0157

*That is always a good thing just to have in the back of your mind. *0172

*Atoms, molecules, and moles -- Atoms are made up of protons and we will call the number of protons the Z-number, so when we write something like a molecule or an atom, (x) is the symbol for it and (z) is the number of protons that goes down here to the bottom and to the left of the element. *0177

*(N) the number of neutrons, they do not have a charge and if neutral, you have Z-electrons; you have one electron with a charge of -1 elementary charge for every proton. *0195

*Now the atomic mass (A) is the number of protons plus neutrons, so you have the protons here and you have the protons plus neutrons here, so if you want to adjust the number of neutrons, take (A) subtract (Z) and you will be left with the number of neutrons you have for your atom molecule element. *0205

*Here we have 2, 4 helium, that means that we have two protons; we have four protons and neutrons, which means we must have two neutrons and one mole of helium is approximately 4 grams (g). *0225

*Here we have Carbon-14 -- the 6 tells you that it has six protons, the 14 means it has 6 protons and neutrons, therefore we must have 8 neutrons and one mole of this material, which has a mass of about 14 g. *0246

*And if we looked at something like oxygen, that has 8 protons, 8 neutrons, and if we looked at one mole of molecular oxygen (O2), it is going to have a mass of about 32 g -- because it is O2, we have two of them there. *0268

*And that is going to have 6.02 × 10 *^{23} molecules. 0288

*All right, so let us see how we can put some of this together. *0298

*How many moles of an ideal gas are equivalent to 3.01 × 10*^{24} molecules? 0301

*Well, let us start with 3.01 × 10*^{24} molecules and I am going to multiply that by one mole over 6.02 × 10 ^{23} molecules in a mole, Avogadro's number. 0307

*And really what I am doing is multiplying by 1 and anything I multiply by 1, I get the same value even if the units are changing. *0328

*One mole and 6.02 × 10 *^{23} molecules are really the same thing, so 1/1 = 1, however, when I do this multiplication, my molecules will cancel out and I will be left with units of moles. 0334

*3.01 × 10*^{24} × 1 divided by 6.02 × 10 ^{23} = 5 moles. 0348

*Let us look at another example -- For moles of carbon dioxide in a bottle, how many moles of gas are present in a 0.3 m*^{3} bottle of carbon dioxide held at a temperature of 320 K and a pressure of 1 million Pa? 0360

*We will use our ideal gas law, PV = NRT, therefore (N) the number of moles, is going to be equal to PV/RT, where our (P) is 1 million or 10*^{6} Pa, and (V) is 0.3 m^{3}. 0376

*(R), our universal gas constant is 8.31 and our (T) is 320 K. *0394

*That gives me about 113 moles. *0402

*Let us take a look at another example -- Pressurized carbon dioxide. *0414

*We have a cubic meter of carbon dioxide gas at room temperature, 300 K, an atmospheric pressure at about 101,325 Pa, and it is compressed into a volume of 0.1 m*^{3} and held at a temperature of 260 K. 0418

*What is the pressure of this compressed carbon dioxide? *0431

*Since the number of moles of gas is a constant here, we can simplify the ideal gas equation into some combined gas law by setting the initial pressure volume and temperature relationship equal to the final pressure volume in temperature relationship. *0435

*If PV = NRT, and we are holding (N) and (R) constant, I could pull (T) over to this side for PV/T = NR, so NR must be constant. *0450

*So I could write this as P1(V1)/T1 = P2(V2)/T2. *0466

*And since I want P2, I can rearrange that and say P2 = P1(V1)T2/T1(V2). *0476

*Now I can substitute in my values -- P2 = 101,325 Pa (P1) 1*^{3} (V1) 260 K (T2)/300 K (T1) 0.1 M^{3} (V2). 0493

*If I do this, I come out with a P2 or final pressure of about 878,000 Pa. *0517

*Let us look at a helium balloon. *0533

*One mole of helium gas is placed inside a balloon. *0535

*What is the pressure -- looking for pressure inside the balloon -- when the balloon rises to a point in the atmosphere where the temperature is -12 ° C and the volume of the balloon is 0.25 m*^{4}?0538

*First thing is to convert this temperature from Celsius to Kelvins. *0551

*Temperature in K is our temperature in ° C + 273.15, so that is going to be -12 ° C + 273.15 or 261.15 K. *0556

*Now, if PV = NRT, then P = NRT/V. *0578

*Well, (N), 1 mole; (R), the gas constant (8.31); our temperature (261.15 K)... *0589

*...and our volume here (0.25 m*^{3}) gives us a pressure of about 8,680 Pa. 0600

*As we talk about the internal energy of an ideal gas, we call it the average kinetic energy of the particles is described by the equation, average kinetic energy is 3/2 times Boltzmann's constant times the temperature in Kelvins. *0618

*Now the total internal energy of the ideal gas can then be found by multiplying the average kinetic energy of the gases particles by the number of particles. *0632

*The total internal energy (U) is going to be the number of particles, (N), times the average kinetic energy, but we can do a little bit of manipulation here. *0641

*The total number of atoms, particles, (N), is going to be equal to the number of moles times Avogadro's number and the average kinetic energy is 3/2 KVT. *0653

*So when I substitute those into my equation, the total internal energy is going to be equal to 3/2 × number of moles (Avogadro's number) × Boltzmann's constant × temperature. *0671

*This implies then, however, Avogadro's number × Boltzmann's constant is our universal gas constant (R), so I am going to take that and replace it with (R) to write that the total internal energy (U) is 3/2(N) -- now I can place my (R) in there -- NRT. *0689

*I have a formula for the total internal energy of an ideal gas. *0712

*Let us see how we can use that. *0719

*Find the internal energy of 5 moles of oxygen at a temperature of 300 K. *0722

*U = 3/2 (NRT), so that's 3/2 × 5 moles × our universal gas constant, 8.31 × 300 K or about 18,700 J or 18.7 kJ.*0729

*Let us do another one.*0759

*What does the temperature of 20 moles of argon with the total internal energy of 100 kJ?*0761

*Well, total internal energy (U) is 3/2 (NRT), therefore temperature equals 2 × the total internal energy divided by 3 × the number of moles × that universal gas constant (R).*0768

*So that's 2 × 100 kJ or 100,000 J divided by 3 × 20 moles × our universal gas constant, 8.31...*0786

*...which gives me about 401 K.*0799

*Great. Let us look a little bit more at the velocity of these particles.*0817

*The root means square velocity or (RMS) velocity is the square root of the average velocity squared for all the molecules in the system.*0822

*You can kind of think of it as a sort of average velocity for molecules when we are using this Maxwell Boltzmann distribution statistics, probabilistic statistics.*0831

*What we have down here, is we have a plot of number of atoms or molecules -- number of particles with some specific velocity for different materials at about 293.15 K -- closing in on room temperature. *0842

*By the way, that C*_{4}H^{10}, that is butane.0860

*You can see that we have different spreads here.*0863

*For butane, we have a peak here at something just shy of 300 m/s*0867

*That is where you are going to have the most particles, but you have a fairly tight distribution around that.*0876

*As we go to something like ammonia, NH*_{3}, we have a much wider distribution and a greater tail down here at the higher velocities.0881

*So, just an idea, giving you a feel for what root means square velocity is and what it means.*0889

*Calculating the root means square velocity -- We are going to start with the average kinetic energy as 3 1/2 × Boltzmann's constant × the temperature in Kelvins.*0896

*What this means then, is average kinetic energy -- is we are taking the average of 1/2MV*^{2} for all those particles and that is equal to 3/2 × Boltzmann's constant × the temperature (T).0905

*But the mass of these particles is constant, so taking the average of it we can pull the (M) out of the average and multiply it and we are done and so can the 1/2, that is a constant too.*0921

*That implies if we pull the M/2 out that M/2 × the average of V*^{2} = 3/2 K_{b}t.0931

*Or if I divide both sides by M/2 or think of it as multiplying both sides by 2/M, the left hand side is just going to be the average of V*^{2} and the right hand side we are going to have 3/M Boltzmann's Constant × (T).0947

*If I take the square root of both sides -- well, this is the definition of the average of the VRMS -- the root means square velocity.*0970

*So VRMS = the square root of 3/MK*_{b} × (T).0981

*But this (M), the mass, we can give another symbol that is often used. *0992

*(M) is often written as the mass of the molecule -- is written as μ.*0998

*So I could write VRMS = 3K*_{b}t/μ square root.1003

*There is another equation for the root means square velocity, but we can take that even further.*1014

*If we start with the root means square velocity equal to the square root of 3 K*_{b}t/μ...1022

*...this implies then, knowing that Boltzmann's Constant, K*_{b} is actually R/Na (Avogadro's number), that we could write VRMS, our root means square velocity as equal to the square root of 3. 1032

*Now we have our R/Na right there and we still our μ down here and we still have our (T). *1055

*So we have 3RT/μ × Na, but even more, μ × Na, the mass of our molecules × Avogadro's number is going to give us what is known as the molar mass, (M) in kilograms per mole. *1067

*So a little bit more we can do here. We could write this then as VRMS = the square root of 3RT/M -- another version for calculating the root means square velocity.*1087

*Let us take a look at an example here.*1113

*An ideal gas is placed in a closed bottle and cooled to half its original temperature. *1114

*What happens to the average speed of the molecules?*1119

*Well, the root means square velocity is the square root of 3RT/M and we are going to cut (T) in half.*1123

*Everything else is going to stay the same, but if we cut (T) in half and (VRMS) is proportional to -- well that is going to be -- if (T) is 1/2 of what it was, that is going be proportional to square root of 1/2 -- what we had for its original velocity.*1139

*Square root of 1/2 is about 0.71, so it is going to be 0.71 of its original velocity or you could write this as the average speed -- if we think of it in terms of average speed -- is going to be about 71% its original value.*1158

*Take its original value multiplied by 0.71.*1177

*Let us do another one. These can be a little bit tricky when you see them the first time.*1180

*The root means square velocity of the molecules of a 300 K gas is 1000 m/s. *1184

*What is the root means square velocity of the molecules at 600 K?*1194

*Well, again we will start with VRMS = square root of 3RT/M.*1199

*Now we are going to double the temperature and when we double the temperature its proportional to the square root of (T), so (VRMS) is proportional to the square root of 2 times the original (RMS) velocity.*1210

*So that is going to be square root of 2 × 1,000 m/s or VRMS = 1.41, the original, which is 1410 m/s. *1228

*You get a 41% increase and the root means square velocity of the molecules and you double the temperature.*1245

*All right, trying another one -- Hydrogen (H*_{2}) and Nitrogen (N_{2}) gas are in thermal equilibrium in a closed box. 1259

*Compare the root means square velocities of the molecules.*1267

*Well, we are going to start by referencing our (VRMS) equation is equal to the square root of 3RT/M.*1271

*Now the (M) of hydrogen is 2 and the (M) of nitrogen is 28. *1282

*That means we have a 14 times difference.*1291

*All right, so when I look at what these are proportional to it is 1/M, so if I were to take a ratio of these two, at the top I would the square root of 1/2 because we have 2 for the (M) of hydrogen compared to 1/square root of 28 for my ratio for nitrogen.*1297

*Which is going give me an (x) factor of 3.74.*1320

*That means the root means square velocity for hydrogen is going to be 3.74 times larger than the root means square velocity for nitrogen.*1326

*That has to be expected; it is a lot smaller.*1339

*Find the number of molecules in 0.4 moles of an ideal gas. *1346

*All right, a conversion problem -- 0.4 moles -- and we want to convert this into molecules.*1350

*I am going to multiply and I want moles to go away, so I will put that in the denominator so they make a ratio of 1 and cancel out.*1360

*I want molecules as my unit and now I need to make sure I am multiplying by a value of 1.*1366

*One mole is equal to 6.02 × (10)*^{23} molecules. 1372

*My moles, units, will make your ratio of 1 or cancel out and I will be left with 0.4 × 6.02 × 10*^{23} molecules, which is 2.4 × (10)^{23} molecules.1378

*All right, let us look at one last problem here.*1401

*The temperature of an ideal gas is doubled. *1405

*What happens to its internal energy?*1408

*The first thing I am going to do is recall that internal energy equation, U 3/2 NRT.*1411

*Now if I double the temperature -- All right if I am doubling the temperature here, I must be doubling the internal energy and I get double the internal energy.*1420

*So the short answer -- internal energy doubles.*1433

*Hopefully that gets you a good start on ideal gases.*1445

*Thank you so much for your time coming to Educator.com.*1450

*Make it a great day everyone!*1454

*Hi everyone and welcome back to Educator.com. *0000

*Today's lesson is on thermodynamics. *0002

*Our objectives are going to be to understand that energy is transferred spontaneously from a higher temperature system to a lower temperature system, to explain the first law of thermodynamics in terms of conservation of energy involving the internal energy of a system, and to represent transfers of energy through work and heat by using PV diagrams. *0006

*Let us begin by talking about the zeroth law of thermodynamics. *0026

*The zeroth law of thermodynamics, they added after some other laws of thermodynamics because they needed it to help make all of their proofs work out. *0034

*It saids if Object (A) is in thermal equilibrium with Object (B), and Object (B) is in thermal equilibrium with Object (C), then Object (A) must be in thermal equilibrium with Object (C). *0041

*Sounds kind of obvious, but just so we have everything in there, that is the zeroth law of thermodynamics. *0052

*The first law of thermodynamics is a little bit more practical for our purposes. *0059

*It says that the change in the internal energy of a closed system is equal to the heat added to the system plus the work done on the system. *0063

*ΔU, change in internal energy is heat added to plus work done on, and those are for the positive values. *0072

*This is really just a restatement of the law of conservation of energy applied in the thermal sense. *0079

*The sign conventions are extremely important. *0085

*Positive heat is heat added to the system; positive work is work done on the system. *0088

*If heat is taken from the system, it is negative and if work is done by the system, the work is negative. *0096

*All right. Let us talk about work done on a gas. *0104

*Typically we will use the first law of thermodynamics to analyze the behavior of ideal gases.*0106

*It may be useful to explore our understandings of the work done on a gas a little bit though. *0111

*If you recall, work is force times the displacement -- and we are going to assume that we have it in the same direction so that we do not have to worry about sines/cosines. *0117

*That is a reasonable assumption as we are talking about thermodynamics, which implies then -- well if we know pressure is force over area, then force must be pressure times area. *0125

*I could rewrite this as work is equal to pressure times area times δr, but we are going to take another step here. *0137

*Change in volume is equal to A(δr) and because we have the convention, that work done on the gas is positive, corresponding to a decrease in volume, we will put a negative sign there, so our sign conventions work out. *0149

*Then we could say that work is equal to -P × δv. *0165

*All right if work is force multiplied by displacement, then work is pressure times area times displacement and negative -- just there for the sign convention -- replace A × δr with δv and we get that work is minus P(δv). *0183

*That is going to be extremely helpful as we start analyzing these gas systems. *0198

*Let us take an example. *0205

*Five thousands joules of heat are added to a closed system which then does 3,000 J of work. *0207

*What is the net change in the internal energy of the system? *0212

*Well, δu is (Q) + (W) -- 5,000 J of heat are added to, added to, so that must be positive, so 5,000 J is positive, which then does 3,000 J of work. *0216

*If the system is doing the work, that is negative, so -3,000 -- our total change in net internal energy, must be 2,000 J. *0232

*Or a second example -- a gas is expanded at atmospheric pressure, 101,325 Pa. *0248

*The volume of the gas was 5 × 10*^{6}m^{3}. 0254

*The volume of the gas is now 5 × 10*^{-3}m^{3}. 0259

*How much work was done in the process? *0263

*Well, work equals -P(δv), so that's (-P) and δ anything is always the final value minus the initial. *0266

*So that is V-final - V-initial; P is 101,325 Pa; V-final is 5 × 10*^{-3}m^{3}... 0291 ...V-initial is 5 × 10^{-6}m^{3}, which implies then that the work is -506 J. 0276

*All right. Let us talk about another useful tool for analyzing gas systems. *0308

*It is called the pressure volume diagram or PV diagram.*0313

*We put pressure on the y-axis, volume on the x-axis and we are going to keep the amount of gas constant, so when we talk about PV = NRT, our ideal gas law, pressures on the graph, volumes on the graph, the amount of gas is constant, so that stays constant, (R) is already a gas constant...*0317

*...we can solve for (T) using the ideal gas law, so a PV diagram shows us pressure, volume, and indirectly temperature, so we can find (T) once we know these other quantities. *0339

*If we transition from state (A) to state (B) on a PV diagram, the volume is increasing, so our pressure is decreasing. *0355

*The work done then is going to be the area under the curve from (A) to (B). *0364

*That area here is going to be our work. *0371

*As the volume expands, the gas is doing work, so (W) would be negative and as the volume compresses, the work is being done on the gas, (W) is positive. *0379

*Also important to note here is that as you move up into the right on the graph, you move to higher temperatures. *0389

*Let us take a look at some analysis using a PV diagram. *0400

*Using the PV diagram below, find the amount of work required to transition from state (A) to state (B) and then the amount of work required to go from state (B) to state (C). *0403

*Well let us start out with the work going from (A) to (B). *0415

*The work in going from state (A) to state (B) is the area under the graph and as we go from (A) to (B), that is just a straight line, there is no area -- no work done. *0418

*How about the work done as we go from (B) to (C)? *0428

*Well that is -P × δV or minus 50,000 Pa -- V, is δV is final minus initial, so that is going to be 4 m*^{3}- 2 m^{3} or -100,000 J. 0433

*Notice that the gas was expanding, the gas was doing work. *0458

*Work is positive if the work is done on the gas since the gas is doing work it makes sense that we get a negative value for the work done in going from (B) to (C).*0461

*There are several different types of PV processes that we ought to point out, special PV processes. *0472

*They have some goofy names and they are kind of vocabulary words, so you really just have to memorize these. *0478

*Adiabatic -- This is when heat (Q) is not transferred into or out of the system; the heat remains constant. *0483

*That is adiabatic and a PV graph for an adiabatic process looks like this here in the light blue -- adiabatic. *0491

*Isobaric -- pressure (P) remains constant and in an isobaric process, since (P) remains constant, you have a horizontal line. *0498

*Isochoric means volume remains constant so that means you have a vertical line and you stay at the same (V). *0508

*An isothermal means temperature (T) remains constant and you get an isotherm that looks like this -- isothermal lines on a PV diagram, we call isotherms. *0514

*And we will dive into these in a little bit more detail right away. *0524

*Adiabatic process -- heat is not transferred into or out of the system -- Q = 0 -- therefore by the first law of thermodynamics, if δU is equal to Q + W, and we know that Q = 0 in an adiabatic process, then the change in internal energy of the gas is the work done on the gas, δU = W. *0529

*Pretty straightforward and the processes have that sort of shape. *0553

*An isobaric process -- pressure remains constant. *0560

*Isobaric -- constant pressure -- the PV diagram shows a horizontal line and if PV = NRT, (P) is constant and then (R) are constant, we can rearrange this to say that V/T = NR/P. *0564

*If all of that is constant, that means that V/T, that ratio remains constant for any gas processes. *0580

*That happens in an isobaric or constant pressure process. *0586

*In an isochoric process, the volume remains constant. *0591

*In an isochoric, we have constant volume or a vertical line and the work done on the gas is 0, because remember work done on a gas is the area under the graph and in a vertical line, you do not have any area under it and if PV = NRT and volume remains constant, well constant P/T = NR/V. *0595

*All of those are constant, so the ratio of P/T remains constant for all of your processes. *0620

*In an isothermal processes where the temperature remains constant, the lines on the PV diagram for these are called isotherms; there is an isothermal process. *0627

*If PV remains constant, the internal energy of the gas must remain constant. *0638

*Let us look at an example for an adiabatic expansion. *0647

*An ideal gas undergoes an adiabatic expansion -- adiabatic -- Q = 0 -- no transfer -- doing 2,000 J of work. *0650

*How much does the gases internal energy change? *0660

*Well, δu = Q + W, but since it is adiabatic, we know that Q = 0, so δu = W, which must be - 2,000 J. *0663

*The biggest trick here is remembering the definitions of these terms. *0681

*Example 5: Removing some heat -- Heat is removed from an ideal gas as its pressure drops from 2,000 Pa to 100,000 Pa. *0686

*The gas then expands from a volume of 0.05 m*^{3} to 0.1 m^{3} as shown in the PV diagram below. 0695

*If curve (AC) represents an isotherm, find the work done by the gas and the heat added to the gas. *0703

*Well, right away the work in going from (A) to (B) is 0, because there is no area under that graph and the work going from (B) to (C) is just -P(δv)... *0709

*... or -100,000 Pa × V-final - V-initial or 0.1 - 0.05, which is -5,000 J. *0721

*That is the work done by the gas, that is why it is negative. *0735

*Now we are on an isotherm going from (A) to (C), so (U) must be constant; our internal energy has to stay the same. *0739

*Δu = 0, which equals Q + W, therefore, Q = -W = 5,000 J.*0746

*You must have added 5,000 J to the gas. *0758

*Our key answers -- find the work done by the gas -- the work done by the gas was 5,000 J and the heat added to the gas -- we added 5,000 J. *0768

*The gas did 5,000 J of work and we added 5,000 J to it. *0782

*Let us take a look at the PV diagram below and answer these questions. *0790

*During which process is the most work done by the gas? *0793

*Well, work done by the gas, that is a negative work or an expanding gas. *0798

*We see that -- that is the area under the graph going to the right here from (A) to (B), so that must be (A) to (B) here. *0803

*Going from (B) to (C) is no work or no area and from (C) to (A), we are compressing the gas, so work is being done on it. *0811

*Again, during which process is the most work done on the gas? *0817

*That must be going from (C) to (A). *0820

*We have the most area going from (C) to (A) and we are compressing the gas, so work is being done to the gas. *0823

*In which state is it the highest temperature? *0829

*Remember temperature gets bigger as you go up into the right, so that must be state (C). *0831

*On to the second law of thermodynamics. *0840

*Heat flows naturally from a warmer object to a colder object and cannot flow from a colder object to a warmer object without doing work on the system. *0842

*Heat energy also cannot be completely transformed into mechanical work or another way to say that is nothing is 100% efficient. *0851

*Now all natural systems tend toward a higher level of disorder or entropy. *0859

*The only way to decrease the entropy of a system is to do work on it. *0864

*An entropy is kind of a state of disorder. *0868

*For example, if I had a really cool Lego castle here right now and I dropped it, it is going to become more messy. *0870

*In the natural state of the world, I am never going to have a bunch of Lego's in all different pieces dropping and then when I look down and go to pick it up, the castle is already built. *0878

*Things do not get more ordered unless you do work on it. *0886

*That is the second law of thermodynamics. *0889

*Now, another way to look at this is in terms of heat engines. *0893

*Heat engines convert heat into mechanical work. *0896

*And the efficiency of a heat engine is the ratio of the energy you get out in the form of work to the energy you put in, so typically how these work... you have a high temp reservoir, a place where you create a lot of heat. *0899

*You use that to do some sort of work. *0913

*If you have heat energy at the high temp reservoir, some of it becomes productive output and some of it goes into the low temp reservoir, where it is not very useful. *0915

*The work that you get out is equal to what you put in minus what is left over -- what goes to that low temp reservoir, and the efficiency of your system is going to be what you wanted to work out divided by what you put in. *0926

*And we will put the absolute value signs around that, just so you do not have to deal with negatives. *0945

*But W = Qh - Qc/Qh, so you could rewrite that if you wanted as 1 - Qc/Qh. *0950

*A couple of key things, but the efficiency is one of the key formulas from this slide. *0962

*Power in heat engines -- Power is the rate at which work is done, work over time. *0969

*We talked about that back in mechanics. *0974

*From a heat engine perspective, though, we can take this a little bit further. *0977

*If efficiency is work over the high temp heat, then we could rewrite that as work is equal to the efficiency times (Qh) or dividing both sides by time -- W/t is efficiency × Qh/t. *0982

*Work over time is power, so since P = W/t, then power on the left hand side becomes efficiency × Qh/t, but let us go another step. *1002

*We just found that efficiency could also be written as 1 - Qc/Qh, therefore, P = 1 - Qc/Qh × Qh/t. *1015

*Well with a little bit more rearrangement and a little more Algebra, P = Qh/t - -- well the Qh's will cancel -- Qc/t. *1037

*A couple of other ways to help you calculate the power from heat engines. *1051

*All right. Heat engines and PV diagrams -- On a PV diagram, a heat engine is a closed cycle. *1058

*For clockwise processes, these are heat engines. *1065

*If you go in the other direction, counter-clockwise processes -- those are refrigerators. *1068

*Now let us talk a little bit about the Carnot engine. *1075

*The Carnot engine is not something that you just go out and buy. *1077

*It is a theoretical model, a theoretical idea of an engine that has the maximum possible efficiency. *1081

*It uses only isothermal and adiabatic processes and Carnot's theorem states that no engine operating between two heat reservoirs can be more efficient than the Carnot engine operating between those same two reservoirs. *1087

*So the Carnot engine is kind of the theoretical model of the maximum efficiency you could get from an engine and the efficiency of the Carnot engine is equal to the temperature of the hot reservoir minus the temperature of the cold reservoir, divided by the temperature of the hot reservoir. *1099

*When you actually utilize this to do calculations, keep a note that the temperature must be in standard SI units or Kelvins. *1115

*Let us take another look at a Carnot engine problem. *1129

*A 35% efficient Carnot engine absorbs 1,000 J of heat per cycle from a high temp reservoir held at 600 K. *1131

*Find the heat expelled per cycle as well as the temperature of the cold reservoir. *1138

*Well, if our efficiency is 35% or 0.35, we also know that our Qh is 1,000 J per cycle and that the temperature of our high temp reservoir is 600 K. *1143

*We could start with efficiency as our high temperature when its our cold temperature divided by our hot temperature for the engine, therefore, efficiency equals 1 - cold temperature/hot temperature or cold temperature/hot temperature is 1 - efficiency. *1163

*Therefore, to find the cold temperature, (TC) is going to be equal to the hot temperature times 1 - the efficiency or 600 K × 1 - 0.35 = 0.65 × 600 or 390 K. *1184

*Now we have the heat expelled per cycle as well as the temperature of the cold reservoir, so if we want E = W/Qh and we want to find what that W is, that is going to be E × Qh or our efficiency 0.35 × the heat on the hot side (1,000 J) or 350 J. *1207

*So then W = Qh - Qc. *1235

*Therefore, Qc = Qh - W or 1,000 - 350 = 650 J. *1242

*Let us look at a maximum efficiency problem. *1262

*Determine the maximum efficiency of a heat engine with a high temperature reservoir of 1200 K and a low temperature reservoir of 400 K. *1265

*Now, this is not really asking for a Carnot efficiency because the most efficiency you can have is the Carnot engine. *1274

*The Carnot efficiency is Th - Tc/Th or 1200 K - 400 K/1200 K = 0.667 or about 66.7%. *1283

*One last problem here -- Which of the following terms best describes a PV process in which the volume of the gas remains constant? *1308

*Constant volume -- So I check on vocabulary words from those PV processes -- Adiabatic, no that is constant (Q); isobaric -- that is constant pressure; isochoric -- that is constant volume, and isothermal of course is constant temperature. *1320

*Our correct answer there must be C. *1337

*Hopefully that will give you a good start in thermodynamics. *1340

*I appreciate your time and thanks for coming to visit us at Educator.com. *1343

*Make it a great day everyone!*1347

*Hi everyone! We are thrilled to have you back with us here at Educator.com.*0000

*Today we are going to start a lesson on electric fields and forces which is the beginning of our unit on electricity and magnetism.*0003

*Our objectives are going to be to calculate the charge on an object and explain the Law of Conservation of Charge, describe differences between conductors and insulators, and explain the difference between conduction and induction.*0010

*We will also use Coulomb's Law to solve for the force on a charged particle due to other point charges, calculate the electric field due to one or more point charges and finally to analyze electric field diagrams.*0023

*Let us start by talking about electric charges.*0034

*As you know matter is made up of atoms, and those atoms even have smaller particles -- subatomic particles such as protons, electrons and neutrons.*0037

*Now protons have a charge of +1 and +1e is a +1 elementary charge, the smallest stable single charge, that is equal to 1.6 × 10*^{-19} coulombs (C), the SI unit of charge.0045

*Electrons on the other hand have a charge of -1 elementary charge and neutrons are neutral.*0060

*Now most atoms are neutral -- they have equal numbers of protons and electrons.*0066

*The positives and the negatives balance out for a net charge of 0; it is neutral.*0070

*But if an atom loses its electrons, loses an electron or two, it is going to become positively charged, or if it gains an electron or two, it is going to become negatively charged.*0075

*We call those charged atoms 'ions'.*0085

*Now the fundamental unit of charge in the SI system is the coulomb (C), and that is a big amount of isolated charge.*0089

*The smallest isolated unit of electric charge is the elementary charge, 1e on a proton, -1e on an electron and it's magnitude is 1.6 × 10*^{-19} C.0098

*Now like charges repel each other, opposites attract, and of course electric charge is conserved -- that is called the Law of Conservation of Charge.*0110

*You cannot spontaneously have just a (+) charge appear.*0119

*If you start off with 0 net charge, later on in a closed system you must still have 0 net charge.*0122

*If you start off with 0 net charge you could have a +1e and a -1e so that your net is still 0, but you cannot spontaneously get a +3e and a -1e for a net of 2 -- charge has to be conserved.*0127

*Now let us take a look at charge on an object.*0143

*Mittens the cat possesses an excess of 6,000,000 electrons.*0146

*Let us find the net charge on Mittens in coulombs SI units.*0150

*Well charge gets the letter (Q) and that is going to be 6 × 10*^{6} electrons, and since they are negative I will put that there so we have -6 × 10^{6} elementary charges...0154

*...and we are going to try and convert that to coulombs and the way we are going to do that is by multiplying by 1 again -- that old trick.*0168

*We want elementary charges to go away so I will put (e) down here and I want units of coulombs, so I will put that up here, and then I have to write numbers in here to make a ratio of 1.*0175

*Well 1 elementary charge = 1.6 × 10*^{-19} C.0186

*Now when I go through and do the math, my elementary charge units are going to cancel out and I will get -6 × 10*^{6} × 1.6 × 10^{-19} C for a total of -9.6 × 10^{-13} C as my answer. Great!0192

*Let us take a look at the charge of an alpha particle.*0216

*An alpha particle, which is also known as a helium nucleus, consists of 2 protons and 2 neutrons, no electrons.*0219

*So what is the charge of an alpha particle?*0225

*Well if an alpha particle has 2 protons, those are the only charged particles, it must have a charge of +2 elementary charges.*0228

*But let us convert that into coulombs -- +2 elementary charges, and we need to multiply that by 1, I will write 1 as 1e = 1.6 × 10*^{-19} C... 0239

*...so I get a charge of 3.2 × 10*^{-19} C when I put that into SI units -- 3.2 × 10^{-19}.0254

*All right. As we talk about materials -- conductors are materials that allow charges to move freely.*0268

*They have a very low resistivity, so charges can go through them very, very easily.*0274

*Insulators, on the other hand, do not allow charges to move freely.*0279

*They have what is known as a very high resistivity, and resistivity is a material property.*0282

*If you look up the resistivity of gold, it has a certain value; if you look up the resistivity of glass, it has a specific value.*0288

*It is a material property measured in ohm meters and it typically gets the symbol ρ, a squirrely little P.*0296

*All right. Materials may be charged by contact known as conduction.*0306

*You might have tried this trick before -- rub a balloon against your hair someday. *0310

*Some electrons from the atoms in your hair get transferred to the balloon.*0313

*The balloon now has a net negative charge and your hair, because it lost some electrons, now has a net positive charge.*0317

*You can also charge conductors by contact.*0324

*For example, if you bring a charge conductor into contact with an identical neutral conductor, you will share the charge across those two conductors.*0327

*Let us take a look and see how that would work.*0335

*A conductor carrying a net charge of 8 elementary charges is brought into contact with an identical conductor with no net charge.*0338

*When they are brought into contact because their conductors and charges can move freely, those charges -- those 8e -- they are all positive; that is a positive charge; they are going to repel; they want to get as far apart from each other as they can.*0346

*So what do they do? They split up so you get 4e on each conductor.*0357

*If those conductors are identical, then you get the exact same charge on each.*0361

*Then if you split them apart, you now have 4 elementary charges on each object.*0365

*So what would the charge be?*0370

*4e on each one, is going to be 4 times the charge on an elementary charge -- 1.6 × 10*^{-19} C, which is 6.4 × 10^{-19} C.0372

*Now on a conductor, the charge is always going to sit on the outside surface.*0387

*So the electric field inside the conductor, as we talk about fields here soon, is always going to be 0.*0392

*Something to remember is that an electric field inside a conductor is 0.*0398

*Let us talk about the electroscope.*0405

*The electroscope is a really cool tool that is used to detect small electric charges based on conduction.*0407

*It consists of a conducting rod in a beaker that is then insulated from the outside world, except for this metal knob that sticks out the top.*0413

*If you were to bring something like a positively charged rod over to it to share the charge, you are going to get a net positive charge on your metal rod and as that is distributed throughout the rod, at the bottom you have these two very thin leaves.*0421

*If you have positive charges on each of the leaves, they are going to repel and you get a spreading of the leaves, which indicates that you have a charge.*0434

*You could have the same basic thing happen if you were trying to do this with a negative charge.*0442

*If we draw just the metal part of our electroscope here and we bring a negatively charged rod near it... *0446

*...well now the electrons that are already on that metal rod without even touching, they are going to be repelled from the negative charge there and they are going to try and hang out as far away as possible from that negative charge -- leaving this end positive.*0456

*Well once again down here on those leaves you have negative charges on each leaves they will repel, so this is really a charge detector.*0470

*You could also charge by induction -- that is charging a conductor without actually coming into contact with another charged object.*0480

*So if we start off with a neutral electroscope here and bring a positive rod near it, we will have the electrons all want to gather near the positive charge -- opposites attract -- leaving a net positive charge down near the leaves of the electroscope; they will spread apart.*0487

*Now what we are going to do though is while we hold that rod in place, we are going to connect this metal bar to ground and by some connection to ground -- to the earth through a conductor -- the earth acts as an infinite sync or source of electrons.*0502

*If you need some electrons to balance things out you could pull them straight from the earth; we have tons of extras. *0517

*If you need to get rid of some, you can throw them into the earth very easily.*0522

*So now when you connect this to ground it sees all the positive charges here and you will start sucking some electrons up from the ground in order to make that balance nice and happy.*0526

*Now if you disconnect the ground connection, those charges are stuck on the metal conductor, you have a net negative charge on the electroscope and of course down here where you once again have negative charges on the leaves, the light charges repel each other and you see the spreading of the scope.*0535

*You have therefore charge to the electroscope by induction -- you have not touched it specifically to the source of charges to that glass rod.*0553

*Let us talk about electrostatic attraction.*0564

*A positively charged glass rod attracts object X. What can you say about the net charge of object X?*0566

*Well before we can answer that, let us think about what could happen here.*0574

*If we have a positively charged glass rod, it is pretty easy to see that if you have some other negatively charged object that it will attract.*0577

*So without a doubt, you could say that you could attract a negatively charged object. *0590

*That is easy, but what if the object is neutral? *0594

*Well if it is a conductor and it is a neutral object, you will actually have some of the electrons in the object who will want to hang out over near the positive charge.*0598

*It remains*