WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I am Dan Fullerton and in this lesson we are going to start to explore circular and relative motion.
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Our objectives include understanding and applying relationships between translational and rotational kinematics which we are going to continue in future lessons.
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Use the right hand rule to determine the direction of the angular velocity vector.
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Describe the meaning of the phrase motion is relative and calculating the velocity of an object relative to various reference frames.
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Let us start by talking about how we measure circular motion in degrees and once around the circle is 360°.
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In radians however once around the circle is 2 π.
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A radians measures the distance around an arc equal to the length of the arcs radius.
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If there is a circle there is our radius it takes 2 π × this value to go once around the circle the circumference.
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We call that distance around the circle δ s or C circumference = 2 π × the radius of that circle.
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It is going to be useful to convert from radians to degrees and back again.
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Especially, in this APC course, we are using both types of measurements depending on the problem.
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If you are using your calculators, be very careful you got them in the right mode that when you are inputting angles
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you know whether you are using degrees or radians in your calculator is also on the same page as you are.
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First let us convert 90° to radians.
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If we start with 90° and we want radians from 90° / 1 we will multiply that there are 2 π radians in one time around the circle or 360°.
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Therefore I can say that we have π / 2 radians is equal to 90° or 1.57 radians if you to put it in decimal form.
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Interesting to know radians is not an actual unit of measure.
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It is helpful for us to use mentally but radians are officially unit less.
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Let us convert 6 radians to degrees going the other way.
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We have 6 radians and we want to convert to degrees.
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I know there are 2 π radians and 360° therefore our radians will make a ratio of 1 and 6 × 360 degrees / 2 π is about 344°.
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Nice and straightforward.
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How do we use these?
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Let start talking about displacements.
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Linear position their displacement we have talked about is being given by δ r or we are also going to see it now especially as we talk about things in circles as δ s.
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Angular position or displacement is given by δ θ.
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How many degrees you have is you are going around the circle.
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S = r × θ if you know your radius and the angle through which you pass you can find how far you have traveled
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this distance along the circumference s = r θ or δ s = r δ θ.
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We can also take a look at that in terms of velocity.
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Linear speed and velocity we have been talking about is being given by V.
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Angular speed or velocity is given by ω that is the curvy w symbol.
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If we look V our average velocity is dr dt or ds dt the time rate of change of position.
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Angular velocity is the time rate of change of θ your angular displacement.
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Know that any point here your velocity vector is constantly changing the pointing in this direction.
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Ω is around the circle your angular velocity as you go around there.
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Now the direction of angular velocity is a little bit tricky because if you look here depending
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on where you are in the circle your linear velocity is constantly changing.
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Trying to describe a constant velocity for something around the circle is very tough in the angular world.
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When we talk about angular velocity to define its direction we are going to use the right hand rule.
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For something that is moving counterclockwise around the circle take the fingers of your right hand and wrap them in that counterclockwise direction.
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Your thumb points in what we call the direction of the angular velocity vector.
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To illustrate that I got a diagram on our next slide here.
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If something is moving in this direction wrap your hands around it and your thumb points in the direction of the angular velocity vector.
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It is very interesting to note that the angular velocity vector does not tell you the direction the object is moving at that specific point in time.
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It is of very nebulous concept the first couple of times you see that.
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We will also see this with angular acceleration and some other things angular momentum coming up as well.
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How do we convert linear to angular velocity?
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Linear velocity was ds dt or dr dt and using those interchangeably at this point.
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S = r θ how we found our linear displacement along the curve from an angular displacement.
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We can write this then it as velocity =dr θ dt.
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But as we go around a circle the radius is a constant that is what makes a circle.
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We can pull that out of the derivation.
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We have V =r d θ dt but we just to find d θ dt as ω.
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ω is d θ dt and we can write that our velocity vector is equal to r × ω vector V= rω.
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We have a conversion method between the 2.
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Let us take an example.
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If we look at Earth's angular velocity find the magnitude of Earth's angular velocity and radians per second.
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Ω angular velocity is angular displacement over some time interval.
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Once around is 2 π radians and how long does it take the Earth once all the way around.
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That is a day or 24 hours so that is going to be π /12 radians / hr.
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radians / hr is kind of goofy metric.
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Let us make that into radians / s to more standard unit.
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I know that 1hr is 3600s.
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My hours make a ratio of 1 and I come up with about 7.27 × 10⁻⁵ radians /s.
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We will look at angular displacement angular velocity what come next? angular acceleration.
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Linear acceleration we said was given by the A vector.
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Angular acceleration is given by the α vector.
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Angular acceleration was a derivative of velocity with respect to time.
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Angular acceleration is the derivative of angular velocity with respect to time.
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How fast something is accelerating as it goes around that circle?
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By accelerating, we are not talking about the centripetal acceleration towards the center of a circle
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that allows it to keep moving a circle or talk about whether it speeding up or slowing down in its path around the circle.
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Looking at centripetal acceleration in much more detail and putting this together.
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We can express the position vector in terms of unit vectors.
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If we say that the position vector r(t) = function x is sum function of time in the i hat direction
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+ y position as function of time in the j hat direction to give us r vector.
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When we do this is we are going around the circle we should be fairly easy to see.
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If we broke this up into components our x position is going to be r cos θ.
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R y is going to be r sin θ.
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x(t) is given by r cos θ or θ is going to be a function of time.
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Ry function, y is a function (t) is r sin θ.
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We could then write that r(t) Position vector is our cos θ in the x direction × I hat + r sin θ in the y direction or × j hat.
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We also know θ = ωt angular displacement is angular velocity × time.
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It is like in the linear world linear displacement is linear velocity × time.
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R(t) = r cos ωt i hat + r sin ωt j hat.
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We have gone that far let us take a look and say what if we want velocity though?
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Velocity was dr dt which is going to be the derivative with respect to time of all the stuff we have up here.
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R cos ωt i hat + r sin ωt j hat.
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Let us take the derivative there then that is going to be equal to the derivative of r cos ω t
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is going to be - ωr sin ωt i hat + derivative of r sin ωt is going to be ωr cos ωt j hat.
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There is our velocity function.
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Just write rv in here so we do not forget.
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There is velocity, let us go to acceleration.
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Let us keep going.
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We know the acceleration is dv dt so that is going to be the derivative with respect to time of r - ωr sin ωt i hat + ωr cos ωt j hat.
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Acceleration, as we take that derivative am going to have -ω² r cos ωt I hat - ω² r sin ωt j hat.
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We can factor an ω² out of that so I could write that as acceleration is - ω² will have r cos ωt i hat + r sin ωt j hat.
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That looks mighty familiar this term right here because we defined that way up here.
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If we look up above we said that r cos ωt I hat + r sin ωt j hat that was our r vector.
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We could then say that acceleration is - ω² r.
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Why that negative sign?
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This is because the acceleration points in the opposite direction of the radius.
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If our radius vector r is going out to that position this implies that acceleration is in the opposite direction towards the center of the circle.
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That is centripetal acceleration.
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We can make it look a little more familiar if you like.
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If you want to look at the magnitude of that centripetal acceleration that is ω² r still not familiar
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but what if they put in our substitutions V = ωr or ω = V / r.
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We can say that the magnitude of a = V² / r² × r or V² / r.
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Centripetal acceleration V² / r it all works out.
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Let us do an amusement park example.
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Riders on a merry go round move in a circle of radius 4m executing 4 revolutions every minute or 1 revolution every 15s.
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Find the angular velocity and the centripetal acceleration of a rider on the merry go round.
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Let us start with the angular velocity that will be angular displacement divided by the time interval
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which is going to be 4 × around in each time around is 2 π radians that takes 60s for all of that which will come out to be 0.419 radians / s.
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If we want centripetal acceleration that is going to be ω² r or 0.419² × r radius 4m or 0.702 m/s².
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Another example, let us look at the moon.
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The Moon revolves around the earth every 27.3 days and the radius of the orbit is 382 mega meters.
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What is the magnitude and direction of the acceleration of the moon relative to earth?
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Couple ways we can do this let us start with the old fashioned way.
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We know that the speed is distance /time.
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It travels one time around 2π r one circumference in the period of t which is going to be 2π × r radius 382,000,000 m
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in our time period is going to be 27.3 days but we want that in seconds.
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Let us do the conversion.
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27.3 days × we know there are 24 hr in one day and we know that there are 3600s in 1hr so why do all that and I come up with about 1018 m/s.
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The centripetal acceleration ac =V² / r which is going to be r 1018 m / s² divided by our radius again 382,000,000m or about 0.00271 m/s².
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That is one way to do it.
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Let us take what we just learned during the different way.
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Our angular velocity is δ θ divided by the time interval below 2π radians in again 27.3 days × 24 hours in the day × 3600 s/hr.
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That comes up with about 2.66 × 10 -6 radians / s.
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If we want the centripetal acceleration that is going to be ω² r which is 2.66 × 10⁻⁶² × r radius 382,000,000m.
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I get the same thing 0.00271 m/s².
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Let us move on and talk about reference frames for couple moments.
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A reference frame describes the motion of an observer watching something else moving.
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Our most common reference frame is earth that is the one we deal with every day.
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We typically measure the motion of things compare to the earth because that is what we typically see is its surroundings.
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Now the laws of physics, we study in this course assume we are in an inertial not accelerating reference frame.
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That is not quite true.
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As we are spinning on the earth we are constantly accelerating
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and have other motion from the universe but it is negligible compared to what we are typically dealing with.
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Let us assume we are going to call the earth an inertial reference frame.
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There is no way to distinguish between motion at a rest and motion at the constant velocity in an inertial reference frame.
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What that means is, let us think about what is going on if you are in an airplane as an example.
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Let us take the airplane and put a wing in here.
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Here we are nice and happy in our airplane.
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Assuming this is an extremely smooth airplane, no turbulence, no bumps, whatsoever.
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You cannot tell the difference whether you are sitting on the runway or whether you are in the air moving 500 miles/hr if all the window shades are down.
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As long as you are moving it that constant speed at the inertial reference frame there is no way to distinguish the difference between the 2.
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That is what we are talking about when we are talking about an inertial reference frame.
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How can you tell if you cannot look out the window?
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There is no experiment you can do to tell the difference between those 2.
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Once we have that settled what we have talked about that reference frames we will talk about motion being relative.
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For example imagine you are sitting in a lawn chair watching a train travel past you.
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It is going to the right at 15m/s.
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From your reference frame if you saw a cup of water through the train’s window it would look to you at your lawn chair like it is moving at 50 m/s.
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If you are sitting on the train right in front of a cup of water though the couple of water would appear to be at rest not moving at all.
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It all depends on your frame of reference.
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Now imagine, instead you are on the trains staring out of the window you are watching some students sitting on a lawn chair for some unknown reason.
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From your reference frame, the cup of water on the train remains still but you see that student as though they are moving to the left that 50 m/s.
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Again, it all depends on your frame of reference.
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What is going on as far as motion and how you describe that?
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When we want to calculate velocities between 2 things, relative velocities, let us consider 2 objects a and b.
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Taking the velocity of a with respect to some reference framed b helps us understand exactly what we are talking about.
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It is pretty straightforward.
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For example you might want to know the speed of a car with respect to the ground.
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Or if you are walking on the train you might want to know the speed of a person with respect to the train.
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It does not matter that the train is speeding along at 40 m/s.
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What you are really concerned about is how fast that person is moving with respect to the rest of the train.
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Let us do some examples here as we talk about strategies.
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When we are calculating relative velocities here is a little trick I use to help keep things straight.
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What we are going to do is say that the velocity with some object a with respect to c is defined as the velocity of a
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with respect to some intermediate object b + the velocity of b with respect to some objects c.
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As long as you change things like this a to b, b to c, c to d.
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As long as you have them all daisy chained you can say that the total is the velocity of the first letter with respect to the last letter.
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In the example we were doing for example.
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The example we are doing for example a redundant again.
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The example we were doing, we will call velocity of b with respect to c, the velocity of the train with respect to the ground.
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We will call object a, that is a is our cup, b is our train, and c is the ground.
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Velocity of a with respect to b then would be the velocity of the cup with respect to the train.
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The velocity of a with respect to c then would be the velocity of the cup with respect to the ground.
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If you want to know the velocity of the cup with respect to the ground you take the velocity of the cup
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with respect to the train and add it to the velocity of the train with respect to the ground.
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And the math will all work out.
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As you look at that pattern here we could have any number of these that we daisy chain
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for example the velocity of some a with respect to e would be the velocity of a with respect to b + the velocity of some b
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with respect to c + the velocity of c with respect to d + the velocity of d with respect to e.
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As we do all these all we have to do is take a look and go with we got a b and b they match.
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C and c they match.
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D and d they match.
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Our total is going to be the first with respect to the last.
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Velocity of a with respect to e.
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It will be a little easier to show you how that is done with an example or 2 that is by just talking about it.
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Let us do a couple.
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Let us say we have a train we will call that a traveling at 60 m/s to the east with respect to the ground.
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We will call the ground c.
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A businessman b on the train runs it 5 m/s to the west with respect to the train.
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Find the velocity of the man b with respect to the ground c.
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We are looking for the velocity of b with respect to c.
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One way I could do that is I can find the velocity of b with respect to a + the velocity of a with respect to c.
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That would work because we have the a's in the middle so we end up with b and c for a total.
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Based on what we are given I think we will be ok there.
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The velocity of b with respect to a would be the velocity of the man with respect to the train.
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It says the man on the train runs 5 m/s to the west with respect to the train.
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Velocity of b with respect to a is 5 to the west.
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Let us call it to the east positive.
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That will be -5 m/s because the man is running to the west with respect to the train + V ac the velocity of the train
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with respect to the ground that is 60 m/s to the east.
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That will be + 60 m/s I will end up with 55 m/s east.
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The velocity of the businessman with respect to the ground is 55 m/s east.
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You probably could have done that in your head on a simple problem like this but
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as we get more involved having a way to keep track like this can be very valuable.
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Let us take a look at an airspeed example.
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An airplane let us call that P flies at 250 m/s to the east with respect to the air.
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The air we will give that an a is moving at 15 m/s to the east with respect to the ground.
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We will call to the east our positive direction again.
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Find the velocity of the plane with respect to the ground.
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We are looking for the velocity of the plane with respect to the ground.
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We are given the velocity of the plane with respect to the air that is 250 m/s and we are given
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the velocity of the air with respect to the ground which is 15 m/s.
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If we want the velocity of the plane with respect to the ground that will be the velocity of the plane
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with respect to the air + the velocity of air with respect to the ground.
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Those matches in the middle are left with P and g that should work.
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So that is going to be 250 m/s + 15 m/s or the velocity of the plane with respect to the ground is 265 m/s.
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This works in multiple dimensions as well.
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An airplane P flies a 250 m/s to the east with respect to the air.
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The air is moving at 35 m/s to the north with respect to the ground.
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Air with respect to the ground is 35 m/s north.
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Find the velocity of the plane with respect to the grounds.
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We want plane with respect to the ground.
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The velocity of the plane with respect to the ground is going to be the velocity of the plane
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with respect to the air + the velocity of the air with respect to the ground.
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It will be Pg Pg that should work.
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What I am going to do is draw this out.
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The velocity of the plane with respect to the air that is 250 m/s to the east.
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The velocity with of air with respect to the ground is 35 m/s to the north and we are going to add that
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so I will line them up to tail to remember as we talked about vectors.
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There is our velocity of air with respect to the ground which is 35 m/s.
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If I want the sum of those the velocity of the plane with respect to the ground I will draw a line
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from the starting point of my first factor to the ending point of my last once they are lined up to tail.
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To add these now in vector form let us find a magnitude and find the speed here first.
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We can do that using the Pythagorean Theorem.
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The magnitude of the velocity of the plane with respect to the ground is going to be √250² + 35² or about 252 m/s.
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If we wanted to know this angle θ we can find that as well.
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Θ equals the inverse tan.
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We know the opposite and the adjacent side.
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Inverse tan of opposite / adjacent which will be the inverse tan of 35 m/s / 250 m/s or about 7.97 degrees.
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My answer the velocity of the plane with respect to the ground is 252 m/s and angle of roughly 8° NE.
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One last 2D problem here with relative velocities.
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An oil tanker let us call it T travels east at 3 m/s with respect to the ground while a tugboat moves north at 4 m/s with respect to the tanker.
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Our tug boat we will call b with respect to our tanker T.
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What is the velocity of our tug boat with respect to the ground?
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We want the velocity of our tug boat with respect to the ground and that is going to be the velocity of our tugboat
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with respect to the tanker + the velocity of the taker with respect to the ground.
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All right drawing these out, we know that we have as I look at the problem tanker travels east that 3 m/s with respect to the ground.
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The tanker with respect to the ground VT g is 3 m/s.
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We also have the tugboat pushes it north or moves forward moves north at 4 m/s with respect to the tanker.
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The velocity of our boat with respect to that tanker is going to be 4 m/s.
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We can add these in any way we want I am just going to drop this way so tip to tail again.
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Velocity of tugboat with respect to the tanker is 4 m/s.
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Find the velocity of the tugboat with respect to the ground.
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To do that I will add these up starting at the starting point of my first going to the ending point of my last.
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And I can do that when my head that is a 345 right triangle.
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The velocity of our tugboat with respect to the ground must be 5 m/s.
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And we can use trigonometry if they wanted to if we needed to know that angle exactly.
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Hopefully that gets you a good start on circular and relative motion.
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Thank you so much for watching www.educator.com and make it a great day everybody.