For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### Center of Mass

- The
*center of mass*(COM) for a system is the point that moves as if all the system's mass were located there and all the forces on the system were applied there. - This means we can describe the motion of a system (to some extent) by describing the motion of the COM.
- We can find the COM of a system by taking the weighted average of the location of every object in the system:
→x

COM= m _{1}→x_{1}+ m _{2}→x_{2}+ …+ m _{n}→x_{n}m_{1}+ m_{2}+ …+ m_{n}. - If we have a homogeneous (uniform mass distribution) object, we can use symmetry to find the COM.
- If we have a void (missing piece) in a homogeneous object, we can still use symmetry: we find the COM for the object
__without__the void, then find the COM for just the void and assign it a__negative__mass. Then we look at the COM for these two objects put together (don't forget the negative mass in the void). - Center of mass and center of gravity are often used interchangeably, but there is a technical difference. The center of gravity is the point we can treat as if the force of gravity were all located there. Generally, this is the same as the center of mass, but for very, very tall objects, they might be slightly different because gravity changes the higher up you go.
- We can expand Newton's Second Law to work on an entire system through the idea of center of mass:

where M is the mass of the entire system.→F

net= M →a

COM,

### Center of Mass

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Introduction to Center of Mass
- Center of Mass
- Definition of Center of Mass
- Example of center of Mass
- Center of Mass: Derivation
- Center of Mass: Formula
- Center of Mass: Formula, Multiple Dimensions
- Center of Mass: Symmetry
- Center of Mass: Non-Homogeneous
- Center of Gravity
- Newton's Second Law and the Center of Mass
- Example 1: Finding The Center of Mass
- Example 2: Finding The Center of Mass
- Example 3: Finding The Center of Mass
- Example 4: A Boy and His Mail

- Intro 0:00
- Introduction to Center of Mass 0:04
- Consider a Ball Tossed in the Air
- Center of Mass 1:27
- Definition of Center of Mass
- Example of center of Mass
- Center of Mass: Derivation
- Center of Mass: Formula
- Center of Mass: Formula, Multiple Dimensions
- Center of Mass: Symmetry
- Center of Mass: Non-Homogeneous
- Center of Gravity 12:09
- Center of Mass vs. Center of Gravity
- Newton's Second Law and the Center of Mass 14:35
- Newton's Second Law and the Center of Mass
- Example 1: Finding The Center of Mass 16:29
- Example 2: Finding The Center of Mass 18:55
- Example 3: Finding The Center of Mass 21:46
- Example 4: A Boy and His Mail 28:31

### High School Physics Online Course

### Transcription: Center of Mass

*Hi, welcome back to educator.com, today we are going to be talking about centre of mass.*0000

*Before we try to explain what the centre of mass is, lets us just consider, if we tossed a ball into the air.*0006

*As we have seen before and learned before, the ball is going to fall a parabolic path, why?*0010

*Because we know that the horizontal movement is constant, it is just going to be a constant linear rate of movement, whereas the vertical motion is going to be based on a parabolic motion, it is going to be based on time squared, t ^{2}.*0015

*It is going to wind up giving us this nice curved parabola of motion.*0028

*You have seen this before, and if you see anything in real life, you toss the ball into the air, you can see it move as a parabola, you see a jet of water, you are going to see a parabola, parabolas are around us because of this.*0032

*But what is the thing that is moving in that parabola?*0042

*In this case, it is the ball that is moving along.*0044

*What if we threw something else, what if we threw a stick?*0047

*We grab one end of the stick, and we chuck the stick.*0050

*As it goes through the air, the stick is going to twirl.*0053

*But the stick as a whole, is still going to fall that parabolic motion.*0056

*If we look at this diagram, it cuts through the same location, a little bit before every point here, it is the same location that falls the parabola.*0060

*What do we call that location, what is that?*0072

*That is the 'centre of mass', this stick is still following a parabolic path.*0076

*It does rotate in the air, it has this centre, this centre of mass.*0080

*That is going to perfectly fall in the path.*0084

*We call this centre the centre of mass, for any system of objects, the centre of mass is going to be the point that move as if all the system's mass was located there, and all forces on the system were applied there.*0087

*This is really powerful, this means that we can have a complicated system that is having multiple things happening to it, so we might have little adjustments like, say, a car driving along on the free way.*0102

*Lots of things are happening to the car.*0112

*If you move around inside the car, and there will be little things happening inside of the car, we can look at the whole entity and say what is the centre of mass of the car, and its motion is going to be effected only by what is happening between the external environment and it.*0114

*We do not have to worry about the internal things.*0129

*So, centre of mass is a really powerful concept.*0131

*Here is an example of centre of mass: A fire works rocket is launched into the sky, and there is no air resistance, and this is important because if there were air resistance, it would completely break this example up, and we will explain why in a few moments.*0134

*The centre of mass of the system will take the same path, whether or not the rocket explodes.*0149

*Here, we shoot the rocket up, it gets to the top of its arc, it starts to fall down, it falls down, unimpressive!*0153

*But over here, we shoot it up, it explodes, we are used to thinking an explosion as, it goes in all different ways.*0161

*It does, but the centre of the mass of the system still continues to being the same.*0169

*Even as the arcs move outward as it falls, the centre of mass of that set of arcs, is going to still wind up being in the same place.*0177

*The centre of the system is always the same, because the explosion is internal to the system.*0187

*Why does it matter if there is no air resistance?*0192

*Because, the air resistance of the rocket on its own is a totally different set of air resistance, that all of the fragments of the rocket.*0194

*The fragments of the rocket are going to experience way more air resistance, and that is why it is going to have a shifting parabola, its arc is going wind up changing because of the change in air resistance once it makes that transition to be exploded.*0201

*But the centre of mass is still going to fall as being the unified thing, it will wind up changing the arc, because now the centre of mass is experiencing a new set of forces, because of the increased air resistance.*0215

*But this stuff about the centre of mass is still true.*0226

*It is easier to understand the idea when you think of no air resistance, because there is clear difference of exploding and non-exploding, having no difference in the way they fall.*0228

*That is surprising, it is not necessarily intuitive, but that is because of the way the centre of mass works.*0235

*Once we include air resistance, it starts to get more complicated, but the important thing to remember, is that the centre of mass, tells us how the system moves as the whole thing.*0240

*No matter what happens to the system internally, so explosion, you moving around inside the car, on a train, you throw up a ball, centre of mass of the whole system is going to be the same thing.*0249

*Let us figure out a way to get what the centre of mass is, as a useful mathematical formula.*0260

*Start off with these two pictures.*0267

*In this one, we have got the same mass here, and here, and so, it is obvious.*0269

*If you got two masses, they are the same thing, the centre of mass of the system is going to be in the middle.*0273

*If you had a stick, that was just equal density throughout, you put the finger to the middle of the stick to hold it up, right?*0277

*It makes sense, the middle is going to be there.*0283

*But what happens if one end of the stick was really heavy.*0286

*You got a big knot on one end of your stick, and at the end we got long and thin and tapered.*0288

*Over here, we are going to have a heavy mass, and over here, we are going to have a small mass, so what is that going to mean?*0294

*The centre of mass of the system is going to be way closer to the heavy part.*0299

*We have to take into account, both the mass of the objects, and how far they are, in terms of some location.*0302

*The distance apart matters, the distance from some reference point is what we will wind up going with, and how heavy each point mass in our system is.*0308

*Centre of mass in a system where we are using point masses, we use point masses, because we know that centre of mass of a circle is just the centre of the circle.*0316

*Centre of mass of the system, we consider the mass and distance together, the weighted average.*0329

*The location of the centre of mass, x = m _{1}x_{1}, (mass of the first one × location of the first one).*0334

*Then, m _{2}x_{1}, and that gives us a way to be able to unify both the idea of mass and its location, which we are to be able to do, to be able to talk about the location of mass.*0345

*Then, in addition to that, we are to be able to deal, with how much each one of those, put into the system.*0357

*If we have two giant masses, but they are equal masses, and we have two small masses, and equal masses, both these cases, we are going to wind up having the same centre of mass between these two systems.*0363

*So, we are going to have a way of dealing with not just, they are going to have very different m's, but those m's will get divided out, so it is what they are in relation to the rest of their system.*0379

*So, that is why m _{1}+m_{2} is on the bottom, because we have to have a way of cancelling out what the mass we are dealing with is.*0390

*So, it i snot just about what the mass and location is, it is what the total masses of the system is, it is weighted average.*0398

*What if we were to do this with an arbitrary number of point masses?*0405

*The exact same method, we just need to expand it.*0408

*Now the centre of mass is going to be, (m _{1}x_{1} + m_{2}x_{2} + .... + m_{N}x_{N})/(m_{1}+m_{2}+....+m_{N}), for N masses.*0410

*One important thing to notice is that it is up to us to define our position system.*0435

*It is going to make things easier often, if we are able to define one of our points as a zero point, to be able to say, "Okay, let us make x _{1} effectively zero", and we will measure everything in reference to that x_{1}, because we will gets distance from object to object when we are measuring things in the real world, but it is up to us, as usual, to impose a coordinate system.*0440

*An easy thing to do is often to say, we will just make one of these zero, and work from there.*0459

*Sometimes, that is not what you want to do, it is going to depend on the problem.*0466

*It is also going to depend on if the coordinate system has already been imposed for you.*0470

*We will have to take it in a case to case basis, but as a general rule, it is going to work some times, if you put one as a zero point.*0473

*If you already have position system, you have to use that one, if you get the chance to impose one of your own, you might want to consider making one of your points the zero point.*0480

*But, you do not always want to do that, as we will see in our fourth and final example for this lesson.*0490

*What happens if we are doing this in multiple dimensions?*0496

*Working with point masses in multiple dimensions, it is just the same, we just throw on vectors.*0498

*The formula holds true in each dimension on its own, the x axis dimension is going to hold true, the y axis dimension is going to hold true, the z axis dimension is going to hold true, so we just expand our formula to vectors.*0502

* x = (m_{1}x_{1} + m_{2}x_{2} + .... + m_{2}x_{2})/(m_{1} + m_{2} + ....+ m_{N}).*0514

*And, x is (x,y) or (x,y,z), so x is not just the x axis, it is now meaning all of the axes as one, we are now saying, our location is called x.*0529

*This just becomes, what your location is for a given thing.*0543

*Another idea: If we have an object that has a uniform (or homogeneous, meaning same throughout) distribution of mass, we can use symmetry to find the centre of mass.*0546

*For example, this square, the centre of this side is here, and the centre of this side is here.*0559

*You put those through, and BOOM!, we see through symmetry, that the centre has to be there.*0565

*We do the same sort of thing here, the centre of this side is here, the centre of this side is here, BOOM!, we find the next one.*0572

*Triangles are little bit more complicated, but here if you want to get very clever, we can figure out that the amount of area above is the same as the amount of area below.*0582

*So, we know that this is our dash line.*0598

*How do we figure out the width?*0600

*That is very easy, we just drop a vertical line from the very centre, since in this case we got an equilateral triangle, and that is how we figure it out.*0605

*That might be a little bit harder to do on our own, but we can at least get an idea of where it approximately is going to be.*0614

*What if we had a torus (torus is a special word for a donut), or a circular disc, that is missing the inner disc?*0619

*In this case, where is the centre of the big circle?*0627

*Centre of the big circle is just the centre of the circle.*0629

*Where is the centre of the small circle?*0632

*Centre of the small circle is also going to be the centre.*0633

*So, we have got this removed centre out of the middle, it is not going to affect the large thing, because the centre from all these points is going to wind up still being here.*0637

*One centre out of another centre, it is still going to wind up being the same thing.*0647

*So, our centre is still just in the middle, because we have got this sort of negative mass, taken out of a solid disc, and both of them wind up having the same centre of mass, so we have got it.*0651

*What do we do if we were given a non-homogeneous thing?*0662

*Something with the mass is not evenly distributed through the object, or what if we have an evenly distributed object, but it is a weird, complicated, non-symmetric shape, like this, how do we find out the centre of mass?*0665

*Well, you do not! We are screwed, I am sorry.*0683

*At the moment, we cannot do that, but that is just for now.*0686

*The reason why we cannot do it, is just because we do not have calculus in our tool box.*0691

*To be honest, the stuff that you will learn in calculus to do this, actually is not that hard, it is totally in the range of doing, once you got little bit of calculus under your belts.*0694

*You will not find it that hard, once you take calculus, but right now we just do not have the ability to talk about it, we have to refer to integrals, and right now, we cannot refer to integrals, because we do not know what calculus is, for this course.*0703

*But, once we do get to calculus in the future, you will totally be able to understand it, but right now we cannot, so we will not be discussing any hard centre of mass problems.*0712

*We still got lots of things that we can do.*0726

*Finally, what is the centre of gravity?*0730

*Often, the centre of mass, the centre of gravity, they get thrown about, completely interchangeably.*0732

*But, there is a very minor distinction between them, what is that distinction?*0737

*Centre of mass is where the aggregate mass of the system is located.*0741

*We can think of that as, where the average, the place where we can consider all the mass effectively is.*0745

*The car might be in different places, but we can treat it as a point mass, by being able to figure out where is its mass located around the centre.*0751

*Like this stick, like the rocket, where is everything sort of located around, that is where the centre of mass is, the aggregate, the sort of average where we can locate the mass as a specific point, even if that specific point is hollow, like in the case of the donut or the torus.*0761

*Centre of gravity on the other hand, is a little bit different.*0775

*Centre of gravity is the place where the force of gravity acts on a system in aggregate, where the average of the centre of gravity is.*0777

*Consider, if we had a super tall building.*0784

*A super tall building is going to have a different gravity here, a than up here.*0788

*They are going to be very similar, but the farther you get away from earth, the lesser the effect of gravity of earth is.*0793

*On a really tall building, hundreds of storeys tall, that building is going to experience slightly less gravity than when you are on the top floor, you will weigh ever so slightly less than when you are on the bottom floor.*0799

*This means that what is up here, is going to be affected less by gravity, than what is down here, which means that the centre of mass might be in the very centre of the building, the centre of gravity is going to be a little bit below that, because more of the bottom is going to be affected by gravity than the top.*0812

*More of the gravity contributes towards where the centre of gravity is located.*0829

*Now, keep in mind, for a small object, these are going to be the same thing.*0833

*These are going to be so minisculely different that we can effectively treat them as the same thing.*0837

*For vey large, very tall objects, the pull of gravity will be slightly different.*0841

*Keep in mind, slightly different in this case means that for the tallest of skyscrapers in the world, we are talking about a matter of 1 mm, and that is like the tallest of skyscrapers.*0847

*For our purposes, we can pretend they are the same thing.*0859

*In result, we will be able to consider that the centre of gravity, and the centre of mass effectively same thing.*0861

*In reality, they are not precisely the same thing, but for our purposes, they are going to be so close, we might as well treat them as synonyms, we might as well treat them as the same thing.*0867

*Newton's second law and the centre of mass:*0876

*We got his great, really useful formula from Newton's second law: F = ma.*0879

*We can expand this idea, to work on an entire system by using the centre of mass.*0885

*The net force on a system, F _{net} = (total mass of the system)×(acceleration of the centre of mass) = Ma_{CM}.*0889

*F, if it is internal, so if we have some system, and some F occurs inside of the system, what do we know from the Newton's third law?*0909

*We know that there is also resisting inner force, that is going to be the same thing.*0916

*If there is an internal force that happens inside of the system, the internal system force is also going to result in another internal system force, equal in magnitude, opposite in direction, equal and opposite reaction force.*0920

*We know that from Newton's third law.*0933

*So, if our force is internal, it is going to get canceled out by its reaction force, so we do not have to worry about having a net force in the end.*0935

*The only forces that we are going to have to worry about, are external forces.*0941

*If it is an external force, we have to care.*0946

*Finally, acceleration of the centre of mass, is just the acceleration of the centre of mass.*0949

*We do not necessarily know anything about the specific objects that make up our system.*0954

*Our system is made up of many different things, many disparate objects.*0959

*It might be the case that the acceleration of the centre of mass is going one way, but one piece of our system, is going in the opposite direction.*0963

*Just because we know something about the centre of mass, does not mean we know everything about the system.*0970

*It just gives us one more piece of information to understand our system.*0974

*It is really useful in many cases, but it will not necessarily tell us everything.*0978

*Once again, you have to pay attention, you have to understand what you are looking at, and really be thinking about the problem you are working on, not just blindly follow formulae.*0981

*We are ready for our examples.*0990

*Nice one to start off with: We have got three balls on a mass less rod.*0992

*Mass less means we do not have to worry about it for our centre of mass, all we have to worry about are the balls on the rod.*0996

*The first ball, is right here, and it weighs 4 kg, the next ball is 1 m away, and it weighs 1 kg, and the next ball is 2 m away, and it weighs 2 kg.*1002

*How do we figure out where the centre of mass is?*1025

*Do we have a coordinate system yet? We do not have a coordinate system yet.*1028

*We know what the distance is, but we have to impose a coordinate system.*1031

*The first thing is to impose a coordinate system.*1034

*I like the 4 kg ball, because it is right on the left, we are used to going right as positive, and the 4 kg, since that is the biggest, most massive, the centre of mass is going to be the closest to 4 kg, it is probably going to suck up most of the centre of mass towards it.*1036

*We will call this (1), x _{1}, and it is going to be zero, it also means m_{1} = 4 kg.*1056

*What is this going to be located at? It will be at a location of 1, because it is 1 m away.*1063

*So, m _{2} is 1 m, m_{2} = 1 kg.*1068

*Finally, x _{3} will be at 3 m, and correlates to m_{3} = 2 kg.*1070

*x _{CM} = m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3})/(m_{1} + m_{2} + m_{3}).*1082

*So, x _{CM} = (4×0 + 1×1 + 2×3)/(4+1+2) = 7/7 = 1 m away, which means that the centre of mass is actually right there, right under that second ball, that is our precise centre of mass.*1095

*Next up: We got three point masses, and this time we are going to do it multi dimensionally.*1128

*m _{1} = 1 kg, m_{2} =5 kg, m_{3} = 2 kg, we got located at point, x_{1} = (0,10,0), x_{2} = (2,1,3), x_{3} = (3,-2,-4).*1142

*What do we have for vectors?*1155

* x = (m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3})/(m_{1} + m_{2} + m_{3}).*1159

*So, (1×(0,10,0) + 5×(2,1,3) + 2×(3,-2,-4))/(1+5+2) = [(0+10+6),(10+5-4),(0+15-8)]/8 = (16,11,7)/8 = (2,11/8,7/8) m, which tells us the position of the centre of mass.*1174

*There is nothing specifically there, but we know that, that is where the aggregate mass is, we will be able to do a force that worked on the whole system, worked on any part of the system in fact, that would effectively work on the whole system, because that is working on the system.*1287

*So we know that the centre of mass is going to be adjusted based on this, so that is where our starting location is.*1301

*Example 3: Now things are starting to get [?].*1306

*Now we are going to have to do some real thinking as opposed to following just formulae.*1309

*To begin with, let us look at a problem: A circle of uniform density with radius 2r has a circle of radius r removed from its far left side.*1312

*Where is the centre of mass?*1320

*This thing is not simply symmetric.*1322

*What we can do, is we can think of it as a trick, we can come up with a trick to think of this, we know what the centre of mass of the whole circle is.*1325

*It is just the centre of the circle.*1334

*We can consider the large circle as a whole circle.*1336

*So, our big circle would have its centre of mass at its centre, at x = 0, let is make here the very centre, x = 0.*1344

*We know what that centre is, we can figure out what its area is, we have a uniform density, we do not know what the mass is precisely, but that is going to work out in the end, because we are going to divide by the mass amount in the end.*1354

*But, we have to also deal with this part that is missing.*1367

*Here is the trick: We can think of the part that is missing not being there, but being there as a negative, we can think of a big circle that had a smaller negative circle, an anti-circle added to it.*1371

*If we put these together, then it is going to equal, a big circle with the negative circle cut out off it, now it kind of look like the Death Star, but that does not matter.*1397

*So, we have got, (big circle) + (negative circle) = (this thing right here).*1412

*How do we do this in terms of math?*1418

*What is the formula for the area of a circle? A = πr ^{2}, so for the big circle, A = 4πr^{2}, and the location of its centre is zero.*1427

*The centre of big circle is, x _{big} = 0.*1448

*What about the nega., what is the area of this one?*1451

*A = πr ^{2}, and x_{nega} = -r, making going to the right as positive, so now we have got a location for this.*1454

*Now, we could figure out what the mass is hypothetically.*1477

*For some amount of area, we get some amount of mass, there is going to be some conversion factor.*1480

*We wind up using this conversion factor both on the top and the bottom because it will be divide by the total mass, so ultimately what we really care about, is the area, because it is going to be a constant conversion factor throughout for both our negative circle, (which is going to be the negative version of that conversion factor), and the big circle.*1486

*So, in the end, we can just turn out our x _{CM} to paying attention to where the area is distributed around these centre points.*1504

*So instead, we can make this x _{CM} = (m_{1}x_{1} + m_{2}x_{2})/(m_{1} + m_{2}).*1510

*In this case, (1) is the big circle, and (2) is the negative circle.*1521

*So, x _{CM}, what is the mass, what is the effective mass of the big circle? The effective mass of the big circle is just how much area it has to sling around.*1537

*That is, 4πr ^{2}.*1546

*But, multiplied by its location, i.e. zero, so it is going to get knocked out.*1550

*How much area does the negative circle has to sling around? It has got -πr ^{2}, negative area because we have to be removing it, we know that the total area of this thing is going to be (big circle) - (negative circle), so it has got to be -πr^{2} is what its real area is, because what it contributes, is how much it takes away, it is a really important thing, I do not want you to be forgetting about that.*1557

*So, 0 - πr ^{2}×(-r) / (4πr^{2} - πr^{2}) = πr^{3}/3πr^{2} = r/3.*1612

*Let us think about that before we quickly assume that it is right.*1661

*That makes sense, we got a big circle which is missing a chunk, so where do you think the centre of mass is going to be?*1664

*Is the centre of mass wind up begin here, closer to the non-existent part?*1671

*No, there is no mass there to pull it there, so it is going to wind up showing over here.*1677

*The centre of mass is going to get shown at r/3, it is going to wind up being pulled over just a little bit, because it is missing that chunk, so it is going to be a little bit over.*1684

*A circle of uniform density, radius 2r, has a circle of radius r removed from its far left side, and we find out that the centre of mass has to be located at r/3 to the right of this place that has not been removed of the larger circle.*1696

*Final example, Example 4: A boy of mass m _{b} = 40 kg, is standing on the right edge of a raft, that is 2 m long, with mass m_{r} = 20 kg.*1712

*A camp counselor comes to hand in a piece of mail from the edge of the dock, but he can only reach out 1 m.*1723

*The boy can reach out 0.5 m, but the left edge of the boat is 0.5 m from the dock.*1728

*Will the boy be able to walk over and get the mail without rowing the raft?*1733

*The first thing to think is, yes, of course he can walk over, he walks over here, and grabs his mail, right?*1735

*Wrong! It just cannot happen like that, he cannot just walk over, because if he walks over, what happens is, there is no friction between the water and boat effectively, it is going to wind up getting moved.*1743

*So there is no external forces over here, we do not expect the water to resist, it does not give us much friction.*1758

*So the boat is going to wind up sliding this way, the centre of mass if the system will be preserved, so by the time the boy gets to the left edge of the boat, he might be only be here, and it might be the case that the boat edge is like this.*1763

*The boat and the boy represent the system, and we know that the centre of mass of that system will be preserved, because there is nothing to keep it in place, there is nothing to hold it in place, there is no external forces acting on the system.*1782

*As the boy moves around, the boat will have to move in response to that.*1792

*The centre of mass of the system has to be preserved.*1796

*If you are an astronaut in space, and you threw a wrench, and the wrench would fly away from you, but in result, you would wind up floating backwards, because (1) we can think about it as the response force of the wrench on you, but we can also think of it, as the centre of mass of the system is preserved.*1799

*They are the same thing, equivalently, the idea that forces come in pairs that cancel one another out, is the same idea as the centre of mass of the system is not affected by the forces, they are equivalent.*1821

*Now, we are going to have to get down and do some maths.*1836

*What is the centre of mass of the system to begin with?*1840

*To begin with, centre of the mass of the raft, it is reasonable to assume that the raft is homogeneous, or at least that it is symmetrical, so we can assume that the raft's centre is in the middle of it.*1843

*The boy is at the edge at 2 m, and we know that x _{r} is going to be at 1.5 m, but we have not discussed yet what our coordinate system is.*1855

*Let us make, to the right positive, but where are we going to make zero?*1864

*In this case, we are going to have shifting motion between our before and after pictures, before the boy starts moving, after he walks over.*1867

*So, it is up to us to figure out what is a good stationary reference frame that we can use.*1874

*In this case, I think the pier is a good place to make x = 0.*1882

*The pier is going to stay there, and that is what we care about.*1886

*We care if the boy can get 1.5 m away from the edge of the pier.*1888

*Camp counselor can reach out 1 m, the boy can reach out 0.5 m, so if the boy can get to x = 1.5 m, or less, he is good, he can get his mail without having to row.*1893

*If he cannot, he has to use the row, and actually move the boat, or swim over.*1902

*At this point, we can solve for what the centre of mass of the system is.*1908

*So, x _{r} = 1.5 m, x_{b} = 2.5 m, the edge of the boat was 0.5, and the boat itself was 2 m long.*1913

*What is x _{CM} going to be?*1924

*x _{CM} = (m_{1}x_{1} + m_{2}x_{2})/(m_{1}+m_{2}) = (20×1.5 + 40×2.5)/(20+40) = 130/60 = 2.17 m, is the centre of mass of the boat-boy system at the beginning.*1929

*If you think about where that is, that winds up being somewhere over here.*1978

*Now, without even doing the rest of the math to figure out where the boy can get to, we can figure out he is not going to be able to do it.*1981

*How can we figure out? We know symmetry is going to work on our end here.*1989

*The boy, if he moves here, is going to be directly over the centre of mass of the system.*1993

*That means that when the boy gets to here, the boat is going to wound up getting here as well, the centre of mass of each of the objects is going to have to line up with the centre of mass of the system.*1997

*Each of the objects is moved to the centre of mass of the system, then the centre of mass of the system is going to be, where the objects are, if the object moves to the centre of mass, then the other object will also has to move to the centre of mass since it is a two object system.*2007

*That means, if the boy is at one extreme here, then he is only going to be able to make that much further and jump out afterwards.*2022

*The boat will wind up making out its centre of mass, will also wind up getting up over here, so the boy to get to the other maximum is only going to here.*2030

*So, the difference between that was less than 0.5 m, to get to here, and then it will be less than 0.5 m, to get here, so this is going to be greater than 1.5 m.*2039

*We do not know the precise numbers right now, but we can at least know that he is not going to make it.*2046

*But, let us figure out exactly what it is, for practice.*2050

*Now we have to think about this again.*2054

*Now we have got our after picture, the counselor is standing here, still, at x = 0.*2055

*But now, the boat, is there, and the boy, is on the left edge.*2062

*Do we know where the boy is located?*2070

*No we do not, that is the whole point if this.*2072

*So, x _{b} = ?*2074

*Do we know where the raft is located?*2077

*If we know where the raft is located, we know where the boy is located, exactly because of that, we know that the two are related, so we can think that, x _{r} = x_{b} + 1, because he is going to be 1 m from the end, because he is going to walk as far as he possibly can.*2079

*We have got, x _{r} = x_{b} + 1, x_{b} is unknown, but we know that x_{CM} is going to remain 2.17 m from what we solved before.*2095

*If we know that x _{CM} = (m_{1}x_{1} + m_{2}x_{2})/(m_{1}+m_{2}).*2104

*2.17 m = (20×(x _{b}+1) + 40×x_{b})/60.*2116

*130 = 20x _{b}+20+40x_{b}, 100 = 60x_{b}, x_{b} = 1.83 m, is the boy's location at the end, so we know he cannot get the mail.*2138

*He cannot get mail without rowing.*2172

*1.83 is where he finishes getting up to, and we know this because we know that the centre of mass of the system is going to have to be the same before and after, because there is no external forces on the system.*2177

*We know that there is a connection between where the boy's location is, and the centre of mass of the raft is, and we can treat that as a point mass for the raft, so that gives us enough information for us to be able to solve for where the boy must land.*2186

*And there you go, we just do the math, and we will be able to figure it out.*2197

*Hope you understood this, hope it made a lot of sense.*2200

*It will be really useful when we will be talking about momentum.*2204

*We will be able to talk about collisions, which will give us pretty much all the understanding we need about kinematics and force.*2209

1 answer

Last reply by: Professor Selhorst-Jones

Mon Oct 21, 2013 9:33 AM

Post by Robert Mills on October 20, 2013

On Example 4 Cont, talking about the boys new position, why is it that xr=xb+1, and not xr=xb+2, because the boat is 2m long? He's starting from the extreme opposite at 2m, walks to the other side, equalling 2 full meters. I'm probably just reading this all wrong, haha!