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Lecture Comments (2)

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!

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:



    + m2

    + …+ mn


    m1 + m2 + …+ mn
  • 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:


    = M


    where M is the mass of the entire system.

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 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

Transcription: Center of Mass

Hi, welcome back to, 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, t2.0015

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 = m1x1, (mass of the first one × location of the first one).0334

Then, m2x1, 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 m1+m2 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, (m1x1 + m2x2 + .... + mNxN)/(m1+m2+....+mN), 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 x1 effectively zero", and we will measure everything in reference to that x1, 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 = (m1x1 + m2x2 + .... + m2x2)/(m1 + m2 + ....+ mN).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, Fnet = (total mass of the system)×(acceleration of the centre of mass) = MaCM.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), x1, and it is going to be zero, it also means m1 = 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, m2 is 1 m, m2 = 1 kg.1068

Finally, x3 will be at 3 m, and correlates to m3 = 2 kg.1070

xCM = m1x1 + m2x2 + m3x3)/(m1 + m2 + m3).1082

So, xCM = (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

m1 = 1 kg, m2 =5 kg, m3 = 2 kg, we got located at point, x1 = (0,10,0), x2 = (2,1,3), x3 = (3,-2,-4).1142

What do we have for vectors?1155

x = (m1x1 + m2x2 + m3x3)/(m1 + m2 + m3).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 = πr2, so for the big circle, A = 4πr2, and the location of its centre is zero.1427

The centre of big circle is, xbig = 0.1448

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

A = πr2, and xnega = -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 xCM to paying attention to where the area is distributed around these centre points.1504

So instead, we can make this xCM = (m1x1 + m2x2)/(m1 + m2).1510

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

So, xCM, 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, 4πr2.1546

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 -πr2, 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 -πr2 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 - πr2×(-r) / (4πr2 - πr2) = πr3/3πr2 = 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 mb = 40 kg, is standing on the right edge of a raft, that is 2 m long, with mass mr = 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 xr 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, xr = 1.5 m, xb = 2.5 m, the edge of the boat was 0.5, and the boat itself was 2 m long.1913

What is xCM going to be?1924

xCM = (m1x1 + m2x2)/(m1+m2) = (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, xb = ?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, xr = xb + 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, xr = xb + 1, xb is unknown, but we know that xCM is going to remain 2.17 m from what we solved before.2095

If we know that xCM = (m1x1 + m2x2)/(m1+m2).2104

2.17 m = (20×(xb+1) + 40×xb)/60.2116

130 = 20xb+20+40xb, 100 = 60xb, xb = 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