AP Physics C: Mechanics Center of Mass

Hello, everyone, and welcome back to www.educator.com.0000

I’m Dan Fullerton and in this lesson we are going to talk about center of mass.0004

Our objectives include identifying by inspection the center of mass of a symmetrical object.0008

Locating the center of mass of a system consisting of two such objects.0014

Using integration to find the center of mass of the thin rod on non uniform density.0018

Applying the relation between center of mass velocity in a linear momentum0025

and between center of mass acceleration and net external force for system particles.0029

Defining center of gravity and using this concept to express the gravitational potential energy of a rigid object0034

in terms of the position of its center of mass.0040

Let us start by talking about what center of mass is.0045

Real objects are more complex than these theoretical particles we been dealing with.0049

It is never just the entire amount of mass at some tiny point and you can treat it that way.0053

Real objects are more irregular, they are more complicated than that.0058

However, what is really nice, is from a physics perspective, we can treat the entire object as0062

if its entire mass are concentrated in a single point that we are going to call the object center of mass.0067

Usually abbreviated CM or C o center of mass.0073

Mathematically speaking, center of mass is the weighted average of the location of mass and object.0080

How will we find center of mass?0086

Well, with some objects we can do it by inspection.0088

For uniform density of objects, the center of mass is going to be the geometric center of that object.0091

For object of multiple parts, you can find the center of mass of each part and treat it as a point and then look at their geometric center.0097

For the irregular objects, one way you can find it experimentally is to suspend the object from two or more points and draw a plumb line.0105

The lines are always going to intersect the center of mass.0112

If you attached it by a couple points, drop a plumb line from it, wherever they cross a right there that would be your center of mass of the object.0115

We can also figure this out by inspection especially if we have a highly symmetric object.0125

If we have something like this with the uniform density even though it is got a complex object it is pretty easy to see0130

that the center of mass is going to be right in the geometric center of our object.0136

That was pretty straightforward.0141

If we have a system of particles, however, we need to get a little bit more detail.0144

The position vector to the center of mass is the sum of all the little objects there mass × position ÷ total mass of your system.0149

Or if you want to look at coordinates, an x center of mass coordinates that would be your first mass × its x position + your second mass × x position and so on.0159

Divided by the sum of the masses.0170

In a similar fashion, if you wanted the center of mass for Y, that would be M1Y1 + M2Y2 and so on divided by the total mass.0173

Or m + m2 + 3 or total mass typically written capital M.0187

Let us do a quick example here.0194

Find the center of mass of an object model as two separate masses on the x axis.0196

Our first mass is2kg in the x coordinate of 2 and the second mass is 6kg here in the x coordinate of 8.0201

To find that, we can go to our function, that x coordinate of center of mass is M1X1 + M2X2 / M0210

which would be M1 2kg × its x position 2m + our second mass 6kg × its x position 8m divided by their total mass 8 kg.0224

4 + 48 ÷ 8 is just going to give us 6.5m.0244

We could treat this system as if its entire mass was one object here at about 6 1/2 with a mass of 8kg.0250

What if it is a continuous system?0268

Find the center of mass of the combination object here below.0271

The density of the object is uniform.0274

What we can do by inspection if this is a 3 kg block, we can see that its center of mass0278

is right in the center at what the position we are calling 00.0283

This object, we can find its center of mass here at 0, 3 on the number line, the center of that object.0286

What we can do is add these up as if the particles.0293

We can pretty easily see that the x center of mass is going to be right on that to x =0 line by symmetry.0297

All we need to figure out is the y center of mass which will be M1Y1 + M2Y2 divided by our total mass M0304

which will be 3kg × its y position 0 + 6kg × this objects y position 3 divided by our total mass 9 kg.0317

Our y position, center of mass is going to be 6 × 3 is 18 ÷ 9 are going to be 2.0333

The center of mass of our system would be at 0, 2 or right about there.0344

Let us do a two dimensional problem, find the coordinates of the center of mass for the system shown below,0358

where we got 3 kg mass at 1, 2, a 4kg mass 5, 3, and a 1kg mass at 7, 1.0364

We can do the x first, the x coordinate for the center of mass is going to be M1X1 3 × 1 + M2X2 4 × 5 + M3X3 1 × 7 divided by our total mass 3 + 4 + 1 or 8.0374

I come up with the value of 3.75.0399

Our y coordinate for the center of mass, we find the same way.0405

We will have 3 × 2 + 4 × 3 + 1 × 1 divided by their total mass 8.0409

6 + 12 =18 + 1= 19/8 or 2.38.0421

Our total, our center of mass of the system of object is going to be at 3.75 up to .38 is going to be right around there.0428

We can treat the entire system as if we had one particle with the mass of 8kg here at 3.75, 2. 38.0440

For more complex objects, you can find the center of mass by summing up all the little pieces a position vectors multiplied by that differential,0459

that little tiny piece of mass and then dividing by the total mass.0466

Just doing the calculus version, the integration version, is to make all of those mass is smaller and smaller.0470

The position vector to the center of mass is equal to the integral of the position vector × all those little tiny small masses divided by the total mass.0477

Let us put that in the play so you can see how that works.0488

We will start by finding the center of mass of a uniform rod of length LMSM.0491

The first thing I'm going to do is I'm going to do a couple definitions here.0496

I'm going to say the linear of mass density which we are going to call λ.0500

The mass in the length λ is M / L.0504

When I go and break this up into little pieces and when you break it up into little pieces along the X axis Dx and V.0510

Linear mass density there is just going to be that little bit of mass DM / dx, the mass in length again.0519

Which implies that just a little tiny piece of mass, DM is going to be our linear mass density × DX.0528

I can go to my formula for the position vector to the center of mass.0539

Our center of mass = the integral of our DM / our total mass which implies then that the vector to the center of mass is equal to the integral.0547

Our r in this case is just going to be our x coordinate.0562

X DM / our total mass but what is this the DM?0567

For that, we have to go up to our definition here, the little bit of mass contain a bit in that piece Dx is λ.0572

The linear mass density × DX.0579

That is going to be equal to the integral of x × differential of mass λ DX all divided by M.0583

We can pull out our constants because it is a uniform rod, λ is constant.0597

Our total mass already is constant.0601

We would have the λ divided by M integral of XDX.0604

We are going to integrate from x = 0 to X = some final value L the length of the rod.0611

This implies then that the vector r CM = we have a λ / M integral of X is x² /2 evaluated from 0 to L which is going to be λ/M L² /2.0619

We also know now that if λ = M/L well M = L λ.0646

We can replace M with L λ then we have λ L² /M which is L λ 2.0657

Our linear mass density cancels out and we end up with just L /2.0667

It is right in the middle.0678

That is common sense right.0678

If it is a uniform mass density rod the center of mass is going to be right in the center.0681

We proved that using calculus, going step by step as an exercise to see if we could do it, to show how you would go doing integration like this.0686

How you would determine the center of mass which is going to be useful when we do our next example problem.0694

Let us find the center of mass, now the non uniform rod.0701

This non uniform rod has length L and mass M, and its density is given by λ = KX.0705

You get the larger and larger values as x gets more and more dense.0711

Right away, just common sense tells me I would expect that we are going to have the center of mass somewhere to the right of center of this object.0716

We will use that as a check when we are all done.0724

To find the total mass that is going to be helpful here.0726

The total mass is going to be the integral from X = 0 to L of our linear mass density DX which is going to be our integral from 0 to L.0729

λ is KXDX, this is going to be KL² /2.0746

Let us go to our function again.0756

The position vector to the center of mass is the integral of r DM divided by our total mass.0758

We know that our R is just x coordinate.0768

We are dealing with one dimension.0773

Now, λ = KX which means that our differential of mass is going to be λ DX but now that is KX DX.0776

Our position vector to the center of mass = the integral of X × DM which is KXDX all divided by our total mass M.0787

Or position vector to the center of mass is we can pull K and M out of there, they are constants K /M integral for x=0 to x=L of we have got an x² in here now DX.0806

That is going to be K/M integral of x² is x³/3 evaluated from 0 to L.0821

We are going to have KL² / 3M.0829

But remember, we said M was KL² / 20836

this implies then that our position vector to the center of mass is going to be, we got KL² / 3 and our M KL² / 2.0844

A little bit of simplification here, L² we got an L there, we have got a K and I end up with 2/3 L.0860

It is somewhere over in this area, just like we predicted, a little bit to the right of center.0875

Let us take a look at some relationships here, if the position vector to the center of mass is 1/the total mass × the sum0883

over all the little tiny pieces of the mass of those pieces × their position vector.0894

That implies that the velocity of the center of mass is 1/ that mass × the sum over all i for all those little pieces of Mi × there individual little velocities.0903

We also know that momentum is mass × velocity so we can write this as the velocity of the center of mass is 1/ M sum0920

overall I, all of those individual little momentum of each of those pieces.0933

Which implies then that the total momentum which is what this is the total momentum = the total mass × the velocity of the center of mass.0941

The total momentum is mass × the velocity of the center of mass.0954

You can find the total momentum by just using that center of mass piece as well.0957

Or total momentum, if that is equal to M × velocity of the center of mass0963

and we know from previous work that force is that derivative of momentum with respect to time.0971

Let us take the derivative of our both sides here.0982

For the left hand side, we have a derivative of the total momentum with respect to T.0986

Which we know is must be our total force and that must be equal to the derivative of the right hand side.0993

The derivative with respect to T of mass × the velocity of the center of mass.1004

Which implies then that our total force must be equal to, the derivative of mass × velocity1011

mass is a constant the derivative of the velocity of the center of mass must be the acceleration of the center of mass.1020

Newton’s second law, the total force is equal to our total mass × the acceleration of the center of mass.1026

With Newton’s second law again, we do not have to worry about individual point particles or irregularly shaped object.1033

We treat the whole thing as if it had its entire mass right at that point the center of mass and we are allowed to do that.1046

Simplifies this up tremendously.1053

Another term you might have heard is called center of gravity.1056

Center of gravity refers to the location at which the force of gravity acts upon the object is if it were point particle with all of its mass at that point.1060

In a uniform gravitational field, the center of gravity and the center of mass are the same.1068

But if you happen to be the non uniform gravitational field they could be different.1074

If you have a humongous object that is so big, that parts that are in different gravitational field strengths where you have to bring that into account.1078

The center of mass and the center of gravity can be different.1088

For the most part, when we are in uniform gravitational fields where1091

we make the estimation like here on the surface of earth with relatively small objects, they are the same.1094

Technically speaking now, they are different quantities.1101

Let us finish up by looking in an old free response problem.1105

We will take the 2004 Mechanics exam free response number 1, you can find the link up here or google it.1108

Take a minute, print it out, and give it a try, then come back here and we will see how you do.1117

Taking a look at this question.1126

We have got someone swinging from a rope, right when they get to the vertical position on the rope they are grabbing that rock,1130

that mass, and then they are flying off as a projectile.1136

Find the speed of a person just before the collision with the object.1140

I would do that by conservation of energy.1143

The gravitational potential energy at position A must be converted into kinetic energy at position B.1146

Which implies that MG L the length of the rope must equal ½ M1 V².1154

I suppose that is M1 as well.1164

Which implies then the V² = 2 GL or V is going to be equal to the √2 GL right before the collision.1167

It also asks us to find the tension in the rope just before the collision with the object.1181

Our free body diagram, we have tension up and we have M1 G down.1186

We know that the net force, this case we will talk about a centripetal force, must be T - M1 G toward the center of the circle the positive direction.1192

And that has to be equal to mass × centripetal acceleration.1204

In this case, that will be M1 V² /L.1208

Therefore, solving for tension T will be M1 V² /L + M1 G.1215

But we just found out that V² = 2 GL right up there.1225

We can write this as T = M1 2 GL ÷ L + M1 G.1233

We got an L so I have 2M1 G + M1 G is just going to be 3 M1 G.1249

Let us move on to part C, find the speed of the person and object right after the collision.1263

That is a conservation of momentum problem.1271

Our initial momentum must equal our final momentum.1273

You can make a momentum table here if you want to.1277

Which implies then the D mass 1 × initial velocity must equal the combined mass after they collide × their final velocity.1280

Or V final is going to be equal to M1/M1 + M2 × the initial.1293

But V initial is √2 GL so that is just going to be M1/M1 + M2 √2 GL.1301

Moving on the part D, find the ratio of the kinetic energy of the person object system before the collision so the kinetic energy after the collision.1325

Let us start off with a kinetic energy before, that is going to be ½ M1 Vb 4² which is ½ M1 V² was 2 GL.1338

That is going to be M1 GL.1353

To find the kinetic energy after, the kinetic energy after will be 1/2 and their combined mass is M1 + M2 VF²1359

which is ½ F M1 + M2 × (M1 / M1) + M2 × √2 GL² our velocity we found in part C.1370

That implies then with a little bit of algebra, kinetic energy after is going to be ½ M1 + M2.1393

We will have M1² / M1 + M2² × 2 GL.1402

We can do little bit of simplification, M1 + M2 ÷ M1 + M2 to give us ½ M1 + M2 that is going to be equal to M1² / M1 + M2 GL.1415

When we take the final ratio, kinetic energy of B / kinetic energy at A, we are going to have kinetic energy at B/A.1447

We will have M1 GL / A.1457

We have got M1² GL M1 + M2 left over.1463

A little bit of simplification GGLL.1472

We got that M1 vs. M1².1477

I come up with M1 + M2 all divided by M 1.1480

We have got a part E in this question here as well.1493

Let us go give ourselves a little more room again.1496

For part E, find the total horizontal displacement x of the person from position A and tell the person object when the water at D.1499

That second part looks like it is a projectile motion problem, something launch horizontally.1510

Let us do that first.1515

We will find out how long they are in the air.1517

We will take a look at the vertical motion, we will call down the positive y direction.1520

V initial vertically is 0.1525

Our δ y is going to be L.1528

Our acceleration is G, the acceleration due to gravity.1531

δ Y = V initial T + ½ AY T².1536

Our V initial is 0 so that term goes away.1543

We could then write that L = ½ GT² or T is going to be equal to √2 L/G.1547

If we want the total horizontal displacement from B to D, δ x from B to D is just going to be the velocity in x × the time it is in the air.1560

We figured out the velocity in the x direction as they went the edge of the cliff previously, that was M1 / M1 + M2 × √2GL.1576

We need to multiply all of that by the time they are in the air √2L/G.1591

√2L √2L = 2L √G √G is a little bit of simplification here then I come up with 2 LM1 / M1 + M2.1598

We are not asked for just that distance, we are asked for the total distance.1617

The total horizontal displacement from position A.1621

Our total displacement is going to be L + what we just found there 2 LM1 / M1 + M2.1625

Or if you want to do a little bit of algebra, I think which you have there perfectly acceptable.1642

But you could go distribute that through a little bit and have L × get a common denominator M1 + M2 / M1 + M2 + 2 L M1 / M1 + M2,1647

which would be L × M1 + M2 + 2 M1 / M1 + M2.1666

Which is equivalent to, we would have L × 3 M1 + M2 / M1 + M2.1678

I'm not sure that is a whole lot prettier than what we have right there but either one of those should work.1690

Hopefully, that gets you a good feel for center of mass and center of gravity.1698

Thanks for watching www.educator.com.1701

We will see you again soon and make it a great day everybody.1703