WEBVTT physics/ap-physics-c-mechanics/fullerton
00:00:00.000 --> 00:00:03.700
Hello, everyone, and welcome back to www.educator.com.
00:00:03.700 --> 00:00:08.000
I’m Dan Fullerton and in this lesson we are going to talk about center of mass.
00:00:08.000 --> 00:00:13.800
Our objectives include identifying by inspection the center of mass of a symmetrical object.
00:00:13.800 --> 00:00:18.300
Locating the center of mass of a system consisting of two such objects.
00:00:18.300 --> 00:00:25.200
Using integration to find the center of mass of the thin rod on non uniform density.
00:00:25.200 --> 00:00:28.800
Applying the relation between center of mass velocity in a linear momentum
00:00:28.800 --> 00:00:34.100
and between center of mass acceleration and net external force for system particles.
00:00:34.100 --> 00:00:40.500
Defining center of gravity and using this concept to express the gravitational potential energy of a rigid object
00:00:40.500 --> 00:00:44.800
in terms of the position of its center of mass.
00:00:44.800 --> 00:00:48.900
Let us start by talking about what center of mass is.
00:00:48.900 --> 00:00:52.700
Real objects are more complex than these theoretical particles we been dealing with.
00:00:52.700 --> 00:00:58.200
It is never just the entire amount of mass at some tiny point and you can treat it that way.
00:00:58.200 --> 00:01:01.900
Real objects are more irregular, they are more complicated than that.
00:01:01.900 --> 00:01:07.100
However, what is really nice, is from a physics perspective, we can treat the entire object as
00:01:07.100 --> 00:01:12.900
if its entire mass are concentrated in a single point that we are going to call the object center of mass.
00:01:12.900 --> 00:01:20.100
Usually abbreviated CM or C o center of mass.
00:01:20.100 --> 00:01:25.800
Mathematically speaking, center of mass is the weighted average of the location of mass and object.
00:01:25.800 --> 00:01:28.000
How will we find center of mass?
00:01:28.000 --> 00:01:30.800
Well, with some objects we can do it by inspection.
00:01:30.800 --> 00:01:36.900
For uniform density of objects, the center of mass is going to be the geometric center of that object.
00:01:36.900 --> 00:01:44.700
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.
00:01:44.700 --> 00:01:52.000
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.
00:01:52.000 --> 00:01:55.000
The lines are always going to intersect the center of mass.
00:01:55.000 --> 00:02:05.500
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.
00:02:05.500 --> 00:02:10.200
We can also figure this out by inspection especially if we have a highly symmetric object.
00:02:10.200 --> 00:02:16.400
If we have something like this with the uniform density even though it is got a complex object it is pretty easy to see
00:02:16.400 --> 00:02:21.000
that the center of mass is going to be right in the geometric center of our object.
00:02:21.000 --> 00:02:24.000
That was pretty straightforward.
00:02:24.000 --> 00:02:28.900
If we have a system of particles, however, we need to get a little bit more detail.
00:02:28.900 --> 00:02:38.900
The position vector to the center of mass is the sum of all the little objects there mass × position ÷ total mass of your system.
00:02:38.900 --> 00:02:50.500
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.
00:02:50.500 --> 00:02:53.100
Divided by the sum of the masses.
00:02:53.100 --> 00:03:07.100
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.
00:03:07.100 --> 00:03:14.200
Or m + m2 + 3 or total mass typically written capital M.
00:03:14.200 --> 00:03:16.300
Let us do a quick example here.
00:03:16.300 --> 00:03:21.400
Find the center of mass of an object model as two separate masses on the x axis.
00:03:21.400 --> 00:03:30.200
Our first mass is2kg in the x coordinate of 2 and the second mass is 6kg here in the x coordinate of 8.
00:03:30.200 --> 00:03:44.300
To find that, we can go to our function, that x coordinate of center of mass is M1X1 + M2X2 / M
00:03:44.300 --> 00:04:04.200
which would be M1 2kg × its x position 2m + our second mass 6kg × its x position 8m divided by their total mass 8 kg.
00:04:04.200 --> 00:04:10.300
4 + 48 ÷ 8 is just going to give us 6.5m.
00:04:10.300 --> 00:04:28.600
We could treat this system as if its entire mass was one object here at about 6 1/2 with a mass of 8kg.
00:04:28.600 --> 00:04:31.400
What if it is a continuous system?
00:04:31.400 --> 00:04:34.500
Find the center of mass of the combination object here below.
00:04:34.500 --> 00:04:37.800
The density of the object is uniform.
00:04:37.800 --> 00:04:42.800
What we can do by inspection if this is a 3 kg block, we can see that its center of mass
00:04:42.800 --> 00:04:46.500
is right in the center at what the position we are calling 00.
00:04:46.500 --> 00:04:53.500
This object, we can find its center of mass here at 0, 3 on the number line, the center of that object.
00:04:53.500 --> 00:04:56.800
What we can do is add these up as if the particles.
00:04:56.800 --> 00:05:04.300
We can pretty easily see that the x center of mass is going to be right on that to x =0 line by symmetry.
00:05:04.300 --> 00:05:16.800
All we need to figure out is the y center of mass which will be M1Y1 + M2Y2 divided by our total mass M
00:05:16.800 --> 00:05:33.000
which will be 3kg × its y position 0 + 6kg × this objects y position 3 divided by our total mass 9 kg.
00:05:33.000 --> 00:05:44.300
Our y position, center of mass is going to be 6 × 3 is 18 ÷ 9 are going to be 2.
00:05:44.300 --> 00:05:58.200
The center of mass of our system would be at 0, 2 or right about there.
00:05:58.200 --> 00:06:04.600
Let us do a two dimensional problem, find the coordinates of the center of mass for the system shown below,
00:06:04.600 --> 00:06:14.400
where we got 3 kg mass at 1, 2, a 4kg mass 5, 3, and a 1kg mass at 7, 1.
00:06:14.400 --> 00:06:39.300
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.
00:06:39.300 --> 00:06:44.800
I come up with the value of 3.75.
00:06:44.800 --> 00:06:49.300
Our y coordinate for the center of mass, we find the same way.
00:06:49.300 --> 00:07:00.800
We will have 3 × 2 + 4 × 3 + 1 × 1 divided by their total mass 8.
00:07:00.800 --> 00:07:08.000
6 + 12 =18 + 1= 19/8 or 2.38.
00:07:08.000 --> 00:07:20.100
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.
00:07:20.100 --> 00:07:39.300
We can treat the entire system as if we had one particle with the mass of 8kg here at 3.75, 2. 38.
00:07:39.300 --> 00:07:46.400
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,
00:07:46.400 --> 00:07:50.100
that little tiny piece of mass and then dividing by the total mass.
00:07:50.100 --> 00:07:56.700
Just doing the calculus version, the integration version, is to make all of those mass is smaller and smaller.
00:07:56.700 --> 00:08:08.000
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.
00:08:08.000 --> 00:08:11.300
Let us put that in the play so you can see how that works.
00:08:11.300 --> 00:08:15.800
We will start by finding the center of mass of a uniform rod of length LMSM.
00:08:15.800 --> 00:08:20.100
The first thing I'm going to do is I'm going to do a couple definitions here.
00:08:20.100 --> 00:08:24.000
I'm going to say the linear of mass density which we are going to call λ.
00:08:24.000 --> 00:08:29.700
The mass in the length λ is M / L.
00:08:29.700 --> 00:08:39.500
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.
00:08:39.500 --> 00:08:48.300
Linear mass density there is just going to be that little bit of mass DM / dx, the mass in length again.
00:08:48.300 --> 00:08:59.500
Which implies that just a little tiny piece of mass, DM is going to be our linear mass density × DX.
00:08:59.500 --> 00:09:06.700
I can go to my formula for the position vector to the center of mass.
00:09:06.700 --> 00:09:21.900
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.
00:09:21.900 --> 00:09:26.800
Our r in this case is just going to be our x coordinate.
00:09:26.800 --> 00:09:31.700
X DM / our total mass but what is this the DM?
00:09:31.700 --> 00:09:39.100
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.
00:09:39.100 --> 00:09:42.700
The linear mass density × DX.
00:09:42.700 --> 00:09:56.700
That is going to be equal to the integral of x × differential of mass λ DX all divided by M.
00:09:56.700 --> 00:10:01.600
We can pull out our constants because it is a uniform rod, λ is constant.
00:10:01.600 --> 00:10:04.100
Our total mass already is constant.
00:10:04.100 --> 00:10:11.400
We would have the λ divided by M integral of XDX.
00:10:11.400 --> 00:10:19.100
We are going to integrate from x = 0 to X = some final value L the length of the rod.
00:10:19.100 --> 00:10:46.300
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.
00:10:46.300 --> 00:10:57.200
We also know now that if λ = M/L well M = L λ.
00:10:57.200 --> 00:11:07.400
We can replace M with L λ then we have λ L² /M which is L λ 2.
00:11:07.400 --> 00:11:17.700
Our linear mass density cancels out and we end up with just L /2.
00:11:17.700 --> 00:11:18.500
It is right in the middle.
00:11:18.500 --> 00:11:20.800
That is common sense right.
00:11:20.800 --> 00:11:26.100
If it is a uniform mass density rod the center of mass is going to be right in the center.
00:11:26.100 --> 00:11:34.500
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.
00:11:34.500 --> 00:11:41.300
How you would determine the center of mass which is going to be useful when we do our next example problem.
00:11:41.300 --> 00:11:45.100
Let us find the center of mass, now the non uniform rod.
00:11:45.100 --> 00:11:51.100
This non uniform rod has length L and mass M, and its density is given by λ = KX.
00:11:51.100 --> 00:11:55.900
You get the larger and larger values as x gets more and more dense.
00:11:55.900 --> 00:12:03.700
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.
00:12:03.700 --> 00:12:06.400
We will use that as a check when we are all done.
00:12:06.400 --> 00:12:09.600
To find the total mass that is going to be helpful here.
00:12:09.600 --> 00:12:26.600
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.
00:12:26.600 --> 00:12:35.900
λ is KXDX, this is going to be KL² /2.
00:12:35.900 --> 00:12:38.500
Let us go to our function again.
00:12:38.500 --> 00:12:48.600
The position vector to the center of mass is the integral of r DM divided by our total mass.
00:12:48.600 --> 00:12:52.800
We know that our R is just x coordinate.
00:12:52.800 --> 00:12:55.700
We are dealing with one dimension.
00:12:55.700 --> 00:13:07.000
Now, λ = KX which means that our differential of mass is going to be λ DX but now that is KX DX.
00:13:07.000 --> 00:13:26.400
Our position vector to the center of mass = the integral of X × DM which is KXDX all divided by our total mass M.
00:13:26.400 --> 00:13:41.200
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.
00:13:41.200 --> 00:13:49.000
That is going to be K/M integral of x² is x³/3 evaluated from 0 to L.
00:13:49.000 --> 00:13:55.700
We are going to have KL² / 3M.
00:13:55.700 --> 00:14:03.900
But remember, we said M was KL² / 2
00:14:03.900 --> 00:14:19.900
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.
00:14:19.900 --> 00:14:34.700
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.
00:14:34.700 --> 00:14:43.500
It is somewhere over in this area, just like we predicted, a little bit to the right of center.
00:14:43.500 --> 00:14:54.400
Let us take a look at some relationships here, if the position vector to the center of mass is 1/the total mass × the sum
00:14:54.400 --> 00:15:02.800
over all the little tiny pieces of the mass of those pieces × their position vector.
00:15:02.800 --> 00:15:20.100
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.
00:15:20.100 --> 00:15:33.500
We also know that momentum is mass × velocity so we can write this as the velocity of the center of mass is 1/ M sum
00:15:33.500 --> 00:15:41.100
overall I, all of those individual little momentum of each of those pieces.
00:15:41.100 --> 00:15:54.300
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.
00:15:54.300 --> 00:15:57.500
The total momentum is mass × the velocity of the center of mass.
00:15:57.500 --> 00:16:03.400
You can find the total momentum by just using that center of mass piece as well.
00:16:03.400 --> 00:16:11.400
Or total momentum, if that is equal to M × velocity of the center of mass
00:16:11.400 --> 00:16:22.300
and we know from previous work that force is that derivative of momentum with respect to time.
00:16:22.300 --> 00:16:25.900
Let us take the derivative of our both sides here.
00:16:25.900 --> 00:16:33.300
For the left hand side, we have a derivative of the total momentum with respect to T.
00:16:33.300 --> 00:16:43.800
Which we know is must be our total force and that must be equal to the derivative of the right hand side.
00:16:43.800 --> 00:16:51.400
The derivative with respect to T of mass × the velocity of the center of mass.
00:16:51.400 --> 00:16:59.700
Which implies then that our total force must be equal to, the derivative of mass × velocity
00:16:59.700 --> 00:17:06.200
mass is a constant the derivative of the velocity of the center of mass must be the acceleration of the center of mass.
00:17:06.200 --> 00:17:13.200
Newton’s second law, the total force is equal to our total mass × the acceleration of the center of mass.
00:17:13.200 --> 00:17:26.600
With Newton’s second law again, we do not have to worry about individual point particles or irregularly shaped object.
00:17:26.600 --> 00:17:33.000
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.
00:17:33.000 --> 00:17:36.400
Simplifies this up tremendously.
00:17:36.400 --> 00:17:39.900
Another term you might have heard is called center of gravity.
00:17:39.900 --> 00:17:48.300
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.
00:17:48.300 --> 00:17:53.800
In a uniform gravitational field, the center of gravity and the center of mass are the same.
00:17:53.800 --> 00:17:58.400
But if you happen to be the non uniform gravitational field they could be different.
00:17:58.400 --> 00:18:07.800
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.
00:18:07.800 --> 00:18:11.500
The center of mass and the center of gravity can be different.
00:18:11.500 --> 00:18:14.600
For the most part, when we are in uniform gravitational fields where
00:18:14.600 --> 00:18:20.800
we make the estimation like here on the surface of earth with relatively small objects, they are the same.
00:18:20.800 --> 00:18:25.200
Technically speaking now, they are different quantities.
00:18:25.200 --> 00:18:28.000
Let us finish up by looking in an old free response problem.
00:18:28.000 --> 00:18:36.700
We will take the 2004 Mechanics exam free response number 1, you can find the link up here or google it.
00:18:36.700 --> 00:18:45.900
Take a minute, print it out, and give it a try, then come back here and we will see how you do.
00:18:45.900 --> 00:18:50.000
Taking a look at this question.
00:18:50.000 --> 00:18:56.500
We have got someone swinging from a rope, right when they get to the vertical position on the rope they are grabbing that rock,
00:18:56.500 --> 00:19:00.200
that mass, and then they are flying off as a projectile.
00:19:00.200 --> 00:19:03.500
Find the speed of a person just before the collision with the object.
00:19:03.500 --> 00:19:06.200
I would do that by conservation of energy.
00:19:06.200 --> 00:19:14.300
The gravitational potential energy at position A must be converted into kinetic energy at position B.
00:19:14.300 --> 00:19:24.600
Which implies that MG L the length of the rope must equal ½ M1 V².
00:19:24.600 --> 00:19:27.500
I suppose that is M1 as well.
00:19:27.500 --> 00:19:40.800
Which implies then the V² = 2 GL or V is going to be equal to the √2 GL right before the collision.
00:19:40.800 --> 00:19:46.100
It also asks us to find the tension in the rope just before the collision with the object.
00:19:46.100 --> 00:19:52.000
Our free body diagram, we have tension up and we have M1 G down.
00:19:52.000 --> 00:20:04.100
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.
00:20:04.100 --> 00:20:07.800
And that has to be equal to mass × centripetal acceleration.
00:20:07.800 --> 00:20:15.500
In this case, that will be M1 V² /L.
00:20:15.500 --> 00:20:25.000
Therefore, solving for tension T will be M1 V² /L + M1 G.
00:20:25.000 --> 00:20:32.700
But we just found out that V² = 2 GL right up there.
00:20:32.700 --> 00:20:49.300
We can write this as T = M1 2 GL ÷ L + M1 G.
00:20:49.300 --> 00:21:03.500
We got an L so I have 2M1 G + M1 G is just going to be 3 M1 G.
00:21:03.500 --> 00:21:10.900
Let us move on to part C, find the speed of the person and object right after the collision.
00:21:10.900 --> 00:21:13.400
That is a conservation of momentum problem.
00:21:13.400 --> 00:21:17.000
Our initial momentum must equal our final momentum.
00:21:17.000 --> 00:21:20.300
You can make a momentum table here if you want to.
00:21:20.300 --> 00:21:32.800
Which implies then the D mass 1 × initial velocity must equal the combined mass after they collide × their final velocity.
00:21:32.800 --> 00:21:41.600
Or V final is going to be equal to M1/M1 + M2 × the initial.
00:21:41.600 --> 00:22:05.400
But V initial is √2 GL so that is just going to be M1/M1 + M2 √2 GL.
00:22:05.400 --> 00:22:17.900
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.
00:22:17.900 --> 00:22:33.500
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.
00:22:33.500 --> 00:22:38.800
That is going to be M1 GL.
00:22:38.800 --> 00:22:50.500
To find the kinetic energy after, the kinetic energy after will be 1/2 and their combined mass is M1 + M2 VF²
00:22:50.500 --> 00:23:12.900
which is ½ F M1 + M2 × (M1 / M1) + M2 × √2 GL² our velocity we found in part C.
00:23:12.900 --> 00:23:22.100
That implies then with a little bit of algebra, kinetic energy after is going to be ½ M1 + M2.
00:23:22.100 --> 00:23:35.200
We will have M1² / M1 + M2² × 2 GL.
00:23:35.200 --> 00:24:06.900
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.
00:24:06.900 --> 00:24:17.200
When we take the final ratio, kinetic energy of B / kinetic energy at A, we are going to have kinetic energy at B/A.
00:24:17.200 --> 00:24:22.800
We will have M1 GL / A.
00:24:22.800 --> 00:24:32.200
We have got M1² GL M1 + M2 left over.
00:24:32.200 --> 00:24:36.900
A little bit of simplification GGLL.
00:24:36.900 --> 00:24:40.600
We got that M1 vs. M1².
00:24:40.600 --> 00:24:53.300
I come up with M1 + M2 all divided by M 1.
00:24:53.300 --> 00:24:56.600
We have got a part E in this question here as well.
00:24:56.600 --> 00:24:59.600
Let us go give ourselves a little more room again.
00:24:59.600 --> 00:25:10.400
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.
00:25:10.400 --> 00:25:15.400
That second part looks like it is a projectile motion problem, something launch horizontally.
00:25:15.400 --> 00:25:17.300
Let us do that first.
00:25:17.300 --> 00:25:20.600
We will find out how long they are in the air.
00:25:20.600 --> 00:25:25.000
We will take a look at the vertical motion, we will call down the positive y direction.
00:25:25.000 --> 00:25:27.800
V initial vertically is 0.
00:25:27.800 --> 00:25:30.700
Our δ y is going to be L.
00:25:30.700 --> 00:25:36.000
Our acceleration is G, the acceleration due to gravity.
00:25:36.000 --> 00:25:42.800
δ Y = V initial T + ½ AY T².
00:25:42.800 --> 00:25:46.900
Our V initial is 0 so that term goes away.
00:25:46.900 --> 00:26:00.600
We could then write that L = ½ GT² or T is going to be equal to √2 L/G.
00:26:00.600 --> 00:26:16.300
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.
00:26:16.300 --> 00:26:30.700
We figured out the velocity in the x direction as they went the edge of the cliff previously, that was M1 / M1 + M2 × √2GL.
00:26:30.700 --> 00:26:38.600
We need to multiply all of that by the time they are in the air √2L/G.
00:26:38.600 --> 00:26:57.100
√2L √2L = 2L √G √G is a little bit of simplification here then I come up with 2 LM1 / M1 + M2.
00:26:57.100 --> 00:27:01.600
We are not asked for just that distance, we are asked for the total distance.
00:27:01.600 --> 00:27:05.300
The total horizontal displacement from position A.
00:27:05.300 --> 00:27:22.100
Our total displacement is going to be L + what we just found there 2 LM1 / M1 + M2.
00:27:22.100 --> 00:27:27.600
Or if you want to do a little bit of algebra, I think which you have there perfectly acceptable.
00:27:27.600 --> 00:27:46.600
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,
00:27:46.600 --> 00:27:58.100
which would be L × M1 + M2 + 2 M1 / M1 + M2.
00:27:58.100 --> 00:28:10.400
Which is equivalent to, we would have L × 3 M1 + M2 / M1 + M2.
00:28:10.400 --> 00:28:17.700
I'm not sure that is a whole lot prettier than what we have right there but either one of those should work.
00:28:17.700 --> 00:28:21.400
Hopefully, that gets you a good feel for center of mass and center of gravity.
00:28:21.400 --> 00:28:23.200
Thanks for watching www.educator.com.
00:28:23.200 --> 00:28:26.000
We will see you again soon and make it a great day everybody.