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Intro to Center of Mass | 14 mins | 0 completed | Learn |

Concept #1: Intro to Center of Mass

**Transcript**

Hey guys! In this video I'm going to talk about center of mass which is a pretty straightforward concept in physics. The idea of center of mass is that when you have a bunch of objects in a system, you can simplify that system and represent it as a single objects. For example if you have thousands or even millions of planets and stars in the galaxy spread all over the place, you could treat the entire system with millions of things as one object. Let's check it out. As I said, sometimes it's useful to simplify a system of objects, a collection of objects, by replacing all of them with a single equivalent object. For example instead of having a bunch of things moving this way, I can simplify this and just say that there's a single object that goes that way. This single object will have mass m equals the summation of all the individual masses. Since you're combining, you add up all the masses and the system will be located at the system's center of mass. Here's a really simple example of that. Let's say I want to combine the system made up of two masses, 2 kilograms and 2 kilograms, into a single object. First of all, the total combined mass will be 4 kilograms. Where would it go? I hope you're thinking that if I wanted to simplify this into one thing, the center of this whole combination of things is actually right down the middle. I would have something like this where this gap here is 5 meters okay and this would be a 4-kilogram object. The reason it's down the middle is because this is a balanced system. The left and the right have exactly the same masses. But if for example I had something like a 2 here and a 10 here, the center of mass would be much closer to the 10. It'd be somewhere over here. That's the idea.

The middle things only works if they are the same. You're not going to get that. You're going to need to use an equation. You're going to need to use the center of mass equation to figure out where the center of this combination of masses is. You're going to use the x position so where along the horizontal the center of mass is. ItÕs going to use this equation: sum of m*x / sum of m. I will explain what that means. I'll give an example where you have let's say 3 objects. Sum of m*x looks like this: m1x1 + m2x2 + m3x3. That's what it means to do the summation of m*x. YouÕre gonna have m1x1 + m2x2 + m3x3. The sum of mÕs is just m1 + m2 + m3. This is in the case like this where both objects are the x-axis. We draw a line between them and the center of mass will be somewhere along that line. But if you have objects at a two-dimensional plane, something like this and notice that I'm intentionally drawing the balls different sizes. If you have a system made up of four objects, the center of mass will be somewhere in the middle of the system. In this case, the left side balances with the right so it's going to be somewhere down the middle. But the bottom is much heavier than the top, so you're not going to have the center of mass be along a line here. It's going to be further down somewhere here, a little bit closer. This is where the center of mass is. I hope you see how this is two dimensions because I have x and y. In that case we're going to use not only the Y equation but you're also going to use in addition to the Y equation, you're going to also use x and y and the equation is the same, sum of m instead of x, sum of m*y divided by sum of m. This looks like this: m1y1 + m2y2 + m3y3 / m1 + m2 + m3. x and y is the x and y position of these objects. Let's do two quick examples to see what this looks like in practice. I have the two masses placed along the x axis. Here's the x axis. One mass is mass A at zero meters, so let's say zero is over here. At zero meters I got A which is 10 kilograms. At 4 meters, I have B which is 20 kilograms. I want to know what is the center of mass. This is x axis only so I'm going to say that the center of mass is the X center of mass and the equation is this. It's m1x1 + m2x2. What I'm going to do is write the masses here, 10X + 20X / 10+20. All I got to do is plug in the numbers. What is the x position of the 10? The 10 is at 0, and the 20 is at 4. I put a 4 here and that's it. Very straightforward. This cancels. I have 80 / 30 and the x position of this thing is 2.67 meters. The middle between 0 & 4 is 2. You should have been expecting that the actual center of mass is somewhere to the right of the middle because the system is heavier on the right side and that's what it is. X center-of-mass, thatÕs what we got, 2.67. If you think about it a little bit, sometimes youÕd be able to sort of guess where the answer will be. I want to make a quick point and then we're going to jump into the next example. There are two terms that are similar. One is center of gravity and the other one is center of mass. In this video, we're only talking about center of mass. We're not talking about center of gravity and we're not really going to do problems with center of gravity. But I do want to clarify that there are two different terms and they mean different things. Without getting into what center of gravity is, I will just tell you that they actually are the same thing if the gravitational field is constant. This is a conceptual point that I want you to remember. If the gravitational field is constant, then center of mass and center of gravity are the same thing. We can use them interchangeably. What does it mean for the gravitational field to be constant? Let me draw something real quick. You don't have to draw. This is just conceptual. But let's say here's the Earth and you're here and your sister is here whatever. You're very close to each other and you form a system. The gravity where you are is let's say 9.8 and it points straight down. Your sister is right next to you so the gravity that she feels is also 9.8 and it's also straight down. If you guys are close enough together, the gravities will be almost identical. In fact, they'll be so close that we can consider them to be the same. Because the two objects feel the same gravity, the gravitational fields is the same for both. The center mass is the same as the center of gravity. That's the idea. If two people on earth are really far apart, they will feel different gravities because even if it's 9.8, one is being pulled this way, the other one is being pulled this way. ItÕs conceptual point for you to know. A lot of professors don't even get into this. If you didn't really hear the distinction between the two, don't even worry about it. It's not that big of a deal.

The last point I wanna make here is that unless otherwise stated, we're just going to assume that this here is the case. We're going to assume that the gravitational field is constant. What does that mean? It means that these two things mean the same, are the same thing. If gravitational field is constant, these two words mean the same. We're going to assume it's constant. We're going to assume that these terms are the same thing and we can use them interchangeably. Again, not going to do problems with center of gravity but I do want to touch up on the conceptual point there. Let me quickly do example 2 and we'll be done with this concept. That's all we're doing. Three masses are placed on the XY plane. I'm going to draw a little XY plane like this, Y X. Mass A is placed at 0,0. 0,0 is right here. Remember guys, coordinate systems are x,y so this means x equals zero and y equals zero. This is object A which has a mass of 10. B is that (0,3) so this is x and y. X equals zero is on this line here and three going up will be somewhere here. This is zero in the x axis and three. Zero in the x axis means you haven't moved left or right, you're in the middle and then three in the y axis means you go up 3. B 8 kilograms. C is that (4,0). The x value is 4, I go 4 this way but I stay on the x axis, I don't go up or down. (4,0) looks like this and this is C which has a mass of 6 kilograms. Here's the diagram. I want to know what is the x and y coordinates of the center of mass of the system. You might be thinking that the center of mass of the system is somewhere here. You can actually think about this in terms of x and y. Let's try to look at this. In the y axis, I have this guy on the y axis and these two guys on the y axis. Notice how this one is 8 and these two here combine to be 16. The y axis is heavier towards the bottom. I'm going to predict that this thing will be somewhere like Ôround 1, not 3, not in the middle but closer to zero. On the x axis, you have these two here and this here. The center of mass will be somewhere in the middle. The left, I have 10 and 8, that's 18. To the right, I have 6 so this thing is much heavier towards the left. Instead of being in the middle between 0 and 4 which would be 2, I'm going to guess it's going to be to the left of 2. I'm going to guess that's going to be around 1. Again, I'm just doing rough estimates so that we can later see if it makes sense with what we expected. We're just going to plug it in and weÕre done. X center of mass is going to be the masses are 10, so IÕm gonna do this, 8 and 6 divided by 10+8+6. Then the Y center of mass is the same thing, 10, 8, and 6 divided by 10+8+6. The only difference is that here I'm going to add X values and then the other one I'm going to add Y values. What's the x value of the 10? Just look right here. It's 0. Of the 8 is 0 and of the 6 is 4. What about the Y value? The Y value of the A is 0, of B is 3 and of C is zero as well. When you do this, you get 6*4 = 24 / 24. ItÕs a coincidence that the numbers happen to divide so neatly, 1 meter. Then here, this cancels and this cancels. I get 8*3 = 24 as well. Again, a coincidence that this happens to be the same as that divided by 24, 1 meter. My rough estimate happened to be dead on. But usually you just know that it's roughly a number. It's (1,1). You can they say that the center of mass of the system is at position (1,1) meter. That's the final answer. That's it for center of mass. Let me know if you have any questions.

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Concept #1: Intro to Center of Mass

Two metal spheres are connected by a light, 15 cm rod. One sphere has a density of 1500 kg/m3 and a radius of 10 cm, the other has a density of 2200 kg/m3 and a radius of 5 cm. What is the center of mass of this system?

A meter stick can be balanced by placing a support at the 50.0-cm mark. If a mass of 50.0 g is attached to the stick at the 90.0-cm mark, the stick can be balanced by placing a support at the 61.3-cm mark. What is the mass of the meter stick? NOTE: 'g' here is grams.
A) 32.6 g
B) 178 g
C) 73.4 g
D) 89.7 g
E) 127 g

Where is the center of mass of the Earth – Sun system located?
A) Just outside the surface of the Sun.
B) Halfway between the Earth and Sun.
C) Inside the Sun, 25 km from the center.
D) Inside the Sun, 7500 km from the center.
E) Inside the Sun, 450 km from the center.

Three balls with masses of 5 g, 1 g, and 6 g, respectively, are connected by massless rods. The balls are located as shown in the figure. What are the coordinates of the center of mass (in units of m)?

The masses of the Earth and Moon are 5.98×1024 kg and 7.35×1022 kg , respectively, and their centers are separated by 3.84×108 m. Where is the CM of this system located?

A croquet mallet balances when suspended from its center of mass, as shown in the figure(above). If you cut the mallet in two at its center of mass, as shown, how do the masses of the two pieces compare?
[A] The masses are equal.
[B] The piece with the head of the mallet has the greater mass.
[C] The piece with the head of the mallet has the smaller mass.
[D] It is impossible to tell.

A projectile fired into the air suddenly explodes into several fragments. What can be said about the motion of the center of mass of the system made up of all the fragments after the explosion?
1. Center of mass of the system moves in the direction opposite to the direction of a fragment with the biggest mass,
2. Center of mass of the system moves in the direction of a fragment with the biggest mass,
3. Center of mass of the system moves in the direction of the biggest fragment,
4. Center of mass does not move,
5. Center of mass of the system follows the same parabolic path the projectile would have followed if there had been no explosion,
6. There is not enough information given.
7. Center of mass of the system moves in the direction opposite to the direction of a projectile just before it exploded,

A uniform piece of sheet metal is shaped as shown in the figure. Compute the x and y coordinates of the center of mass of the piece.

(a) Find the x-coordinate xcm of the system's center of mass. Give your answer in terms of m1, m2, x1, and x2.(b) If m2 >> m1 , then the center of mass is located:A. Left of m1 at a distance much greater than x2 - x1.B. Left of m1 at a distance much less than x2 - x1.C. Right of m1 at a distance much less than x2 - x1.D. Right of m2 at a distance much greater than x2 - x1.E. Right of m2 at a distance much less than x2 - x1.F. Left of m2 at a distance much less than x2 - x1.

The blocks shown can only move along the x-axis.(a) Their velocities at a certain moment are v1 and v2 . Find the velocity of the center of mass vcm at that moment. Keep in mind that, in general: vx = dx/dt. Give your answer in terms of m1, m2, v1, and v2.(b) The blocks' momenta at a certain moment are P1 and P2 . Find the x component of the velocity of the center of mass vcm at that moment. Give your answer in terms of m1, m2, P1, and P2.(c) The blocks' accelerations at a certain moment are a1 and a2 . Find the acceleration of the center of mass acm at that moment. Keep in mind that, in general, ax = dvx/dt. Give your answer in terms of m1, m2, a1, and a2.

(a) How far is the center of mass of the Earth-Moon system from the center of the Earth? (Appendix C gives the masses of the Earth and the Moon and the distance between the two.)(b) Express the answer to (a) as a fraction of the Earth's radius. (Rcm / Rearth) =

Consider a system of two blocks that have masses m1 and m2 . Assume that the blocks are point-like particles and are located along the x axis at the coordinates x1 and x2 as shown (Figure 1) . In this problem, the blocks can only move along the x axis.Part ASuppose that v⃗cm=0 . Which of the following must be true?A) |p1x|=|p2x|B) |v1x|=|v2x|C) m1=m2D) none of the abovePart B(acm)x = F1x+F2xm1+m2Under what condition would the acceleration of the center of mass be zero? Keep in mind that F1x and F2x represent the components, of the corresponding forces.A)F1x=−F2xB) F1x=F2xC) m1=m2D) m1≪m2Part CConsider the same system of two blocks. Now, there are two internal forces involved. An internal force F⃗12 is applied to block m1 by block m2 and another internal force F⃗21 is applied to block m2 by block m1 (Figure 4) . Find the x component of the acceleration of the center of mass (acm)x of the system.Express your answer in terms of the x components F12x and F21x of the forces, m1 and m2 .

Find the coordinates of the center of mass. Find the x-coordinate.Express your answer to two significant figures and include the appropriate units.

Find the y-coordinate.Express your answer to two significant figures and include the appropriate units.

Calculate the center of mass of the object below. Assume the mass is uniformly distributed.

Consider a system consisting of three particles: m1 = 2 kg, v1 < 10, -5, 12 > m/s m2 = 9 kg, v2 <- 14, 3, -3>m/s m3 = 2 kg, v3 <- 23, 37, 20 > m/s What is the kinetic energy of this system relative to the center of mass? Krel = JOne way to calculate Krel is to calculate the velocity of each particle relative to the center of mass, by subtracting the center-of-mass velocity from the particle's actual velocity to get the particle's velocity relative to the center of mass, then calculating the corresponding kinetic energy, then adding up the three relative kinetic energies However, there is a much simpler way to determine the specified quantity, without having to do all those calculations; think about what you know about the relationships among the various kinds of kinetic energy in a multiparticle system. (If you wish, you can check your result by doing the complicated calculation just described.)The total kinetic energy of the system is 3530J and the translational kinetic energy of the system is J.

Consider a system consisting of three particles: m1 = 2 kg, v1 < 10, -5, 12 > m/s m2 = 9 kg, v2 <- 14, 3, -3>m/s m3 = 2 kg, v3 <- 23, 37, 20 > m/s vCM = (-11.7, 7.0, 2.85) m/sWhat Is the translational kinetic energy of this system? Ktrans = J

Consider a system consisting of three particles: m1 = 2 kg, v1 < 10, -5, 12 > m/s m2 = 9 kg, v2 <- 14, 3, -3>m/s m3 = 2 kg, v3 <- 23, 37, 20 > m/s What Is the total momentum of this system? ptot = kg m/s

The three masses shown in the figure are connected by massless, rigid rods. Assume that m1 = 170 g and m2 = 310 g.A.) What is the x-coordinate of the center of mass?B.) What is the y-coordinate of the center of mass?

Consider a system consisting of three particles: m1 = 2 kg, v1 < 10, -5, 12 > m/s m2 = 9 kg, v2 <- 14, 3, -3>m/s m3 = 2 kg, v3 <- 23, 37, 20 > m/s What Is the velocity of the center of mass of thls system? vcm = m/s

Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates: (1) 0.300kg , (0.300m , 0.400m ); (2) 0.400kg , (0.100m , -0.400m ); (3) 0.200kg , (-0.300m , 0.600m ).Part A: Find the x-coordinate of the center of mass of the system of three chocolate blocks.Part B: Find the y-coordinate of the center of mass of the system of three chocolate blocks.

Find the x coordinate x_cm of the center of mass of the system of particles shown in the figure.

A. Find the coordinates of the center of mass. Find the x-coordinate. Find the y-coordinate.B. Find the moment of inertia about a diagonal axis that passes through masses B and D.

What is the x coordinate xcm of the center of mass of the system shown in the figure? described in Part D?Express your answer in terms of L.

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