# Homework Help: Angular Momentum and Pulleys

1. Aug 24, 2013

### postfan

1. The problem statement, all variables and given/known data

A block of mass m1 is attached to a block of mass m2 by an ideal rope passing over a pulley of mass M and radius R as shown. The pulley is assumed to be a uniform disc rotating freely about an axis passing through its center of mass (cm in the figure). There is no friction between block 2 and the surface. Assume that the pulley rotates counterclockwise as shown with an angular speed ω and that the rope does not slip relative to the pulley, and that the blocks move accordingly and do not topple or rotate.

Consider the system to be formed by the pulley, block 1, block 2 and the rope.

Calculate the magnitude of the angular momentum of the system about the center of mass of the pulley. Express your answer in terms of some or all of the variables m1, m2, M, R, ω and g. Type omega for ω

2. Relevant equations

3. The attempt at a solution

Used formula L=Iw I=.25MR^2 then multiply by w so L=1/4*(M*R^2)*w. What am I doing wrong?

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2. Aug 24, 2013

### TSny

Is there any way you can post the figure?

Note that they are asking for the angular momentum of the system, not just the pulley.

I don't think the 1/4 factor for the moment of inertia of the pulley is correct.

3. Aug 24, 2013

### postfan

I attached the image. If so don't we need to know the distances between the blocks and the pulley?

4. Aug 24, 2013

### TSny

You don't need those distances. Treat each block as a point particle located at the center of each block. Think about how to calculate the angular momentum of each block relative to the center of the pulley. You can use the definition of the angular momentum of a point particle relative to an origin.

5. Aug 24, 2013

### postfan

The formula is mr^2, right, so you still need the distances?

6. Aug 24, 2013

### TSny

That's not the formula for angular momentum. That's the moment of inertia of a single particle moving in a circle of radius r. The blocks are moving along straight lines. You should have covered the formula for the angular momentum of a point particle. It was probably at the very beginning of your study of angular momentum.

7. Aug 24, 2013

### postfan

The 2 formula I know are m*r*v and I*w, we don't know both r and w, so how do we do it?

8. Aug 24, 2013

### TSny

OK, the m*r*v formula can be used for a particle moving along a straight line with speed v. r is then the perpendicular distance from the origin to the straight line.

The speed v of one of the blocks is related to the angular speed of ω of the pulley. So in the formula m*r*v you can express v in terms of ω.

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9. Aug 24, 2013

### postfan

Ok so the formula is m*r*(w*r)=m*r^2*w, right?

10. Aug 24, 2013

### TSny

Yes. That's it.

11. Aug 24, 2013

### postfan

OK so the total angular momentum for the system is (M+m1+m2)*r^2*w, right?

12. Aug 24, 2013

### TSny

No, the part that deals with the pulley is not correct. You had the right approach for the pulley in your first post, but you didn't quite have the right expression for the moment of inertia, I.

13. Aug 24, 2013

### postfan

Ok, so the angular momentum is M*r^2*w, right?

14. Aug 25, 2013

### TSny

No, look at a table of moments of inertia.

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• ###### Rot_Inertia_table.gif
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15. Aug 25, 2013

### mshmsh_2100

i met this problem once.... but still cannot figure out the final formula of the angular momentum!!

16. Aug 25, 2013

### mshmsh_2100

i think the final formula for the angular momentum is (m1+m2)*v*R+M*omega*R^2, right?

17. Aug 25, 2013

### postfan

OK, so according to the diagram the answer is .5*M*r^2*w, right?

18. Aug 25, 2013

### TSny

Yes, for the pulley.

19. Aug 25, 2013

### postfan

Do we need to add anything for the blocks?

20. Aug 25, 2013

### TSny

Yes. See post #9 for the angular momentum of each block.