# Pulley and angular momentum

1. Nov 24, 2006

### candyq27

Hi. I have this problem and I really need help with it. I'm not even sure where to start. Here is the problem:
A horizontal solid disk of mass M and radius R rotates at an angular velocity of w with respect to an axis perpendicular to the disk at its center. Assume that the axis is perfectly frictionless, so that the disk rotates freely.
a. The moment of inertia for a solid disk is .5MR^2. What is the angular momentum of the disk?
From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of R/2 from the axis. The total mass of the sand deposited is M/2.
b. After all the sand is in place, wha is the final angular velocity of the disk? Express your answer in terms of the initial angular velocity w.
c. Calculate the initial and final values of the kinetic energy in the system. Is kinetic energy conserved in this situation? Explain why or why not.

Please help me get started on each step of this problem. I'm not even sure where to begin. Thank you!

2. Nov 24, 2006

### Staff: Mentor

Please give the problems a try and show your work in order to get help. To start with, look up the definitions of angular momentum, moment of inertia, and rotational kinetic energy. For part (b) consider what might be conserved.

3. Nov 24, 2006

I suggest you start with the definition of angular momentum and the law of its conservation.

Edit: late again.

4. Nov 24, 2006

### candyq27

Thanks for replying. I'm still not sure even where to begin but here's a try...
I know angular momentum is L=Iw, momentum is p=mv, moment of inertia for the disk is 1/2MR^2, and rotational kinetic energy is K=1/2Iw^2.
So for part (a) I know the moment of inertia is 1/2MR^2 and I need to find out the angular momentum. So since angular momentum is Iw, then...? Is the angular momentum just (1/2MR^2)w?

Now for part (b)...
The initial angular velocity is w. The sand is R/2 and M/2. I need to know the final angular velocity...so I don't know where to go from there. Do I have to use one of the formulas such as w=wo +at (a as in alpha)?

Thanks again.

5. Nov 24, 2006

### Staff: Mentor

You got it.

Big hint: Angular momentum the system (sand + disk) is conserved. But what's the moment of inertia of the system after the sand is added?

6. Nov 24, 2006

### candyq27

Well since we're adding sand that adds another M/2 and R/2 so would the moment of inertia after the sand is added be 1/2((3/2)Mx(3/2)R)^2?

7. Nov 25, 2006

### Staff: Mentor

No. What's the moment of inertia of the sand by itself? (It's not a disk of sand, but a ring of sand.) The total moment of inertia of the system is just the sum of the parts:
$$I_{total} = I_{disk} + I_{sand}$$

8. Nov 25, 2006

### candyq27

Ok, so the Itotal= (1/2MR^2) + (MR^2)
but since it tells me that the ring of sand is at M/2 and R/2 that part is (M/2xR/2^2)?
Then I'm not sure where to go from there
So I know the total I and I need to know the final angular w for the disk...do I have to figure out the torques?

9. Nov 25, 2006

### Staff: Mentor

Right.

Refer to my hint in post #5.

10. Nov 25, 2006

### candyq27

I don't get it...
I just found the I of the system, but now I have to compare the angular momentums? So the L of the disk was part (a) but now with the sand it's....? ((1/2MR^2)+((M/2)(R/2)^2))w?

11. Nov 25, 2006

### Staff: Mentor

You calculated the angular momentum of the system in part a--that has not changed (the sand dropped onto the disk with zero speed, so it added no angular momentum of its own). Conservation of angular momentum tells you:
$$I_1 \omega_1 = I_2 \omega_2$$

You have the new moment of inertia, now calculate the new angular speed.

12. Nov 25, 2006

### candyq27

oooh, ok so then i get...
(1/2MR^2)(w) = ((1/2MR^2)+((M/2)(R/2)^2))w2
so w2= [(1/2MR^2)(w)]/[(1/2MR^2)+((M/2)(R/2)^2)]

13. Nov 25, 2006

### candyq27

So for part (c) I'm not sure what to do either...I know the kinetic energy for rotational motion is KE=1/2 Iw^2
So for the initial kinetic energy it would be 1/2(1/2MR^2)w^2?
and for the final kinetic energy it would be 1/2[(1/2MR^2)+((M/2)(R/2)^2)]w^2?
Then would the kinetic energy not be conserved because of friction?

14. Nov 25, 2006

### Staff: Mentor

Please simplify that expression for $\omega_2$ before my head explodes.

Rotational KE is simply $1/2 I \omega^2$. (Calculate it twice: before and after the sand falls.) The sand hitting the moving disk is an inelastic collision; mechanical energy is "lost" to thermal energy.

Last edited: Nov 25, 2006
15. Nov 25, 2006

### candyq27

I'm afraid to simplify incase it's wrong but is it just w2=wo/2? no that doesnt look right
for the KE was what i said right for the formulas? thanks so much!

16. Nov 25, 2006

### candyq27

Ok I simplified the equation for part (b) and i got w2= 4wo/5
That makes more sense now.
For part (c) I got KEi=1/4MR^2w^2 and KEf=5/16MR^2w^2

17. Nov 26, 2006

### Staff: Mentor

That looks good.
Careful how you write these terms. In KEi, w is the original speed; but in KEf, you are using w to mean the final speed--confusing! Rewrite KEf in terms of the original w so that the variables all mean the same thing. (Note that 5/16 > 1/4, so left as is it looks like the system gained energy!)

18. Nov 26, 2006

### candyq27

Ok so for the final kinetic energy using the original w I get 1/5MR^2w^2. That makes more sense because 1/5 is less than 1/4. Thanks so much!