Angular momentum and a turntable

1. May 4, 2006

williams31

Angular momentum....

A uniform disk has a mass of 3.7 kg and a radius of 3.8 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 20 rpm. A thing walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for .20 s until it acquires the same angular velocity as the turntable. The final angular momentum of the system is closest to:
A) 1.96 kg m^2/s
B) 1.12
C) .56
D) .80
E) 1.68

2. May 4, 2006

Staff: Mentor

Hint: Does the angular momentum of the system change when the cylinder is dropped onto the disk?

3. May 4, 2006

Hootenanny

Staff Emeritus
HINT: The angular momentum of a closed system (which this is) must remain constant.

~H

4. May 4, 2006

williams31

im guessing no by the hint that was given after yours...is all the information given in the problem relevant...or is there some extra stuff in there

5. May 4, 2006

Hootenanny

Staff Emeritus
This information is superfluous.

And apologies to Doc Al, I didn't see your post.

~H

6. May 4, 2006

arildno

Not quite.
It also depends upon whether the point you are calculating the angular momentum with respect to is moving or not.

7. May 4, 2006

Hootenanny

Staff Emeritus
Sorry I assumed it was the axis of rotation, as they both have the same one, which is not moving relative to both the disk and cylinder. I missed it though, good point.

~H

8. May 4, 2006

arildno

That would be the only sensible choice of course.

9. May 4, 2006

williams31

now im confused

10. May 4, 2006

Hootenanny

Staff Emeritus
What's confusing you my friend?

~H

11. May 4, 2006

williams31

just the whole problem...im not too smart when it comes to physics...every problem creates mess within my brain

12. May 4, 2006

Hootenanny

Staff Emeritus
Okay, all you basically need to know is that because no external torques are acting on the system and you are taking moments about the same stationary axis, angular momentum is conserved. In otherwords, the intial angular momentum of the system is equal to the final angular momentum of the system.

~H

13. May 4, 2006

williams31

do i use the equation L= r X mv???

14. May 4, 2006

Hootenanny

Staff Emeritus
This equation is only valid when the object which is rotating is a particle (point mass). However, you have a rotating disk, so you must use the equation $L = I\omega$, where I is the moment of inertia of the disk and $\omega$ is the angular velocity of the rotation.

~H

15. May 4, 2006

williams31

for some reason i am coming up with answers that are nowhere near close to the choices

16. May 4, 2006

Hootenanny

Staff Emeritus
Could you show your working? I'll try and point you in the right direction.

~H

17. May 4, 2006

williams31

for moment of inertia...i use the equation for a thin-walled hollow cylinder?

18. May 4, 2006

Hootenanny

Staff Emeritus
No, you should use it for a flat disk, as your are calculating the intial angular momentum. Remember Inital angular momentum = final angular momentum.

~H

19. May 4, 2006

williams31

i dont even see an equation in my book for a flat disk...i dunno what to do

20. May 4, 2006

Hootenanny

Staff Emeritus