What is the Final Angular Momentum of a Turntable and Hollow Cylinder System?

[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

Im trying to figure out where to start with this problem.

 

[Doc Al]

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

 

[Hootenanny]

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

~H

 

[williams31]

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

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

 

[Hootenanny]

The hollow cylinder slips on the turntable for .20 s until it acquires the same angular velocity as the turntable.

This information is superfluous.

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

~H

 

[arildno]

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

~H

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

 

[Hootenanny]

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

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

 

[arildno]

That would be the only sensible choice of course.

 

[williams31]

now I am confused

 

[Hootenanny]

now I am confused

What's confusing you my friend?

~H

 

[williams31]

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

 

[Hootenanny]

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

 

[williams31]

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

do i use the equation L= r X mv?

 

[Hootenanny]

do i use the equation L= r X mv?

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

 

[williams31]

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

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

 

[Hootenanny]

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

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

~H

 

[williams31]

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

 

[Hootenanny]

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

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

 

[williams31]

i don't even see an equation in my book for a flat disk...i don't know what to do

 

[Hootenanny]

The moment of inertia of a flat disk of uniform density rotatied about its centre is given by;

I = \frac{1}{2}mr^2

A useful resource is located at http://hyperphysics.phy-astr.gsu.edu/Hbase/mi.html#mi

~H

 

[williams31]

so it should be 1/2(3.7)(.38)^2 x 20?

 

[Hootenanny]

so it should be 1/2(3.7)(.38)^2 x 20?

Note quite. r = 3.8 not 0.38. Also angular velocity is given by; \omega = 2\pi f where f is the frequency (revolutions per second).

Do you follow?

~H

 

[williams31]

i follow but i don't understand why r= 3.8 instead of .38

 

[Hootenanny]

A uniform disk has a mass of 3.7 kg and a radius of 3.8 m.

You stated in your question that the radius was 3.8m therefore, r = 3.8.

~H

 

[williams31]

i did not realize that...sorry..it is supposed to be .38m

but i figured out the answer thanks to your help