Releasing a hollow cylinder on a rotating disk

[archaic]

Homework Statement

A uniform disk has a mass of 3.8 kg and a radius of 0.52 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 50 rpm. A thin-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 0.20 s until it acquires the same final angular velocity as the turntable. What is the final angular momentum of the system?

Relevant Equations

##L=I\omega##

There is no net external torque since the cylinder is slipping (no friction), so the angular momentum should be conserved.
$$L_f=\frac 12MR^2\omega_i=\frac 12\times3.8\times0.52^2\times50\times\frac{2\pi\times0.52}{60}$$

 

[haruspex]

I think you mean "since the disk is slipping (no friction)".
Your reasoning is right, but you seem to have an extra 0.52 in there.

 

[archaic]

I think you mean "since the disk is slipping (no friction)".
Your reasoning is right, but you seem to have an extra 0.52 in there.

Right, I should have the velocity in radians per second. Thank you.

 

[archaic]

I think you mean "since the disk is slipping (no friction)".

The hollow cylinder slips on the turntable

 

[haruspex]

The hollow cylinder slips on the turntable

Yes, but with friction, or they would not converge to the same angular velocity.

 

[haruspex]

Right, I should have the velocity in radians per second. Thank you.

I'm not sure how that is a response to my post. What do you have as the answer now?

 

[archaic]

I'm not sure how that is a response to my post. What do you have as the answer now?

$$\frac 12\times3.8\times0.52^2\times50\times\frac{2\pi}{60}$$

Yes, but with friction, or they would not converge to the same angular velocity.

This actually bothered me a bit. I went with conservation of angular momentum since I saw no other way of doing the problem. Doesn't friction imply a net external torque on the disk and the cylinder?

 

[kuruman]

Doesn't friction imply a net external torque on the disk and the cylinder?

The system, the angular momentum of which is conserved, is the turntable plus the hollow cylinder. There is no external torque acting on the system; friction is internal. This is the rotational equivalent of a one dimensional inelastic collision where two blocks collide and stick together after the collision.