Turntable, block, and rotation angular velocity

[abeltyukov]

Homework Statement

A 200g, 40-cm-diameter turntable rotates on frictionless bearings at 60 rpm. A 20g block sits at the center of the turntable. A compressed spring shoots the block radially outward along a frictionless groove in the surface of the turntable. What is the turntable's rotation angular velocity when the block reaches the outer edge?

Homework Equations

1/2Iw^2
I = 1/2MR^2

The Attempt at a Solution

I did:
1/2Iw^2 = 1/2Iw^2
1/2(1/2(0.200)(0.20)^2)(2pi)^2 = 1/2(1/2(0.200)+0.020)(wf)

wf = 32.898 rad/sec


Does that seem right?



Thanks!

 

[radou]

You have to use conservation of angular momentum. I'm not sure what you did in the equation 1/2(1/2(0.200)(0.20)^2)(2pi)^2 = 1/2(1/2(0.200)+0.020)(wf). The angular momentum of the turntable and the block in the center must equal the angular momentum of the turtable with the block on the edge. Watch out for the moment of inertia.

 

[abeltyukov]

I did rotational kinetic energy of the turntable = rotational kinetic energy of the turntable and block. So KE = 1/2Iw^2. Can I do that?

Thanks!