Angular Velocity of a turntable after weights are added

[doneganstm]

A 2.0 {\rm kg}, 20-cm-diameter turntable rotates at 100 {\rm rpm} on frictionless bearings. Two 500 {\rm g} blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?

I'm not quite sure on how to set this one up or what equation to use, can anyone give me a push in the right direction?

 

[tiny-tim]

Welcome to PF!

Hi doneganstm! Welcome to PF!

Hint: conservation of angular momentum.

 

[kerimek]

This scenario can be solved using the principle of conservation of angular momentum. The initial angular momentum of the turntable is given by L1 = I1*ω1, where I1 is the moment of inertia of the turntable and ω1 is its initial angular velocity. After the blocks are added, the moment of inertia of the turntable changes to I2 = I1 + 2mr^2, where m is the mass of the blocks and r is the distance from the center of the turntable to the point of impact. The final angular velocity of the turntable, ω2, can be found using the equation L1 = L2, which implies I1*ω1 = I2*ω2. Plugging in the values, we get:

I1*ω1 = (I1 + 2mr^2)*ω2

Solving for ω2, we get:

ω2 = (I1*ω1)/(I1 + 2mr^2)

Substituting the values given in the question, we get:

ω2 = (1/2*2*π*100)/(1/2*2*π*100 + 2*0.5*0.2^2) = 83.33 rpm

Therefore, the turntable's angular velocity just after the event is 83.33 rpm.