A 2 kg, 35 cm diameter turntable rotates at 150 RPM on frictionless bearings. Tw

Question

A 2 kg, 35 cm diameter turntable rotates at 150 RPM on frictionless bearings. Two 400g blacks fall from above, hit the turntable simultaneously at opposite ends of a diagnal and stick. What is the turntable's angular speed, in RPM, just after the event?

Wf =( Ii / If) x Wi

Rotating cylindar for Ii = 1/2 mr^2

Wf = [(1/2) (2kg) (0.75m)^2 / If ] x 150 RPM

Question::???::: How do I find the value of I final? I have a note that it is mr^2.

The answer should be 83.3, but I do not know how to solve this from here.

Explanation / Answer

Initial I =mr^2/2 = 2*(0.35)^2/2=(0.35)^2

Intial w = 150 rpm

Since,2 blacks fall so,

mass of black =400 g =0.4 kg

Final I= Initial I + 2 *(mr^2) =(0.35)^2 +2*0.4*(0.35)^2

Since, angular momentum will remain conserved,

Initial I*Intial w = Final I * Final W

Final W= (Initial I*Intial w)/ Final I

Wf=((0.35)^2*150)/((0.35)^2 +2*0.4*(0.35)^2)

Final angular speed=83.3 rpm

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