Final Angular Velocity. Need help w/ part 4

Question

A phonograph turntable driven by an electric motor accelerates at a constant rate (i.e., constant angular acceleration) from 0 to 33.33 revolutions per minute in 2 minutes. The turntable is a uniform disc of mass 1.5 kg and radius R = 13 cm, as indicated in the figure. The radius of the driving wheel is r = 2.0 cm.

(i)What is the torque required to drive the turntable at this angular acceleration?

(ii)If the driving wheel makes contact with with the turntable at its outer rim, what is the force exerted? (i.e., the F vector shown in the figure).

(iii)What is the final angular velocity of the driving wheel? Express your answer in “revolutions per minute”.

(iv)Now consider that the driving wheel is disengaged, and the larger turntable coasts (without friction) at the 33.33 rpm when a stack of 10 phonograph records suddenly drops on it. Assuming no slippage between the records and the turntable or between the records, what is the final angular velocity of the turntable/records combination? Each record has a mass 0.17 kg and a radius of 0.152m.

Answer 1

Angular momentum must be conserved.  So the momentum of the spinning turntable before the records are added must equal the momentum of the turntable and the records together after they are added.

 

Angular momentum L=Iω

ω₁ = 33.33 rpm or 3.49031 rad/s

moment of inertia for a disk is I=1/2*m*r²

I₁ = 1/2*1.5kg*(.13m)² = .012675 kg*m²

L₁ = I₁*ω₁ = (.012675 kg*m²)*(3.49031 rad/s) = .044239679 kg*m²/s

 

L₂ = I₂*ω₂

L₁ = L₂  -->  ω₂ = L₁/I₂

I₂ = I₁ + 10(1/2*.17kg*(.152m)² = .0323134 kg*m²

ω₂ = 1.369 rad/s or 13.07 rpm