A potters wheel is a uniform solid disk of diameter 0.72m and a mass of 60.0 kg.

Question

A potters wheel is a uniform solid disk of diameter 0.72m and a mass of 60.0 kg. the wheel is initially spinning at a rate of 85 rev/min and the potter wants to bring it down to 45 revs/min with her hands within 15.0 revolutions,

a) What is the moment of inertia of the disk?

b) What is the angular acceleration that will bring down the rotational speed under the stated conditions?

c) What minimum total friction force must the potter supply to the disk if her hands are acting at the edge of the disk?

Explanation / Answer

Part A)

I = .5mr^2

I = (.5)(60)(.36)^2

I = 3.89 kg m^2

Part B)
Apply wf^2 = wo^2 + 2a(theta)

wo = 85 rev/min which is 8.90 rad/s

15 rev = 94.2 rad, so

0 = 8.90^2 + 2(a)(94.2)

a = -.420 rad/s^2 (Negative since it is slowing doen - sign convention only)

Part C)

Fr = Ia

F(.36) = (3.89)(-.42)

F = -4.54 N (Again the negative is sign convention, you can ignore it if your answer key wants you to)

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