A potter%u2019s wheel is a uniform solid disk of diameter .72 m and a mass of 60

Question

A potter%u2019s wheel is a uniform solid disk of diameter .72 m and a mass of 60 kg. The wheel is initially spinning at a rate of 85 revs/min, and the potter wants to bring it down to 45 revs/min with her hands within 15 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

A) mR^2/2 = 60*(0.36*0.36)/2 = 3.888 kgm^2

B) wf^2 = wi^2+2*a*@

a = (45^2-85^2)/(2*15) = -177.77 rev/min^2

a =173.33*2*pi/3600 = 0.3025 rad/sec^2

C)

F=I*a = 3.888*0.3025 = 1.176 N

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