A 45 cm -diameter potter\'s wheel with a mass of 21 kg is spinning at 180 rpm. U

Question

A 45 cm -diameter potter's wheel with a mass of 21 kg is spinning at 180 rpm. Using her hands, a potter forms a 14 cm-diameter pot that is centered on and attached to the wheel. The pot's mass is negligible compared to that of the wheel. As the pot spins, the potter's hands apply a net frictional force of 1.3 N to the edge of the pot.

If the power goes out, so that the wheel's motor no longer provides any torque, how long will it take the wheel to come to a stop? You can assume that the wheel rotates on frictionless bearings and that the potter keeps her hands on the pot as it slows.

Explanation / Answer

Use eqn,

Torque by friction = I*

-F*r= I*

-F*r= I*(wf-wi)/t

t= I*(wf-wi)/(-F*r)

Plug values.

F+1.3N, r=0.07m, I=1/2mr^2=1/2*21*0.225^2 = 0.53, wf=0rad/s, wi=18.85rad/s

t= [0.53(0-18.85)]/(1.3*0.07)

t= 109.8 s

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