Question

A 60 cm diameter potter's wheel with a mass of 30 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 wheels 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.

Answer 1

Answer:

It will take the wheel 278.9 s to come to a stop

Explanation:

Mass of the potter's wheel, M = 30 kg

Diameter of the potter's wheel, d₁ = 60 cm = 0.6 m

Radius, r₁ = d/2 = 0.6/2

r₁ = 0.3 m

The moment of inertia of the wheel, $I = 0.5Mr_1^{2}$

$I = 0.5*30*0.3^{2}\\I = 1.35 kg.m^2$

d₂ = 14 cm = 0.14 m

r₂ = 0.14/2 = 0.07 m

Angular velocity, $\omega = 180 rpm$

$\omega = \frac{180*2\pi }{60} \\\omega = 18.85 rad/s$

Frictional Force, F = 1.3 N

The torque generated:

$\tau = F*r_{2}\\\tau = 1.3*0.07\tau = 0.091 Nm$

Torque can also be calculated as:

$\tau = I \alpha\\\tau = I \frac{\omega }{t} \\0.091 = 1.35*\frac{18.8 }{t} \\t = (18.8*1.35)/0.091\\t = 278.9 s$