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:
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, [tex]I = 0.5Mr_1^{2}[/tex]
[tex]I = 0.5*30*0.3^{2}\\I = 1.35 kg.m^2[/tex]
d₂ = 14 cm = 0.14 m
r₂ = 0.14/2 = 0.07 m
Angular velocity, [tex]\omega = 180 rpm[/tex]
[tex]\omega = \frac{180*2\pi }{60} \\\omega = 18.85 rad/s[/tex]
Frictional Force, F = 1.3 N
The torque generated:
[tex]\tau = F*r_{2}\\\tau = 1.3*0.07\tau = 0.091 Nm[/tex]
Torque can also be calculated as:
[tex]\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[/tex]