Question

A potter's wheel is initially at rest. A constant external torque of 74.0 N·m is applied to the wheel for 15.0 s, giving the wheel an angular speed of 490 rev/min.

1)What is the moment of inertia of the wheel?

2)The external torque is then removed, and a brake is applied. If it takes the wheel 210 s to come to rest after the brake is applied, what is the magnitude of the torque exerted by the brake?

Answer 1

Initial angular speed of the potter's wheel = ₁ = 0 rad/s (At rest)

Rotational speed of the potter's wheel after 15 sec = N₂ = 490 rev/min

Angular speed of the potter's wheel after 15 sec = ₂

₂ = 51.313 rad/s

Angular acceleration of the potter's wheel = ₁

External torque applied for accelerating the potter's wheel = ₁ = 74 Nm

Time period the torque accelerates the potter's wheel = T₁ = 15 sec

Moment of inertia of the potter's wheel = I

₂ = ₁ + ₁T₁

51.313 = 0 + ₁(15)

₁ = 3.421 rad/s²

I₁ = ₁

I(3.421) = 74

I = 21.631 kg.m²

Torque of the brake applied = ₂

Time period the brake is applied for = T₂ = 210 sec

Final angular speed of the potter's wheel = ₃ = 0 rad/s (Comes to rest)

Angular acceleration of the potter's wheel when the brake is applied = ₂

₃ = ₂ + ₂T₂

0 = 51.313 + ₂(210)

₂ = -0.2443 rad/s²

I₂ = ₂

(21.631)(-0.2443) = ₂

₂ = -5.284 Nm

Negative as the torque is opposing the motion of the wheel.

We are asked for the magnitude of the braking torque therefore it will be positive.

1) Moment of inertia of the wheel = 21.631 kg.m²

2) Magnitude of the torque applied by the brake = 5.284 Nm