A potter is shaping a bowl on a potter's wheel rotating at a constant angular velocity of 1.6 rev/s (Fig. 8–48). The frictional force between her hands and the clay is 1.5 N. (a) How large is her torque on the wheel if the diameter of the bowl is 9.0 cm? (b) How long would it take for the potter's wheel to stop if the only torque acting on it is due to the potter's hands? The moment of inertia of the wheel and the bowl is \(0.11\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

FIGURE 8-48

Problem 40

The constant angular speed means that the net torque on the wheel is zero. Due to the frictional torque, the pot slows down when there is no other torque on the wheel.

The constant angular velocity of the wheel is \(\omega = 1.6\;{\rm{rev/s}} = \left( {1.6 \times 2\pi } \right)\;{\rm{rad/s}}\).

The diameter of the bowl is \(d = 9.0\;{\rm{cm}} = 0.09\;{\rm{m}}\).

The moment of the wheel and the bowl is \(I = 0.11\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

The frictional force between the hand and the clay is \(F = 1.5\;{\rm{N}}\).

Let it take \(\Delta t\) time to come to rest when the torque is only due to the frictional force on the potter's hand.

You can assume that the axis of rotation is through the center of the bowl, and the potter is applying the force perpendicular to the radius of the bowl.

Part (a)

The radius of the bowl is \(r = \frac{d}{2}\).

Now, the potter is applying the force at a distance r from the rotating axis.

As the wheel is rotating with a constant angular velocity, the torque on the wheel is equal to the torque by the frictional force of the potter.

Therefore, the torque is

\(\begin{align}\tau &= Fr\\ &= F \times \frac{d}{2}\\ &= \left( {1.5\;{\rm{N}}} \right) \times \frac{{0.09\;{\rm{m}}}}{2}\\ &= 0.068\;{\rm{N}} \cdot {\rm{m}}\end{align}\).

Hence, the magnitude of the torque by the potter is \(0.068\;{\rm{N}} \cdot {\rm{m}}\).

Part (b)

When the torque is only by the potter, the wheel comes to rest, and the angular acceleration of the wheel is

\(\begin{align}\alpha &= \frac{{0 - \omega }}{{\Delta t}}\\ &= - \frac{\omega }{{\Delta t}}\end{align}\).

The torque due to the frictional force against the motion is negative as it reduces the angular velocity. Also, the acceleration due to the torque for the frictional force is \(\alpha = - \frac{\tau }{I}\).

Now, comparing the values of the angular acceleration,

\(\begin{align}\frac{\omega }{{\Delta t}} &= \frac{\tau }{I}\\\Delta t &= \frac{{I\omega }}{\tau }\\ &= \frac{{\left( {0.11\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right) \times \left( {1.6 \times 2\pi } \right)\;{\rm{rad/s}}}}{{0.068\;{\rm{N}} \cdot {\rm{m}}}}\\ &= 16.25\;{\rm{s}}\end{align}\).

Hence, it takes \(16.25\;{\rm{s}}\) to stop the wheel.