91Ó°ÊÓ

Torque is the product of the moment of inertia and the angular acceleration. It would help if you considered the ball as a rod that is rotating about its one end.

The mass of the ball is \(m = 350\;{\rm{g}} = 0.350\;{\rm{kg}}\).

The radius of the circular path is \(r = 1.2\;{\rm{m}}\).

The force by the air resistance is \(f = 0.020\;{\rm{N}}\).

The ball is moving with a constant angular velocity. You can consider the ball as a point mass.

Part (a)

The moment of inertia of the ball about the center of the circle is

\(\begin{align}I &= m{r^2}\\ &= \left( {0.350\;{\rm{kg}}} \right) \times {\left( {1.2\;{\rm{m}}} \right)^2}\\ &= 0.50\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{align}\).

Hence, the moment of inertia of the ball about the center of the circle is \(0.50\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

Part (b)

The ball is moving at a constant angular velocity, i.e., the net torque on the ball is zero.

Therefore, the torque by the frictional force is equal to the applied torque.

Therefore, the torque on the ball against the motion by the resistive force of air is

\(\begin{align}\tau &= fr\\ &= \left( {0.020\;{\rm{N}}} \right) \times \left( {1.2\;{\rm{m}}} \right)\\ &= 0.024\;{\rm{N}} \cdot {\rm{m}}\end{align}\).

Hence, for a constant angular velocity, the value of the torque should be \(0.024\;{\rm{N}} \cdot {\rm{m}}\).