Question

A small 350-gram ball on the end of a thin, light rod is rotated in a horizontal circle of a radius of 1.2 m. Calculate (a) the moment of inertia of the ball about the center of the circle and (b) the torque needed to keep the ball rotating at a constant angular velocity if the air resistance exerts a force of 0.020 N on the ball. Ignore the air resistance on the rod and its moment of inertia.

Short answer

The applied torque should be \(0.024\;{\rm{N}} \cdot {\rm{m}}{\rm{.}}\)

Step-by-step solution

Concept

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.

Given data

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.

Calculation for part (a)

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 \(0.50\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

Calculation for part (b)

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 \(0.024\;{\rm{N}} \cdot {\rm{m}}\).