A softball player swings a bat, accelerating it from rest to 2.6 rev/s in a time of 0.20 s. Approximate the bat as a 0.90-kg uniform rod of length 0.95 m, and compute the torque the player applies to one end of it.

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

The mass of the bat is \(m = 0.90\;{\rm{kg}}\).

The length of the bat is \(L = 0.95\;{\rm{m}}\).

The initial angular velocity of the bat is \({\omega _1} = 0\).

The final angular velocity of the bat is \({\omega _2} = 2.6\;{\rm{rev/s}} = \left( {2.6 \times 2\pi } \right)\;{\rm{rad/s}}\).

The ball takes the time \(\Delta t = 0.20\;{\rm{s}}\) to gain the final speed.

Let \(\tau \) be the torque applied by the player.

You can consider the bat a uniform rod rotating about its one end.

Now, the angular acceleration of the bat is \(\alpha = \frac{{{\omega _2} - {\omega _1}}}{{\Delta t}}\).

The moment of inertia of the bat about the rotating axis is \(I = \frac{1}{3}m{L^2}\).

Therefore the torque on the bat is

\(\begin{align}\tau &= I\alpha \\ &= \frac{1}{3}m{L^2} \times \frac{{{\omega _2} - {\omega _1}}}{{\Delta t}}\\ &= \frac{1}{3} \times \left( {0.90\;{\rm{kg}}} \right) \times {\left( {0.95\;{\rm{m}}} \right)^2} \times \frac{{\left( {2.6 \times 2\pi } \right)\;{\rm{rad/s}} - 0}}{{0.20\;{\rm{s}}}}\\ &= 22\;{\rm{N}} \cdot {\rm{m}}\end{align}\).

Hence, the torque applied by the player is \(22\;{\rm{N}} \cdot {\rm{m}}\).