Chapter 8, Problem 35P

To determine

To find: The torque that the player applies at one end.

Answer to Problem 35P

The value is $\tau =62.3\text{N}\cdot \text{m}$ is the required torque.

Explanation of Solution

Given information:

A softball player swings a bat, accelerating it from rest to $3.0\text{rev/s}$ in a time of $0.20\text{s}$ . Approximate the bat as a $2.2\text{kg}$ uniform rod of length $0.95\text{m}$ .

Formula used:

  $I=\frac{1}{3}M{L}^2$

Here,

  $I$ is the moment of inertia,

  $M$ is the mass of the rod,

  $L$ is the length of rod.

Calculation:

The mass of bat $M=2.2\text{kg}$

The length of bat $L=0.95\text{m}$

Player hits the softball with bat,

Hence the axis of rotation will be at one end of the bat.

For this, Moment of Inertia $I=\frac{1}{3}M{L}^2$

The ball accelerates from rest to $3.0\text{rev/s}$ in time $0.20\text{s}$

The angular acceleration $\alpha =\frac{{\omega }_f-{\omega }_i}{t}$

But ${\omega }_i=0$

And ${\omega }_f=3\times 2\pi \text{rad/s}$

Therefore,

  $\begin{array}{l}\alpha =\frac{6\pi -0}{0.2}{\text{rad/s}}^{\text{2}}\\ \alpha =94.2{\text{rad/s}}^{\text{2}}\end{array}$

The Torque $\tau =I\alpha =\frac{1}{3}M{L}^2\alpha$

  $=\left(\frac{1}{3}\times 2.2\times {\left(0.95\right)}^2\times 94.2\right)\text{N}\cdot \text{m}$

Hence, $\tau =62.3\text{N}\cdot \text{m}$ is the required torque.

Conclusion:

The value is $\tau =62.3\text{N}\cdot \text{m}$ is the required torque.