Chapter 9.1, Problem 7PP

To determine

Angular velocity of ball and average torque exerted on ball by pitcher.

Answer to Problem 7PP

Angular velocity is , $\omega =62.8\text{ rad/s}$ and average torque is $\tau =0.033\text{ N-m}$ .

Explanation of Solution

Given :

Angular speed, $N=6\times {10}^2=600\text{ rpm}$

Moment of inertia, $I=8\times {10}^{-5}{\text{ kg-m}}^2$

Time interval, $\Delta t=0.15\text{ s}$

Formula used:

Angular velocity is given as

  $\omega =\frac{2\pi N}{60}$  (1)

Here, $\omega$ in angular velocity in rad/s and N is angular speed in rpm.

First equation of motion (rotational).

  ${\omega }_f={\omega }_i+\alpha t$  (2)

Here, ${\omega }_f$ is final angular velocity, ${\omega }_i$ is initial angular velocity, $\alpha$ is angular acceleration and t is time.

Also,

  $\tau =I\alpha$  (3)

Here, $\tau$ is torque, Iis moment of inertia and $\alpha$ is angular acceleration

Calculation:

Now, substitute the value of Nin equation (1).

  $\begin{array}{c}\omega =\frac{2\times 3.14\times 600}{60}\\ =62.8\text{ rad/s}\end{array}$

As the ball starts motion from rest. Therefore, initial angular speed of ball, ${\omega }_i=0\text{ rad/s}$

Substitute the value of ${\omega }_i\text{,}{\omega }_f\text{ and }t$ in equation (2).

  $\begin{array}{c}62.8=0+\alpha \times 0.15\\ 0.15\alpha =62.8\\ \alpha =\frac{62.8}{0.15}\\ =418.67{\text{ rad/s}}^2\end{array}$

Now, substitute the value of Iand $\alpha$ in equation (3).

  $\begin{array}{c}\tau =8\times {10}^{-5}\times 418.67\\ =0.033\text{ N-m}\end{array}$

Conclusion:

Therefore, angular velocity of ball thrown is 62.8 rad/s and average torque is $0.033\text{ N-m}$