Chapter 9.1, Problem 8PP

To determine

Angular impulse and average torque.

Answer to Problem 8PP

Angular impulse $L=2.45{\text{ kg-m}}^{\text{2}}\text{/s}$ and average torque $\tau =1.02\text{ N-m}$

Explanation of Solution

Given:

Angular velocity, $\omega =70\text{ rad/s}$

Moment of inertia, $I=0.0350{\text{ kg-m}}^2$

Time interval, $\Delta t=2.4\text{ s}$

Formula used:

Angular impluse is given as,

  $L=I\omega$  (1)

Here, L is angular impulse, I is moment of inertia and $\omega$ is angular velocity.

Also,

  $L=\tau \Delta t$  (2)

Here, $\tau$ is torque and $\Delta t$ is time interval

Calculation:

Substitute the value of I and $\omega$ in equation (1).

  $\begin{array}{c}L=0.0350\times 70\\ =2.45{\text{ kg-m}}^{\text{2}}\text{/s}\end{array}$

Now, substitute the value of L and $\Delta t$ in equation (2).

  $\begin{array}{c}2.45=\tau \times 2.4\\ \tau =\frac{2.45}{2.4}\\ =1.02\text{ N-m}\end{array}$

Conclusion:

Therefore, angular impulse is equal to $\text{2}{\text{.45 kg-m}}^{\text{2}}\text{/s}$ and average torque is $1.02\text{ N-m}$ .