Chapter 8.1, Problem 8SSC

To determine

The angular acceleration of the drum.

Answer to Problem 8SSC

  $-8.3{\text{rad/s}}^{\text{2}}$

Explanation of Solution

Given:

The drum was turning at $635\text{rev/min}$ .

It requires $8.0\text{s}$ to slow to a stop.

Formula used:

The angular acceleration can be obtained by:

  $\alpha =\frac{\Delta \omega }{\Delta t}=\frac{{\omega }_f-{\omega }_i}{t_f-t_i}$

Where ${\omega }_f$ is the final angular velocity, ${\omega }_i$ is the initial angular velocity, and the $\Delta t$ is the time taken.

Calculation:

Initial angular velocity of the drum is,

  $\begin{array}{c}{\omega }_i=635\text{rev/min}\\ \text{= 635}\text{rev/min}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\text{min}}{60\text{s}}\\ =66.46\text{rad/s}\end{array}$

Finally the drum stops, thus

  ${\omega }_f=0\text{rad/s}$

Time taken to stop is, $\Delta t=8.0\text{s}$

Then, the angular acceleration will be,

  $\begin{array}{c}\alpha =\frac{{\omega }_f-{\omega }_i}{\Delta t}\\ =\frac{0-66.46}{8}\\ =-8.3{\text{rad/s}}^{\text{2}}\end{array}$

Negative sign shows that the angular velocity decreases with time.

Conclusion:

Thus, angular acceleration of the drum is $-8.3{\text{rad/s}}^{\text{2}}$ .