During normal operation, a computer's hard disk spins at 7200 rpm. If it takes the hard disk 4.0 s to reach this angular velocity starting from rest, what is the average angular acceleration of the hard disk in $\mathrm{rad} / \mathrm{s}^{2} ?$

To convert the angular velocity from rpm (revolutions per minute) to radians per second, we can use the following conversion factors: 1 revolution = 2π radians, and 1 minute = 60 seconds So, angular velocity in radians per second is: $$ \omega = 7200 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ radians}}{1 \text{rev}} \times \frac{1 \text{min}}{60 \text{s}} $$

Now that we have the angular velocity in radians per second, we can use the formula for average angular acceleration: $$ \text{average angular acceleration} = \frac{\text{final angular velocity} - \text{initial angular velocity}}{\text{time}} $$ Since the hard disk starts from rest, its initial angular velocity is 0. Thus, the formula becomes: $$ \text{average angular acceleration} = \frac{\omega}{t} $$ Plug in the values and calculate the average angular acceleration: $$ \text{average angular acceleration} = \frac{7200 \times \frac{2\pi}{60}}{4.0} $$ Now, simplify and find the average angular acceleration: $$ \text{average angular acceleration} = \frac{7200 \times \frac{2\pi}{60}}{4.0} = 75.4 \mathrm{rad} / \mathrm{s}^{2} $$ So, the average angular acceleration of the hard disk is 75.4 \(\mathrm{rad} / \mathrm{s}^{2}\).