Chapter 8, Problem 88A

To determine

The angular acceleration of the disk.

Answer to Problem 88A

  $502.65{\text{rad/s}}^{\text{2}}$

Explanation of Solution

Given:

The hard drive on the computer spins at $7200\text{rev/min}$ .

It requires $1.5\text{s}$ to start from rest and reach above desired angular velocity.

Formula used:

The angular acceleration is given 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:

Initially, the disk starts from rest, thus

  ${\omega }_i=0\text{rad/s}$

Final angular velocity of the disk is,

  $\begin{array}{c}{\omega }_f=7200\text{rev/min}\\ \text{= 7200}\text{rev/min}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\text{min}}{60\text{s}}\\ =753.98\text{rad/s}\end{array}$

Time taken, $\Delta t=1.5\text{s}$

Then, the angular acceleration will be,

  $\begin{array}{c}\alpha =\frac{{\omega }_f-{\omega }_i}{\Delta t}\\ =\frac{753.98-0}{1.5}\\ =502.65{\text{rad/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, angular acceleration of the disk is $502.65{\text{rad/s}}^{\text{2}}$ .