Question

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 411.5 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds , what is the magnitude of the average angular acceleration of the disk

Answer 1

Answer:

$723.2 rad/s^2$

Explanation:

The angular acceleration of an object in rotation is equal to the rate of change of its angular velocity.

Mathematically:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\alpha$ is the angular acceleration

$\omega_i$ is the initial angular velocity

$\omega_f$ is the final angular velocity

t is the time elapsed

For the optical disk in this problem, we have:

$\omega_i = 411.5 rad/s$ is the initial angular velocity

$\omega_f=0$ is the final angular velocity (because the disk comes to rest)

t = 0.569 s is the time elapsed

So, the angular acceleration is

$\alpha=\frac{0-411.5}{0.569}=-723.2rad/s^2$

And since we are only interested in the magnitude, then the magnitude is $723.2 rad/s^2$