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

`723.2 rad/s^2`

Step-by-step explanation:

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

Mathematically:

`\alpha = (\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=(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`