Question

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is spun at 910.0 rad/s910.0 rad/s while it is being read, and then is allowed to come to rest over 0.167 seconds0.167 seconds , what is the magnitude of the average angular acceleration of the disk?

Answer 1

To solve this problem we must keep in mind the concepts related to angular kinematic equations. For which the angular velocity is defined as

`\omega_f =\omega_i-\alpha t`

Where

`\omega_f =` Final angular velocity

`\omega_i =` Initial angular velocity

`\alpha =`Angular acceleration

t= time

In this case we do not have a final angular velocity, then

`\omega_i = \alpha t`

Re-arrange for
`\alpha`

`\alpha= (\omega_i)/(t)`

`\alpha = (910)/(0.167)`

`\alpha = 5449.1 rad\s^2`

Therefore the mangitude of the angular aceleration is 5449.1rad/s²