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 $\alpha$

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

$\alpha = \frac{910}{0.167}$

$\alpha = 5449.1 rad\s^2$

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