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 734.1 rad/s 734.1 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds0.569 seconds , what is the magnitude of the average angular acceleration of the disk

Answer 1

Answer:

$1290.16\ \text{rad/s}^2$

Explanation:

$\omega_i$ = Initial angular velocity = 734.1 rad/s

$\omega_f$ = Final angular velocity = 0

t = Time = 0.569 seconds

$\alpha$ = Angular acceleration

From the kinematic equations of rotational motion we have

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-734.1}{0.569}\\\Rightarrow \alpha=-1290.16\ \text{rad/s}^2$

The magnitude of the average angular acceleration of the disk is $1290.16\ \text{rad/s}^2$.