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 235.6 rad/s while it is being read, and then is allowed to come to rest over 0.167 seconds , what is the magnitude of the average angular acceleration of the disk? average angular acceleration: rad s 2 If the disk is 0.12 m in diameter, what is the magnitude of the tangential acceleration of a point 1 / 3 of the way out from the center of the disk? tangential acceleration

Answer 1

Answer:

Part(a): The angular acceleration is $\bf{1410.78~rad/s^{2}}$.

Part(b): The linear acceleration is $\bf{28.21~m/s^{2}}$.

Explanation:

Given:

The angular velocity, $\omega = 235.6~rad/s$

Time taken, $t = 0.167~s$

Diameter of the disk, $d = 0.12~m$

Radius of the concerned point, $r_{p} = \dfrac{1}{3}(d/2)$.

Part(a):

The angular acceleration ($\alpha$) is given by

$\alpha &=& \dfrac{\omega}{t}\\&=& \dfrac{235.6~rad/s}{0.167~s}\\&=& 1410.78~rad/s^{2}$

Part(b):

The radius ($r_{p}$) of the concerned point is given by

$r_{p} &=& \dfrac{1}{3}\dfrac{0.12~m}{2}\\&=& 0.02~m$

Linear acceleration ($a$) is give by

$a &=& \alpha \times r_{p}\\&=& (1410.78~rad/sec)(0.02~m)\\&=& 28.21~m/s^{2}$