Question

A disk, initially rotating at 146 rad/s, is slowed down with a constant angular acceleration of magnitude 4.55 rad/s2. (a) How much time does the disk take to stop? (b) Through what angle (rad) does the disk rotate during that time?

Answer 1

a) 32.1 s

b) 2342 rad

Explanation:

a)

To solve this problem, we can use the equivalent of the suvat equations for a rotational motion.

In fact, the motion of the disk is a rotational motion with unifom angular acceleration.

So we can use the following suvat equation:

$\omega_f = \omega_i + \alpha t$

where:

$\omega_i$ is the initial angular velocity

$\omega_f$ is the final angular velocity

$\alpha$ is the angular acceleration

t is the time elapsed

In this problem:

$\omega_i = 146 rad/s$ is the initial angular velocity

$\omega_f=0$, since the disk comes to a stop

$\alpha = -4.55 rad/s^2$ (negative since the disk is slowing down)

Therefore, the time taken to stop is

$t=\frac{\omega_f - \omega_i}{\alpha}=\frac{0-146}{-4.55}=32.1 s$

b)

To solve this part of the problem, we can use another suvat equation for the rotational motion, which is:

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

where

$\omega_i$ is the initial angular velocity

$\alpha$ is the angular acceleration

t is the time elapsed

$\theta$ is the angular displacement covered

For the disk in this problem:

$\omega_i = 146 rad/s$ is the initial angular velocity

$\alpha = -4.55 rad/s^2$ (negative since the disk is slowing down)

t = 32.1 s (time elapsed, found in part a)

Substituting, we find the angle through which the disk has rotated in this time:

$\theta = (146)(32.1)+\frac{1}{2}(-4.55)(32.1)^2=2342 rad$