Question

An electric motor rotating a workshop grinding wheel at a rate of 1.31 ✕ 102 rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude 3.40 rad/s2. (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in (a)?

Answer 1

(a) 4.03 s

The initial angular velocity of the wheel is

$\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s$

The angular acceleration of the wheel is

$\alpha = -3.40 rad/s^2$

negative since it is a deceleration.

The angular acceleration can be also written as

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

where

$\omega_f = 0$ is the final angular velocity (the wheel comes to a stop)

t is the time it takes for the wheel to stop

Solving for t, we find

$t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s$

(b) 27.6 rad

The angular displacement of the wheel in angular accelerated motion is given by

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

where we have

$\omega_i=13.7 rad/s$ is the initial angular velocity

$\alpha = -3.40 rad/s^2$ is the angular acceleration

t = 4.03 s is the total time of the motion

Substituting numbers, we find

$\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad$