Question

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time interval, the wheel rotates through 62.4 rad. What is the angular speed of the wheel at the end of the 4.20-s interval?

Answer 1

We use rotational kinematic equations,

$\theta =\theta _{0}  +  \omega_{0}  t+ \frac{1}{2}\alpha t^2$             (A)

$\omega^2=  \omega^2_{0} +2\alpha\theta$                                     (B)                                          

Here, $\theta$ and $\theta _{0}$ are final and initial angular displacements respectively,  $\omega$ and $\omega_{0}$ are final and initial angular speed and $\alpha$ is the angular acceleration.

Given,  $\alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad$ and $t = 4 .20 \ s$.

Substituting these values in equation (A), we get

$62.4 \ rad = 0 +  \omega_{0}  4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\  \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s$

Now from equation (B),

$\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s$

Thus, the angular speed of the wheel at the end of the 4.20-s interval is 36.7  rad/s.