Question

During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 33.3 rad/s. Find the angular displacement of the tub during a spin of 77.5 s, expressed both in radians and in revolutions.

Answer 1

Answer:

The angular displacement of the tub will be 2580.75 rad or 411 revolutions.

Explanation:

We know the relation between angular velocity ($\omega$), the angular displacement ($\theta$) and time ($t$) is given by

$\omega = \dfrac{\theta}{t}$

Given $\omega$ = $rad~s^{-1}$ and the time ($t$) required to complete a spin is 77.5 s.

Therefore, the required angular displacment ($\theta$) is

$&& \theta = \omega \times t\\&or,& \theta = 33.3~rad~s^{-1} \times 77.5~s = 2580.75~rad$

Also we know that,

$&& 2 \pi~rad = 1~revolution\\&or,& 580.75~rad = \dfrac{2580.75}{2~\pi} \approx 411~revolutions$

So, the angular displacement of the tub will be 2580.75 rad or 411 revolutions.