Question

The dryer of a washing machine goes into its spin cycle from rest and gains a speed of 5.0 rev/s in 3.0s. When the lid is opened, the dryer slows down and finally stops in the 7.0s. Compute the angular acceleration during the first 3.0s and the last 7.0s. Through how many revolutions does the dryer turn before coming to stop?

Answer 1

Given:

$n_1=\rm 0\:rev/s$

$n_2=\rm 5.0\:rev/s$

$\Delta t_1=\rm 3.0\: s$

$\Delta t_2=\rm 7.0\:s$

$n_3=\rm 0\:rev/s$

The angular acceleration by definition

$$\alpha=\frac{\Delta \omega}{\Delta t}=2\pi \frac{\Delta n}{\Delta t}$$

Hence,

$$\alpha_1=2\pi \frac{n_2-n_1}{\Delta t_1}=2\pi \frac{5.0-0}{3.0}=10\rm\: rad/s^2$$

$$\alpha_2=2\pi \frac{n_3-n_2}{\Delta t_2}=2\pi \frac{0-5.0}{7.0}=-4.5\rm\: rad/s^2$$

The numbers of revolutions

$$N=N_1+N_2=\frac{\theta_1}{2\pi}+\frac{\theta_2}{2\pi}\\ =\frac{n_1+n_2}{2}\Delta t_1+\frac{n_2+n_3}{2}\Delta t_2$$

$$N=\frac{0+5.0}{2}3.0+\frac{5.0+0}{2}7.0=25\:\rm rev$$