Question

The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 6.00 s, at which time it is turning at 4.00 rev/s. At this point, the lid of the washing machine is opened, and a safety switch turns it off. The tub then smoothly slows to rest in 15.0 s. Through how many revolutions does the tub rotate while it is in motion?_____________ rev

Answer 1

Answer:

$N = 42 rev$

Explanation:

As we know that initial angular speed of the tub was zero and then it increases uniformly to 4 rev/s in t = 6.00 s

so we will have

$\theta_1 = \frac{\omega_f + \omega_i}{2} t$

$\theta_1 = \frac{2\pi f_2 + 2\pi f_1}{2}(t)$

$\theta_1 = \frac{2\pi \times 4 + 0}{2}(6)$

$\theta_1 = 75.4 rad$

Now when the tub will comes to rest uniformly after opening the lid in time interval of t = 15 s

then we have

$\theta_2 = \frac{\omega_f + \omega_i}{2} t$

$\theta_2 = \frac{2\pi f_2 + 2\pi f_1}{2}(t)$

$\theta_2 = \frac{2\pi \times 4 + 0}{2}(15)$

$\theta_2 = 188.5 rad$

Now total angular displacement of the tub is given as

$\theta = \theta_1 + \theta_2$

$\theta = 75.4 + 188.5$

$\theta = 263.9 rad$

so number of revolutions is given as

$\N = \frac{\theta}{2\pi}$

$N = 42 rev$