Question

Ask Your Teacher The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 7.0 s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 13.0 s. Through how many revolutions does the tub turn during this 20 s interval?

Answer 1

Answer:

$\theta = 20\ rev$

Explanation:

Case 1

Given,

initial angular speed = 0 rev/s

final angular speed = 2 rev/s

time, t = 7 s

angular acceleration of the washer

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

$\alpha = \dfrac{2-0}{7}$

$\alpha = 0.286\ rev/s^2$

using equation of rotational motion

$\omega_f^2 = \omega_i^2 + 2 \alpha \theta_1$

$2^2 = 0^2 + 2\times 0.286 \times \theta_1$

$\theta_1 = 7 rev$

Case 2

Given,

initial angular speed = 2 rev/s

final angular speed = 0 rev/s

time, t = 12 s

angular acceleration of the washer

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

$\alpha = \dfrac{0-2}{13}$

$\alpha = -0.154\ rev/s^2$

using equation of rotational motion

$\omega_f^2 = \omega_i^2 + 2 \alpha \theta_2$

$0^2 = 2^2 + 2\times (-0.154) \times \theta_2$

$\theta_2 = 13 rev$

total revolution in this case

$\theta = \theta_1 + \theta_2$

$\theta =7 +13$

$\theta = 20\ rev$

total revolution of the washer is equal to 20 rev.