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

`\theta = 20\ rev`

Step-by-step 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 = (\omega_f-\omega_i)/(t)`

`\alpha = (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* 0.286 * \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 = (\omega_f-\omega_i)/(t)`

`\alpha = (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* (-0.154) * \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.