Question

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 9.0 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 15.0 s. Through how many revolutions does the tub turn during this 24 s interval? Assume constant angular acceleration while it is starting and stopping.

Answer 1

Answer

given,

angular speed of the tub = 5 rev/s

time = 9 s

he tub slows to rest  =  15.0 s

the angular acceleration

$\omega_f - \omega_i = \alpha t$

$\alpha = \dfrac{5-0}{9}$

$\alpha = 0.556 rev/s^2$

angular displacement

$\theta_1 = \omega_i \ t + \dfrac{1}{2}\alpha t^2$

$\theta_1 = \dfrac{1}{2}\times 0.556 \times 9^2$

$\theta_1 = 22.52 rev$

$\theta_1 = 23 rev$

case 2

now,

$\omega_i = 5 rev/s$

$\omega_f = 0 rev/s$

time = 15 s

the angular acceleration

$\omega_f - \omega_i = \alpha t$

$\alpha = \dfrac{0-5}{15}$

$v\alpha =-0.333 rev/s^2$

angular displacement

$\theta_2 = \omega_i \ t + \dfrac{1}{2}\alpha t^2$

$\theta_2 =5\times 15 -\dfrac{1}{2}\times 0.333 \times 15^2$

$\theta_2 = 37.875 rev$

$\theta_2 =38 rev$

total revolution in 24 s

= 23 + 38

= 62 revolution