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

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 = (5-0)/(9)`

`\alpha = 0.556 rev/s^2`

angular displacement

`\theta_1 = \omega_i \ t + (1)/(2)\alpha t^2`

`\theta_1 = (1)/(2)* 0.556 * 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 = (0-5)/(15)`

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

angular displacement

`\theta_2 = \omega_i \ t + (1)/(2)\alpha t^2`

`\theta_2 =5* 15 -(1)/(2)* 0.333 * 15^2`

`\theta_2 = 37.875 rev`

`\theta_2 =38 rev`

total revolution in 24 s

= 23 + 38

= 62 revolution