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 8.0 s. At this oint, the person doing the laundry open the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Thru how many revolutions does the tub trun during the entire 20-s interval? Assume constant angular acceleration while it is starting and stopping.

Answer 1

The total number of revolution is 50 rev.

Step-by-step explanation:

Given that,

Angular speed = 5.0 rev/s

Time = 8.0 s

We need to calculate the angular acceleration

Using equation of angular motion

`\omega_(f)-\omega_(i)=\alpha t`

Put the value into the formula

`5.0-0=\alpha*8.0`

`\alpha=(5.0)/(8.0)`

`\alpha=0.625\ rev/s^2`

We need to calculate the angular displacement

Using equation of angular motion

`\theta=\omega_(i)t+(1)/(2)\alpha t^2`

Put the value into the formula

`\theta=0+(1)/(2)*0.625*(8.0)^2`

`\theta=20\ rev`

Now, The washer coming to rest from top spin

We need to calculate the angular acceleration

Using equation of angular motion

`\omega_(f)-\omega_(i)=\alpha t`

`\alpha=(\omega_(f)-\omega_(i))/(t)`

`\alpha=(0-5)/(12)`

`\alpha=−0.4167\ rev/s^2`

We need to calculate the angular displacement

Using formula of displacement

`\theta'=\omega_(i)t+(1)/(2)\alpha t^2`

Put the value into the formula

`\theta'=5*12+(1)/(2)*(-0.4167)*12^2`

`\theta'=30\ rev`

We need to calculate the total number of revolution

`\theta''=\theta+\theta'`

`\theta''=20+30`

`\theta''=50\ rev`

Hence, The total number of revolution is 50 rev.