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

Answer:

The total number of revolution is 50 rev.

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

$5.0-0=\alpha\times8.0$

$\alpha=\dfrac{5.0}{8.0}$

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

We need to calculate the angular displacement

Using equation of angular motion

$\theta=\omega_{i}t+\dfrac{1}{2}\alpha t^2$

Put the value into the formula

$\theta=0+\dfrac{1}{2}\times0.625\times(8.0)^2$

$\theta=20\ rev$

Now, The washer coming to rest from top spin

We need to calculate the angular acceleration

Using equation of angular motion

$\omega_{f}-\omega_{i}=\alpha t$

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}$

$\alpha=\dfrac{0-5}{12}$

$\alpha=−0.4167\ rev/s^2$

We need to calculate the angular displacement

Using formula of displacement

$\theta'=\omega_{i}t+\dfrac{1}{2}\alpha t^2$

Put the value into the formula

$\theta'=5\times12+\dfrac{1}{2}\times(-0.4167)\times12^2$

$\theta'=30\ rev$

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.