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.

Respuesta :

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

[tex]\omega_{f}-\omega_{i}=\alpha t[/tex]

Put the value into the formula

[tex]5.0-0=\alpha\times8.0[/tex]

[tex]\alpha=\dfrac{5.0}{8.0}[/tex]

[tex]\alpha=0.625\ rev/s^2[/tex]

We need to calculate the angular displacement

Using equation of angular motion

[tex]\theta=\omega_{i}t+\dfrac{1}{2}\alpha t^2[/tex]

Put the value into the formula

[tex]\theta=0+\dfrac{1}{2}\times0.625\times(8.0)^2[/tex]

[tex]\theta=20\ rev[/tex]

Now, The washer coming to rest from top spin

We need to calculate the angular acceleration

Using equation of angular motion

[tex]\omega_{f}-\omega_{i}=\alpha t[/tex]

[tex]\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}[/tex]

[tex]\alpha=\dfrac{0-5}{12}[/tex]

[tex]\alpha=−0.4167\ rev/s^2[/tex]

We need to calculate the angular displacement

Using formula of displacement

[tex]\theta'=\omega_{i}t+\dfrac{1}{2}\alpha t^2[/tex]

Put the value into the formula

[tex]\theta'=5\times12+\dfrac{1}{2}\times(-0.4167)\times12^2[/tex]

[tex]\theta'=30\ rev[/tex]

We need to calculate the total number of revolution

[tex]\theta''=\theta+\theta'[/tex]

[tex]\theta''=20+30[/tex]

[tex]\theta''=50\ rev[/tex]

Hence, The total number of revolution is 50 rev.