Question

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 11.0 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 9.0 s. Through how many revolutions does the tub turn during this 20 s interval? Assume constant angular acceleration while it is starting and stopping.

Answer 1

Total angular displacement will be 19.998 radian

Step-by-step explanation:

It is given that the washer starts from the rest and reach reach the speed of 2 rev/sec in 11 sec

So initial angular velocity
`\omega _i=0rad/sec`

And final angular velocity
`\omega _f=11rad/sec`

Time t = 11 sec

So angular acceleration
`\alpha =(\omega _f-\omega _i)/(t)=(2-0)/(11)=0.1818rad/sec^2`

So angular displacement in this 11 sec

`\Theta =\omega _it+(1)/(2)\alpha t^2=0* 11+(1)/(2)* 0.1818* 11^2`

`\Theta =\omega _it+(1)/(2)\alpha t^2=0* 11+(1)/(2)* 0.1818* 11^2=10.99radian`

Now the washer slows down and stops in 9 sec

So final angular velocity = 0 rad/sec

So angular acceleration
`\alpha =(0-2)/(9)=-0.222rad/sec^2`

So angular displacement
`\Theta =2* 9-(1)/(2)* 0.222* 9^2=8.991radian`

So total displacement in 20 sec =
`=10.999+8.999=19.998radian`