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

Answer:

Total angular displacement will be  19.998 radian  

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 $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{2-0}{11}=0.1818rad/sec^2$

So angular displacement in this 11 sec

$\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times 11+\frac{1}{2}\times 0.1818\times 11^2$

$\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times 11+\frac{1}{2}\times 0.1818\times 11^2=10.99radian$

Now the washer slows down and stops in 9 sec

So final angular velocity = 0 rad/sec

So angular acceleration $\alpha =\frac{0-2}{9}=-0.222rad/sec^2$

So angular displacement $\Theta =2\times 9-\frac{1}{2}\times 0.222\times 9^2=8.991radian$

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