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 8.0 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 12.0 s. Through how many revolutions does the tub turn during this 20-s interval? Assume constant angular acceleration while it is starting and stopping. rev

Answer 1

Very clearly there are two different accelerations, so we need to use two seperate intervals.

While spinning we must find the angular acceleration.

Using w = a * t for an object at rest that spins under constant angular acceleration, a [alpha if you prefer], to a final angular velocity of w in time t. We see that t = w/a = 2 / 8 rev/s^2 (Note: You can do these operations with radians or revolutions. It only is a factor of 2 pi anyways.)

Then while it slows down, we can use w2 = w1 + at, which is the general form of the 1st equation for constant acceleration from angular speed w1 to w2. Here, w1 = 2 rev/s and w2 = 0.
So we get a = -w1/t = -2/12rev/s^2.

So for the 1st time, we can use an equation theta2 = theta1 + w1*t + 0.5 a*t^2. But instead of angles, theta, we can use revolutions -- call this r. And by choice we can make r1 (theta1) be 0.

From 0 to 13s, r2 = 0.5 * a* t^2, since it starts with 0 angular velocity.
From 13 to 20s, r2 = 5*t + 0.5* a*t^2. Here though, t is not 20 but instead 12 since it is really a difference in time under which the acceleration is being applied. Remember that a is negative.



So the total revolution will be the sum of the two values