The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev?

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 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 15.0 s. Through how many revolutions does the tub turn during this 26 s interval? Assume constant angular acceleration while it is starting and stopping.

Answer 1

Assuming constant angular acceleration while it is starting and stopping.

SPEED UP

Velocity (vi) = 0 rev/s      

Velocity (vf) = 5 rev/s

Time (t) = 11s

Δx = 1/2 ( vi + vf ) * Δt

Δx = 1/2 ( 0 + 5 ) * 11

Δx = 27.5 rev

SLOW DOWN

Velocity (vi) = 5 rev/s

Velocity (vf) =  0 rev/s

Time (t) = 15s

Δx = 1/2 ( vi + vf ) * Δt

Δx = 1/2 ( 0 + 5 ) * 11

Δx = 37.5 rev

Total revolutions in 26 seconds:

27.5 + 37.5 = 65.0 rev

Answer 2

Since the speeding up and slowing down is uniform, we can say:-

Average angular speed during spin up  =  (0 + 5)/2  =  2.5 rev/s

Average speed during slowdown  =  (5  +  0)/2   =   2.5rev/s

In fact, during the whole action, the average speed is 2.5 rev/s

It turned at an average of 2.5rev/s for 26 sec.

No. of rotations  =   26 * 2.5   =   65 revs.