Question

The tub of a clothes washer goes into its spin-dry cycle starting from rest and reaches an angular speed of 4.70 rev/s in 7.03 seconds. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub then slows to a stop in 20.04 seconds. Assuming a constant angular acceleration, while the tub is starting and stopping, how many revolutions does the tub turn during this 27.07 second interval?

Answer 1

Case 1: until the washer reaches its top spin

Initial angular speed ωi = 0 rev /s

Final angular speed ωf = 4.70 rev /s

Time t = 7.03 s

The angular acceleration is

   ωf - ωi = α t

   α = 4.7 - 0 / 7.03

= 0.668563 rev/s^2

The angular displacement

   θ1 = ωi t + (1/2) α t^2

= 0 + (1/2) (0.668563)(7.03)^2

= 16.5205 rev

Case II: the washer coming to rest from top spin

Initial angular speed ωi = 4.70 rev /s

Final angular speed ωf = 0 rev /s

Time t = 20.04 s

The angular acceleration is

   ωf - ωi = α t

   α = 0 - 4.7 / 20.04

= - 0.234531 rev/s^2

The angular displacement

   θ2 = ωi t + (1/2) α t^2

= 4.7(20.04) + (1/2) ( -0.234531)(20.04)^2

= 47.094 rev

Total number of revolutions

θ1 + θ2 = 16.5205 rev + 47.094 rev

   = 63.61 rev

or approximately 63 revolutions