The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 4.0 rev/s in 10.0 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 19 s interval

Total number of revolutions = 38 rev

Explanation:

Case I:

Initial angular velocity, ω1 = 0 rev/s

Final angular velocity, ω2 = 4 rev/s

time, t1 = 10 s

Case II:

Initial angular velocity, ω1' = 4 rev/s

Final angular velocity, ω2' = 0 rev/s

Time, t2 = 9 s

Solution:

Case I:

From the equations of motion,

θ1 = [ ( ω1 + ω2 ) / 2 ] * t1

   = [ ( 0 + 4 ) / 2 ] * 10

    = 20 rev

Case II:

From the equations of motion,

θ2 = [ ( ω1' + ω2' ) / 2 ] * t2

   = [ ( 4 + 0 ) / 2 ] * 9

    = 18 rev

Total number of revolutions = θ1 + θ2

                                         = 20 + 18

                                         = 38 rev

Total number of revolutions = 38 rev

v=ir = 3x15 =45 v

the maxium height is 34m/s