Question

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 6.0 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 14.0 s. Through how many revolutions does the tub turn during this 25 s interval? Assume constant angular acceleration while it is starting and stopping.

Answer 1

given,

ω₁ = 0 rev/s

ω₂ = 6 rev/s

t = 11 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

11 α = 6 - 0

= 0.545 rev/s²

The angular displacement

θ₁= ωi t + (1/2) α t²

θ₁= 0 + (1/2) (0.545)(11)^2

θ₁= 33 rev

case 2

ω₁ = 6 rev/s

ω₂ = 0 rev/s

t = 14 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

14 α = 0 - 6

= - 0.428 rev/s²

The angular displacement

θ₂= ωi t + (1/2) α t²

θ₂= 6 x 14 + (1/2) (-0.428)(14)^2

θ₂= 42 rev

total revolution in 25 s is equal to

θ = θ₁ + θ₂

θ = 33 + 42

θ = 75 rev