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

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

the force required to move the cart is divided by three or 33.33 n.