The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 8.00 s, at which time it is turning at 5.00 rev/s. at this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. the tub smoothly slows to rest in 12.0 s. through how many revolutions does the tub turn while it is in motion?

50

Explanation:

ωo = 0, t = 8 s, f = 5 rev/s, ω = 2 x 3.14 x f = 2 x 3.14 x 5 = 31.4 rad/s

Initially when the washer is accelerating

Let α be the angular acceleration

ω = ωo + α t

31.4 = 0 + α x 8

α = 3.925 rad/s^2

Let the number of revolutions be θ.

θ = ωo x t + 1/2 α x t^2

θ = 0 + 0.5 x 3.935 x 8 x 8 = 125.6 rad

Number of revolutions, n1 = θ / 2π = 125.6 / (2 x 3.14) = 20 revolutions

When the washer is switch off

ω = 0, t = 12 s, ωo = 31.4 rad/s

Let α be the angular acceleration

ω = ωo + α t

0 = 31.4 + α x 12

α = - 2.62 rad/s^2

Let the number of revolutions be θ.

θ = ωo x t + 1/2 α x t^2

θ = 31.4 x 12 - 0.5 x 2.62 x 12 x 12 = 188.16 rad

Number of revolutions, n2 = θ / 2π = 188.16 / (2 x 3.14) = 30 revolutions

Total number of revolutions, n = n1 + n2 = 20 + 30 = 50

the answer to this question is 60 n

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