Question

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 4.7 rev/s in 8.3 s. You lift the lid of the washer and notice that the tub decelerates and comes to a stop in 17.5 s. Assuming that the tub rotates with constant angular acceleration while it is starting and stopping, determine the total number of revolutions undergone by the tub during this entire time interval.

Answer 1

Number of revolutions = Δθ = 60.63 revs

Step-by-step explanation:

The angular acceleration of the washer is given by

α = (ωf - ωi)/Δt

Where ωf is the initial angular speed, ωi is the final angular speed and Δt is the interval of time during this acceleration.

α = (4.7 - 0)/8.3

α = 0.56 rev/s²

As we know from the equation of kinematics,

2αΔθ = ωf² - ωi²

Where Δθ is the change in angular displacement of the washer.

Δθ = (ωf² - ωi²)/2α

Δθ = (4.7² - 0²)/2*0.56

Δθ = 19.72 revs

Now the tub decelerates and comes to a stop in 17.5 s

α = (ωf - ωi)/Δt

α = (-4.7 - 0)/17.5

α = -0.27 rev/s²

the corresponding change in angular displacement of the washer is

Δθ = (ωf² - ωi²)/2α

Δθ = (0² - 4.7²)/2*-0.27

Δθ = 40.91 revs

Therefore, the total number of revolutions undergone by the tub during this entire time interval is

Δθ = 19.72 + 40.91

Δθ = 60.63 revs