ASK YOUR TEACHER 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 4.00 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 13.0 s. Through how many revolutions does the tub turn while it is in motion?

Explanation:

Given that,

The tub starts from rest

Then, it initial angular velocity is zero

ωi = 0 rev/s

After t = 8 seconds, it angular speed is

ωf = 4 rev /s

The lid was open at the point and the tub slows to rest after t = 13 sec

How many revolution does the tub turn while in motion.

We need to know the constant angular for the first case when it angular velocity moves from 0 rev/s to 4rev/s

Then, angular acceleration can be determine using

α = ∆ω/∆t

α = (ωf - ωi) / (t2-t1)

α = (4—0) / (8—0)

α = 4/8 = 0.5rev/s²

Using the angular motion equation

ωf² = ωi² +2αθ

4² = 0² + 2 × 0.5× θ

16 = 0 + θ

θ = 16 rev

Second case

When the tub safety switch was on and the tub is decelerating,

So we need to find the deceleration, from 4rev/s to 0rev/s in 13 seconds

α = ∆ω/∆t

α = (ωf - ωi) / (t2-t1)

α = (0—4) / (13—0)

α = -4/13 = —0.308 rev/s²

Also, to find the revolution

Using the angular motion equation

ωf² = ωi² +2αθ

0² = 4² + 2× -0.308 × θ

0 = 16 — 0.6154•θ

0.6154 θ = 16

θ = 16 / 0.6154

θ = 26 rev

Then, the total revolution is the addition of the first case revolution and the second case

θ = 26 + 16

θ = 42 resolutions

the components of a vector are the parts of the vectors that are to each other.

b perpendicular