Answer:
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
