Aflywheel turns through 28 rev as it slows from an angular speed of 5.4 rad/s to a stop. (a) assuming a constant angular acceleration, find the time for it to come to rest. (b) what is its angular acceleration? (c) how much time is required for it to complete the first 14 of the 28 revolutions? (a) number enter your answer for part (a) in accordance to the question statement units choose the answer for part (a) from the menu in accordance to the question statement

a) 65.2 s

b) -0.083 rad/s/s

c) 19 s

Explanation:

a) Angular displacement = θ = 28 rev = 28 × 2 π =175.9 rad

Initial angular velocity = ω₁ = 5.4 rad/s

Final angular velocity = ω₂ = 0, since it comes to rest

Angular acceleration = α = ( ω₂² -  ω₁² ) / 2θ = - 0.083 rad/s/s

Time taken to come to rest = t =  (ω₁ - ω₁ ) / α = 65.2 s

b) Angular acceleration = α = - 0.083 rad/s/s

c) Angular displacement = θ ' = 14 rev = 87.96 rad

θ'= ω₁  t  + 0.5 α t²

Angular acceleration is said to be constant.

⇒ 87.96 = 5.4 t + 0.5 ( - 0.083 ) t²

⇒ 0.0415 t² - 5.4 t + 87.96 = 0

Solving this quadratic equation for t,

⇒ t =111 s or 19 s

(Since time for 28 rev is 65.2 s, for 14 rev , it cannot be greater than that so 19 s is chosen out of the two roots for t)

the total distance traveled by the car is

30km + 49km = 79 km .