Revolutions and Acceleration?

Question

The tires of a car make 65 revolutions as the car reduces its speed from 100km/hr to 50km/hr. The tires have a diameter or .80m A) what was the angular acceleration? b) if the car continues to decelerate at this rate, how much more time is it required to stop?

Answer 1

A) ωf² = ωi² + 2α∆θ

∆θ = 65 revolutions x 2 x = 410 radians

ωi = (100 km/hr x (1000/3600)) / 0.80 m = 35 rad/s

ωf = (50 km/hr x (1000/3600)) / 0.80 m = 17 rad/s

α = (ωf² - ωi²) / 2∆θ

= ((17 rad/s)² - (35 rad/s)²) / (2 x 410 rad)

= - 1.1 rad/s²

B) ωf = ωi + αr

t = (ωf - ωi) / α

= (0 rad/s - 17 rad/s) / (- 1.1 rad/s²)

= 15 s

Answer 2

First, convert your linear velocity (v) into angular velocity (omega), v = r * w, and multiply your revolutions by 2 * (pi) to get your angular displacement (theta).  Then you solve for your angular acceleration (alpha), (omega)_f ^2 = (omega)_i ^2 + 2 (alpha) (theta).  Once you have your angular acceleration, plug that into the equation (omega)_f = (omega)_i + (alpha) * t, where (omega)_f is now your angular velocity at 0 kph and (omega)_i is your angular velocity at 50 kph.  Solve for t and you have the time required to stop.