Angular velocity

Question

During a very quick stop a car decelerates at 10 m/s² before coming to stop. If the tires have a 0.45m radius and assuming they do not slip on the pavement: how many revolutions do the tyres make given the initial angular velocity is 105.0 rad/s?

Answer 1

The tangential velocity of the tire is equal to the velocity of the car, and the tangential acceleration of the tire is also equal to the acceleration of the car.

With this in mind, the angular acceleration of the tire is given by α = a / r where a is the tangential acceleration of the tire, which is equal to the acceleration of the car, - 10 m/s^2.

Therefore α = -10 m/s^2 / 0.45 m = -22.2 rad/s^2. Since the initial angular velocity of the tire is given as 105.0 rad/s, we can use the equation ω²_final = ω²_initial – 2 α θ, where θ is the total angular displacement. Solving for θ, we get θ = - ω²_initial / -2α = 105.0^2 rad^2/s^2 / 2 x 22.2 rad/s^2 = 248 rad, remembering that ω_final = 0.

Since one revolution goes through 2π radians, the number of revolutions of the tire before stopping is 248 rad / 2π rad/rev. = 39.5 revolutions.

Please check my math.