During a very quick stop, a car decelerates at 7.6 m/s2. Assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the pavement).
(a) What is the angular acceleration of its tires in rad/s2, assuming they have a radius of 0.26 m and do not slip on the pavement?
(b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95 rad/s ?

24.57 revolutions

Explanation:

(a) If they do not slip on the pavement, then the angular acceleration is

(b) We can use the following equation of motion to find out the angle traveled by the wheel before coming to rest:

where v = 0 m/s is the final angular velocity of the wheel when it stops, = 95rad/s is the initial angular velocity of the wheel, is the deceleration of the wheel, and is the angle swept in rad, which we care looking for:

As each revolution equals to 2π, the total revolution it makes before stop is

154.375 / 2π = 24.57 revolutions

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