A car has wheels (including the tire) of radius OA= 45.0 cm. It accelerates from

Question

A car has wheels (including the tire) of radius OA= 45.0 cm. It accelerates from rest to 40.0 mph = 17.9 m/s in 10 seconds. (i) Calculate the angular acceleration of the wheels while the car is accelerating, assuming that the wheels rotate without slipping on the pavement. (ii) Calculate the number of revolutions through which the wheels rotate while the car is accelerating. (iii) The car comes to rest from 17.9 m/s with a uniform deceleration. During this deceleration, the wheels are observed to rotate through exactly 80 revolutions. How long does it take for the car to come to rest?

Explanation / Answer

Here,
OA = 45 cm

OA = 0.45 m

v = 17.9 m/s

t = 10 s

a) let the angular acceleration is a

using first equation of motion

a = (v/(OA * t))

a = 17.9/(0.45 * 10)

a = 3.98 rad/s^2

the angular acceleration of the wheels while the car is accelerating is 3.98 rad/s^2

b)

angle rotated = 0.5 * a * t^2

angle rotated = 0.5 * 3.98 * 10^2

angle rotated = 198.9 radian

angle rotated = 198.9/(2pi) rev

angle rotated = 31.7 revs

the number of revolutions through which the wheels rotate while the car is accelerating is 31.7

c)

let the angular acceleration is a

Using third equation of motion

17.9^2 - 0^2 = - 2 * (80 * 2pi/(OA)) * a

17.9^2 - 0^2 = - 2 * (80 * 2pi/(0.45)) * a

solving for a

a = -0.143 rad/s^2

time taken to stop = v/(a * OD)

time taken to stop = 17.9/(0.143 * 0.45)

time taken to stop = 278.2 s

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