Question

A 1200 kg car is practicing on a flat test track.   The car begins moving in a 100 m radius circle at a speed of 20 m/s.   Viewed above, it is traveling around the circle clockwise, beginning from the top of the circle.

(A) What is the centripetal acceleration of the car?
(B) How much force is required for this?
(C) What is the coefficient of friction between the tires and the track?
Three quarters of the way around the track the car hits a 4m patch of oil, which reduces friction of the tires to zero.
(D) Which way does the car move, with what acceleration?
The driver now applies the brakes, aiming the car straight ahead.
(E) If the force stopping the car is the same as your answer from part (C), how much time does it take to stop the car.

Answer 1

a) v = 20m/s

R = 100m

centripetal acceleration = v²/R = (20)²/100 = 4m/s²

b) centripetal force = mv²/R = 1200x4 = 4800N

c)Let the coefficient of friction be f

mgf = mv²/R

f = v²/gR

= (20x20)/(9.8x100) = 0.408

d) The required coefficient of friction would be same because it is independent of mass of vehicle

e) Minimum speed should be such that the centripetal force is equal to weight of car

mg = mv²/R

v = sqrt(gR)

= sqrt(9.8x100)

= 31.3 m/s

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