A car of mass m is travelling at a speed v around a circular track of radius r banked at an angle θ. (a) What is the centripetal acceleration of the car? (b) What is the normal force acting on the car? (c) If θ = 45°, r = 1 km what is the maximum speed?

(a) The formula for the centripetal acceleration of an object undergoing circular motion at a radius r and speed v is a = v^2/r so as F = ma the centripetal force is F = mv^2/r. 2 marks By drawing a diagram, labelling the forces including weight (W = mg) and the normal force (N) and knowing that these forces must add to the centripetal force (F = mv^2/r) horizontally and must cancel vertically (as the car is not accelerating vertically there can be no net force) we can first show that the weight must balance the vertical component of N. So N cosθ = mg therefore N = mg/ cosθ. 6 marks By balancing the horizontal component of the normal force N sinθ and the centripetal force F = mv^2/r we can show that N sinθ = mv^2/r as N = mg/ cosθ then mg tanθ = mv^2/r cancelling the masses on both sides of the equation we show that v^2 = g r tanθ so the maximum speed is v = √ (g r tanθ) as g = 10 m/s^2, r = 1 km = 1000 m and tan45° = 1. The maximum speed v = √ (1000 × 10 × 1) = √ 10000 = 100 m/s. [6 marks]If the car went any faster than 100 m/s the horizontal component of the normal force would not be large enough to keep the car travelling in circular motion so the car would come off the track unless another force besides the normal force was present. One force that can do this is the friction between the tyres and the track which we have neglected.