A track consists of two circular parts ABC and CDE of equal radius 100 m , joined smoothly. Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track. Determine the normal force between the road and the cycle just before and just after the cycle crosses C . Take g = 10 m/s ^2 .

Step-by-step solution

Given v = 18 km/h = 5 m/s , m = 100 kg , R = 100 m . The track consists of two circular arcs, each subtending 90^ at their respective centres. By symmetry, the radius vector at point C makes an angle = 45^ with the vertical. Just before crossing C , the cycle is on the convex part of the track. The centripetal force is provided by the component of weight towards the centre minus the normal reaction. mg 45^ - N₁ = mv^2 R N₁ = mg 45^ - mv^2 R Substituting the values: N₁ = 100 10 1 2 - 100 (5)^2 100 N₁ = 500 2 - 25 70