Applied Maths - Mechanics 41

Let the railway line be banked up at an angle of θ to the horizontal.
Let the mass of the truck be m.
Let v be the speed of the truck.
Let r be the radius of curvature wrt the centre of mass of the truck.
When a truck is stationary at that point, there is a lateral thrust equal to mgsinθ to the inner rail. When the truck is in motion, this thrust will decrease due to the curvature of the railway line. The reduction of thrust is given by mv2cosθ/r. When v increases to the point, its effect is more than sufficient to cancel out the effect of the weight of the truck, and there will be a net thrust on the outer rail.
It is given that mgsinθ – mv12cosθ/r = mv22cosθ/r – mgsinθ
(v12 + v22)cosθ/r = 2gsinθ
(v12 + v22)/(2gr) = tanθ
At certain speed v, the effect of weight and the effect of circular motion just cancel out each other. This happens at v such that mgsinθ – mv2cosθ/r = 0
mgsinθ – mv2cosθ/r = 0 =>
v2 = grsinθ/cosθ
= gr(v12 + v22)/(2gr)
= (v12 + v22)/2
v =  √[(v12 + v22)/2]