What should be the elevation of outer track of the train to move in a circular path of radius R, width of the track is w (< < R) and speed of the train is v ? (Neglect friction)

Answer :

Solution :

$Nsin\mathrm{\theta <!--\; \theta \; -->}=\frac{m{v}^2}{R}$ Ncosθ = mg $tan\mathrm{\theta <!--\; \theta \; -->}=\frac{v^2}{Rg}=\frac{h}{w}$ $\therefore <!--\; \therefore \; -->h=\frac{v^{2}w}{Rg}$