# Obtain an expression for the angle of lean when a cyclist takes a curved path

Consider a cyclist of negotiating a curve of radius r with velocity 'v'
Weight of the cyclist, W=m g
In order to provide the necessary centripetal force, the cyclist leans through angle $\theta^{\circ}$ in inward direction as shown in figure above.
The various forces acting on the cyclist are:
(i) Weight Mg acting vertically downward at the centre of gravity of cycle and the cyclist.
(ii) The reaction R of the ground onto the cyclist, acting along a line, making angle $\theta^{\circ}$ with the vertical.

The cyclist while taking the turn is in equilibrium, therefore the vertical component Rcos$\theta$ of the normal reaction R will balance the weight of the cyclist.
The horizontal component Rsin$\theta$ will provide the necessary centripetal force to the cyclist.

$\begin{array}{l} R \cos \theta=M g \\ R \sin \theta=\frac{M v^{2}}{r} \end{array}$

Dividing \ (2) \ by \ (1),\ we \ have \ \\ \\ \begin{aligned} \tan \theta &=\frac{v^{2}}{r g} \\ \theta &=\tan ^{-1}\left(\frac{v^{2}}{r g}\right) \end{aligned}

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