A cyclist is travelling with velocity v on a | vedclass

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(D) When a cyclist turns on a curved road,they must lean inwards to maintain balance against the centrifugal force.Let $m$ be the mass of the cyclist,

Vedclass · Accepted Answer

$	an 	heta = \frac{v^2}{Rg}$

Vedclass · Answer

(D) When a cyclist turns on a curved road,they must lean inwards to maintain balance against the centrifugal force.Let $m$ be the mass of the cyclist,$v$ be the velocity,and $R$ be the radius of the curve.The vertical component of the normal force $N$ balances the weight of the cyclist: $N \cos 	heta = mg$.The horizontal component of the normal force provides the necessary centripetal force: $N \sin 	heta = \frac{mv^2}{R}$.Dividing the second equation by the first,we get: $\frac{N \sin 	heta}{N \cos 	heta} = \frac{mv^2 / R}{mg}$.Therefore,$	an 	heta = \frac{v^2}{Rg}$.

$A$ cyclist is travelling with velocity $v$ on a curved road of radius $R$. The angle $\theta$ through which the cyclist leans inwards is given by

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