Question

a cyclist of mass M negotiates A curve of radius R with uniform speed V the contact force exerted by the ground on the cyclist is​

Answer 1

Answer:

$\sqrt{(Mg)^2-(\frac{MV^2}{R})^2}$

Explanation:

The centripetal force acting on the cyclist will be

$M\frac{V^2}{R}$

If the angle made by the cyclist is θ then the vertical component will be Mgsinθ and the horizontal component will be Mgcosθ

The centripetal force which is horizontal will be balanced by the horizontal component

i.e. $Mg\cos\theta=M\frac{V^2}{R}$

$\implies \cos\theta=\frac{V^2}{Rg}$

$\implies \sin\theta=\sqrt{1-(\frac{V^2}{Rg})^2}$

The normal reaction will be

$N=Mg\sin\theta$

$N=Mg\sqrt{1-(\frac{V^2}{Rg})^2}$

$\implies N=\sqrt{(Mg)^2-(\frac{MV^2}{R})^2}$

Hope this helps.