Solve the following :

Solution:

(a)

At highest point,

$N+\frac{m v^{2}}{R}=m g$

For maximum speed, $\mathrm{N}=0$

$\frac{m v_{\max }^{2}}{R}=m g$

$V_{\max }=\sqrt{(R g)}$

(b)

$V=\frac{V_{\max }}{\sqrt{2}}=\sqrt{\frac{R g}{\sqrt{2}}}$

Let at angle $\boldsymbol{\Phi}$ it loses its contact with road

$N^{\prime}=0$

$\operatorname{Cos} \Phi=\frac{1}{2}$

$\Phi=60^{\circ}$ or ${ }^{\frac{\pi}{3}} \mathrm{rad}$

So it loses contact after a distance $\pi R / 3$ along the bridge from the highest point.

(c)

Normal contact at any angle $\boldsymbol{\Phi}$

$\mathrm{N}^{\prime}=\operatorname{mgcos} \boldsymbol{\Phi} \frac{m v^{2}}{R}$

So, as motorcycle moves along the curve $\boldsymbol{\Phi}$ increases, so $\cos \boldsymbol{\Phi}$ decreases and hence N' decreases.

Normal contact will be minimum at the extreme point.

At extreme point

$\boldsymbol{\Phi}=\frac{\theta}{2}=\frac{L}{2 R}=$ and $\mathrm{N}^{\prime}=0$

From equation (1),

$0=\operatorname{mgcos}\left(\frac{L}{2 R}\right)-\frac{m v^{2}}{R}$

$V=\sqrt{\mathrm{g} \operatorname{Rcos}\left(\frac{L}{2 R}\right)}$

 

$\mathrm{N}^{\prime}+\frac{m V^{2}}{R}=\operatorname{mgcos} \boldsymbol{\Phi}$