Question

A motorcycle has to move with a constant speed on an over bridge which is in the form of a circular arc of radius R and has a total length L. Suppose the motorcycle starts from the highest point.(a) What can its maximum velocity be for which the contact with the road is not broken at the highest point? (b) If the motorcycle goes at speed 1/√2 times the maximum found in part (a), where will it lose the contact with the road? (c) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge?

Answer 1

According to the question The chances of losing contact is maximum at the end of the bridge for which α =L/2R Read more on Sarthaks.com - https://www.sarthaks.com/43475/a-motorcycle-has-to-move-with-a-constant-speed-on-an-overbridge-which-is-in-the-form-of

Answer 2

(a) $\begin{equation}v^{2}=\sqrt{R g}$

(b) Hence it losses contact at $\begin{equation}\frac{\pi \mathrm{R}}{3}$ distance from the highest point  

(c) $\begin{equation}v=\sqrt{R g \cos \alpha}$

Explanation:

According to the figure we can see that,

R is the radius of the bridge and L is the total length of the bridge  

(a) Now at highest point ,

$\begin{aligned}&m g=\frac{m v^{2}}{R}\\&m g R=m v^{2}\\&\frac{m g R}{m}=v^{2}\\&v^{2}=\sqrt{R g}\end{aligned}$

(b) If the motorcycle goes at a speed $\frac{1}{\sqrt{2}}$ times the maximum found in part.

Now for this case $v^{2}=\frac{1}{\sqrt{2}} \sqrt{R g}$ is given.

Let it loses the contact at point p,

Therefore at point p  

$mgcos$$\theta=\frac{m v^{2}}{R}$

$\begin{equation}\mathrm{v}^{2}=\mathrm{Rg} \cos \theta$

$(\frac{1}{\sqrt{2}} \sqrt{R g})^{2}=Rgcos \theta$

$\frac{Rg}{2} = Rgcos\theta$

$cos\theta = 1/2 = 60^{\circ}$

$=\pi / 3$

Now we can say that $\theta=\frac{1}{r}=R \theta=\frac{\pi R}{3}$

Hence it losses contact at  $\frac{\pi R}{3}$ distance from the highest point.

(c)  let the uniform speed of bridge be v

The chances of losing contact is maximum at the end of the bridge for which  $\begin{equation}\alpha=\frac{L}{2 R}$

So,

$\begin{equation}\begin{aligned}&\frac{m v^{2}}{R}=m g \cos \alpha\\&m v^{2}=R m g \cos \alpha\\&v^{2}=\frac{R m g \cos \alpha}{m}\\&v^{2}=R g \cos \alpha\\&v=\sqrt{R g \cos \alpha}\end{aligned}$