Question

A motor cyclist moving with a velocity of $72\text{km/h}$ on a flat road takes a turn on the road at a point where the radius of curvature of the road is $20\text{m}$. The acceleration due to gravity is $10{\text{ms}}^{-2}$. In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than

• A. $\theta ={\tan }^{-1}\left(6\right)$
• B. $\theta ={\tan }^{-1}\left(2\right)$
• C. $\theta ={\tan }^{-1}\left(0.75\right)$
• D. $\theta ={\tan }^{-1}\left(4\right)$

Answer 1

The correct option is B $\theta ={\tan }^{-1}\left(2\right)$

Speed of motor cyclist is,

$v=72{\text{kmh}}^{-1}=72\times \frac{5}{18}=20{\text{ms}}^{-1}$

Let the maximum angle bend by motor cyclist with the vertical be $\theta$.


From vertical equilibrium,

$N\cos \theta =mg$ ...........$\left(1\right)$

From circular dynamics equation,

$N\sin \theta =\frac{m{v}^2}{R}$ ..........$\left(2\right)$

After diving above two equations, we get

$\tan \theta =\frac{v^2}{gR}$

$\tan \theta =\frac{(20{)}^2}{10\times 20}=2$

$\therefore \theta ={\tan }^{-1}\left(2\right)$

This is the maximum angle with which cyclist can bend without skidding.

$\begin{array}{c}\text{Why this Question}?\\ \text{Tip: For maximum angle of bend, the}\\ \text{skidding will not happen if component}\\ \text{of normal reaction}N\sin \theta \text{is able to}\\ \text{provide required centripetal force for the}\\ \text{given speed (v).}\end{array}$