Question

A motor cyclist moving with a velocity of $72\text{km/h}$ on a flat frictionless 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 $10{\text{m/s}}^2$. In order to avoid skidding, he must bend with respect to the vertical plane by an angle

• A. ${\tan }^{-1}\left(2\right)$
• B. ${\cos }^{-1}\left(0.3\right)$
• C. ${\tan }^{-1}\left(3\right)$
• D. ${\sin }^{-1}\left(0.2\right)$

Answer 1

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


Given $v=72\text{km/h}=72\times \frac{5}{18}\text{m/s}=20\text{m/s}$
$R$ is the total reaction force.
Let the motorcyclist bend w.r.t vertical plane by an angle $\theta$, then
$R\cos \theta =mg..\left(1\right)$ and
$R\sin \theta =\frac{m{v}^2}{r}..\left(2\right)$
Dividing (2) by (1) gives $\tan \theta =\frac{v^2}{rg}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{v^2}{rg}\right)$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{{20}^2}{20\times 10}\right)$
$\Rightarrow \theta ={\tan }^{-1}\left(2\right)$