Question

A motorcyclist moving with a velocity of $72\,kmh{r^{ - 1}}$ on a flat road takes a turn on the road at a point where the radius of curvature of the road is $20\,m$. The acceleration due to gravity is $10\,m{s^{ - 2}}$ In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than
A. $\theta = {\tan ^{ - 1}}(6)$
B. $\theta = {\tan ^{ - 1}}(2)$
C. $\theta = {\tan ^{ - 1}}(25.92)$
D. $\theta = {\tan ^{ - 1}}(4)$

Answer 1

Hint: Using the basic relationship between banking angle, velocity, and radius of the circular path for a body to turn safely, we will solve for the smallest angle at which the bicycle should turn to prevent skipping before deciding on the best course of action.

Formula used:
If $\theta $ is the angle required for a body to turn safely in the circular path of radius r and if v is the velocity of the body then the relation between banking angle, radius and velocity is,
$\tan \theta \leqslant \dfrac{{{v^2}}}{{rg}}$
where g is the acceleration due to gravity and tan is the tangent of the angle $\theta $

Complete step by step solution:
According to the question, we have given that the velocity of the cyclist is $v = 72\,kmh{r^{ - 1}} = 20\,m{s^{ - 1}}$ and the radius of the curvature of the road is $r = 20\,m$ and \[g = 10\,m{s^{ - 2}}\].

So, using the relation $\tan \theta \leqslant \dfrac{{{v^2}}}{{rg}}$ and solving for $\theta $ we get,
$\tan \theta \leqslant \dfrac{{{{(20)}^2}}}{{(20)(10)}} \\
\Rightarrow \tan \theta \leqslant 2 \\
\therefore \theta \leqslant {\tan ^{ - 1}}(2)$
So, the maximum angle at which a cyclist can turn without skidding is $\theta = {\tan ^{ - 1}}(2)$.

Hence, the correct answer is option B.

Note: It should be remembered that, while riding on a curved road, the cyclist bends slightly inwards to provide the necessary centripetal force, which is focussed towards the centre and aids in turning around a bend.