Question

A motorcycle moving with a uniform speed of $72{\text{kmh}}^{-1}$ 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. $\theta ={\tan }^{-1}\left(2\right)$
• B. $\theta ={\tan }^{-1}\left(6\right)$
• C. $\theta ={\tan }^{-1}\left(4\right)$
• D. $\theta ={\tan }^{-1}\left(25.92\right)$

Answer 1

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

Applying the condition that bending from vertical will provide the component of normal reaction, so that it is able to sustain the required centripetal acceleration at bend.

By the free body diagram and writing the force equation we get,

$N\cos \theta =mg...\left(i\right)$
$N\sin \theta =\frac{m{v}^2}{r}...\left(ii\right)$
Now by dividing both the above equations, we get,

$\theta ={\tan }^{-1}\left(\frac{v^2}{rg}\right)$

where, $r=20m$

$v=72{\text{kmh}}^{-1}$

$=72\times \frac{5}{18}=20{\text{ms}}^{-1}$
$\therefore \theta ={\tan }^{-1}\left(\frac{20\times 20}{20\times 10}\right)$
or $\theta ={\tan }^{-1}\left(2\right)$