Question

A cyclist riding the bicycle at a speed of $14\sqrt{3}\text{m/s}$ takes a turn around a circular road of radius $20\sqrt{3}\text{m}$ without skidding. What is his inclination to the vertical?
[Take $g=9.8{\text{m/s}}^2$]

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. $0^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D ${60}^{\circ }$


Let the cyclist bend w.r.t the vertical plane by an angle $\theta$, then
To avoid skidding, $\frac{v^2}{rg}\ge \tan \theta$
For the given value of $v$ and $r$,
$\theta ={\tan }^{-1}\left(\frac{v^2}{rg}\right)$
$={\tan }^{-1}\left(\frac{(14\sqrt{3}{)}^2}{20\sqrt{3}\times 9.8}\right)$
$={\tan }^{-1}\left(\sqrt{3}\right)$
$={60}^{\circ }$