Q. A cyclist riding the bicycle at a speed of $14\sqrt{3}\ ms^{- 1}$ takes a turn around a circular road of radius $20\sqrt{3}\ m$ without skidding. Given $g = 9.8\ ms^{- 2}$, what is his inclination to the vertical
A) $30^{o}$
B) $90^{o}$
C) $45^{o}$
D) $60^{o}$
Explanation
(d) $\theta = \tan^{- 1}\left( \frac{v^{2}}{rg} \right) = \tan^{- 1}{\left\lbrack \frac{\left( 14\sqrt{3} \right)^{2}}{20\sqrt{3} \times 9.8} \right\rbrack =}\tan^{- 1}{\lbrack\sqrt{3}\rbrack} = 60^{o}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.