Question

A cyclist riding the bicycle at a speed of $14\sqrt{3} \, m \, s^{- 1}$ takes a turn around a circular road of radius $20\sqrt{3} \, m$ without skidding. Given $g=9.8 \, m \, s^{- 2}$ What is his inclination to the vertical ?

• A. $30^\circ $
• B. $90^\circ $
• C. $45^{o}$
• D. $60^\circ $

Answer 1

Answer: (D)

$\theta=\tan ^{-1}\left(\frac{v^{2}}{r g}\right)=\tan ^{-1}\left[\frac{(14 \sqrt{3})^{2}}{20 \sqrt{3} \times 9.8}\right]=\tan ^{-1}[\sqrt{3}]=60^{\circ}$