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 $g=9.8m{s}^{-2}$ What is his inclination to the vertical ?

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

Answer 1

Answer: (D)

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