Question

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

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

Answer 1

Answer: (D)

${60}^o$

Given,

$v=14\sqrt{3}m/s$

$r=20\sqrt{3}m$

$g=9.8m/s^2$

The inclination to the vertical is given by,

$tan\theta =\frac{v^2}{rg}=\frac{(14\sqrt{3}{)}^2}{20\sqrt{3}\times 9.8}$

$tan\theta =\sqrt{3}$

$\theta =ta{n}^{-1}\left(\sqrt{3}\right)$

$\theta ={60}^0$

The correct option is D.