Question

A cyclist riding at a speed of $9.8\, m/s$ takes a turn around a circular road of radius $19.6\, m$. What is his inclination to the vertical?

• A. $ 30^{\circ}$
• B. $45^{\circ}$
• C. $10.5^{\circ}$
• D. $26.5^{\circ}$

Answer 1

Answer: (D)

Angle of incline $\tan \theta=\frac{v^{2}}{r g}$
where $v$ is velocity, $r$ the radius of circular path and $g$ the acceleration due to gravity.
Given, $v=9.8\, m/s,\, r=19.6\, m,\, g=9.8\, m/s ^{2}$
$\therefore \tan \theta=\frac{9.8 \times 9.8}{19.6 \times 9.8}=\frac{1}{2}$
$\Rightarrow \theta=26.5^{\circ}$