Question

A motor cyclist rides on a cylindrical wall of a velodrome (hollow metal cylinder kept with the axis vertical) of radius $r$ rotating about its axis at the rate $f\frac{rev}{sec}$. The least coefficient of friction necessary so that he does not fall down is

• A. $\frac{g}{\left(4{\pi }^2{f}^{2}r\right)}$
• B. $\frac{\left(4{\pi }^2{f}^{2}r\right)}{g}$
• C. $\frac{\left(4{\pi }^2{f}^2\right)}{gr}$
• D. $\frac{g}{\left(4{\pi }^{2}r\right)}$

Answer 1

The correct option is A $\frac{g}{\left(4{\pi }^2{f}^{2}r\right)}$

Normal reaction from wall = centrifugal force

$N=mr{\omega }^2=mr{\left(2\pi f\right)}^2$

Friction force, $F=\mu N=\mu \left(mr{\omega }^2\right)=\mu mr{\left(2\pi f\right)}^2$ (vertically upward)

Weight of motor cycle, $W=mg$ (vertically downward)

For non-skidding condition

$F\ge \text{W}$

$\mu mr{\left(2\pi f\right)}^2\ge mg$

$\mu \ge \frac{g}{4{\pi }^2{f}^{2}r}$

Hence, coefficient of friction should be greater than $\frac{g}{4{\pi }^2{f}^{2}r}$