Question

A motorcyclist wants to drive on the vertical surface of a wooden well of radius $5\text{m}$, with a minimum speed of $5\sqrt{5}{\text{ms}}^{-1}$. The minimum value of coefficient of friction between the tyres and the wall of the well must be
$($Take, $g=10{\text{ms}}^{-2})$

• A. $0.10$
• B. $0.20$
• C. $0.30$
• D. $0.40$

Answer 1

The correct option is D $0.40$

F.B.D of motorcyclist is as shown below


For minimum value of coefficient of friction, the friction will act at its maximum value, so,

$f=\mu N$

For vertical equilibrium of motorcyclist in wooden well,

$f=mg$

$\therefore \mu N=mg..........\left(1\right)$

Using equation of circular dynamics,

$N=\frac{m{v}^2}{R}...........\left(2\right)$

From eqs. $\left(1\right)\&\left(2\right)$,

$\mu \frac{m{v}^2}{R}=mg\Rightarrow {\mu }_{min}=\frac{Rg}{v^2}$

${\mu }_{min}=\frac{5\times 10}{(5\sqrt{5}{)}^2}=0.40$

Hence, option $\left(d\right)$ is correct.