Question

The coefficient of friction between the tyres and the road is $0.1$ . The maximum speed with which a cyclist can take a circular turn of radius $3\,m$ without skidding is? (Take $g = 10\,m{s^{ - 2}}$).
A) $\sqrt {15} \,m{s^{ - 1}}$
B) $\sqrt 3 \,m{s^{ - 1}}$
C) $\sqrt {30} \,m{s^{ - 1}}$
D) $\sqrt {10} \,m{s^{ - 1}}$

Answer 1

The cyclist can take a turn of the circular path, so that the cyclist will get some centripetal force. And the tyre and the road are having some frictional force, by equating these two forces, the velocity or the maximum speed of the cyclist will be determined.

Useful formula:
The frictional force between the road and tyre is given by,
$F = \mu mg$
Where, $F$ is the frictional force between the tyre and the road, $\mu$ is the coefficient of the friction, $m$ is the mass of the both cycle and cyclist and $g$ is the acceleration due to gravitational force.
The centripetal force of the cycle is given by,
$F = \dfrac{{m{v^2}}}{r}$
Where, $F$ is the centripetal force, $m$ is the mass of the both cycle and cyclist, $v$ is the velocity of the cycle or cyclist and $r$ is the radius of the circular path.

Complete step by step solution:
Given that,
The coefficient of friction between the tyres and the road is $0.1$.
The radius of the circular path is, $r = 3\,m$
The acceleration due to gravitational force is, $g = 10\,m{s^{ - 2}}$
Now,
The frictional force between the road and tyre is given by,
$F = \mu mg\,...............\left( 1 \right)$
The centripetal force of the cycle is given by,
$F = \dfrac{{m{v^2}}}{r}\,...................\left( 2 \right)$
By equating the equation (1) and equation (2), then
$\mu mg = \dfrac{{m{v^2}}}{r}$
By cancelling the mass on both sides, then
$\mu g = \dfrac{{{v^2}}}{r}$
By keeping the velocity in one side and the other terms in the other side, then the above equation is written as,
${v^2} = \mu gr$
By taking square root on both sides, then
$v = \sqrt {\mu gr}$
Now substituting the coefficient of friction, acceleration due to gravity and radius in the above equation, then
$v = \sqrt {0.1 \times 10 \times 3}$
On multiplying the above equation, then
$v = \sqrt 3 \,m{s^{ - 1}}$
Thus, the above equation shows the maximum speed.

Hence, the option (B) is the correct answer.

Note: Here there are two forces acting, one is centripetal force because the cycle travels in the circular path and the other force is the frictional force. According to the law of conservation of energy, when the two forces acting on the system are equal.