Question

A cyclist moving with a speed of $4.9\, m/s$ on a level road can take a sharp circular turn of radius $4\,m$, then coefficient of friction between the cycle tyres and road is:

• A. 0.71
• B. 0.61
• C. 0.31
• D. 0.81

Answer 1

Answer: (B)

Centripetal force is provided by the frictional force between the tyres and the road. A body performing a circular motion is acted upon by a force which is always directed towards the center of the circle, This force is called Centripetal centripetal force. If m is mass, $v$ is velocity, $r$ is radius then
$F=\frac{m v^{2}}{r}$ ...(i)
Also when the cyclist takes a turn on the road, the centripetal force is provided by the frictional force between the tyres and road.
$\therefore F=\mu m g$ ...(ii)
Equating Eqs. (i) and (ii), we get
$\mu=\frac{v^{2}}{r g}$
Putting the numerical values from the question, we have
$\therefore \mu=\frac{(4.9)^{2}}{4 \times 9.8}=0.61$