question_answer

If a cyclist moving with a speed of 4,9 m/s on a level road can take a sharp circular turn of radius 4m. Then coefficient of friction between the cycle tyre and the road is:

A) 0.71                                       

B) 0.61

C) 0.51                                       

D) 0.81

Correct Answer: B

Solution :

Key Idea: Body performing circular motion is acted upon by a force directed towards its centre provided by force of friction in our case. When the cyclist moves with a uniform speed v on a circular path of radius r, it has centripetal acceleration\[\left( \frac{{{v}^{2}}}{r} \right),\] hence cyclist is acted upon by a centripetal force directed towards its centre. \[F=\frac{m{{v}^{2}}}{r}\]where m is mass. Also, from Newton's laws \[F=\mu mg\]where \[\mu \]is coefficient of friction. \[\therefore \]      \[\frac{m{{v}^{2}}}{r}=\mu mg\]\[\Rightarrow \]   \[\mu =\frac{{{v}^{2}}}{rg}\] Putting the numerical values from the question, we have  \[\mu =\frac{{{(4.9)}^{2}}}{4\times 9.8}=0.61\]