Question

For a bicycle of mass $m$ moving with a speed $v$ in a circular motion of radius $r$ on a road with friction coefficient $\mu$, which of the following relations is correct.

• A. $\frac{m{v}^2}{r}=mg$
• B. $\frac{m{v}^2}{r}\le \mu mg$
• C. $\frac{m{v}^2}{r}\ge \mu mg$
• D. $\frac{v^2}{r}=mg$

Answer 1

The correct option is B $\frac{m{v}^2}{r}\le \mu mg$

When a bicycle moves in a circular path, the centripetal force is provided by the friction force.

Now we know that friction force acting on a body of mass $m$ can range from $zero$ to $\mu mg$.

The centripetal force required for the motion is $\frac{m{v}^2}{r}$. And for the bicycle to stay in circular motion, the force of friction should be equal to the centripetal force.

Hence $\frac{m{v}^2}{r}\le \mu mg$