Question

A bicycle travelling at speed v requires frictional force F to turn on an unbanked road. How much frictional force would it require to make the same turn at a speed of 2v?a. F/4b. F/8c. 4Fd. F/2e. 2Ff. 8Fg. F

Answer 1

Given:

The required frictional force to turn on a road is,

$F$

while the bicycle is moving at a speed of

$v$

To find:

The frictional force required to make the same turn at a speed of 2v

Explanation:

The required frictional force should be equal to the centrifugal force. So, we can write,

$F=\frac{mv^2}{r}$

Now, for the speed 2v, the frictional force is,

$\begin{gathered} F^{\prime}=\frac{m(2v)^2}{r} \\ =\frac{m\times4v^2}{r} \\ =\frac{4mv^2}{r} \\ =4F \end{gathered}$

Hence, the required force is 4F.