Question

A cyclist rides along the circumference of a circular horizontal plane of radius R, the friction coefficient being dependent only on distance r from the centre O of the plane as k = k0(1 - r/R), where k0 is a constant. Find the radius of the circle with the centre at the point along which the cyclist can ride with the maximum velocity. What is this velocity?

Answer 1

Solution

As given in question

The cyclist moves in circular path and centripetal force is given by frictional force so,

$f=m\omega$

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

$k_0(1-\frac{r}{R}) g=\frac{v^2}{r}$

$v=\sqrt{k_0(r-\frac{r^2}{R}) g}$ . ............. eq. 1

For maximum velocity

$\frac{dv}{dr}=0$

So we get

$r=\frac{R}{2}$

So velocity becomes as

$v=\frac{\sqrt{k_0gR}}{2}$