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 $k=k_0(1-\frac{r}{R})$, where $k_0$ 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

According to the question, the cyclist moves along the circular path and the centripetal force is provided by the frictional force. Thus from the equation $F_n=m{w}_n$
$fr=\frac{m{v}^2}{r}$ or $kmg=\frac{m{v}^2}{r}$
or $k_0(1-\frac{r}{R})g=\frac{v^2}{r}$ or $v^2=k_0(r-\frac{r^2}{R})g$ (1)
For $v_{max}$, we should have $\frac{d(r-\frac{r^2}{R})}{dr}=0$
or, $1-\frac{2r}{R}=0$, so $r=\frac{R}{2}$
Hence $v_{max}=\frac{1}{2}\sqrt{k_{0}gR}$