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 $k={{k}_{0}}\left( 1-\dfrac{r}{R} \right)$, 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

Hint: Due to the presence of friction, we have centripetal force which acts to the centre of the plane and because of this force the cyclist does not slide. Find the centripetal force and the frictional force on the cyclist and then equate them to find the maximum velocity.

Complete step-by-step answer:
When a cyclist moves through a circular path, the cyclist does not slide on the rod because of the centripetal force given by the frictional force.
So, we can equate the frictional force to the centripetal force to find the maximum velocity of the cyclist.
Now, the centripetal force is given by, ${{f}_{c}}=\dfrac{m{{v}^{2}}}{r}$
The frictional force is given as, ${{f}_{f}}=kN=kmg$
Where, m is the mass of the system, v is the velocity of the cyclist, r is the distance from the centre of the plane, k is the frictional coefficient and g is the acceleration due to gravity.
The frictional coefficient is given as, $k={{k}_{0}}\left( 1-\dfrac{r}{R} \right)$
Equating the two forces we get,
$\begin{align}
  & \dfrac{m{{v}^{2}}}{r}=kmg \\
 & \dfrac{{{v}^{2}}}{r}=kg \\
\end{align}$
Putting the value of the coefficient of friction,
$\begin{align}
  & \dfrac{{{v}^{2}}}{r}={{k}_{0}}\left( 1-\dfrac{r}{R} \right)g \\
 & {{v}^{2}}={{k}_{0}}\left( r-\dfrac{{{r}^{2}}}{R} \right)g \\
\end{align}$
Now, the velocity of the cyclist will be maximum if the space derivative of the velocity is zero.
$\begin{align}
  & \dfrac{dv}{dr}=\dfrac{d{{v}^{2}}}{dr}=0 \\
 & {{k}_{0}}g\left( 1-\dfrac{2r}{R} \right)=0 \\
 & r=\dfrac{R}{2} \\
\end{align}$
Now,
$v=\sqrt{{{k}_{0}}\left( r-\dfrac{{{r}^{2}}}{R} \right)g}$
So, the maximum velocity will be,
${{v}_{\max }}=\dfrac{1}{2}\sqrt{{{k}_{0}}gR}$

Note: Whenever we move in a circular path, we feel a force towards the centre of the circular path. This force is called the centripetal force. In this case the centripetal force is given by the frictional force between the wheel of the cycle of and the road.