Concepts of Physics

IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

Direction of the Frictional Forces on the Bicycle Wheels during Paddling

An interesting problem was asked in IIT JEE 1990. During paddling of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts,

  1. in the backward direction on the front wheel and in the forward direction on the rear wheel.
  2. in the forward direction on the front wheel and in the backward direction on the rear wheel.
  3. in the backward direction on both the front and the rear wheels.
  4. in the forward direction on both the front and the rear wheels.

Disc This problem is explored in detail for conceptual clarity. Let a non-rotating ($\omega_0=0$) disc of radius $r$ having initial velocity $u$ be gently placed on a frictional surface (see figure). Initial velocity of the contact point P, $\vec{u}_{P}=\vec{u}_C+\vec{\omega}_0\times\vec{r}_\text{CP}=\vec{u}_C$ is same as that of the centre C. Thus, P moves towards right relative to the surface. To oppose this, frictional force at P acts towards left (see figure). The frictional force retards the velocity of C i.e., $v < u$. The torque about C due to frictional force is $\tau=fr$ (clockwise). This torque gives clockwise angular acceleration ($\tau=I\alpha$) and disc starts rotating clockwise. If the coefficient of friction is sufficiently large then retardation and angular acceleration continue till $v=\omega r$. At this instant, velocity of P relative to the surface becomes zero making $f=0$. After it the disc continues to roll without slipping.

Disc Now, let disc with non-zero angular velocity $\omega_0$ and zero linear velocity ($u=0$) be gently placed on a frictional surface. Initial velocity of the contact point P, $\vec{u}_{P}=\vec{u}_C+\vec{\omega}_0\times\vec{r}_\text{CP}=\omega_0 r$, leftward relative to the surface. To oppose this, frictional force at P acts towards right (see figure). The frictional force increases the velocity of C. The torque about C due to frictional force is anti-clockwise. This torque gives anti-clockwise angular acceleration i.e., $\omega<\omega_0$. If coefficient of friction is sufficiently large then acceleration and angular retardation continue till $v=\omega r$. At this instant, velocity of P relative to the surface becomes zero making $f=0$. After it the disc continues to roll without slipping.

Bicycle Now, consider the bicycle. The front wheel is connected to rest of the bicycle by a rod passing through its centre (axle). The torque on the wheel about its centre by the force coming from rest of the bicycle is zero. Thus, paddling can give linear velocity to the front wheel but cannot rotate it. The situation is similar to the first case discussed above and frictional force acts in backward direction. The situation of rear wheel is different. The rear wheel is connected to rest of the bicycle by a rod passing through its centre and a chain connected to the paddles (see figure). Pressing the paddle increases tension in upper portion of chain. This tension gives rise to clockwise torque and wheel starts rotating in clockwise direction. Thus, situation of rear wheel is similar to the second case discussed above and frictional force acts in forward direction.