Question

A cylindrical drum, pushed along by a board rolls forward. There is no slipping at any contact. Find the distance moved by the man who is pushing the board, when the axis of the cylinder covers a distance L.

Answer 1

Rolling motion is that basic blend of rotational and translational movement that we see all over the place, each day. Consider the various circumstances of wheels proceeding onward a vehicle along a thruway, or wheels on a plane arrival on a runway, or wheels on a mechanical pioneer on another planet. Understanding the powers and forces engaged with rolling motion is a vital factor in various sorts of circumstances.
The velocity of the point of contact between the cylinder and the board, i.e., the point instantaneously at the top of the cylinder is equal to twice that of the axis of the cylinder. Therefore, the man moves a distance of 2L.

Complete step by step solution

Let ${{v}_{0}}$ be the linear speed of the axis of the cylinder and $\omega$ be its angular speed about the axis. As it does not slip on the ground hence $\omega =\dfrac{{{v}_{0}}}{R}$ , where R is the radius of the cylinder.
Speed of the topmost point is
$\begin{aligned} & v={{v}_{0}}+\omega R \\\ & v=2{{v}_{0}} \\\ \end{aligned}$
Time taken by the axis to move a distance L is equal to
$t=\dfrac{L}{{{v}_{0}}}$
In the same interval of the time distance moved by the topmost point is:
$\begin{aligned} & s=2{{v}_{0}}\times \dfrac{L}{{{v}_{0}}} \\\ & s=2L \\\ \end{aligned}$
As there is no slipping between any points of contact.
Hence, distance moved by the man is 2L.

Note
Rolling without slipping is a blend of translation and rotation where the purpose of contact is immediately very still. At the point when an object encounters unadulterated translational motion, the entirety of its focus moves with the very speed as the center of mass; that is in a similar direction and with a similar speed.