Question

A cylinder A rolls without slipping on a plank B. The velocities of the centre of the cylinder and that of the plank are \[4\,m/s\] and \[2\,m/s\] respectively in the same direction with respect to the ground. Find the angular velocity ( in rad /sec) of the cylinder if its radius is 1m.

Answer 1

Hint:Frictional force is a special type of force which is usually against the direction of motion as opposed. In the above problem, the frictional force is between the cylinder A horizontal plank B. For rolling without slipping, the velocity of the cylinder at the point of contact with the plank will be zero.

Formula Used:
Frictional force,
\[f = \mu ma\]
where \[\mu \]= coefficient of friction, m = mass of the solid and a = acceleration.
\[v = \omega R\]
where v is the linear velocity, R is the radius of the solid and \[\omega \] is the angular velocity.

Complete step by step solution:
Given: Given that cylinder rolls on the ground without slipping. Let the velocity of the centre of cylinder be \[{v_{CM}}\]= \[4m/s\] and velocity of plank be \[{v_B}{ = _{}}2m/s\]. Let the radius of the cylinder be \[R = 1cm\]. Let the velocity of the cylinder be \[{v_C}\]. Let the angular velocity of the centre of the cylinder be \[\omega \] (in rad/sec) which is to be calculated.

According to the question, the velocity of the cylinder at the point of contact with the plank is zero.
So, \[{v_C} = {v_{CM}} - \omega R\]---(1)
And, \[{v_C} - {v_B} = 0\]----(2)
From equations (1) and (2), we get,
\[{v_{CM}} - \omega R - {v_B} = 0\]
\[\Rightarrow {v_{CM}} - {v_B} = \omega R\]
\[\Rightarrow \omega = \dfrac{{{v_{CM}} - {v_B}}}{R}\]
Applying values as per the question,
\[\therefore \omega = 2\,rad/s\]

Hence the angular velocity of the cylinder is 2 rad/sec.

Note: Since the cylinder rolls without slipping on the plank, the velocity of the cylinder at the point of contact with the plank is zero. This means that there is no translational motion of the cylinder on the plank, only purely rotational and hence pure rolling. For pure rolling, the rotational acceleration and the acceleration of the centre of mass should have an equivalence of the form \[R\alpha = {a_{CM}}\]. In order that the motion occurs as desired, the two opposing forces should balance out each other.