Role of friction during pure rolling

Question

I am confused about the role of friction during pure rolling. I tied to analyse these two cases:

Case 1: An object is rotated rapidly and put on a frictionless (smooth) table, will it execute pure rolling?

Case 2: The same object is rotated rapidly, kept on the smooth table and slightly pushed, will it now execute pure rolling?

For case 1, because there's no friction on the table, I feel that the object will keep sliding on the table but what about the rotation? Will that be hindered or not?

For second case, I feel that whatever be the magnitude of the push force, the same thing will happen i.e. the object will slide.

I am very unsure about my analysis. Also, there's this article in my book which has puzzled me more:

In accelerated pure rolling if $a = R\alpha$, then there's no need of friction and force of friction $f=0$.

Different people on different sites like Quora, give different contradicting answers.

Answer 1

There is no rolling in either case. A slight push will lead to a slight velocity for the object. You can sort of simulate the effect of rolling by accelerating it in such a way that $a=R\alpha$ always, and then it will look like it's rolling. Since in that case even with a non-zero coefficient of friction, the force of friction will be zero.

Answer 2

I assume that by pure rolling you mean that there is no slipping between the object and the surface over which it is rolling?

If this is the case then there are three equivalent relationships between the horizontal distance moved by the object $x$ and its angle of rotation $\theta$ during the same time which are satisfied where $r$ is the radius of the object.

$x=r\,\theta \;\;\; \dot x=r\,\dot \theta \;\;\; \ddot x=r\,\ddot \theta$

A way to decide whether of not a frictional force needs to act is to consider the motion of the rotating abject as the sum of the translational motion of its centre of mass and the rotational motion about its centre of mass.

Assume that the system is the rolling object.

On a horizontal surface if $\dot x = r \dot \omega$ then to maintain a constant speed horizontal motion no external horizontal force is acting on the object and so there is no frictional force.

On this horizontal surface if you want the centre of mass of the object undergo a horizontal acceleration there must be an horizontal external force acting on the object and suppose that the horizontal external force acts through the centre of mass of the object.
That horizontal force external applies no torque about the centre of mass of the object so without any other force acting the translational speed of the object $\dot x$ would increase and yet the rotational speed $\dot \theta$ would stay constant.
This would result in relative motion between the object and the surface at the point of contact and the pure rolling condition would no longer be satisfied.
So there must also be a horizontal frictional force acting in such a direction as to try and make sure that the condition $\ddot x = r \ddot \theta$ is satisfied.
If the horizontal external force causing the centre of mas of the object to accelerate is increasing the translational speed of the object then to maintain the pure rolling condition the horizontal frictional force must act in such a way as to try and increase the rotational speed of the object (and reduce the increase in translational speed of the centre of mass of the object) and so the frictional force will act in the opposite direction to the horizontal external force.

Similar arguments apply if the object starts of slipping relative to the surface at the point of contact.

To get to the pure rolling condition $\dot x = r \dot \theta$ the translational speed $\dot x$ must increase/decrease whilst at the same time the rotational speed $\dot \theta$ must decease/increase.

With no horizontal forces acting both the translational speed and the rotational speed cannot change and so with no horizontal frictional force acting the pure rolling condition cannot be reached.

If you start with $\dot x > r\dot \omega$ then the object is not rotating fast enough or you could say that the translational speed is too high for the pure rolling condition to be satisfied.
In this case a horizontal frictional force at the point of contact must act in the opposite direction to the direction of translational motion of the object.
That horizontal frictional force will reduce the translational speed of the centre of mass of the object whilst at the same time will apply a torque about the centre of mass of the object which will increase its rotational speed.

$\dot x \downarrow \ne r\dot \theta \uparrow \;\;\;\; \Rightarrow\;\;\;\; \dot x = r\dot \theta \;\;\;\;$ which is the pure rolling condition.

Answer 3

Let's try!!

Case 1 When rotated, the body will acquire angular velocity ω. As soon as it touches the ground, the lowermost point of contact will have velocity Rω. Since the surface is frictionless, there will be no torque to produce angular deceleration and no force to produce linear acceleration so that after a time the relation V=Rω is satisfied. So the sphere will not exhibit pure rolling.

Case 2 Here, as you are gently pushing the sphere the lowermost point of contact will acquire both linear and angular velocity. If by chance the linear velocity(V) comes out to be equal to Rω and that too both of them are opposite in direction only then the sphere will exhibit pure rolling. If friction had been present it would try to make both the velocities equal. So in your case there will be no evidence of pure rolling.

In case you have any problem understanding my answer please leave a comment.