Rolling Motion made Easy

Q. Many of us we think that Rolling Motion is tough?
The answer to the above question is "BIG NO" things are only tough till you understand them completely.

So, I am here to guide you on the above topic. Leave comments on any Doubts faced by you recently.

When we talk about pure Rotational Motion we imagine a fixed axis about which the all the particles of the Rigid Body are moving in a Circular Motion. Hence if the body moves with angular Velocity 'omega'
(w) than every particle on the periphery will have linear velocity as v =wr ( where r is the radius of the particle from its axis ).

In the above Diagram we see 3 figures:
1st figure shows when the body is in pure translational motion only, every particle of the rigid body will have same velocity ( same Direction ) as visible.
2nd figure shows when the body is in pure rotational motion as discussed above particle will have same speed but different velocity about fixed axis O.
3rd figure shows when both are combined i.e ( rotational + translational ) than body undergoes the Pure Rolling Motion "provided there is sufficient friction".

What happens on an Inclined Plane without friction

Suppose that I have some frictionless block/sphere on an inclined plane. The block/sphere can only accelerate in the direction along the plane. This means that if I put the x-axis in this direction of plane, the net forces in the x-direction will be mass*acceleration and the net forces in the y-direction will be zero/balanced. The only force acting in the x-direction is a component of the gravitational force. This means that the forces in the x-direction will be:
Now the question is asked many times when we release solid cylinder, solid sphere, hollow sphere, ring from the top of Inclined plane which of them will reach first on ground ??

Ans to this is all of them are rolling on frictionless surface hence will undergo pure translational motion and hence will be pulled by "gsinQ". So all will reach the same time.

**Also we can apply work-energy theorem to above problem. 1/2mv^2 = mgh for ALL

When there is Sufficient Friction on the Inclined Plane  

When there is sufficient friction on the Inclined plane than it means it will perform Pure Rolling in that case there are two Kinetic Energies 

For Example let's drop a Solid Cylinder from the top of Inclined plane

 As we see in Work Energy theorem we have 2 types of KE

1. Translational KE = 1/2mv^2

2. Rotational KE = 1/2Iw^2

3. I =1/2mr^2

4. PE = mgh