# Curiosity on conservation of energy

Imagine a ball rolling down a hill or inclined plane without slipping from a particular height.
The conservation of energy law says that the final energy remains the same as the initial energy.
But, when a ball is rolling without slipping, there must be a friction.
So, what I think is
E = E'
mgh + Ffriction s = (1/2) mv^2 + (1/2) Iω^w

But, in every book I read says that
E = E'
mgh = (1/2) mv^2 + (1/2) Iω^w

Why is there no work done by friction ?
I questioned this since if the ball is not rolling, it must be mgh + Ffriction s = (1/2)mv^2
But, when it's rolling, there is no work done by friction.. Why?

A.T.
...rolling without slipping...

Why is there no work done by friction ?
no slipping -> static friction -> displacement of force application point is zero (in the ground frame) -> no work done

• terryds
The model is assuming that the motion is exhibiting pure roll and that there is no rolling friction (caused by hysteresis). Under these idealized conditions no work is done by friction or internal forces.

Static friction is necessary to develop rolling motion (otherwise it would just slide), but does not transfer any energy.

no slipping -> static friction -> displacement of force application point is zero (in the ground frame) -> no work done
The model is assuming that the motion is exhibiting pure roll and that there is no rolling friction (caused by hysteresis). Under these idealized conditions no work is done by friction or internal forces.

Static friction is necessary to develop rolling motion (otherwise it would just slide), but does not transfer any energy.

Why the displacement of frictional force is zero ?
Is the tangential velocity at the point where the ball touches the ground zero ?
Why ? It rolls and makes some distance, right ? The tangential velocity is α r , right ? It's not static, right ?

A.T.
Why the displacement of frictional force is zero ?
If the contact point doesn't slip, then it must be have the same horizontal velocity as the surface.

Is the tangential velocity at the point where the ball touches the ground zero ?
The tangential velocity is not the total velocity of the contact point in the ground frame. But it is In the frame of the wheel's center, where static friction is indeed doing work.

• terryds
Doc Al
Mentor
I questioned this since if the ball is not rolling, it must be mgh + Ffriction s = (1/2)mv^2
Just for the record, this remains valid. A more general theorem, of which this is an example, can be written as (at least for constant forces):
$$F_{net}\Delta x_{cm}=\Delta (\frac{1}{2}m v_{cm}^2)$$
Just do not confuse this as a statement of energy conservation, which it is not. Here's something that I wrote in an older thread about this concept, which you might find helpful:
If you take a net force acting on an object (like friction) and multiply it by the displacement of the object's center of mass, you get a quantity that looks like a work term but is better called pseudowork (or "center of mass" work)--what it determines is not the real work done on the object, but the change in the KE of the center of mass of the object. This is usually called the "Work-Energy" theorem:

$$F_{net}\Delta x_{cm}=\Delta (\frac{1}{2}m v_{cm}^2)$$
Despite the name, this is really a consequence of Newton's 2nd law, not a statement of energy conservation.

"Real" work is the work that appears in the first law of thermodynamics (conservation of energy) and depends on the details of how the force is applied and the movement of the point of contact.