# I Why is there no virtual work done by a rolling friction?

1. Mar 1, 2016

### Happiness

Why is there no virtual work done by a rolling friction?

In pure rolling, is the virtual displacement parallel or perpendicular to the surface? I believe it can't be perpendicular because the object is not allowed to lose contact with the surface. But if it's parallel, then there must be a non-zero virtual work done since the rolling friction is also parallel to the surface. The book says the point of contact is momentarily at rest, does that mean that the virtual displacement must be zero (but the actual infinitesimal displacement is parallel to the surface)?

Last edited: Mar 2, 2016
2. Mar 2, 2016

### drvrm

The principle of virtual work states that in equilibrium the virtual work of the forces applied to a system is zero. Newton's laws state that at equilibrium the applied forces are equal and opposite to the reaction, or constraint forces. This means the virtual work of the constraint forces must be zero as well.

the virtual displacements are consistent with the forces of constraint on the system and in that case the displacements will be perpendicular to the force of constraint. In the case of rolling the constraining forces are the component of the weight W in the direction of perpendicular to the inclined plane- actually the friction force is generated by this constraint otherwise it will not be there.
moreover the motion of the body in contact with the plane will be zero-momentarily as the plane is at rest . the real turning moment is being provided by the force in the direction parallel to the plane.

3. Mar 2, 2016

### Happiness

You must have assumed the virtual work done by the applied forces is zero to conclude that the virtual work done by the constraint forces must be zero as well. This is not true in the case of a body experiencing sliding friction and moving at a constant velocity, where the virtual work done by the applied forces is $+W$ and the virtual work done by the constraint forces is $-W$. This does not violate the principle of virtual work since the total virtual work is still zero.

4. Mar 2, 2016

### drvrm

In classical mechanics the concept of a virtual displacement, related to the concept of virtual work, is meaningful only when discussing a physical system subject to constraints on its motion.

A special case of an infinitesimal displacement (usually notated dr), a virtual displacement (denoted δr) refers to an infinitesimal change in the position coordinates of a system such that the constraints remain satisfied.a book kept on the table has a constraint that it is in contact with table- a displacement consistent with the constraint will be normally be perpendicular to the force of constraint.
since 'virtual displacement' is consistent with forces also the principal of virtual work holds true;

For example, if a bead is constrained to move on a hoop, its position may be represented by the position coordinate θ, which gives the angle at which the bead is situated. Say that the bead is at the top. Moving the bead straight upwards from its height z to a height z + dz would represent one possible displacement but would violate the constraint. The only possible virtual displacement would be a displacement from the bead's position, θ to a new position θ + δθ (where δθ could be (+ve) or negative).
fo more details see <https://en.wikipedia.org/wiki/Virtual_displacement> [Broken]

Last edited by a moderator: May 7, 2017
5. Mar 2, 2016

### Happiness

What you wrote here is true but it does not answer the question! A displacement consistent with the constraint is usually perpendicular to the force of constraint but it is not always perpendicular.

Last edited by a moderator: May 7, 2017
6. Mar 2, 2016

### drvrm

I think you are making a conceptual error- The Virtual displacements are not to be seen as equivalent real Physical displacements
- so its a geometric change of arrangement during a time interval which is zero.thats why it is called 'virtual'.
The principal of virtual work or D'Alembert's principle is invoked to get to a form of equations of motion such that Knowledge of nature of forces acting on the system may not be needed/required a priory.

7. Mar 2, 2016

### Happiness

As posted with the original question:

The book clearly says there are systems for which the net virtual work of the forces of constraint is non-zero and there are those for which it is zero, and rolling friction belongs to the latter class.

8. Mar 2, 2016

### drvrm

how come i read differently the same quote shown above - it does say that work done in virtual displacement is zero even when in real /actual displacements it may not vanish...also it says that rolling friction obeys the principle of virtual work.

9. Mar 2, 2016

### Happiness

Read it again? "The net virtual work of force of constraint is zero." "this condition holds true for rigid bodies .... the virtual work vanishes." "This is no longer true if sliding friction forces are present"

Basically, your argument is that virtual work always vanishes, so the virtual work for rolling friction must vanish too. But virtual work does not always vanish. The virtual work for rolling friction indeed vanishes but your argument is incorrect.

10. Mar 2, 2016

### drvrm

then read the line that these will not be part of this formulation <We must exclude such systems from our formulation>

11. Mar 2, 2016

### Happiness

Ermm.. I know we must exclude them. This wasn't my question!

12. Mar 2, 2016