# Virtual work and kinetic energy

1. Apr 8, 2012

### dEdt

Suppose we have a system with N degrees of freedom, and hence N generalized coordinates ${q_i}, \ i=1...N$. A virtual displacement is defined as an infinitesimal change in a generalized coordinate without producing a change in the generalized velocities. Virtual work is defined as the work done by the acting forces as a result of any virtual displacement.

Suppose the total virtual work is $\delta W$ and the virtual displacements are $\delta q_i, \ i=1...N$. I would expect then that
$$\delta W = \delta T = \sum_i \frac{\partial T}{\partial q_i}\delta q_i$$
where $\delta T$ is the infinitesimal change in kinetic energy produced by the virtual work.

However, this is wrong. It's obviously incorrect in the cases that T doesn't depend on $q_i$, but only on $\dot{q_i}$, because then the right hand side of the above equation would be zero. So, what gives?

2. Apr 9, 2012

### Hassan2

But $\dot{q_i}$ and $q_i$ are dependent through variable t.

3. Apr 9, 2012

### dEdt

They're treated as independent variables in analytical mechanics.

4. Apr 9, 2012

### Hassan2

Thanks,

It's a confusion thing for me. I remember I had a discussion about this with my lecturer and in the end, I confessed that I have not understood " partial derivatives" properly. Now the misunderstanding is back again!

Ok. I think you should correct the equations as follows;

That is:

$T=T(q_{1},q_{2},...,q_{n},\dot{q}_{1},\dot{q}_{2},...,\dot{q}_{n})$

so

$\delta T = \sum\frac{\partial T}{\partial q} \delta q +\sum\frac{\partial T}{\partial \dot{q}} \delta \dot{q}$

5. Apr 9, 2012

### dEdt

Thing is, $\delta\dot{q_i}=0$ for a virtual displacement :/. That's why I'm confused: if T doesn't depend on $q_i$, then $\delta T$ is automatically zero.

Last edited: Apr 9, 2012
6. Apr 9, 2012

### Hassan2

That means the work done on the system goes to potential energy then.

7. Apr 9, 2012

I'm defining $\delta W[\itex] to be the total work done on the system, including the work done by any conservative forces. 8. Apr 9, 2012 ### AlephZero Yes, I think your basic mistake is assuming that the virtual work all goes into kinetic energy. For example the principle of virtual work also applies to statics problems where the KE is zero by definition. In that case the virtual work is balanced to the change in internal energy of the system (e.g internal force or stress) and the work done by external forces against the virtual displacements (including what you might describe as "gravitational potential energy", etc). 9. Apr 9, 2012 ### dEdt As above, I'm defining [itex]\delta W$ to include work done by conservative forces.

For clarity, I'll give a example.

Suppose we have a particle falling near the surface of the Earth. The three generalized coordinates will be the coordinates of the particle's position, x, y, z, in a coordinate system such that x is parallel to the surface, y is perpendicular, etc. The only force acting on the particle is Earth's gravity, $\vec{F}=m\vec{g}=-mg\hat{y}$. So, the virtual work done by a virtual displacement is $\delta W = -mg\delta y$.

Now, $T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)$. There is no dependence on the generalized coordinates, so $\delta T=\frac{\partial T}{\partial x}\delta x+\frac{\partial T}{\partial y}\delta y+\frac{\partial T}{\partial z}\delta z=0$. If $\delta W=\delta T$, it's easy to see that we have a contradiction.

10. Apr 9, 2012

### Studiot

11. Apr 9, 2012

### dEdt

12. Apr 9, 2012

### cepheid

Staff Emeritus
I think he/she is saying that the work in the case of a falling particle clearly cannot be virtual, because $\delta \dot{y} \neq 0$ in this case.

But then what do I know...I remember very little from Lagrangian mechanics.

13. Apr 9, 2012

### Studiot

Yes exactly. A falling particle is undergoing real y displacements.

A virtual displacement of such a particle would be δx or δz.

14. Apr 9, 2012

### dEdt

I don't think I understand what your claim is.

By definition, a virtual displacement $\delta q$ is never followed by a change in generalized velocity $\delta\dot{q}$. When we make a virtual displacement, what we're doing is freezing the system at some instant in time, and manually making some infinitesimal changes in the generalized coordinates PERIOD, without any changes in generalized velocities. This is true in the falling particle example as well: the virtual displacement $\delta y$ is, by definition, not accompanied by a change in the $\dot{y}$.

Now, it IS true that if the system, running on its own accord, were to produce changes in the generalized coordinates, then $\delta\dot{y}\neq 0$. But that's different than a virtual displacement.

15. Apr 9, 2012

### Studiot

All virtual displacements have to satisfy the condition of system compatibility, in addition to whichever mechanics laws you employ. Your proposed one does not.

16. Apr 9, 2012

### dEdt

Could you clarify what this means?

Or, better yet, give me a system that does satisfy the 'condition of system compatibility' and I'll show you that $\delta W = \delta T$ doesn't work. I've tried it on other systems, including systems given in my textbook where virtual work and virtual displacements were used, and outcome was still wrong.

17. Apr 9, 2012

### Studiot

18. Apr 9, 2012

### Hassan2

For the falling object in your example, you need to apply a force in order to freeze the system. Shouldn't you take the force into account?