Register to reply 
Virtual work and kinetic energy 
Share this thread: 
#1
Apr812, 09:53 PM

P: 282

Question about analytical mechanics.
Suppose we have a system with N degrees of freedom, and hence N generalized coordinates [itex]{q_i}, \ i=1...N[/itex]. 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 [itex]\delta W[/itex] and the virtual displacements are [itex]\delta q_i, \ i=1...N[/itex]. I would expect then that [tex]\delta W = \delta T = \sum_i \frac{\partial T}{\partial q_i}\delta q_i[/tex] where [itex]\delta T[/itex] 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 [itex]q_i[/itex], but only on [itex]\dot{q_i}[/itex], because then the right hand side of the above equation would be zero. So, what gives? 


#2
Apr912, 01:59 AM

P: 409




#3
Apr912, 09:51 AM

P: 282




#4
Apr912, 11:00 AM

P: 409

Virtual work and kinetic energy
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: [itex] T=T(q_{1},q_{2},...,q_{n},\dot{q}_{1},\dot{q}_{2},...,\dot{q}_{n})[/itex] so [itex]\delta T = \sum\frac{\partial T}{\partial q} \delta q +\sum\frac{\partial T}{\partial \dot{q}} \delta \dot{q}[/itex] 


#5
Apr912, 11:31 AM

P: 282




#6
Apr912, 11:46 AM

P: 409

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



#7
Apr912, 12:00 PM

P: 282




#8
Apr912, 12:05 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 7,160

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
Apr912, 02:01 PM

P: 282

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, [itex]\vec{F}=m\vec{g}=mg\hat{y}[/itex]. So, the virtual work done by a virtual displacement is [itex]\delta W = mg\delta y[/itex]. Now, [itex]T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)[/itex]. There is no dependence on the generalized coordinates, so [itex]\delta T=\frac{\partial T}{\partial x}\delta x+\frac{\partial T}{\partial y}\delta y+\frac{\partial T}{\partial z}\delta z=0[/itex]. If [itex]\delta W=\delta T[/itex], it's easy to see that we have a contradiction. 


#10
Apr912, 02:07 PM

P: 5,462

There is nothing virtual about your δW for a falling particle.



#11
Apr912, 02:26 PM

P: 282




#12
Apr912, 02:37 PM

Emeritus
Sci Advisor
PF Gold
P: 5,196

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


#13
Apr912, 03:29 PM

P: 5,462

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


#14
Apr912, 03:53 PM

P: 282

By definition, a virtual displacement [itex]\delta q[/itex] is never followed by a change in generalized velocity [itex]\delta\dot{q}[/itex]. 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 [itex]\delta y[/itex] is, by definition, not accompanied by a change in the [itex]\dot{y}[/itex]. Now, it IS true that if the system, running on its own accord, were to produce changes in the generalized coordinates, then [itex]\delta\dot{y}\neq 0[/itex]. But that's different than a virtual displacement. 


#15
Apr912, 04:01 PM

P: 5,462

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
Apr912, 05:26 PM

P: 282

Or, better yet, give me a system that does satisfy the 'condition of system compatibility' and I'll show you that [itex]\delta W = \delta T[/itex] 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
Apr912, 06:29 PM

P: 5,462



#18
Apr912, 09:36 PM

P: 409




Register to reply 
Related Discussions  
Work Energy Theorem with Kinetic Friction and External Work  Introductory Physics Homework  7  
Misc Questions: Virtual work, RMS, Friction on incline, and Virtual Displacement  Classical Physics  2  
Conservation of Energy and Virtual Work  General Physics  25  
WorkEnergy Theorum: Spring potential energy vs Kinetic Energy  Introductory Physics Homework  4  
Kinetic energy change in virtual photon exchange  Quantum Physics  22 