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dEdt
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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?
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?