Is the Lagrange Equation Valid for All Holonomic Systems?

[Petar015]

Homework Statement

Show that for an arbitrary ideal holonomic system (n degrees of freedom)

$$\frac{1}{2} \frac{\partial \ddot T}{\partial\ddot q_j} - \frac{3}{2} \frac{\partial T}{\partial q_j} = Q_j$$

where T is kinetic energy and q_j generalized coordinates.[/B]

Homework Equations

Lagrange's equation
$$\frac{d}{dt} \frac{\partial T}{\partial\dot q_j} - \frac{\partial T}{\partial q_j} = Q_j$$[/B]

The Attempt at a Solution

We know that $$T(q_1,...q_n,\dot q_1,...,\dot q_n,t)$$
The idea is to express $$\dot T$$ and $$\ddot T$$ and then plug it into initial equation in order to obtain equivalence with Lagrange's equation.

So we write

$$\frac {dT}{dt}=\dot T=\frac{\partial T}{\partial \dot q_j} \ddot q_j + \frac{\partial T}{\partial q_j} \dot q_j + \frac{\partial T}{\partial t}$$

So I figure that I should express $$\ddot T$$
in the same manner, but I'm stuck at doing the chain rule for the first 2 terms.
[/B]

 

[Dr.D]

As you say, T = T(q1,q2,...,q1d,q2d,...) so the partial of T with respect to qidd is necessarily zero.

I really don't believe the result that you are trying to prove. I can say with confidence that in almost 60 years of doing dynamics, I have never seen this expression anywhere, and it looks entirely bogus to me. I look forward to whatever light others may bring to this issue. Maybe there is something new under the sun after all!