I have recently started reading “ Classical Dynamics of Particles and Systems “ by Marion and Thornton. Page 258 gives us a general expression for kinetic energy. The end result of the derivation is given by:
Kinetic energy =$ \sum_{a} \sum_{i,j,k} \frac{1}{2}m_{a} \frac{dx_{a,i}}{dq_{j}} \frac{dx_{a,i}}{dq_{k}} \frac{dq_{j}}{dt} \frac{dq_{k}}{dt} + \sum_{a} \sum_{i,j} m_{a} \frac{dx_{a,i}}{dq_{j}}\frac{dx_{a,i}}{dt} \frac{dq_{j}}{dt} + \sum_{a} \sum_{i} \frac{1}{2} m_{a} \left( \frac{dx_{a,i}}{dt}\right)^{2}$
Which can be expressed as
Kinetic energy= $ \sum_{j,k}a_{j,k} \frac{dq_{j}}{dt} \frac{dq_{k}}{dt} + \sum_{j}b_{j}\frac{dq_{j}}{dt} +c. $
Recently, I tried to test the equation using a self-made example.
Consider two particles $m_{1}$ and $m_{2}$ undergoing independent, uniform circular motion of radius r and R respectively.
Let $q_{1}= \theta_{1} (t), q_{2}= \theta_{2}(t) $ be our generalized coordinates.
Let $ x_{1,1}= Rcos(\theta_{1}(t)), x_{1,2}= Rsin(\theta_{1}(t))$ and let $ x_{2,1}= rcos(\theta_{2}(t)), x_{2,2}= rsin(\theta_{2}(t).)$
Using the above expression, I found the correct expression for kinetic energy that one would find using the general method.
However, using this equation, I found that the system is scleronomic, ie $b_{j}=0, c=0$.
The book states that a system is scleronomic if time does not appear explicitly in the equations of transformation. But doesn’t time appear explicitly in our transformation equations since $\theta_{1}(t)= \omega_{1}t$ and $\theta_{2}(t)= \omega_{2}t$?
I would appreciate if anyone could clarify why this system is scleronomic, and perhaps provide an example of a non- scleronomic system. Thank you.