A When KE is a function of position

[Trying2Learn]

TL;DR

Question about the Lagrangian

Hi all

In the Lagrangian, we have L = KE - PE

In most cases, I have seen KE as a function of q and q-dot (using the generic symbols).

However I first learned how KE = 0.5 m * v-squared.

Later, I used generalized coordinates and THAT is when KE became a function of q.

I get all that. However, I am still wondering WHY I have the feeling that, in most cases, KE is only a function of the parameter's first time derivative and ONLY involves the parameter in certain cases.

Maybe what I am asking is: "why did they first formulate L(q, q-dot) = KE(q, q-dot) and PE(q)

(I get the P only a function of q, by the way. That is not an issue for me.)

 

[ergospherical]

It's perhaps easiest to see through an example; consider a particle moving in $\mathbf{R}^2$, and let the generalised co-ordinates be plane polar co-ordinates $q = (r,\theta)$. The kinetic energy is\begin{align*}
T(q, \dot{q}) = \dfrac{1}{2} \dot{r}^2 + \dfrac{1}{2} r^2 \dot{\theta}^2
\end{align*}Not only does $T$ depend on $\dot{r}$ and $\dot{\theta}$, but it also depends on $r$. In the general case, the kinetic energy is a quadratic form whose coefficients $a_{ij}(q)$ depend on the co-ordinates,\begin{align*}
T(q,\dot{q}) = \dfrac{1}{2} a_{ij}(q) \dot{q}^i \dot{q}^j
\end{align*}where summation over repeated indices is implicit. For the previous example, $a_{rr}(q) = 1$ and $a_{\theta \theta}(q) = r^2$, whilst the mixed coefficients are zero.

 

[Trying2Learn]

Oh... yes Now I see. That was obvious.

Then can you say that tried to keep the form as general as possible? Is that it?

Thank you.