Expression of kinetic energy in polar coordinates Expression for kinetic energy in Cartesian coordinate:

Expression for kinetic energy in polar coordinate (applying the transformation of coordinates):

Why can't we express it in the following terms by taking the time derivative of each degree of freedom:

 A: If you express velocity in polar (planar) coordinates you get:$$\mathbf v = \dot r \hat r + +r\dot \theta \hat \theta,$$
so a correct expression for the kinetic energy would be:$$T=\frac{m}{2}(\dot r ^2 + r^2\dot \theta ^2).$$

To find the expression of the velocity in polar coordinates you can work in different ways. I'll suggest you one, very straightforward in my point of view.
First of all, as you noted, we have $$\mathbf r = r(\cos \theta, \sin \theta)$$
(in the first part of the post I'm simply defining $\hat r:=(\cos \theta, \sin \theta)$). One differentiation yelds:$$\dot {\mathbf r} = \dot r (\cos \theta, \sin \theta) + r \dot \theta (- \sin \theta,\cos \theta),$$
and here we call $$\hat \theta=(-\sin \theta,\cos \theta).$$
You can easily check that $\hat\theta$ is perpendicular to $\hat r$. Also note that the norm of both $\hat r$ and $\hat \theta$ is $1$, hence the norm of $\dot {\mathbf r}$ is:$$|\dot {\mathbf r}|=(\dot r ^2 + r^2 \dot \theta ^2)^\frac{1}{2}.$$

I wish I was able to add also a geometric derivation of the result, it's very easy and nice to compare with the one above. Surely you'll be able to find one on some good mechanic's book.
A: If the particle is near the origin, say r=1, then a change in $\alpha$ corresponds to a small physical distance.   If the particle is far away, r=1000, then the same change in $\alpha$ corresponds to a much larger change in physical location, about 1000 times as large.  A term like ${1 \over 2}m \dot\alpha^2$ doesn't account for that.  But one like ${1 \over 2}m (r\dot\alpha)^2$ does.
