homework and exercises - Finding Lagrangian of a Spring Pendulum

I'm trying to understand Morin's example of a spring pendulum. What I don't get is his expression for $T$. I can understand the $\dot x^2$ term in the brackets. But I don't understand the $(l + x)^2\dot \theta^2$.

Also, it seems rather strange to break up Kinetic Energy into tangential and radial components when it is a scalar.

Answer

$\newcommand{\er}{\hat e_r} \newcommand{\et}{\hat e_\tau} \newcommand{\d}{\dot} \newcommand{\m}{\frac{1}{2}m}$ In radial coordinates, $\d\er=\d\theta \et$, and (useless here) $\d\et= -\d r \er$. $\er,\et$ are unit vectors in radial and tangential directions respectively. Due to this mixing of unit vectors (they move along with the particle), things get a little more complicated than plain 'ol cartesian system, where the unit vectors are constant.

For your particle, writing $x+l\to r$, the position vector is: $$\vec p= r\er$$ $$\therefore \vec v=\d{\vec p}= \d r\er + r\d\er=\d r \er + r\d\theta\et$$ $$\therefore v^2= \vec v\cdot\vec v= \d r^2+r^2\d\theta^2$$

Substituting back the value of $r=x+l,\d r=\d x$ (and mutiplying by $\m$, we get the above expression?

As you can see in my expression for $\vec v$, I had two components of velocity--radial and tangential. Since they are perpendicular, I can just square and add, akin to $T=\m\left(\d x^2 +\d y^2\right)$.

The point is, it may be a scalar, but it contains a vector in its expression: $$T=\m v^2=\m|v|^2=\m \vec v\cdot \vec v=\m(\dot x^2+\dot y^2)$$