Why do some Lagrangians have both a rotational and a non-rotational part of their kinetic energy?

I was going through a derivation and I came across a term in the kinetic energy of the Lagrangian that had both a velocity term and a rotational term. So I looked it up and I found these two links

https://hepweb.ucsd.edu/ph110b/110b_notes/node19.html

https://hepweb.ucsd.edu/ph110b/110b_notes/node20.html

In the first link, they say

For a rigid body, we will find in the equations that the motion can be separated into the motion of the center of mass and the rotation around the center of mass.

[…]

For the purposes of calculation, we will assume that the body is made up of a set of discrete masses, labeled by an index $alpha$. In any inertial frame, the velocity of one of those masses is
$$vec{v}_alpha = vec{V} + vec{omega} times vec{r}_alpha$$

In this equation, the $vec{v}_alpha$ is given in the inertial frame, $vec{V}$ is the velocity of the center of mass of the object, and $vec{r}_alpha$ is the position of the mass in the body frame in which it is at rest.

In the second link, they then show the following

From Lagrangian dynamics, we know that we can extract the physics if we know the kinetic and potential energy, $T$ and $U$. For now we will not be applying any potential so we only have the kinetic energy.
begin{align}
T&=sumlimits_alpha T_alpha = {1over 2}sum_alpha m_alpha v_alpha^2 = {1over 2}sum_alpha m_alpha left(vec{V} + vec{omega}timesvec{r}_alpharight)^2
T &= {1over 2}sumlimits_alpha m_alpha left(V^2 + 2vec{V}cdot(vec{omega}timesvec{r}_alpha)+(vec{omega}timesvec{r}_alpha)^2right)
T &= V^2{1over 2}sumlimits_alpha m_alpha + vec{V}cdotleft(vec{omega}timessum_alpha m_alpha vec{r}_alpharight)+{1over 2}sumlimits_alpha m_alpha(vec{omega}timesvec{r}_alpha)^2
end{align}

[Note that]
$sum_alpha m_alpha vec{r}_alpha = 0$, [so then]
begin{align}
T &= {1over 2}M V^2 + {1over 2}sumlimits_alpha m_alpha(vec{omega}timesvec{r}_alpha)^2
T& = T_{CM} + T_{rot}
end{align}

I’m confused why the kinetic energy has to be broken up this way. For example, consider the example with a pendulum.

Intuitively looking at it, I would say there is a torque from gravity on the pendulum causing it to move. So the force should be accounted for in this "rotational" part of the kinetic energy. However, I understand the standard derivation of a pendulum to be as follows:

The only force acting on the pendulum is gravity. So the potential energy is

$$V = mgr_y = -mgellcos(theta)$$

The kinetic energy is

$$
begin{align}
T &= frac{1}{2} m lVert mathbf{v}rVert^2
&= frac{1}{2} m (v_x^2 +v_y^2)
end{align}
$$

Note,

$$
begin{align}
v_x &= frac{partial r_x}{partial theta} frac{partial theta}{partial t}
&= frac{partial}{partial theta}(ellsin(theta)) dot{theta}
&= ellcos(theta) dot{theta}
v_x^2&= ell^2cos^2(theta) dot{theta}^2
end{align}
$$
and
$$
begin{align}
v_y &= frac{partial r_y}{partial theta} frac{partial theta}{partial t}
&= frac{partial}{partial theta}(-ellcos(theta)) dot{theta}
&= ellsin(theta) dot{theta}
v_y^2&= ell^2sin^2(theta) dot{theta}^2
end{align}
$$

Thus,
begin{align}
v_x^2+v_y^2&= ell^2cos^2(theta) dot{theta}^2+ ell^2sin^2(theta) dot{theta}^2
&= ell^2 dot{theta}^2 (cos^2(theta)+sin^2(theta))
&= ell^2 dot{theta}^2
end{align}

So then, the kinetic energy is
$$
T =frac{1}{2} m (v_x^2 +v_y^2) = frac{1}{2} m ell^2 dot{theta}^2
$$

So then the Lagrangian is given by

$$begin{align}
L & = T – V
&= frac{1}{2} m ell^2 dot{theta}^2 – (-mgellcos(theta))
&= frac{1}{2} m ell^2 dot{theta}^2 + mgellcos(theta)
end{align}$$

which yields

$$ellddot{theta}+gsin(theta)=0$$

Notice in my derivation however, the kinetic energy term for the pendulum was stated using $(v_x^2 +v_y^2)$ which seems analogous to the $T_{CM}$ term from earlier and not $T_{rot}$.

My Question

Why is velocity broken up into $vec{v}_alpha = vec{V} + vec{omega} times vec{r}_alpha$? Why doesn’t the derivation of a pendulum use $vec{omega} times vec{r}_alpha$ given that the mass rotates around a point?