# 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&=\sum\limits_\alpha T_\alpha = {1\over 2}\sum_\alpha m_\alpha v_\alpha^2 = {1\over 2}\sum_\alpha m_\alpha \left(\vec{V} + \vec{\omega}\times\vec{r}_\alpha\right)^2\\ T &= {1\over 2}\sum\limits_\alpha m_\alpha \left(V^2 + 2\vec{V}\cdot(\vec{\omega}\times\vec{r}_\alpha)+(\vec{\omega}\times\vec{r}_\alpha)^2\right)\\ T &= V^2{1\over 2}\sum\limits_\alpha m_\alpha + \vec{V}\cdot\left(\vec{\omega}\times\sum_\alpha m_\alpha \vec{r}_\alpha\right)+{1\over 2}\sum\limits_\alpha m_\alpha(\vec{\omega}\times\vec{r}_\alpha)^2 \end{align}

[Note that] $$\sum_\alpha m_\alpha \vec{r}_\alpha = 0$$, [so then] \begin{align} T &= {1\over 2}M V^2 + {1\over 2}\sum\limits_\alpha m_\alpha(\vec{\omega}\times\vec{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

$$V = mgr_y = -mg\ell\cos(\theta)$$

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}(\ell\sin(\theta)) \dot{\theta}\\ &= \ell\cos(\theta) \dot{\theta}\\ v_x^2&= \ell^2\cos^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}(-\ell\cos(\theta)) \dot{\theta}\\ &= \ell\sin(\theta) \dot{\theta}\\ v_y^2&= \ell^2\sin^2(\theta) \dot{\theta}^2 \end{align}

Thus, \begin{align} v_x^2+v_y^2&= \ell^2\cos^2(\theta) \dot{\theta}^2+ \ell^2\sin^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 - (-mg\ell\cos(\theta)) \\ &= \frac{1}{2} m \ell^2 \dot{\theta}^2 + mg\ell\cos(\theta) \\ \end{align}

which yields

$$\ell\ddot{\theta}+g\sin(\theta)=0$$

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?

• What would you do in the case that the bob is not point-like, e.g. it could be more like your illustration where the bob has some non-zero radius? Jul 26 '20 at 16:39
• Why do you think it has to be one way or the other? It is purely your choice as to whether you work it as translation and rotation or of pure rotation. The reference you quotes say "motion can be separated", not "must be". Normally you choose the most convenient method. Jul 27 '20 at 7:22

Assuming the string remains taut, I think you can still treat the simple pendulum as a rigid body, consisting of one point mass. The centre of mass of such a body is located at the point mass (e.g. the bob in this case) itself so $$\mathbf{r} = \mathbf{0}$$ since this must be measured from the centre of mass of the rigid body.

Therefore, we already see that $$\mathbf{v}_{\alpha} = \mathbf{V} + \boldsymbol{\omega} \times \mathbf{r}_{\alpha}$$ reduces to $$\mathbf{v} = \mathbf{V}$$ - i.e. we just need to find the velocity of the bob (centre of mass).

Why doesn't the derivation of a pendulum use $$\boldsymbol{\omega} \times \mathbf{r}_{\alpha}$$ given that the mass rotates around a point?

I believe this has now been answered since the $$\boldsymbol{\omega} \times \mathbf{r}_{\alpha}$$ term actually vanishes. However, if we measure the location of the bob from the pivot point itself, $$\mathbf{r'}$$, we can deduce $$\mathbf{V}$$ using the same cross product (since it comes from the same mathematics). Using cylindrical polar coordinates:

$$\mathbf{r'} = l \mathbf{\hat{r}'}$$ and $$\boldsymbol{\omega} = \dot{\theta} \mathbf{\hat{z}}$$.

Therefore, $$\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r'} = l \dot{\theta} (\mathbf{\hat{z}} \times \mathbf{\hat{r}'}) = l \dot{\theta} \boldsymbol{\hat{\theta}}$$

We then see that $$v^2 = \mathbf{v} \cdot \mathbf{v} = l^2 \dot{\theta}^2$$, which is consistent with your expression.

You can totally use $$\vec{\omega} \times \vec{r}$$ to derive the kinetic energy of a pendulum. Note that $$\vec{\omega} \perp \vec{r}$$, so $$||v||=||\vec{\omega} \times \vec{r}|| = ||\vec{\omega}|| \cdot ||\vec{r}||$$, and since $$||\vec{\omega}||=\dot{\theta}$$ and $$||\vec{r}|| = \ell$$, we have that the kinetic energy $$T = \frac{1}{2} m v^2 = \frac{1}{2} m \dot{\theta}^2 \ell^2$$, as desired.

More generally, the reason for breaking things up in any particular way is always convenience. When you separate out the center of mass motion of a rigid body, you cut the number of degrees of freedom that remain in half (3 instead of 6). The remaining motion of every point on the body will be highly constrained (it will lie on a sphere centered at the center of mass of the body), and easier to solve analytically than if you hadn't removed the CM motion. By breaking up a problem like this, it just makes it easier to understand and solve.

It's reasonable to ask "Why not do the same thing for a pendulum as for a rigid body?" The answer to this is more profound: It is because the rigid body has symmetries that the pendulum does not. In particular, the rigid body has translation invariance, while the pendulum does not because it is anchored to a wall. Separation of CM motion and rotational motion for rigid bodies only strictly works when you have this translational symmetry: If the rigid body is moving in non-uniform potential, it will generally experience a torque caused by its CM motion, so the CM and rotational degrees of freedom become coupled. This is a generic phenomenon--in the presence of symmetry, certain degrees of freedom will tend to decouple, and if you choose coordinates which respect the symmetry of the problem, it will make it easier to solve.