Rigid body

A rigid body is a solid body that cannot be deformed. Given any two points on the object, the distance in between them remains the same at all times. The Center of mass of a rigid body always remains in the same place with respect to the body's points. A rigid body can be either a system of particles or a continuum of mass.

If the body is a system of particles, we can describe a rigid body as NN point masses bound by the condition ∣ri−rj∣=const\lvert \mathbf{r}_{i}-\mathbf{r}_{j} \rvert=\text{const}, for any i,j=1,…,Ni,j=1,\ldots,N representing a point mass.

Analytical mechanics

We can treat a rigid body in a more formal, accurate manner through analytical mechanics. Firstly, the condition that the distance between points cannot change is a constraint. Secondly, a rigid body has exactly six degrees of freedom: three translational and three rotational. The former describe the position of (a point of) the body with respect to the frame of reference in use, whereas the latter describe the orientation of the body, defined as the difference between a Cartesian triad attached to the object (a moving frame) and the reference triad.

The change in position is pretty straight-forward. The change of orientation is a little more complicated. In an infinitesimal time δt\delta t, the moving triad {e1,e2,e3}\{ \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3} \} is subject to an infinitesimal rotation Ω\Omega. For any given vector u\mathbf{u} moving alongside the body, its variation in time due to the rotation is

δu=Ωuδt\delta \mathbf{u}=\Omega \mathbf{u}\delta t

Since Ω\Omega is an antisymmetric matrix, this operation is equivalent to claiming that there exists some vector ω\boldsymbol{\omega} such that δu=ω×u δt\delta \mathbf{u}=\boldsymbol{\omega}\times \mathbf{u}\ \delta t. By """dividing""" through with δt\delta t, we can find the time derivative of u\mathbf{u}:

u˙=ω×u\dot{\mathbf{u}}=\boldsymbol{\omega}\times \mathbf{u}

The vector ω\boldsymbol{\omega} is, of course, interpreted as the angular velocity of rotation. Since u\mathbf{u} is any vector attached to the object, the Basis {e1,e2,e3}\{ \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3} \} also changes according to this equation

e˙i=ω×ei,i=1,2,3\dot{\mathbf{e}}_{i}=\boldsymbol{\omega}\times \mathbf{e}_{i},\quad i=1,2,3

In fact, if we know the changes in unit vectors, we can even invert this relation to find ω\boldsymbol{\omega}:

ω=12∑i=13ei×e˙i\boldsymbol{\omega}=\frac{1}{2}\sum_{i=1}^{3} \mathbf{e}_{i}\times \dot{\mathbf{e}}_{i}

Knowing ω\boldsymbol{\omega}, we can move on to the angular momentum of the body:

L0=∑i=1Nmiri×vi=∑i=1Nmiri×(ω×ri)≡Iω\mathbf{L}_{0}=\sum_{i=1}^{N}m_{i}\mathbf{r}_{i}\times \mathbf{v}_{i}=\sum_{i=1}^{N} m_{i}\mathbf{r}_{i}\times(\boldsymbol{\omega}\times \mathbf{r}_{i})\equiv \mathcal{I}\boldsymbol{\omega}

where the 00 on L0\mathbf{L}_{0} denotes that the angular momentum is for a point attached to the object, v=r˙=ω×r\mathbf{v}=\dot{\mathbf{r}}=\boldsymbol{\omega}\times \mathbf{r}. The last value I\mathcal{I} is something we just defined: we call it the inertia tensor and it acts as a sort of "rotational mass". In fact, just like the linear momentum is mass times linear velocity, the angular momentum is "rotational mass" (i.e. inertia tensor) times angular velocity: p=mv→L=Iω\mathbf{p}=m\mathbf{v}\to \mathbf{L}=\mathcal{I}\boldsymbol{\omega}. Note however, that I\mathcal{I} is not a Scalar, it is a tensor, a three-dimensional second-order one to be specific. In other words, it is a 3×33\times 3 matrix, with nine components, split across the axes.

We can use this knowledge to describe rotational motion. If the axis of rotation is constant, say e3=ez\mathbf{e}_{3}=\mathbf{e}_{z}, our angular velocity becomes

ω=12(e1×e˙1+e2×e˙2)\boldsymbol{\omega}=\frac{1}{2}(\mathbf{e}_{1}\times \dot{\mathbf{e}}_{1}+\mathbf{e}_{2}\times \dot{\mathbf{e}}_{2})

since e˙3=0\dot{\mathbf{e}}_{3}=0. The basis vectors are

e1=(cos⁡θsin⁡θ),e2=(−sin⁡θcos⁡θ)\mathbf{e}_{1}=\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix},\qquad \mathbf{e}_{2}=\begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix}

with derivatives

e˙1=θ˙(−sin⁡θcos⁡θ)=θ˙e2,e˙2=−θ˙(cos⁡θsin⁡θ)=−θ˙e1\dot{\mathbf{e}}_{1}=\dot{\theta}\begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix}=\dot{\theta}\mathbf{e}_{2},\qquad \dot{\mathbf{e}}_{2}=-\dot{\theta}\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}=-\dot{\theta}\mathbf{e}_{1}

Hence

ω=12θ˙(e1×e2−e2×e1)=θ˙e1×e2=θ˙e3\boldsymbol{\omega}=\frac{1}{2}\dot{\theta}(\mathbf{e}_{1}\times \mathbf{e}_{2}-\mathbf{e}_{2}\times \mathbf{e}_{1})=\dot{\theta}\mathbf{e}_{1}\times \mathbf{e}_{2}=\dot{\theta}\mathbf{e}_{3}

But e3\mathbf{e}_{3} is constant, so the direction of the angular velocity does not change (of course). With this, we can find the kinetic energy of rotation:

T=12ω⋅Iω=12θ˙2e3⋅Ie3=12I3θ˙2T=\frac{1}{2}\boldsymbol{\omega}\cdot \mathcal{I}\boldsymbol{\omega}=\frac{1}{2}\dot{\theta}^{2}\mathbf{e}_{3}\cdot \mathcal{I}\mathbf{e}_{3}=\frac{1}{2}I_{3}\dot{\theta}^{2}

Mix it with a potential energy of choice and you have a Lagrangian to solve for motion.

(TODO: Finish this, lesson 30/04/2025)