2024-03-19 Rotating Rigid Bodies

See video: Ellipsoids and the Bizarre Nature of Rotating Bodies Very good resource on the weirdness of rotating bodies

Continued Rigid Body and Inertia Tensors

Uniform Solid Cube Example: Given length $a$, mass $M$, and the corner of the cube at the origin of the system

figure 1 here

Kronecker delta review

$$I_{ij}={\int }_{v}p(\vec{r})\left({\delta }_{ij}\sum _k{x}_k^2-x_i{x}_j\right)dV$$

Finding $I_{11}$

$$I_{11}={\int }_V\rho (\vec{r})(y^2+z^2)dV$$

We can pull out $\rho$ because the square is uniform

$$I_{11}=\rho \underset{0}{\overset{a}{\iiint }}(y^2+z^2)dxdydz$$

Performing the integration:

$$=\rho \left[\underset{0}{\overset{a}{\int }}dx\underset{0}{\overset{a}{\int }}{y}^{2}dy\underset{0}{\overset{a}{\int }}dz+\underset{0}{\overset{a}{\int }}dx\underset{0}{\overset{a}{\int }}dy\underset{0}{\overset{a}{\int }}{z}^{2}dz\right]$$ $$\rho \left[a\left(\frac{1}{3}{y}^3{|}_0^a\right)+a\left(\frac{1}{3}{z}^3{|}_0^a\right)\right]$$ $$\rho a\left[\frac{1}{3}{a}^{3}a+\frac{1}{3}{a}^{3}a\right]=\frac{2}{3}\rho a^5$$

Because $M=\rho a^3$

$$I_{11}=\frac{2}{3}M{a}^2$$

Now finding the $I_{22}$ component

$$I_{22}=\underset{V}{\int }\rho (x^2+z^2)dV$$ $$I_{22}=I_{33}=\frac{2}{3}M{a}^2$$

This can be proven with symmetry or math

$$I_{13}=\underset{0}{\overset{a}{\iiint }}\rho xzdxdydz$$ $$-\rho \underset{0}{\overset{a}{\int }}xdx\underset{0}{\overset{a}{\int }}dy\underset{0}{\overset{a}{\int }}zdz=\rho \left(\frac{1}{2}{x}^2{|}_0^a\right)a\left(\frac{1}{2}{z}^2{|}_0^a\right)$$ $$=-\rho \frac{1}{2}{a}^2\cdot a\cdot \frac{1}{2}{a}^2=-\frac{1}{4}\rho a^5$$ $$I_{31}=-\frac{1}{4}M{a}^2$$ $$I_{31}=I_{13}=I_{23}=I_{32})$$ $$\{\mathbf{I}\}=\left[\begin{array}{ccc}\frac{2}{3}M{a}^2& -\frac{1}{4}M{a}^2& -\frac{1}{4}M{a}^2\\ -\frac{1}{4}M{a}^2& \frac{2}{3}M{a}^2& -\frac{1}{4}M{a}^2\\ -\frac{1}{4}M{a}^2& -\frac{1}{4}M{a}^2& \frac{2}{3}M{a}^2\end{array}\right]$$

Or simplified:

$$\{\mathbf{I}\}=\frac{1}{12}M{a}^2\left[\begin{array}{ccc}8& -3& -3\\ -3& 8& -3\\ -3& -3& 8\end{array}\right]$$

Angular Momentum

Previously in physics 1, angular momentum

$$\vec{L}=I\vec{\omega }$$

and Angular kinetic energy

$$T=\frac{1}{2}I{\omega }^2$$

In the past, we have treated the moment of inertia as a scalar, but that is only true in a special case where $\vec{L}\parallel \vec{\omega }$

In general:

$$\vec{L}=\{\mathbf{I}\}\vec{\omega }$$

or

$$L_i=\sum _j{I}_{ij}{\omega }_j$$ $$\left(\begin{array}{c}{L}_1\\ L_2\\ L_3\end{array}\right)=\left(\begin{array}{ccc}{I}_{11}& I_{12}& I_{13}\\ I_{21}& I_{22}& I_{23}\\ I_{31}& I_{32}& I_{33}\end{array}\right)\left(\begin{array}{c}{\omega }_1\\ {\omega }_2\\ {\omega }_3\end{array}\right)$$ $$\begin{array}{cc}{L}_1=I_{11}{\omega }_1+I_{12}{\omega }_2& +I_{13}{\omega }_3\\ L_2=I_{21}{\omega }_1+I_{22}{\omega }_2& +I_{23}{\omega }_3\\ L_1=I_{31}{\omega }_1+I_{32}{\omega }_2& +I_{33}{\omega }_3\end{array}$$

Angular momentum of a rigid body with respect to the origin of rotating coordinate system.

$$\vec{L}=\sum _{\alpha }{\vec{r}}_{\alpha }\times {\vec{P}}_{\alpha }$$ $$\vec{L}=\sum _{\alpha }{m}_{\alpha }({\vec{r}}_{\alpha }\times (\vec{\omega }\times {\vec{r}}_{\alpha }))$$ $$=\sum _{\alpha }{m}_{\alpha }[r_{\alpha }^2\vec{\omega }-{\vec{r}}_{\alpha }({\vec{r}}_{\alpha }\cdot \vec{\omega })]$$

Where

$${\vec{P}}_{\alpha }=m_{\alpha }{\vec{v}}_{\alpha }=m_{\alpha }\vec{\omega }\times {\vec{r}}_{\alpha }$$

and

$${\vec{r}}_{\alpha }=\sum _K{x}_{\alpha k}^2$$ $${\omega }_i=\sum _j{\omega }_j{\delta }_{ij}$$

the $i^{th}$ component only remains due to the Kronecker delta

$$L_i=\sum _{\alpha }{m}_{\alpha }\left[{\omega }_i\sum _k{x}_{\alpha k}^2-x_{\alpha i}\sum _j{x}_{\alpha j}{\omega }_j\right]$$ $$=\sum _{\alpha }{m}_{\alpha }\sum _j\left({\omega }_j{\delta }_{ij}\sum _k{x}_{\alpha k}^2-{\omega }_j{x}_{\alpha i}{x}_{\alpha j}\right)$$ $$=\sum _j{I}_{ij}{\omega }_j$$ $$L_i=\sum _j{I}_{ij}\omega j$$

Multiply by 1… ($\frac{1}{2}{\omega }_i$ on both sides)

$$\frac{1}{2}{L}_i{\omega }_i=\frac{1}{2}\sum _{ij}{I}_{ij}{\omega }_j{\omega }_i=T_{\text{rot}}$$ $$T_{\text{rot}}=\frac{1}{2}\vec{\omega }\{\mathbf{I}\}\vec{\omega }$$

Principle Axis of Inertia

For the case when none of the components of $\{\mathbf{I}\}$ are 0 then the calculation of energy and angular momentum can be annoying/complicated. (annoying if simple, etc)

Instead, if we assume our inertia tensor only has non-zero diagonal elements such that

$$\{I\}=\left(\begin{array}{ccc}{I}_{11}& 0& 0\\ 0& I_{22}& 0\\ 0& 0& I_{33}\end{array}\right)$$ $$I_{ij}=I_{ij}{\delta }_{ij}$$

Much easier. Would be nice if this is the case. Examples of how nice everything is:

$$L_i=\sum _j{I}_{ij}{\omega }_j\to \sum _j{I}_{ij}{\delta }_{ij}{\omega }_j=I_{ii}{\omega }_i$$ $$\left(\begin{array}{c}{L}_1\\ L_2\\ L_3\end{array}\right)=\left(\begin{array}{ccc}{I}_{11}& 0& 0\\ 0& I_{22}& 0\\ 0& 0& I_{33}\end{array}\right)\left(\begin{array}{c}{\omega }_1\\ {\omega }_2\\ {\omega }_3\end{array}\right)=\left(\begin{array}{c}{I}_{11}{\omega }_1\\ I_{22}{\omega }_2\\ I_{33}{\omega }_3\end{array}\right)$$ $$T_{\text{rot}}=\frac{1}{2}\sum _{ij}{I}_{ij}{\omega }_i{\omega }_j\to \frac{1}{2}\sum _{ij}{\delta }_{ij}{\omega }_i{\omega }_j=\frac{1}{2}\sum _i{I}_{ii}{\omega }_i^2$$

How do we get here?

The diagonal elements are the principal moments of inertia and the principle axes of inertia are the coordinates (axes) used such that $\{\mathbf{I}\}$ is diagonal.

Since we can pick the coordinate system and axes, we can make selections such that we diagonalize $\{\mathbf{I}\}$

There are a few ways to go about this:

Method 1: Simply choose the right axes such that there is symmetry about the axes. Just literally pick better axes.

Method 2: If no obvious symmetries exist, pick arbitrary coordinate system, then diagonalize $\{\mathbf{I}\}$

The eigenvalues of $\{\mathbf{I}\}$ in the arbitrary coordinate system are the principal moments of inertia and the corresponding eigenvectors are the principal axes. The inertia tensor should always have real eigenvalues

For a matrix $A$:

$$A\vec{v}=\lambda \vec{v}$$

Where $\lambda$ are the eigenvalues and $\vec{v}$ are eigenvectors

for an $n\times n$ matrix, has n eigenvalues $({\lambda }_n)$ and eigenvectors ${\vec{v}}_n$

1. So $(A-\lambda \mathbf{I})\vec{v}=0$ So for a non-trivial this is true when $\det (A-\lambda \mathbf{I})=0$ Use this to solve for ${\lambda }_n$

$$(A-{\lambda }_n\mathbf{I}){\vec{v}}_n=0$$

In which $\mathbf{I}$ is the identity matrix 2) Use this to construct the matrix $P$ from the eigenvectors

$$P=\left(\begin{array}{ccc}{v}_{1x}& v_{2x}& v_{3x}\\ v_{1y}& v_{2y}& v_{3y}\\ v_{1z}& v_{2z}& v_{3z}\end{array}\right)$$

3. Find $P^{-1}:P{P}^{-1}=\mathbf{I}$
4. Diagonalize A: $B=P^{-1}AP$