2024-03-15 Rigid Bodies Cont.

Rotational Kinetic Energy

$$T_{\text{rot}}=\frac{1}{2}\sum _{\alpha }{m}_{\alpha }(\vec{\omega }\times {\vec{r}}_{\alpha }{)}^2\text{\hspace{1em}(1)}$$

Where

$$\begin{array}{c}{\vec{r}}_{\alpha }=(x_{\alpha 1},x_{\alpha 2},x_{\alpha 3}\dots )\\ \vec{\omega }=({\omega }_1,{\omega }_2,{\omega }_3)\end{array}$$

and

$$(\vec{A}\times \vec{B}{)}^2=A^2{B}^2-(\vec{A}\cdot \vec{B}{)}^2$$

Using these two relations we can rewrite (1) as the following

$$=\frac{1}{2}\sum _{\alpha }{m}_{\alpha }[{\omega }^2{r}_{\alpha }^2-(\vec{\omega }\cdot {\vec{r}}_{\alpha }^2)]$$

Expanding this

$$T_{\text{rot}}=\frac{1}{2}\sum _{\alpha }{m}_{\alpha }\left[\left(\sum _i{\omega }_i^2\right)\left(\sum _k{x}_{a,k}^2\right)-\left(\sum _i{\omega }_i{x}_{\alpha ,i}\right)\left(\sum _j{\omega }_j{x}_{\alpha ,j}\right)\right]$$

Using Kronecker delta to simplify this:

$$\frac{1}{2}\sum _{\alpha }\sum _{i,j}{m}_{\alpha }\left[{\omega }_i{\omega }_j{\delta }_{ij}\left(\sum _k{x}_{\alpha k}^2\right)-{\omega }_i{\omega }_j{x}_{\alpha ,i}{x}_{\alpha ,j}\right]$$ $$=\frac{1}{2}\sum _{ij}{\omega }_i{\omega }_j\sum _{\alpha }{m}_{\alpha }\left[{\delta }_{ij}\sum _k{x}_{\alpha ,k}^2-x_{\alpha ,i}{x}_{\alpha ,j}\right]$$

Content after second summation is called the Inertia tensor

For a discrete distribution of particles, the inertia tensor looks like the following:

$$I_{ij}=\sum _{\alpha }{m}_{\alpha }\left({\delta }_{ij}\sum _k{x}_{\alpha ,k}^2-x_{\alpha ,i}{x}_{\alpha ,j}\right)$$

Continuous Distribution

$$I_{ij}={\int }_{v}p(\vec{r})\left({\delta }_{ij}\sum _k{x}_{\alpha k}^2-x_{\alpha i}{x}_{\alpha j}\right)dV$$

Example: Consider 3 masses (all with mass m) $\alpha$ will sum from 1→3 3D problem. They have the following positions:

$$\begin{array}{c}{\vec{x}}_1=a,0,0\\ {\vec{x}}_2=0,a,2a\\ {\vec{x}}_3=0,2a,a\end{array}$$

$$I_{11}=\sum _{\alpha }^3{m}_{\alpha }(x_{{\alpha }_2}^2+x_{{\alpha }_3}^2)$$ $$=m[(x_{1,2}^2+x_{1,3}^2)+(x_{2,2}^2+x_{2,3}^2)+(x_{3,2}^2+x_{3,3}^2)]$$ $$=m[(0^2+0^2)+(a^2+2a^2)+((2a{)}^2+a^2)]$$ $$=m[a^2+4a^2+4a^2+a^2]=10m{a}^2$$

Finding $I_{22}$

$$\sum _{\alpha }{m}_{\alpha }(x_{\alpha ,1}^2+x_{\alpha ,3}^2)$$ $$m[(x_{1,1}^2+x_{1,3}^2)+(x_{2,1}^2+x_{2,3}^2)+(x_{3,1}^2+x_{3,3}^2)]$$ $$m[a^2+0^2+0^2+4a^2+0^2+a^2]$$ $$I_{22}=6m{a}^2$$ $$I_{33}=m[a^2+a^2+4a^2]=6m{a}^2$$ $$I_{12}=-m[(a\times 0)+(0\times a)+(0\times 2a)]=0=I_{21}$$ $$I_{13}=-m[(a\times 0)+(0\times 2a)+(0\times a)]=I_{3,1}$$ $$I_{23}=-m[(0\times 0)+(a\times 2a)+(2a\times a)]=-4m{a}^2=I_{32}$$ $$\{I\}=\left(\begin{array}{ccc}10m{a}^2& 0& 0\\ 0& 6m{a}^2& -4m{a}^2\\ 0& -4m{a}^2& 6m{a}^2\end{array}\right)$$

Example: Rectangular plate of uniform density $\rho$ with total mass M, length a, width b and height c

• Put the origin at the center of the plate

$$\begin{array}{c}-\frac{a}{2}\le x_1\le \frac{a}{2}\\ -\frac{b}{2}\le x_2\le \frac{b}{2}\\ -\frac{c}{2}\le x_3\le \frac{c}{2}\end{array}$$ $$\rho =\frac{M}{abc}$$ $$I_{ij}={\int }_v\rho (r)\left({\delta }_{ij}\sum _k{x}_{\alpha k}^2-x_{\alpha i}{x}_{\alpha j}\right)dV$$ $$I_{ii}=\rho {\int }_V(x_2^2+x_3^2)dV=\rho \left[{\int }_V{x}_2^{2}dV+{\int }_V{x}_3^{2}dV\right]$$ $$\rho \left[\underset{-\frac{a}{2}}{\overset{a/2}{\int }}\underset{-\frac{b}{2}}{\overset{b/2}{\int }}\underset{-\frac{c}{2}}{\overset{c/2}{\int }}{x}_2^{2}d{x}_{1}d{x}_{2}d{x}_3+\underset{-\frac{a}{2}}{\overset{a/2}{\int }}\underset{-\frac{b}{2}}{\overset{b/2}{\int }}\underset{-\frac{c}{2}}{\overset{c/2}{\int }}{x}_3^{2}d{x}_{1}d{x}_{2}d{x}_3\right]$$ $$=\rho \left[ac\frac{1}{3}{x}_2^3{|}_{-b/2}^{b/2}+ab\frac{1}{3}{x}_3^3{|}_{-c/2}^{c/2}\right]$$ $$=\rho \left[ac\frac{1}{3}\frac{b^3}{8}+ac\frac{b^3}{8}\frac{1}{3}+ab\frac{c^3}{8}\frac{1}{3}+ab\frac{c^3}{8}\frac{1}{3}\right]$$ $$\rho \left[\frac{a{b}^{3}c}{12}+\frac{ab{c}^3}{12}\right]=\frac{1}{12}\frac{M}{abc}[a{b}^{3}c+ab{c}^3]=\frac{1}{12}M[b^2+c^2]$$ $$I_{22}=\frac{1}{12}M[a^2+c^2]$$ $$I_{33}=\frac{1}{12}M(a^2+b^2)$$ $$I_{12}={\int }_V\rho (-x_{{\alpha }_1}{x}_{{\alpha }_2})dV-\int \rho x_1{x}_{2}dV$$ $$=-\rho \left[\underset{-c/2}{\overset{c/2}{\int }}d{x}_3\underset{-b/2}{\overset{b/2}{\int }}{x}_{2}d{x}_2\underset{-a/2}{\overset{a/2}{\int }}{x}_{1}d{x}_1\right]$$ $$=\rho c\left(\frac{1}{2}{x}_2^2{|}_{-b/2}^{b/2}\right)\left(\frac{1}{2}{x}_1^2{|}_{-a/2}^{a/2}\right)$$ $$=-\rho c\left[\text{cancelto}0\frac{1}{2}\frac{b^2}{4}-\frac{1}{2}\frac{b^2}{4}\right]=\left[\text{cancelto}0\frac{1}{2}\frac{a^2}{4}-\frac{1}{2}\frac{a^2}{4}\right]=0=I_{21}$$

Similarly

$$I_{13}=I_{31}=0$$ $$I_{23}=I_{32}=0$$ $$\{I\}=\frac{1}{12}M\left(\begin{array}{ccc}{b}^2+c^2& 0& 0\\ 0& a^2+c^2& 0\\ 0& 0& a^2+b^2\end{array}\right)$$ $$T_{rot}=\frac{1}{2}\sum _{ij}2{\omega }_i{\omega }_j$$

Particular case where $\vec{\omega }=(0,0,{\omega }_3)$

$$=\frac{1}{2}\sum _{i=1}^3\sum _{j=1}^3{\omega }_i{\omega }_j{I}_{ij}$$ $$=\frac{1}{2}\sum _{i=1}^3{\omega }_i[\text{cancelto}0{\omega }_1{I}_{i1}+\text{cancelto}0{\omega }_2{I}_{i2}+{\omega }_3{I}_{i3}]$$ $$=\frac{1}{2}{I}_{33}{\omega }_3^2$$

The dzhanibhanibekov effect or tennis racket theorem arises from this ! very cool videos.