<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/7453ccbe-f81e-4ead-9971-cf103e932a47/Rigid_bodies.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/7453ccbe-f81e-4ead-9971-cf103e932a47/Rigid_bodies.png" width="40px" /> Rigid body: a system of particles in which the distance between them does not change regardless of the force acting.
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$$ \vec r^{(k)}=r_i^{(k)}\,\hat e_i $$
where $k=1\cdots N$ labels to the particles.
$$ 3 \;\text{translations} \quad ; \quad 3 \;\text{rotations} $$
$$ \begin{aligned} M&= \sum^N_{k=1}m_k=\text{total mass} \\ \vec R&= \frac 1M\sum^N_{k=1}m_k \vec r^{(k)}=\text{position vector of the CoM} \\ \vec P_\text{tot}&= \sum^N_{k=1}m_k\dot{\vec r}^{(k)}=\text{total linear momentum} \\ \vec L_\text{tot}&=\sum^N_{k=1}m_k\left (\vec r^{(k)}\times \dot{\vec r}^{(k)}\right )=\text{total angular momentum} \\ T_\text{tot}&=\frac 12 \sum^N_{k=1}m_k\left |\dot{\vec r}^{(k)}\right |^2=\text{total kinetic energy} \end{aligned} $$
These can all be turned into continuous limits as follows:
$$ \begin{aligned} M&= \int\rho\,\text dV \\ \vec R&= \frac 1M \int \rho\vec r\,\text dV \\ \vec P_\text{tot}&= \int \rho \dot{\vec r}\,\text dV\\ \vec L_\text{tot}&= \int \rho(\vec r \times \dot{\vec r})\,\text dV\\ T_\text{tot}&= \frac 12\int\rho|\dot{\vec r}|^2\,\text dV \end{aligned} $$
🧠 Remember: $\left . \dot{\vec r}^{(k)} \right |S=\left . \dot{\vec r}^{(k)} \right |{S'}+\vec \omega \times \vec r^{(k)}$
If $S'$ is the frame of the rigid body and $\dot{\bold r}^{(k)} |{S'}\equiv0$ and hence $\dot{\bold r}^{(k)} |{S}= \vec \omega \times \vec r^{(k)}$ We get:
$$ \begin{aligned} \vec L^{(\text{tot})} &=\sum^N_{k=1}m_k\left [ |\vec r^{(k)} |^2 \vec \omega -\left ( \vec r^{(k)}\cdot\vec \omega \right )\vec r^{(k)} \right]\\ L^{(\text{tot})}{i} &=\sum^N{k=1}m_k\left [ |\vec r^{(k)} |^2 \omega_i - r^{(k)}j \omega_j \,\vec r^{(k)}i \right]=I{ij}\omega{j} \end{aligned} $$
where $I$ is the moment of inertia matrix about the origin
$$ I_{ij}=\sum^N_{k=1}m_k\left [ |\vec r^{(k)} |^2 \delta_{ij} - r^{(k)}_j r^{(k)}i \right ]=\int\rho(r^2\delta{ij}-r_ir_j)\,\text dV $$
🧠 Remember: $\vec r^{(k)}=\vec R+{\vec r'}^{(k)}$
If $\vec R$ is the position of the CoM and ${\vec r'}^{(k)}$ is the position of the $k^{\text{th}}$ particle with respect to the CoM which must satisfy $\sum^N_{i=1}m_k{\vec r'}^{(k)}\equiv 0$ . Then we get the following expressions
$$ \begin{aligned} \vec P_\text{tot}&=M\dot{\vec R} \equiv M\vec V \equiv \vec P_\text{CoM} \\ \vec L_\text{tot}&=\vec R\times M\dot{\vec R}+\sum^N_{k=1}m_k {\vec r'}^{(k)}\times \left ( \vec \omega \times {\vec r'}^{(k)} \right )=\vec L_\text{CoM}+\vec L_\text{rot}
\end{aligned} $$
where $L_i^{(\text{CoM})}=M\epsilon_{ijk} R_j \dot R_k=\epsilon_{ijk} P^{(\text{CoM})}k$ is the angular momentum of the CoM and $L_i^{(\text{rot})}=I'{ij}\omega_j$ is the rotational angular momentum where $I'{ij}=\sum^N{k=1} m_k\left [ |{\vec r'}^{(k)} |^2 \delta_{ij} - {r'}^{(k)}_j {r'}^{(k)}i \right ]\ne I{ij}$
$$ \begin{aligned} T_\text{tot}&=\frac 12 M|\dot{\vec R}|^2+\frac 12 \sum^N_{k=1}m_k|\vec \omega\times {\vec r'}^{(k)}|^2=T_{\text{CoM}}+T_\text{rot} \end{aligned} $$
where $T_\text{CoM}=\frac 12 M|\dot{\vec R}|^2$ is the kinetic energy of the CoM and $T_\text{rot}=\frac 12\sum^N_{k=1}m_k\left ( |\vec \omega|^2|{\vec r'}^{(k)}|^2-|{\vec r'}^{(k)}\cdot\omega|^2 \right )=\frac 12\omega_i I'_{ij}\omega_j$ is the rotational kinetic energy
$$ I_{ij}=\int\text dM (r^2\delta_{ij}-r_ir_j)=\int\text dM\begin{bmatrix} y^2+z^2 & -xy & -xz \\ -yx & z^2+x^2 & -yz \\ -zx & -zy & x^2+y^2 \end{bmatrix} $$
Consider a cylinder of radius $R$ and height $h$ about its center of mass
If we have a solid cylinder with uniform density where
$$ \text dM=\rho\,\text dV=\rho r\,\text dr\, \text d\theta \,\text dz \quad ; \quad \rho = \frac{M}{\pi R^2h} $$
$$ \begin{aligned} I_{ij}&=\rho\int^R_0 r\,\text dr \int^{2\pi}0\text d\theta \int^{\frac 12 h}{- \frac 12 h}\text dz \begin{bmatrix} r^2\sin^2\theta+z^2 &-r^2\cos\theta\sin\theta & -rz\cos\theta \\ -r^2\cos\theta\sin\theta & r^2\cos^2\theta+z^2 &-rz\sin\theta \\ -rz\cos\theta & -rz\sin\theta & r^2
\end{bmatrix} \\ &=\frac{M}{\pi R^2h}\int^R_0 r\,\text dr \begin{bmatrix} \pi r^2h+\frac 16 \pi h^3 & 0 & 0 \\ 0 & \pi r^2h+\frac 16 \pi h^3 & 0 \\ 0 & 0 & 2\pi r^2 h \end{bmatrix} \\ &=M\begin{bmatrix} \frac 14 R^2 + \frac 1 {12} h^2 & 0 & 0 \\ 0 & \frac 14 R^2 + \frac 1{12} h^2 & 0 \\ 0 & 0 & \frac 12 R^2 \end{bmatrix}
\end{aligned} $$
If we have a cylindrical shell with uniform density where
$$ \text dM=\sigma \,\text dA=R\,\text d\theta \,\text dz \quad ;\quad \sigma =\frac{M}{2\pi Rh} $$
$$ \begin{aligned} I_{ij}&=\sigma R\int^{2\pi}0\text d\theta \int^{\frac 12 h}{- \frac 12 h}\text dz \begin{bmatrix} R^2\sin^2\theta+z^2 &-R^2\cos\theta\sin\theta & -Rz\cos\theta \\ -R^2\cos\theta\sin\theta & R^2\cos^2\theta+z^2 &-Rz\sin\theta \\ -Rz\cos\theta & -Rz\sin\theta & R^2
\end{bmatrix} \\ &=\frac{M}{2\pi h}\int^{\frac 12 h}_{- \frac 12 h}\text dz \begin{bmatrix} \pi R^2+2\pi z^2 & 0 & 0 \\ 0 & \pi R^2+2\pi z^2 & 0 \\ 0 & 0 & 2\pi R^2 \end{bmatrix} \\ &=M\begin{bmatrix} \frac 12 R^2+\frac 1{12} h^2 & 0 & 0 \\ 0 & \frac 12 R^2+\frac 1{12} h^2 & 0 \\ 0 & 0 & R^2 \end{bmatrix}
\end{aligned} $$