Motion of Rigid Bodies

$$\begin{array}{rl}\overrightarrow{L}& ={\overrightarrow{L}}_i\\ & ={\overrightarrow{r}}_i\times {\overrightarrow{p}}_i\\ \frac{d}{dt}\overrightarrow{L}& =\frac{d}{dt}({\overrightarrow{r}}_i\times {\overrightarrow{p}}_i)\\ & ={\overrightarrow{r}}_i\times {\dot{\overrightarrow{p}}}_i\\ & ={\overrightarrow{r}}_i\times {\overrightarrow{F}}_i^{(e)}.\\ \dot{\overrightarrow{L}}& ={\overrightarrow{\tau }}^{(e)}\end{array}$$

Thus, the internal force does not contribute to the time-change of Angular momentum.

Write ${\overrightarrow{r}}_i=\overrightarrow{R}+{\overrightarrow{r}}_i^{\prime }$ and ${\overrightarrow{v}}_i=\overrightarrow{V}+{\overrightarrow{v}}_i^{\prime }$.

$$\begin{array}{rl}\overrightarrow{L}& =m_i(\overrightarrow{R}+{\overrightarrow{r}}_i^{\prime })\times (\overrightarrow{V}+{\overrightarrow{v}}_i^{\prime })\\ & =\overrightarrow{R}\times m_i\overrightarrow{V}+{\overrightarrow{r}}_i^{\prime }\times m_i{\overrightarrow{v}}_i^{\prime }+m_i{\overrightarrow{r}}_i^{\prime }\times \overrightarrow{V}+\overrightarrow{R}\times m_i{\overrightarrow{v}}_i^{\prime }\\ & =\overrightarrow{R}\times m_i\overrightarrow{V}+{\overrightarrow{r}}_i^{\prime }\times m_i{\overrightarrow{v}}_i^{\prime }\\ & =\overrightarrow{R}\times M\overrightarrow{V}+{\overrightarrow{r}}_i^{\prime }\times m_i{\dot{\overrightarrow{r}}}_i^{\prime }\\ & ={\overrightarrow{L}}_{CM}+{\overrightarrow{L}}^{\prime }.\end{array}$$

Then,

$$\begin{array}{rl}\dot{\overrightarrow{L}}& ={\dot{\overrightarrow{L}}}_{CM}+{\dot{\overrightarrow{L}}}^{\prime }\\ & =\frac{d}{dt}(\overrightarrow{R}\times \overrightarrow{P})+\frac{d}{dt}(\overrightarrow{L}-{\overrightarrow{L}}_{CM})\\ & =\overrightarrow{R}\times \dot{\overrightarrow{P}}+{\overrightarrow{\tau }}^{(e)}-{\dot{\overrightarrow{L}}}_{CM}\\ & =\overrightarrow{R}\times {\overrightarrow{F}}_{total}^{(e)}+{\overrightarrow{\tau }}^{(e)}-{\dot{\overrightarrow{L}}}_{CM}\\ & ={\overrightarrow{\tau }}^{(e)}={\overrightarrow{r}}_i\times {\overrightarrow{F}}_i^{(e)}.\end{array}$$

Note,

$$\begin{array}{rl}{\dot{\overrightarrow{L}}}^{\prime }& =\frac{d}{dt}(\overrightarrow{L}-{\overrightarrow{L}}_{CM})\\ & =({\overrightarrow{r}}_i-\overrightarrow{R})\times {\overrightarrow{F}}_i^{(e)}={\overrightarrow{r}}_i^{\prime }\times {\overrightarrow{F}}_i^{(e)}.\end{array}$$

$$\begin{array}{rl}T& =\frac{1}{2}{m}_i{\overrightarrow{v_i}}^2\\ & =\frac{1}{2}{m}_i(\overrightarrow{V}+{\overrightarrow{v}}_i)\cdot (\overrightarrow{V}+{\overrightarrow{v}}_i)\\ & =\frac{1}{2}M{\overrightarrow{V}}^2+\frac{1}{2}{m}_i{\overrightarrow{v}}_i^{\prime }.\end{array}$$

Thus, the kinetic energy can be computed from the center of mass motion and the relative motion of each particles wrt CM.

Consider two different coordinate systems that share the same origin.

$$\begin{array}{rlr}\overrightarrow{v}& ={\overrightarrow{v}}_i^0{\hat{e}}_i^0& ={\overrightarrow{v}}_i^1{\hat{e}}_i^1\\ \frac{d}{dt}{\left(\overrightarrow{v}\right)}_{inertial}& =\frac{d}{dt}\left({\overrightarrow{v}}_i^0{\hat{e}}_i^0\right)& =\frac{d}{dt}\left({\overrightarrow{v}}_i^1{\hat{e}}_i^1\right)\\ & =\frac{d}{dt}\left({\overrightarrow{v}}_i^0\right){\hat{e}}_i^0& =\frac{d}{dt}\left({\overrightarrow{v}}_i^1\right){\hat{e}}_i^1\end{array}$$

For the other frame, consider if it was rotating relative to the first,

$$\begin{array}{rl}\overrightarrow{v}& ={\overrightarrow{v}}_i{\hat{e}}_i^1\\ \frac{d}{dt}{\left(\overrightarrow{v}\right)}_{inertial}& =\frac{d}{dt}\left({\overrightarrow{v}}_i^1{\hat{e}}_i^1\right)\\ & =\frac{d}{dt}\left({\overrightarrow{v}}_i^1\right){\hat{e}}_i^1+{\overrightarrow{v}}_i^1\frac{d}{dt}\left({\hat{e}}_i^1\right)\\ & ={\left(\frac{d\overrightarrow{v}}{dt}\right)}_{body}+\overrightarrow{\omega }\times \overrightarrow{v}.\end{array}$$

Hence we have the inertial frame and the body-fixed axes. Note that $\overrightarrow{\omega }$ and $\overrightarrow{v}$ are measured in the inertial frame.

Consider an inertial observer which measures the motion of a point B on the rigid body with respect to A. Then, $\overrightarrow{R}={\overrightarrow{R}}_A-{\overrightarrow{R}}_B$. So,

$$\begin{array}{rl}\frac{d}{dt}{\left({\overrightarrow{R}}_B\right)}_{inertial}& ={\left(\frac{d{\overrightarrow{R}}_A}{dt}\right)}_{body}+{\overrightarrow{\omega }}_B\times \overrightarrow{R}.\end{array}$$

But we also have,

$$\begin{array}{rl}\frac{d}{dt}{\left({\overrightarrow{R}}_A\right)}_{inertial}& ={\left(\frac{d{\overrightarrow{R}}_B}{dt}\right)}_{body}+{\overrightarrow{\omega }}_A\times (-\overrightarrow{R}).\end{array}$$

Adding these two,

$$\begin{array}{rl}0& =({\overrightarrow{\omega }}_B-{\overrightarrow{\omega }}_A)\times \overrightarrow{R}\Rightarrow {\overrightarrow{\omega }}_A={\overrightarrow{\omega }}_B.\end{array}$$

Consider a rolling cylinder with no slippage. Then, at the bottom $0+\overrightarrow{\omega }\times \overrightarrow{r}=\overrightarrow{v}$ hence, $\omega r=v$.

Hence, when our body-fixed coordinate is rotating with the object, for some $i$,

$$\begin{array}{rl}\frac{d}{dt}{\left({\overrightarrow{r}}_i\right)}_{inertial}& ={\left(\frac{d{\overrightarrow{i}}_i}{dt}\right)}_{body}+\overrightarrow{\omega }\times {\overrightarrow{r}}_i\\ & =\overrightarrow{\omega }\times {\overrightarrow{r}}_i.\end{array}$$

$$\begin{array}{rl}\overrightarrow{L}& ={\overrightarrow{r}}_i\times {\overrightarrow{p}}_i\\ & ={\overrightarrow{r}}_i\times m_i{\overrightarrow{v}}_i\\ & =m_i{\overrightarrow{r}}_i\times (\overrightarrow{\omega }\times {\overrightarrow{r}}_i)\\ & =m_i[\overrightarrow{\omega }{r}_i^2-{\overrightarrow{r}}_i({\overrightarrow{r}}_i\cdot \overrightarrow{\omega })],\\ & =L_x\hat{i}+L_y\hat{j}+L_z\hat{k},\\ L_x& =m_i[{\omega }_x{r}_i^2-r_{ix}({\overrightarrow{r}}_i\cdot \omega )]\\ & =m_i[{\omega }_x{r}_i^2-r_{ix}(r_{ic}{\omega }_c)]\\ & =m_i[{\omega }_x{r}_i^2-x_i(x_i{\omega }_x+y_i{\omega }_y+z_i{\omega }_z)]\\ & =m_i[{\omega }_x(r_i^2-x_i^2)-x_i{y}_i{\omega }_y-x_i{z}_i{\omega }_z],\\ L_y& =m_i[{\omega }_y(r_i^2-y_i^2)-y_i{x}_i{\omega }_x-y_i{z}_i{\omega }_z],\\ L_z& =m_i[{\omega }_z(r_i^2-z_i^2)-z_i{x}_i{\omega }_x-z_i{y}_i{\omega }_y].\end{array}$$

Hence, we can get an interial tensor,

$$\begin{array}{rl}L=\left(\begin{array}{c}{L}_x\\ L_y\\ L_z\end{array}\right)& =\left(\begin{array}{ccc}{m}_i(r_i^2-x_i^2)& -m_i{x}_i{y}_i& -m_i{x}_i{z}_i\\ -m_i{y}_i{x}_i& m_i(r_i^2-y_i^2)& -m_i{y}_i{z}_i\\ -m_i{z}_i{x}_i& -m_i{z}_i{y}_i& m_i(r_i^2-z_i^2)\end{array}\right)\left(\begin{array}{c}{\omega }_x\\ {\omega }_y\\ {\omega }_z\end{array}\right)=I\omega .\end{array}$$

The kinetic energy,

$$\begin{array}{rl}T& =\frac{1}{2}{m}_i{\overrightarrow{v}}_i^2\\ & =\frac{1}{2}{m}_i{\overrightarrow{v}}_i\cdot (\overrightarrow{\omega }\times {\overrightarrow{r}}_i)\\ & =\frac{1}{2}{m}_i\overrightarrow{\omega }\cdot ({\overrightarrow{r}}_i\times {\overrightarrow{v}}_i)\\ & =\frac{1}{2}\overrightarrow{\omega }\cdot ({\overrightarrow{r}}_i\times {\overrightarrow{p}}_i)\\ & =\frac{1}{2}\overrightarrow{\omega }\cdot {\overrightarrow{L}}_i\\ & =\frac{1}{2}\overrightarrow{\omega }\cdot \overrightarrow{L}\\ & =\frac{1}{2}{\omega }^{T}I\omega .\end{array}$$

Recall for a collection of particles, $\frac{d\overrightarrow{L}}{dt}={\overrightarrow{\tau }}^{(e)}$ and $\frac{d{\overrightarrow{L}}_{CM}}{dt}={\overrightarrow{\tau }}_{CM}^{(e)}$.

For our body-fixed,

$$\begin{array}{rl}{\left(\frac{d\overrightarrow{L}}{dt}\right)}_{inertial}& ={\left(\frac{d\overrightarrow{L}}{dt}\right)}_{body}+\overrightarrow{\omega }\times \overrightarrow{L}={\overrightarrow{\tau }}^{(e)}.\end{array}$$

For a principle axes (which is a special body fixed axes) the inertia matrix is diagonal.

Then for principle axes,

$$\begin{array}{rl}T& =\frac{1}{2}{\omega }^{T}I\omega \\ & =\frac{1}{2}{\omega }_i^2{I}_{ii}.\end{array}$$

And,

$$\begin{array}{rl}{T}_s^{(e)}& ={\tau }^{(e)}\cdot {\hat{e}}^{(s)}.\end{array}$$

Is the $s-$th principle axes’ torque. Similarly,

$$\begin{array}{rl}{\overrightarrow{L}}_s& =\overrightarrow{L}\cdot {\hat{e}}^{(s)},\\ {\overrightarrow{\omega }}_s& =\overrightarrow{\omega }\cdot {\hat{e}}^{(s)}.\end{array}$$

We will derive Euler’s equation from this.

Consider a cue ball on a pool table (one dimensional problem) with an initial velocity and no slippage. How long does it take to stop rolling? $R,M,\mu$ parameters.

$I_{sphere}=\frac{2}{5}M{R}^2$.

Initial conditions: $X(0)=0,\varphi (0)=0,\dot{X}(0)=v_0\ne 0$. If the ball is struck so it translates before rolling, then $\dot{\varphi }(0)=0$.

The body fixed axis is then ${\hat{e}}_i$ and the center of mass is ${\hat{e}}_i^0$. The frictional force is, ${\overrightarrow{F}}_f=-\mu gM{\hat{e}}_2^0$. Then, $\ddot{X}=-\mu g$. So, $x(t)=v_{0}t-\frac{1}{2}\mu g{t}^2$.

We also know, ${\omega }_1={\omega }_2=0$. So, $I_3\frac{d{\omega }_3}{dt}=-F_{f}R=-\mu gMR$. $I_3=\frac{2}{5}M{R}^2$. So, $I_3\ddot{\varphi }=-F_{f}R$. Then, $R\ddot{\varphi }=\frac{\mu gMR}{\frac{2}{5}MR}=\frac{5}{2}\mu g$.

Let $y$ be the vertical position. Let $x^{\prime }$ and $y^{\prime }$ axes be the center of mass fixed axes.

The ball starts to roll when the contact point at the bottom has zero velocity. $I_3\frac{d{\omega }_3}{dt}={\gamma }^{(e)}=\overrightarrow{r}\times \overrightarrow{F}\Rightarrow I_3\ddot{\varphi }=-F_{f}R$. Then, $I_3=\frac{2}{5}M{R}^2$. So, $\ddot{x}=-\mu g$ with this, $R\ddot{\varphi }=\frac{5}{2}\mu g$. $\dot{x}=v_0-gt=-a\dot{\varphi }=-\frac{5}{2}\mu gt$. So, $t$ can be determined to be when those two are equal.

When rotating about the highest moment of inertia, then the stability of the rotation is maximized.

About the principle axes, assume we have the rotation, $\overrightarrow{\omega }={\omega }_1{\hat{e}}_1+{\lambda }_2{\hat{e}}_2+{\lambda }_3{\hat{e}}_3$. Then,

$$\begin{array}{rl}I\overrightarrow{\omega }& =\left(\begin{array}{c}{I}_1{\omega }_1\\ I_2{\lambda }_2\\ I_3{\lambda }_3\end{array}\right).\end{array}$$

For a pure rotation:

$$\begin{array}{rl}\mathcal{L}& =\frac{1}{2}(I_1^2{\dot{\varphi }}_1^2+I_2^2{\dot{\varphi }}_2^2+I_3^2{\dot{\varphi }}_3^2).\end{array}$$

Then,

$$\begin{array}{rl}0=\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial {\dot{\varphi }}_i}\right)-\frac{\partial \mathcal{L}}{\partial {\varphi }_i}& =I_i^2{\ddot{\varphi }}_i\end{array}$$

Then, ${\varphi }_i(t)=A\exp (\pm \frac{t}{I_i})$. ${\dot{\varphi }}_1(0)={\omega }_1$. ${\dot{\varphi }}_i(0)={\lambda }_i$. Then,

$$\begin{array}{rl}{\lambda }_i& =\frac{A_i}{I_i}-\frac{(-A_i)}{I_i}\\ A_i& =\frac{1}{2}{\lambda }_i{I}_i.\end{array}$$

So,

$$\begin{array}{r}{\varphi }_1(t)=\frac{{\omega }_1{I}_1}{2}(\exp \left(\frac{t}{I_1}\right)-\exp (-\frac{t}{I_1})),\\ {\varphi }_2(t)=\frac{{\lambda }_2{I}_2}{2}(\exp \left(\frac{t}{I_2}\right)-\exp (-\frac{t}{I_2})),\\ {\varphi }_3(t)=\frac{{\lambda }_3{I}_3}{2}(\exp \left(\frac{t}{I_3}\right)-\exp (-\frac{t}{I_3})),\end{array}$$

Alternatively,

Suppose $I_3>I_2>I_1$. $I_1\frac{d{\omega }_1}{dt}={\lambda }_2{\lambda }_3(I_2-I_3)\Rightarrow I_1\frac{d{\omega }_1}{dt}\approx =0\Rightarrow {\omega }_1=\mathbb{C}$.

$I_2\frac{d{\lambda }_2}{dt}={\omega }_1{\lambda }_3(I_3-I_1)$.

$I_3\frac{d{\lambda }_3}{dt}={\omega }_1{\lambda }_2(I_1-I_2)$.

$\Rightarrow {\dot{\lambda }}_2=\left(\frac{I_3-I_1}{I_2}{\omega }_1\right){\lambda }_3,{\dot{\lambda }}_3=\left(\frac{I_1-I_2}{I_3}\right)\lambda$. ${\dot{\lambda }}_2+\frac{(I_1-I_3)(I_1-I_2)}{I_2{I}_3}{\lambda }_2=0$.

Then, $\lambda =\exp (\pm i\mathrm{\Omega }t)$.

Suppose $I_3>I_1>I_2$. ${\dot{\lambda }}_2+\frac{(I_1-I_3)(I_1-I_2)}{I_2{I}_3}{\lambda }_2=0$. Then, $\lambda =\exp (\pm \mathrm{\Omega }t)$.

First, rotate along ${\hat{e}}_3^0$ by an angle $\alpha$ then by ${\hat{e}}_2={\hat{e}}_{line-of-note}$ by an angle $\beta$ and finally by ${\hat{e}}_3$ by $\gamma$.

So,

$$\begin{array}{r}\overrightarrow{\omega }=\dot{\alpha }{\hat{e}}_{\alpha }+\dot{\beta }{\hat{e}}_{\beta }+\dot{\gamma }{\hat{e}}_{\gamma }\\ \overrightarrow{\omega }=\dot{\alpha }{\hat{e}}_3^0+\dot{\beta }{\hat{e}}_{line-of-note}+\dot{\gamma }{\hat{e}}_3.\end{array}$$

For body-fixed principle axes, ${\hat{e}}_{\gamma }={\hat{e}}_3$ will be the body fixed. Then,

$$\begin{array}{rl}{\hat{e}}_{\alpha }={\hat{e}}_3^0& =\cos \beta {\hat{e}}_3+\sin \gamma \sin \beta {\hat{e}}_2-\cos \gamma \sin \beta {\hat{e}}_1\end{array}$$ $$\begin{array}{rl}{\hat{e}}_{\beta }={\hat{e}}_2& =\left|\begin{array}{ccc}{\hat{e}}_1& {\hat{e}}_2& {\hat{e}}_3\\ 0& 0& 1\\ -\cos \gamma \sin \beta & \sin \gamma \sin \beta & \cos \beta \end{array}\right|\\ & =-\sin \gamma \sin \beta {\hat{e}}_1+\cos \gamma \sin \beta {\hat{e}}_2,\\ & =-\sin \gamma {\hat{e}}_1+\cos \gamma {\hat{e}}_2.\end{array}$$

${\hat{e}}_{\beta }=\sin \gamma {\hat{e}}_1+\cos \gamma {\hat{e}}_2$.

Then, $\overrightarrow{\omega }=\dot{\alpha }{\hat{e}}_{\alpha }+\dot{\beta }{\hat{e}}_{\beta }+\dot{\gamma }{\hat{e}}_{\gamma }$. So, ${\omega }_1=-\dot{\alpha }\sin \beta \cos \gamma +\dot{\beta }\sin \gamma$. So, ${\omega }_2=\dot{\alpha }\sin \beta \sin \gamma +\dot{\beta }\cos \gamma$. So, ${\omega }_3=\dot{\alpha }\cos \beta \dot{\gamma }$.

Consider a football with a body fixed principle axes ${\hat{e}}_i$ and a fixed coordinate system ${\hat{e}}_i^0$. Suppose we have no torque on the system. Then, we have Euler angles, $\alpha$, $\beta$, $\gamma$ to transform to the body-fixed axes – our generalized coordinates. Then, we have ${\omega }_i$ for the body-fixed axes.

A football is a symmetric top so $I_1=I_2\ne I_3$.

Then, $\hat{\mathcal{L}}=\frac{1}{2}(I_i{\omega }_i^2)=\frac{1}{2}{I}_1({\omega }_1^2+{\omega }_2^2)+\frac{1}{2}{I}_3{\omega }_3^2$. So,

$$\begin{array}{rl}\mathcal{L}& =\frac{1}{2}{I}_1({\dot{\alpha }}^2{\sin }^2\beta +{\dot{\beta }}^2)+\frac{1}{2}{I}_3{(\dot{\alpha }\cos \beta +\dot{\gamma })}^2.\end{array}$$ $$\begin{array}{r}\frac{d}{dt}(I_1\dot{\alpha }{\sin }^2\beta +I_3\cos \beta (\dot{\alpha }\cos \beta +\dot{\gamma }))=0\\ P_{\alpha }=I_1\dot{\alpha }{\sin }^2\beta +I_3\cos \beta (\dot{\alpha }\cos \beta +\dot{\gamma })={\mathbb{C}}_1\\ \frac{d}{dt}(I_3(\dot{\alpha }\cos \beta +\dot{\gamma }))=0\\ P_{\gamma }=I_3(\dot{\alpha }\cos \beta +\dot{\gamma })=I_3{\omega }_3={\mathbb{C}}_2\\ P_{\gamma }& =I_1\dot{\beta },\\ P_i& =\overrightarrow{L}\cdot {\hat{e}}_i.\end{array}$$

Thus, the angular velocity about the third axes is constant - precession?

Choose ${\hat{e}}_3^0$ such that it is aligned with the angular momentum. So, $P_{\alpha }=|\overrightarrow{L}|={\mathbb{C}}_1$ and $P_{\gamma }=|\overrightarrow{L}|\cos \beta ={\mathbb{C}}_2$, hence $\dot{\beta }=0$ – Precession. Thus (looking back at our equations and seeing the new constant variables), $\dot{\alpha }={\mathbb{C}}_3\Rightarrow \ddot{\alpha }=0$ – Precession. Further, then $\dot{\gamma }={\mathbb{C}}_4$. Thus, it precesses at a constant rate at a fixed angle and spins at a constant rate.