# Motion of Rigid Bodies

Consider a system of $$N$$ particles. The position and mass of each particle is $$\vec{r}_i,m_i$$ respectively. Let $$\vec{R}_{CM}$$ be the position of the center of mass, the weighted position, $$\vec{R}_{CM} = \vec{R} = \frac{m_i\vec{r}_i}{m_i}$$ (See Eisenstein Summation Notation).

The $$\dot{\vec{p}}_i = \vec{F}_i^{(e)} + \vec{F}_{ji}^{(interaction)}$$. Write $$\vec{F}_{ji}^{(interaction)} = \vec{F}_{ji}$$. Note that $$i=j\Rightarrow\vec{F}_{ji} = \vec{0}$$. $$\vec{F}_{ij} = -\vec{F}_{ji}$$.

The rate of change of the total momentum is: $$\dot{\vec{P}} = \frac{d}{dt}\left(m_i\vec{v}_i\right) = \vec{F}_i^{(e)} + \vec{F}_{ji} = \frac{d^2}{dt^2}\left(m_i\vec{r}_i\right) = \vec{F}_i^{(e)} = \vec{F}_{total}^{(e)}$$. Thus, the total momentum only changes from external forces. Hence, $$\frac{d^2}{dt^2}M\vec{R} = M\ddot{\vec{R}} = \vec{F}_{total}^{(e)}$$.

## Angular Momentum

\begin{align*} \vec{L} &= \vec{L}_i \\ &= \vec{r}_i\times\vec{p}_i \\ \frac{d}{dt}\vec{L} &= \frac{d}{dt}\left(\vec{r}_i\times\vec{p}_i\right) \\ &= \vec{r}_i\times\dot{\vec{p}}_i \\ &= \vec{r}_i\times\vec{F}_i^{(e)}. \\ \dot{\vec{L}} &= \vec{\tau}^{(e)} \end{align*}

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

Write $$\vec{r}_i = \vec{R} + \vec{r}_i'$$ and $$\vec{v}_i = \vec{V} + \vec{v}_i'$$.

\begin{align*} \vec{L} &= m_i(\vec{R} + \vec{r}_i')\times(\vec{V} + \vec{v}_i') \\ &= \vec{R}\times m_i\vec{V} + \vec{r}_i'\times m_i\vec{v}_i' + m_i\vec{r}_i'\times\vec{V} + \vec{R}\times m_i\vec{v}_i' \\ &= \vec{R}\times m_i\vec{V} + \vec{r}_i'\times m_i\vec{v}_i' \\ &= \vec{R}\times M\vec{V} + \vec{r}_i'\times m_i\dot{\vec{r}}_i' \\ &= \vec{L}_{CM} + \vec{L}'. \end{align*}

Then,

\begin{align*} \dot{\vec{L}} &= \dot{\vec{L}}_{CM} + \dot{\vec{L}}' \\ &= \frac{d}{dt}\left(\vec{R}\times\vec{P}\right) + \frac{d}{dt}\left(\vec{L} - \vec{L}_{CM}\right) \\ &= \vec{R}\times\dot{\vec{P}} + \vec{\tau}^{(e)} - \dot{\vec{L}}_{CM} \\ &= \vec{R}\times\vec{F}_{total}^{(e)} + \vec{\tau}^{(e)} - \dot{\vec{L}}_{CM} \\ &= \vec{\tau}^{(e)} = \vec{r}_i\times\vec{F}_i^{(e)}. \end{align*}

Note,

\begin{align*} \dot{\vec{L}}' &= \frac{d}{dt}\left(\vec{L} - \vec{L}_{CM}\right) \\ &= (\vec{r}_i-\vec{R})\times\vec{F}_i^{(e)} = \vec{r}_i'\times\vec{F}_i^{(e)}. \end{align*}

## Kinetic Energy

\begin{align*} T &= \frac{1}{2}m_i\vec{v_i}^2 \\ &= \frac{1}{2}m_i\left(\vec{V}+\vec{v}_i\right)\cdot\left(\vec{V}+\vec{v}_i\right) \\ &= \frac{1}{2}M\vec{V}^2 + \frac{1}{2}m_i\vec{v}_i'. \end{align*}

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

## Axes of Motion

Consider two different coordinate systems that share the same origin.

\begin{align*} \vec{v} &= \vec{v}_i^0\hat{e}_i^{0} &= \vec{v}_i^1\hat{e}_i^{1} \\ \frac{d}{dt}\left(\vec{v}\right)_{inertial} &= \frac{d}{dt}\left(\vec{v}_{i}^{0}\hat{e}_{i}^{0}\right) &= \frac{d}{dt}\left(\vec{v}_{i}^{1}\hat{e}_{i}^{1}\right) \\ &= \frac{d}{dt}\left(\vec{v}_{i}^{0}\right)\hat{e}_{i}^{0} &= \frac{d}{dt}\left(\vec{v}_{i}^{1}\right)\hat{e}_{i}^{1} \\ \end{align*}

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

\begin{align*} \vec{v} &= \vec{v}_i\hat{e}_i^{1} \\ \frac{d}{dt}\left(\vec{v}\right)_{inertial} &= \frac{d}{dt}\left(\vec{v}_{i}^{1}\hat{e}_{i}^{1}\right) \\ &= \frac{d}{dt}\left(\vec{v}_{i}^{1}\right)\hat{e}_{i}^{1} + \vec{v}_{i}^{1}\frac{d}{dt}\left(\hat{e}_{i}^{1}\right) \\ &= \left(\frac{d\vec{v}}{dt}\right)_{body} + \vec{\omega}\times\vec{v}. \end{align*}

Hence we have the inertial frame and the body-fixed axes. Note that $$\vec{\omega}$$ and $$\vec{v}$$ 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, $$\vec{R} = \vec{R}_A - \vec{R}_B$$. So,

\begin{align*} \frac{d}{dt}\left(\vec{R}_B\right)_{inertial} &= \left(\frac{d\vec{R}_A}{dt}\right)_{body} + \vec{\omega}_B\times\vec{R}. \end{align*}

But we also have,

\begin{align*} \frac{d}{dt}\left(\vec{R}_A\right)_{inertial} &= \left(\frac{d\vec{R}_B}{dt}\right)_{body} + \vec{\omega}_A\times(-\vec{R}). \end{align*}

\begin{align*} 0 &= (\vec{\omega}_B - \vec{\omega}_A)\times\vec{R} \Rightarrow\vec{\omega}_A=\vec{\omega}_B. \end{align*}

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

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

\begin{align*} \frac{d}{dt}\left(\vec{r}_i\right)_{inertial} &= \left(\frac{d\vec{i}_i}{dt}\right)_{body} + \vec{\omega}\times\vec{r}_i \nonumber \\ &= \vec{\omega}\times\vec{r}_i. \end{align*}

## Total Angular Momentum

\begin{align*} \vec{L} &= \vec{r}_i\times \vec{p}_i \\ &= \vec{r}_i\times m_i\vec{v}_i \\ &= m_i\vec{r}_i\times (\vec{\omega}\times\vec{r}_i) \\ &= m_i\left[\vec{\omega}r_i^2 - \vec{r}_i(\vec{r}_i\cdot\vec{\omega})\right], \\ &= L_x\hat{i} + L_y\hat{j} + L_z\hat{k}, \\ L_x &= m_i[\omega_xr_i^2 - r_{ix}(\vec{r}_i\cdot\omega)] \\ &= m_i[\omega_xr_i^2 - r_{ix}(r_{ic}\omega_c)] \\ &= m_i[\omega_xr_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_iy_i\omega_y - x_iz_i\omega_z], \\ L_y &= m_i[\omega_y(r_i^2 - y_i^2) - y_ix_i\omega_x - y_iz_i\omega_z], \\ L_z &= m_i[\omega_z(r_i^2 - z_i^2) - z_ix_i\omega_x - z_iy_i\omega_y]. \end{align*}

Hence, we can get an interial tensor,

\begin{align*} L=\begin{pmatrix} L_x \\ L_y \\ L_z \end{pmatrix} &= \begin{pmatrix} m_i(r_i^2 - x_i^2) & -m_ix_iy_i & -m_ix_iz_i \\ -m_iy_ix_i & m_i(r_i^2 - y_i^2) & -m_iy_iz_i \\ -m_iz_ix_i & -m_iz_iy_i & m_i(r_i^2 - z_i^2) \end{pmatrix}\begin{pmatrix} \omega_x \\ \omega_y \\ \omega_z \end{pmatrix} = I\omega. \end{align*}

The kinetic energy,

\begin{align*} T &= \frac{1}{2}m_i\vec{v}_i^2 \\ &= \frac{1}{2}m_i\vec{v}_i\cdot(\vec{\omega}\times\vec{r}_i) \\ &= \frac{1}{2}m_i\vec{\omega}\cdot(\vec{r}_i\times\vec{v}_i) \\ &= \frac{1}{2}\vec{\omega}\cdot(\vec{r}_i\times\vec{p}_i) \\ &= \frac{1}{2}\vec{\omega}\cdot\vec{L}_i \\ &= \frac{1}{2}\vec{\omega}\cdot\vec{L} \\ &= \frac{1}{2}\omega^T I\omega. \end{align*}

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

For our body-fixed,

\begin{align*} \left(\frac{d\vec{L}}{dt}\right)_{inertial} &= \left(\frac{d\vec{L}}{dt}\right)_{body} + \vec{\omega}\times\vec{L} = \vec{\tau}^{(e)}. \end{align*}

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

Then for principle axes,

\begin{align*} T &= \frac{1}{2}\omega^T I\omega \\ &= \frac{1}{2}\omega_i^2I_{ii}. \end{align*}

And,

\begin{align*} T^{(e)}_s &= \tau^{(e)}\cdot\hat{e}^{(s)}. \end{align*}

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

\begin{align*} \vec{L}_s &= \vec{L}\cdot\hat{e}^{(s)}, \\ \vec{\omega}_s &= \vec{\omega}\cdot\hat{e}^{(s)}. \end{align*}

We will derive Euler’s equation from this.

### Case 1

For a Rb that moves with respect to a fixed internal point. The inertia tensor is time-independent and the omega has all the time dependence: $$\vec{L}(t) = I\omega(t)$$.

### Case 2

For a RB that moves without a fixed point. Then, fix the center of mass coordinates as the same directions as an inertial frame. The body-fixed coordinates may be rotating relative to the center of mass, but shares the same origin. Then, $$\left(\frac{d\vec{L}'}{dt}\right)_{CM} = \left(\frac{d\vec{L}'}{dt}\right)_{body} + \vec{\omega}\times\vec{L}' = \vec{\tau}^{(e)'}$$. Hence, the CM coordinates are a non-inertial frame.

If we choose the Body-Fixed axes to be the principle axes, we get a diagonal Inertia Tensor. So, $$\vec{L}'(t) = I\omega(t) = I_1\omega_1(t)\hat{e}_1 + I_2\omega_2(t)\hat{e}_2 + I_3\omega_3(t)\hat{e}_3$$, so the time dependence is only in the omega. Hence, $$\left(\frac{d\vec{L}}{dt}\right)_{cm} = (I_1\frac{d\omega_1}{dt}(t) + (\omega_2 L'_3 - \omega_3 L'_2))\hat{e}_1 + (I_2\frac{d\omega_2}{dt}(t) - (\omega_3 L'_1 - \omega_1 L'_3))\hat{e}_2 + (I_3\frac{d\omega_3}{dt}(t) + (\omega_1 L'_2 - \omega_2 L'_1))\hat{e}_3$$. So, $$\left(\frac{d\vec{L}}{dt}\right)_{CM} = I_1\frac{d\omega_1}{dt}\hat{e}_1 + I_2\frac{d\omega_2}{dt}\hat{e}_2 + I_3\frac{d\omega_3}{dt}\hat{e}_3 + \omega_2\omega_3(I_3-I_2)\hat{e}_1 + \omega_1\omega_3(I_1 - I_3)\hat{e}_2 + \omega_1\omega_2(I_2 - I_1)\hat{e}_3$$. Hence the relative torque is,

\begin{align*} \vec{\tau}^{(e)} &= \begin{pmatrix} I_1\frac{d\omega_1}{dt} + \omega_2\omega_3(I_3-I_2) \\ I_2\frac{d\omega_2}{dt} + \omega_1\omega_3(I_1-I_3) \\ I_3\frac{d\omega_3}{dt} + \omega_2\omega_1(I_2-I_1) \end{pmatrix}. \end{align*}

## Example

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.

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

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

The body fixed axis is then $$\hat{e}_i$$ and the center of mass is $$\hat{e}_i^0$$. The frictional force is, $$\vec{F}_f = -\mu gM\hat{e}_2^0$$. Then, $$\ddot{X} = -\mu g$$. So, $$x(t) = v_0t - \frac{1}{2}\mu g t^2$$.

We also know, $$\omega_1 = \omega_2 = 0$$. So, $$I_3\frac{d\omega_3}{dt} = -F_fR = -\mu gM R$$. $$I_3 = \frac{2}{5}MR^2$$. So, $$I_3\ddot{\phi} = -F_f R$$. Then, $$R\ddot{\phi} = \frac{\mu gM R}{\frac{2}{5}MR} = \frac{5}{2}\mu g$$.

Let $$y$$ be the vertical position. Let $$x'$$ and $$y'$$ axes be the center of mass fixed axes.

The ball starts to roll when the contact point at the bottom has zero velocity. $$I_3 \frac{d\omega_3}{dt} = \gamma^{(e)} = \vec{r}\times\vec{F} \Rightarrow I_3\ddot{\phi} = -F_f R$$. Then, $$I_3 = \frac{2}{5}MR^2$$. So, $$\ddot{x} = -\mu g$$ with this, $$R\ddot{\phi} = \frac{5}{2}\mu g$$. $$\dot{x} = v_0 - gt = -a\dot{\phi} = -\frac{5}{2}\mu g t$$. 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, $$\vec{\omega} = \omega_1\hat{e}_1 + \lambda_2\hat{e}_2 + \lambda_3\hat{e}_3$$. Then,

\begin{align*} I\vec{\omega} &= \begin{pmatrix} I_1\omega_1 \\ I_2\lambda_2 \\ I_3\lambda_3 \end{pmatrix}. \end{align*}

For a pure rotation:

\begin{align*} \mathcal{L} &= \frac{1}{2}(I_1^2\dot{\phi}_1^2 + I_2^2\dot{\phi}_2^2 + I_3^2\dot{\phi}_3^2). \end{align*}

Then,

\begin{align*} 0=\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{\phi}_i}\right) - \frac{\partial\mathcal{L}}{\partial\phi_i} &= I_i^2\ddot{\phi}_i \\ \end{align*}

Then, $$\phi_i(t) = A\exp\left(\pm \frac{t}{I_i}\right)$$. $$\dot{\phi}_1(0) = \omega_1$$. $$\dot{\phi}_i(0) = \lambda_i$$. Then,

\begin{align*} \lambda_i &= \frac{A_i}{I_i} - \frac{(-A_i)}{I_i} \\ A_i &= \frac{1}{2}\lambda_iI_i. \end{align*}

So,

\begin{align*} \phi_1(t) = \frac{\omega_1I_1}{2}\left(\exp\left(\frac{t}{I_1}\right) - \exp\left(-\frac{t}{I_1}\right)\right), \\ \phi_2(t) = \frac{\lambda_2I_2}{2}\left(\exp\left(\frac{t}{I_2}\right) - \exp\left(-\frac{t}{I_2}\right)\right), \\ \phi_3(t) = \frac{\lambda_3I_3}{2}\left(\exp\left(\frac{t}{I_3}\right) - \exp\left(-\frac{t}{I_3}\right)\right), \end{align*}

Alternatively,

Suppose $$I_3 > I_2 > I_1$$. $$I_1\frac{d\omega_1}{dt} = \lambda_2\lambda_3(I_2 - I_3)\Rightarrow I_1\frac{d\omega_1}{dt}\approx = 0\Rightarrow\omega_1=\mathbb{C}$$.

$$I_2\frac{d\lambda_2}{dt} = \omega_1\lambda_3(I_3 - I_1)$$.

$$I_3\frac{d\lambda_3}{dt} = \omega_1\lambda_2(I_1 - I_2)$$.

$$\Rightarrow \dot{\lambda}_2 = \left(\frac{I_3-I_1}{I_2}\omega_1\right)\lambda_3, \dot{\lambda}_3 = \left(\frac{I_1 - I_2}{I_3}\right)\lambda$$. $$\dot{\lambda}_2 + \frac{(I_1-I_3)(I_1-I_2)}{I_2I_3}\lambda_2 = 0$$.

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

Suppose $$I_3 > I_1 > I_2$$. $$\dot{\lambda}_2 + \frac{(I_1-I_3)(I_1-I_2)}{I_2I_3}\lambda_2 = 0$$. Then, $$\lambda = \exp(\pm \Omega t)$$.

## Rotations

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{align*} \vec{\omega} = \dot{\alpha}\hat{e}_\alpha + \dot{\beta}\hat{e}_{\beta} + \dot{\gamma}\hat{e}_\gamma \\ \vec{\omega} = \dot{\alpha}\hat{e}_3^0 + \dot{\beta}\hat{e}_{line-of-note} + \dot{\gamma}\hat{e}_3. \end{align*}

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

\begin{align*} \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{align*} \begin{align*} \hat{e}_\beta = \hat{e}_2 &= \begin{vmatrix} \hat{e}_1 & \hat{e}_2 & \hat{e}_3 \\ 0 & 0 & 1 \\ -\cos\gamma\sin\beta & \sin\gamma\sin\beta & \cos\beta \end{vmatrix} \\ &= -\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{align*}

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

Then, $$\vec{\omega} = \dot{\alpha}\hat{e}_\alpha + \dot{\beta}\hat{e}_\beta + \dot{\gamma}\hat{e}_\gamma$$. 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 $$I_1 = I_2\neq I_3$$.

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{align*} \mathcal{L} &= \frac{1}{2}I_1 (\dot{\alpha}^2\sin^2\beta + \dot{\beta}^2) + \frac{1}{2}I_3\left(\dot{\alpha}\cos\beta + \dot{\gamma}\right)^2. \end{align*} \begin{align*} \frac{d}{dt}\left(I_1\dot{\alpha}\sin^2\beta + I_3\cos\beta(\dot{\alpha}\cos\beta + \dot{\gamma})\right) = 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}\left(I_3(\dot{\alpha}\cos\beta + \dot{\gamma})\right) = 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 &= \vec{L}\cdot\hat{e}_i. \end{align*}

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, $$P_\alpha = |\vec{L}| = \mathbb{C}_1$$ and $$P_\gamma = |\vec{L}|\cos\beta = \mathbb{C}_2$$, 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.

Created: 2024-05-30 Thu 21:17

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