Rigid Body Kinetics

Euler’s 1st law is the balance of linear momentum extended to rigid bodies. It says that the sum of external forces acting on the rigid body equals the change in linear momentum of the rigid body:

\begin{align} \mathbf{F} = \dot{\mathbf{G}} \end{align}

Equivalently, if the mass of the system is conserved, then we have

\begin{align} \mathbf{F} = m\mathbf{a}_C \end{align}

where \(\mathbf{a}_C\) is the acceleration of the center of mass.

Euler’s 2nd law is the balance of angular momentum extended to rigid bodies. It has the following three equivalent forms:

\begin{align} \mathbf{M}^O &= \dot{\mathbf{H}}^O \\ \mathbf{M}^C &= \dot{\mathbf{H}}^C \\ \mathbf{M}^P &= \dot{\mathbf{H}}^P + (\mathbf{v}_P-\mathbf{v}_C)\times\mathbf{G} = \dot{\mathbf{H}}^C + (\mathbf{r}_C-\mathbf{r}_P)\times m\mathbf{a}_C \end{align}

where \(O\) is some fixed material point, \(C\) is the center of mass, and \(P\) is any material point on the rigid body.

For the case of planar motion, when \(\mathbf{\omega} = \dot{\theta}\mathbf{E}_z\), the right hand side of these equations expands to become:

\begin{align} \dot{\mathbf{H}}^{O} &= (I_{xz}^{O}\dot{\omega}-I_{yz}^{O}\omega^{2})\mathbf{e}_{x}+(I_{yz}^{O}\dot{\omega}+I_{xz}^{O}\omega^{2})\mathbf{e}_{y}+I_{zz}^{O}\dot{\omega}\mathbf{E}_{z} \\ \dot{\mathbf{H}}^{C} &= (I_{xz}^{C}\dot{\omega}-I_{yz}^{C}\omega^{2})\mathbf{e}_{x}+(I_{yz}^{C}\dot{\omega}+I_{xz}^{C}\omega^{2})\mathbf{e}_{y}+I_{zz}^{C}\dot{\omega}\mathbf{E}_{z} \\ \dot{\mathbf{H}}^{P} &= (I_{xz}^{P}\dot{\omega}-I_{yz}^{P}\omega^{2})\mathbf{e}_{x}+(I_{yz}^{P}\dot{\omega}+I_{xz}^{P}\omega^{2})\mathbf{e}_{y}+I_{zz}^{P}\dot{\omega}\mathbf{E}_{z}+\frac{d}{dt}\left((\mathbf{r}_{C}-\mathbf{r}_{P})\times m\mathbf{v}_{P}\right) \end{align}

The linear impulse-linear momentum equation for rigid bodies is:

\begin{align} \boxed{\int_{t_1}^{t_2} \mathbf{F}\text{ d}t = \mathbf{G}(t_2) - \mathbf{G}(t_1)} \end{align}

There are two forms of the angular impulse-angular momentum equation for rigid bodies, one around a fixed point \(O\), and the other around the center of mass \(C\):

\begin{align} \boxed{\int_{t_1}^{t_2} \mathbf{M}^O\text{ d}t = \mathbf{H}^O(t_2) - \mathbf{H}^O(t_1)} \\ \boxed{\int_{t_1}^{t_2} \mathbf{M}^C\text{ d}t = \mathbf{H}^C(t_2) - \mathbf{H}^C(t_1)} \end{align}

We can view the kinetic energy of a rigid body as the sum of kinetic energies of each infinitesimal mass on the rigid body:

\begin{align} T = \int_B \frac{1}{2}\mathbf{v}\cdot\mathbf{v}\text{ d}m \notag \end{align}

We know that \(\mathbf{v}-\mathbf{v}_C = \mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\), so we can rewrite this integral as:

\begin{align} T &= \frac{1}{2} \int_B \left[ \mathbf{v}_C + \mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\right] \cdot \left[ \mathbf{v}_C + \mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\right]\text{ d}m \notag \\ &= \frac{1}{2}\int_B\mathbf{v}_C\cdot\mathbf{v}_C + 2\mathbf{v}_C\cdot\left(\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\right)+\left[\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\right]\cdot\left[\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\right]\text{ d}m \notag \\ &= \frac{1}{2}m\mathbf{v}_C\cdot\mathbf{v}_C + \mathbf{v}_C\cdot\mathbf{\omega}\times\int_B \mathbf{r}-\mathbf{r}_C\text{ d}m + \frac{1}{2}\int_B\left[\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\right]\cdot\left[\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\right]\text{ d}m \notag \\ &= \frac{1}{2}m\mathbf{v}_C\cdot\mathbf{v}_C + \frac{1}{2}\int_B\left[\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\right]\cdot\left[\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_C)\right]\text{ d}m \notag \end{align}

It can be shown that \((\mathbf{a} \times \mathbf{b})\cdot(\mathbf{c}\times\mathbf{d}) = (\mathbf{a}\cdot\mathbf{c})(\mathbf{b}\cdot\mathbf{d})-(\mathbf{a}\cdot\mathbf{d})(\mathbf{b}\cdot\mathbf{c})\), so

\begin{align} T &= \frac{1}{2}m\mathbf{v}_C \cdot\mathbf{v}_C + \frac{1}{2}\int_B\mathbf{\omega}\cdot\mathbf{\omega}\|\mathbf{r}-\mathbf{r}_C\|^2 - \left[\mathbf{\omega}\cdot(\mathbf{r}-\mathbf{r}_C)\right]\left[\mathbf{\omega}\cdot(\mathbf{r}-\mathbf{r}_C)\right]\text{ d}m \notag \\ &= \frac{1}{2}m\mathbf{v}_C \cdot\mathbf{v}_C + \frac{1}{2}\mathbf{\omega} \cdot \int_B \|\mathbf{r}-\mathbf{r}_C\| \mathbf{1} - (\mathbf{r}-\mathbf{r}_C)\otimes(\mathbf{r}-\mathbf{r}_C)\text{ d}m \mathbf{\omega} \notag \end{align}

Simplifying, we get that the kinetic energy of a rigid body is:

\begin{align} \boxed{T = \frac{1}{2}m\mathbf{v}_C\cdot\mathbf{v}_C + \frac{1}{2}\mathbf{I}^C\mathbf{\omega}\cdot\mathbf{\omega} = \frac{1}{2}m\mathbf{v}_C\cdot\mathbf{v}_C + \frac{1}{2}\mathbf{H}^C\cdot\mathbf{\omega}} \end{align}

If there is a fixed material point \(O\), then this simplifies to:

\begin{align} \boxed{T = \frac{1}{2} \mathbf{I}^O\mathbf{\omega}\cdot\mathbf{\omega} = \frac{1}{2}\mathbf{H}^O\cdot\mathbf{\omega}} \end{align}

Taking the derivative of the kinetic energy, we get the work-energy theorem for rigid bodies:

\begin{align} \dot{T} = \mathbf{F}\cdot\mathbf{v}_C + \mathbf{M}^C\cdot\mathbf{\omega} \end{align}