### Conservative forces Conservative forces result in no net work when a particle is moved in a closed path. Conservative forces are store potential energy: $ \delta W_{C} = \sum \vec{f}_{k}\cdot d\vec{r}_{k} =-\delta \mathcal{V} $ More generally, $ f = -\frac{ \partial \mathcal{V} }{ \partial x } $ Including nonconservative forces gives us the relation $ \Delta \mathcal{T}+\Delta \mathcal{V}=W_{NC} $ ### Gravitational Potential Energy $\mathcal{V}=Mgz_{C}+c$ ### Kinetic Energy $\mathcal{T} =\frac{1}{2}M|\vec{v}_{C}|^{2}+\frac{1}{2}\vec{\omega}_{E}^{T}~I_{C}~\vec{\omega}_{E}$ ### Generalized Forces Rewriting the work equation in terms of generalized coordinates gives us generalized forces corresponding to each generalized coordinate: $ \delta W_{NC} = \sum \vec{f}_{k} \cdot\, \delta \vec{r}_{k} = \sum Q_{j} \, \delta q_{j} $ $\delta W_{NC} =$ variational work done by nonconservative forces $\vec{f}_{k} =$ nonconservative force $\delta \vec{r}_{k} =$ displacement resulting from ${} \vec{f}_{k} {}$ $Q_{j} =$ generalized force responsible for $\delta q_{j}$ $\delta q_{j} =$ variation in a generalized coordinate Another way to write this is: $ Q_{j} = \sum_{k}\vec{f}_{k}\cdot \frac{ \partial \vec{r}_{k} }{ \partial q_{j} } $ ### Euler-Lagrange Equations When there are a large number of rigid bodies but well understood degrees of freedom, Euler-Lagrange equations are a way to skip internal forces and derive the equations of motion for the degrees of freedom of concern. $ Q_{j} = \frac{d}{dt}\left( \frac{ \partial \mathcal{T} }{ \partial \dot{q}_{j} } \right) - \frac{ \partial T }{ \partial q_{j} } +\frac{ \partial \mathcal{V} }{ \partial q_{j} } $ #### Derivation Angular momentum is left out because it's exactly the same process 1. Start with conservation of momentum for $i$ rigid bodies $\vec{f}_{i}= m_{i}\dot{\vec{v}}_{i}$ 2. Dot both sides by $\delta \vec{r}_{i}$, the variation in CoM of body $i$ due to $\vec{f}_{i}$, and sum together the equations for all rigid bodies $\sum\vec{f}_{i}\cdot \delta \vec{r}_{i}=\sum m_{i}\dot{\vec{v}}_{i}\cdot\delta \vec{r}_{i}$ 3. Let $\vec{r}'_{ij}=\frac{ \partial \vec{r}_{i} }{ \partial q_{j} }$ and rewrite in terms of the variations of generalized coordinates $\delta q_{j}$ $ \sum_{k} \vec{f}_{k}\cdot \vec{r}'_{kj}\delta q_{j} = \sum_{i} m_{i}\dot{\vec{v}}_{i}\cdot \vec{r}'_{ij}\delta q_{j} $ 4. The conservative forces on the left side follow $dW = \sum_{k}\vec{f}_{k}\cdot d\vec{r}_{k}=-d\mathcal{V}$ , so $ \sum \vec{f}_{k}\cdot \vec{r}_{kj}'\delta q_{j}=-\frac{ \partial \mathcal{V} }{ \partial q_{j} }\delta q_{j} $ 5. The nonconservative forces on the left side by definition sum to $Q_{j}\delta q_{j}$ 6. Repeated product and chain rules allow us to rewrite the right side as $ m \dot{\vec{v}}\cdot \vec{r}'\delta q = \left( \frac{d}{dt}\left( \frac{ \partial \mathcal{T} }{ \partial \dot{q} } \right) - \frac{ \partial \mathcal{T} }{ \partial q } \right) \delta q $ 7. Since $\delta q_{j}$ are independant, we can "turn them off" arbitrarily to split the resulting equation into the Euler-Lagrange equations: $ Q_{j} = \frac{d}{dt}\left( \frac{ \partial \mathcal{T} }{ \partial \dot{q} } \right)-\frac{ \partial T }{ \partial q_{j} } -\frac{ \partial \mathcal{V} }{ \partial q_{j} } $