With d’Alembert’s principle we leave the realm of statics and enter the realm of dynamics.”
We extend the principle of virtual work, which is for statics, to In that sense, one might say that principle of virtual work is just a special case of d’Alembert’s principle for static systems.
Consider a system with \(N\) mass particles. \[ 0 = \sum_{i=1}^{N} ( \mathbf{F}_i - \dot{\mathbf{p}}_i ) \cdot \delta \mathbf{r}_i \;\;\;\;\;\;\; \sum_{i=1}^{N} \mathbf{F}_i \cdot \delta \mathbf{r}_i = \sum_{j=1}^{n} \bigg( \Xi_j - \frac{\partial V}{\partial q_j} \bigg) \delta q_j \]
The term that was omitted in class was: \[ \begin{align*} \sum_{i=1}^{N} \dot{\mathbf{p}}_i \cdot \delta \mathbf{r}_i &= \sum_{i=1}^{N} m_i \ddot{\mathbf{r}}_i \cdot \delta \mathbf{r}_i = \sum_{i=1}^{N} m_i \ddot{\mathbf{r}}_i \cdot \bigg( \sum_{j=1}^{n}\frac{\partial \mathbf{r}_i}{\partial q_j } \bigg)\delta q_j = \sum_{j=1}^{n} \bigg( \sum_{i=1}^{N} m_i \ddot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j}\bigg) \delta q_j \\ \sum_{i=1}^{N} m_i \ddot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j} &= \frac{d}{dt} \bigg( \sum_{i=1}^{N} m_i \dot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j} \bigg) - \sum_{i=1}^{N} m_i \dot{\mathbf{r}}_i \frac{d}{dt}\bigg( \frac{\partial \mathbf{r}_i}{\partial q_j }\bigg) \end{align*} \] There are two terms that we need to derive further. For the first one, we use the so called cancellation of dots: \[ \begin{align*} \dot{\mathbf{r}}_i &= \sum_{j=1}^{n}\frac{\partial \mathbf{r}_i}{\partial q_j} \dot{q}_j + \frac{\partial \mathbf{r}_i}{\partial t} ~~~~~~~~~ \frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j} ~~~~~~~~ \text{Cancellation of Dots}\\ \sum_{i=1}^{N} m_i\dot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j} &= \sum_{i=1}^{N} m_i\dot{\mathbf{r}}_i \cdot \frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j} = \frac{\partial }{\partial \dot{q}_j} \bigg( \frac{1}{2}\sum_{i=1}^{N} m_i( \dot{\mathbf{r}}_i\cdot \dot{\mathbf{r}}_i) \bigg) = \frac{\partial T}{\partial\dot{q}_j} \end{align*} \] where \(T\) is the total kinetic energy of the system. For the second term, we have: \[ \begin{align*} \frac{d}{dt}\bigg( \frac{\partial \mathbf{r}_i}{\partial q_j } \bigg) &= \sum_{k=1}^{n} \frac{\partial }{\partial q_k} \bigg( \frac{\partial \mathbf{r}_i}{\partial q_j} \bigg) \dot{q}_k + \frac{\partial }{\partial t}\bigg( \frac{\partial \mathbf{r}_i}{\partial q_j} \bigg) \\ \dot{\mathbf{r}}_i &= \sum_{k=1}^{n}\frac{\partial \mathbf{r}_i}{\partial q_k} \dot{q}_k + \frac{\partial \mathbf{r}_i}{\partial t} ~~~~~~~~~~ \frac{\partial \dot{\mathbf{r}}_i}{\partial q_j} = \sum_{k=1}^{n}\frac{\partial }{\partial q_j}\bigg( \frac{\partial \mathbf{r}_i}{\partial q_k}\bigg) \dot{q}_k + \frac{\partial}{\partial q_j}\frac{\partial \mathbf{r}_i}{\partial t} \\ & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{d}{dt}\bigg( \frac{\partial \mathbf{r}_i}{\partial q_j } \bigg) = \frac{\partial \dot{\mathbf{r}}_i}{\partial q_j} \end{align*} \] where we have used the Schwarz’s theorem to change the order of partial derivatives. Hence, we have: \[ \begin{align*} \sum_{i=1}^{N} m_i \dot{\mathbf{r}}_i \cdot \frac{d}{dt}\bigg( \frac{\partial \mathbf{r}_i}{\partial q_j }\bigg) = \sum_{i=1}^{N} m_i \dot{\mathbf{r}}_i \cdot \frac{\partial \dot{\mathbf{r}}_i}{\partial q_j} = \frac{\partial }{\partial q_j} \bigg( \frac{1}{2}\sum_{i=1}^{N} m_i( \dot{\mathbf{r}}_i\cdot \dot{\mathbf{r}}_i) \bigg) = \frac{\partial T}{\partial q_j} \end{align*} \] Finally, we have the Euler-Lagrange Equation: \[ \sum_{i=1}^{N} m_i \ddot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j} = \frac{d}{dt} \bigg( \sum_{i=1}^{N} m_i \dot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j} \bigg) - \sum_{i=1}^{N} m_i \dot{\mathbf{r}}_i \cdot \frac{d}{dt}\bigg( \frac{\partial \mathbf{r}_i}{\partial q_j }\bigg) = \frac{d}{dt}\bigg( \frac{\partial T }{\partial \dot{q_j}}\bigg) - \frac{\partial T}{\partial q_j} \] \[ \sum_{j=1}^{n} \bigg( \frac{d}{dt}\bigg( \frac{\partial T }{\partial \dot{q_j}}\bigg) - \frac{\partial T}{\partial q_j} \bigg) \delta q_j = \sum_{j=1}^{n} \bigg( \Xi_j - \frac{\partial V}{\partial q_j} \bigg) \delta q_j \]