D'Alembert Principle and Euler-Lagrange. Virtual displacement

Question

I have a little trouble with d'Alembert Principle and with virtual displacement. Imagine that with the d'Alembert Principle: $$ \sum_i \boldsymbol{\mathrm{F_i}} \; \cdot \delta \boldsymbol{\mathrm{r_i}} = 0 $$

You arrive to the following: $$ (\boldsymbol{\mathrm{F}} - m\boldsymbol{\mathrm{a}}) \; \delta \boldsymbol{\mathrm{r}} = 0 \\ \delta \boldsymbol{\mathrm{r}} \neq0 \\ \therefore \boldsymbol{\mathrm{F}} = m\boldsymbol{\mathrm{a}} $$ Most simple equation you could arrive to, but I don't completely understand $\delta \boldsymbol{\mathrm{r}}$. If you make an small variation to your system, how are you able to get the right equations for the system that has no variation. It feels contradictory.

Same happens to the way you are able to arrive to the Euler-Lagrange equations; how on earth you arrive to the correct equations by varying (it feels like you are changing a little bit the system and you still find a solution). Could someone explain me what $\delta \boldsymbol{\mathrm{r}}$ exactly means? Or in general, why varying and then seeking for a stationary variation gives good solutions?

Hope my explanation is not confuse. If so, I'll try to explain it better.

Answer 1

Starting with Newton second law

$$m_i\,\ddot{\vec{r}}_i=\vec F_i$$

where $~\vec F_i~$ is the applied force $~\vec F^A_i~$ and the constraint force $~\vec F^C_i~$. thus $$m_i\,\ddot{\vec{r}}_i=\vec F^A_i+\vec F^C_i$$

from here you obtain the constraint froce

$$ \vec F^C_i=m_i\,\ddot{\vec{r}}_i-\vec F^A_i\tag 1$$

you can eliminate the constraint force from equation (1) by taking the dot product with the virtual displacements $~\delta\vec{r}_i $ you obtain

$$\sum_{i=1}^N\left(m_i\,\ddot{\vec{r}}_i-\vec F^A_i\right)\dot\,\delta\vec{r}_i =0\tag 2$$

notice that $$~\sum_{i=1}^N\left(\vec F^C_i\,\cdot \delta\vec{r}_i \right)=0$$

The equations of motion

with the holonomic constraint equations you obtain the generalized coordinates $~q=(q_1(t)~,q_2(t)~,\ldots~,q_f(t))~$ thus:

$$\vec r_i=\vec r_i(q)\\ \dot{\vec{r}}_i=\dot{\vec{r}}_i(q,\dot q)\\ \ddot{\vec{r}}_i=\ddot{\vec{r}}_i(q,\dot q\,\ddot q)\\ \delta\vec{r}_i=\sum_{i=1}^f\frac{\partial \vec r_i}{\partial q_j}\delta q_j$$

Example Pendulum

generalized coordinate is $~\varphi~$

$$\vec r=\left[ \begin {array}{c} l\sin \left( \varphi \right) \\ -l\cos \left( \varphi \right) \end {array} \right]~, \dot{\vec{r}}= \left[ \begin {array}{c} l\cos \left( \varphi \right) \dot\varphi \\ l\sin \left( \varphi \right) \dot\varphi \end {array} \right] ~, \vec a=\left[ \begin {array}{c} l\cos \left( \varphi \right) \ddot \varphi -l\sin \left( \varphi \right) {\dot\varphi }^{2} \\ l\sin \left( \varphi \right) \ddot\varphi +l \cos \left( \varphi \right) {\dot\varphi }^{2}\end {array} \right]\\ \delta\vec r= \left[ \begin {array}{c} l\cos \left( \varphi \right) \delta \varphi \\ l\sin \left( \varphi \right) \delta \varphi \end {array} \right]\\ \vec F^A= \left[ \begin {array}{c} 0\\ -mg\end {array} \right] $$

equation (2)

$$\delta \varphi\,l \,m \left( l\ddot\varphi +\sin \left( \varphi \right) g \right) =0$$ and because $~\delta \varphi\ne 0\quad \Rightarrow$

$$\ddot\varphi+\frac gl\sin(\varphi)=0$$

d'Alembert Principle is equivalent to Euler-Langrage equations. you have holonomic constraint equations and the applied Forces are conservative

$$T=\frac m2\,\dot{\vec{r}}\cdot\,\dot{\vec{r}}=\frac m2\,l^2\,\dot\varphi^2\\ U=-m\,g\,l\,\cos(\varphi)$$