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# 6.7: Applications to unconstrained systems

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Although most dynamical systems involve constrained motion, it is useful to consider examples of systems subject to conservative forces with no constraints . For no constraints, the Lagrange-Euler equations $$(6.6.1)$$ simplify to $$\Lambda _{j}L=0$$ where $$j=1,2,..n,$$ and the transformation to generalized coordinates is of no consequence.

Example $$\PageIndex{1}$$: Motion of a free particle, $$U=0$$

The Lagrangian in cartesian coordinates is $$L= \frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}).$$ Then

\begin{aligned} \frac{\partial L}{\partial \dot{x}} &= m\dot{x} \\[4pt] \frac{\partial L}{\partial \dot{y}} &= m\dot{y} \\[4pt] \frac{\partial L}{\partial \dot{z}} &= m\dot{z} \\[4pt] \frac{\partial L}{\partial x} &= \frac{\partial L}{\partial y}=\frac{ \partial L}{\partial z}=0\end{aligned}

Insert these in the Lagrange equation gives

\begin{align*} \Lambda _{x}L &= \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x} \\[4pt] &=\frac{d}{dt}m\dot{x}-0=0 \end{align*}

Thus

\begin{align*} \mathit{\ }p_{x} &= m\dot{x}=constant \\[4pt] p_{y} &= m\dot{y}=constant \\[4pt] \mathit{\ }p_{z} &= m\dot{z}=constant\end{align*}

That is, this shows that the linear momentum is conserved if $$U$$ is a constant, that is, no forces apply. Note that momentum conservation has been derived without any direct reference to forces.

Example $$\PageIndex{2}$$: Motion in a uniform gravitational field

Consider the motion is in the $$x-y$$ plane. The kinetic energy $$T= \frac{1}{2}m\left( \overset{.}{x}^{2}+\overset{.}{y}^{2}\right)$$ while the potential energy is $$U=mgy$$ where $$U(y=0)=0.$$ Thus

$L=\frac{1}{2}m\left( \overset{.}{x}^{2}+\overset{.}{y}^{2}\right) -mgy \nonumber$

Using the Lagrange equation for the $$x$$ coordinate gives

\begin{align*} \Lambda _{x}L &= \frac{d}{dt}\frac{\partial L}{\partial \overset{.}{x}}-\frac{ \partial L}{\partial x} \\[4pt] &=\frac{d}{dt}m\overset{.}{x}-0 \\[4pt] &=0 \end{align*}

Thus the horizontal momentum $$m\dot{x}$$ is conserved and $$\overset{..}{x}=0.$$ The $$y$$ coordinate gives

\begin{align*} \Lambda _{y}L &= \frac{d}{dt}\frac{\partial L}{\partial \overset{.}{y}}-\frac{ \partial L}{\partial y} \\[4pt] &=\frac{d}{dt}m\overset{.}{y}+mg \\[4pt] &=0 \end{align*}

Thus the Lagrangian produces the same results as derived using Newton’s Laws of Motion.

$\ddot{x}=0 \nonumber$

$y=-g \nonumber$

The importance of selecting the most convenient generalized coordinates is nicely illustrated by trying to solve this problem using polar coordinates $$r,\theta ,$$ where $$r$$ is radial distance and $$\theta$$ the elevation angle from the $$x$$ axis as shown in the adjacent figure. Then

$T=\frac{1}{2}m\overset{.}{r}^{2}+\frac{1}{2}m\left( r\overset{.}{\theta } \right) ^{2} \nonumber$

$U=mgr\sin \theta \nonumber$

Thus

$L=\frac{1}{2}m\overset{.}{r}^{2}+\frac{1}{2}m\left( r\dot{\theta}\right) ^{2}-mgr\sin \theta \nonumber$

$$\Lambda _{r}L=0$$ for the $$r$$ coordinate

$r\dot{\theta}^{2}-g\sin \theta -\ddot{r}=0 \nonumber$

$$\Lambda _{\theta }L=0$$ for the $$\theta$$ coordinate

$-gr\cos \theta -2r\dot{r}\dot{\theta}-r^{2}\ddot{\theta}=0 \nonumber$

These equations written in polar coordinates are more complicated than the result expressed in Cartesian coordinates. This is because the potential energy depends directly on the $$y$$ coordinate, whereas it is a function of both $$r,\theta .$$ This illustrates the freedom for using different generalized coordinates, plus the importance of choosing a sensible set of generalized coordinates.

Example $$\PageIndex{3}$$: Central forces

Consider a mass $$m$$ moving under the influence of a spherically-symmetric, conservative, attractive, inverse-square force. The potential then is

$U=- \frac{k}{r} \nonumber$

It is natural to express the Lagrangian in spherical coordinates for this system. That is,

$L=\frac{1}{2}m\dot{r}^{2}+\frac{1}{2}m\left( r\dot{\theta}\right) ^{2}+\frac{ 1}{2}m(r\sin \theta \dot{\phi})^{2}+\frac{k}{r} \nonumber$

$$\Lambda _{r}L=0$$ for the $$r$$ coordinate gives $m\ddot{r}-mr[\dot{\theta}^{2}+\sin ^{2}\theta \dot{\phi}^{2}]=\frac{k}{r^{2}} \nonumber$

where the $$mr\sin ^{2}\theta \dot{\phi}^{2}$$ term comes from the centripetal acceleration.

$$\Lambda _{\phi }L=0$$ for the $$\phi$$ coordinate gives

$\frac{d}{dt}\left( mr^{2}\sin ^{2}\theta \dot{\phi}\right) =0 \nonumber$

This implies that the derivative of the angular momentum about the $$\phi$$ axis, $$\dot{p}_{\phi }=0$$ and thus $$p_{\phi }= mr^{2}\sin ^{2}\theta \dot{\phi}$$ is a constant of motion.

$$\Lambda _{\theta }L=0$$ for the $$\theta$$ coordinate gives

$\frac{d}{dt}(mr^{2}\dot{\theta})-mr^{2}\sin \theta \cos \theta \dot{\phi} ^{2}=0 \nonumber$

That is, $\dot{p}_{\theta }=mr^{2}\sin \theta \cos \theta \dot{\phi}^{2}=\frac{p_{\phi }^{2}\cos \theta }{2mr^{2}\sin ^{3}\theta } \nonumber$

Note that $$p_{\theta }$$ is a constant of motion if $$p_{\phi }=0$$ and only the radial coordinate is influenced by the radial form of the central potential.

This page titled 6.7: Applications to unconstrained systems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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