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Physics LibreTexts

6.7: Applications to unconstrained systems

( \newcommand{\kernel}{\mathrm{null}\,}\)

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 ΛjL=0 where j=1,2,..n, and the transformation to generalized coordinates is of no consequence.

Example 6.7.1: Motion of a free particle, U=0

The Lagrangian in cartesian coordinates is L=12m(˙x2+˙y2+˙z2). Then

L˙x=m˙xL˙y=m˙yL˙z=m˙zLx=Ly=Lz=0

Insert these in the Lagrange equation gives

ΛxL=ddtL˙xLx=ddtm˙x0=0

Thus

 px=m˙x=constantpy=m˙y=constant pz=m˙z=constant

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 6.7.2: Motion in a uniform gravitational field

6.7.1.PNG
Figure 6.7.1: Motion in a gravitational field

Consider the motion is in the xy plane. The kinetic energy T=12m(.x2+.y2) while the potential energy is U=mgy where U(y=0)=0. Thus

L=12m(.x2+.y2)mgy

Using the Lagrange equation for the x coordinate gives

ΛxL=ddtL.xLx=ddtm.x0=0

Thus the horizontal momentum m˙x is conserved and ..x=0. The y coordinate gives

ΛyL=ddtL.yLy=ddtm.y+mg=0

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

¨x=0

y=g

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

T=12m.r2+12m(r.θ)2

U=mgrsinθ

Thus

L=12m.r2+12m(r˙θ)2mgrsinθ

ΛrL=0 for the r coordinate

r˙θ2gsinθ¨r=0

ΛθL=0 for the θ coordinate

grcosθ2r˙r˙θr2¨θ=0

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,θ. This illustrates the freedom for using different generalized coordinates, plus the importance of choosing a sensible set of generalized coordinates.

Example 6.7.3: Central forces

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

U=kr

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

L=12m˙r2+12m(r˙θ)2+12m(rsinθ˙ϕ)2+kr

ΛrL=0 for the r coordinate gives m¨rmr[˙θ2+sin2θ˙ϕ2]=kr2

where the mrsin2θ˙ϕ2 term comes from the centripetal acceleration.

ΛϕL=0 for the ϕ coordinate gives

ddt(mr2sin2θ˙ϕ)=0

This implies that the derivative of the angular momentum about the ϕ axis, ˙pϕ=0 and thus pϕ=mr2sin2θ˙ϕ is a constant of motion.

ΛθL=0 for the θ coordinate gives

ddt(mr2˙θ)mr2sinθcosθ˙ϕ2=0

That is, ˙pθ=mr2sinθcosθ˙ϕ2=p2ϕcosθ2mr2sin3θ

Note that pθ is a constant of motion if pϕ=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.

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