Talk:PlanetPhysics/Example of Generalized Coordinates for Constrained Motion on a Horizontal Circle

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: example of generalized coordinates for constrained motion on a horizontal circle %%% Primary Category Code: 45.20.Jj %%% Filename: ExampleOfGeneralizedCoordinatesForConstrainedMotionOnAHorizontalCircle.tex %%% Version: 2 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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\begin{document}

Let a \htmladdnormallink{particle}{http://planetphysics.us/encyclopedia/Particle.html} of \htmladdnormallink{mass}{http://planetphysics.us/encyclopedia/Mass.html} $m$, constrained to move on a smooth horizontal circle of radius $a$, be given an initial \htmladdnormallink{velocity}{http://planetphysics.us/encyclopedia/Velocity.html} $V$, and let it be resisted by the air with a \htmladdnormallink{force}{http://planetphysics.us/encyclopedia/Thrust.html} proportional to the \htmladdnormallink{square}{http://planetphysics.us/encyclopedia/PiecewiseLinear.html} of its velocity.

Here we have one degree of freedom. Let us take as our coordinate $q_{1}$ the angle $\theta$ which the particle has described about the center of its path in the time $t$.

For the \htmladdnormallink{kinetic energy}{http://planetphysics.us/encyclopedia/KineticEnergy.html} $$ T=\frac{m}{2}a^{2}\dot{\theta}^{2}, $$

and we have

$$ \frac{\partial T}{\partial \dot{\theta}}=ma^{2}\dot{\theta}. $$

\quad Our \htmladdnormallink{differential equation}{http://planetphysics.us/encyclopedia/DifferentialEquations.html} is

$$ ma^{2}\ddot{\theta}\delta\theta=-ka^{2}\dot{\theta}^{2}a\delta\theta, $$

which reduces to

$$ \ddot{\theta}+\frac{k}{m}a\dot{\theta}^{2}=0, $$

or

$$ \frac{d \dot{\theta}}{dt}+\frac{k}{m}a\dot{\theta}^{2}=0. $$

Separating the variables,

$$ \frac{d \dot{\theta}}{\dot{\theta}^{2}}+\frac{k}{m}a dt=0. $$

Integrating,

$$ -\frac{1}{\dot{\theta}}+\frac{k}{m} a t= C =-\frac{a}{V}. $$

$$ \frac{1}{\dot{\theta}}=\frac{ma+k V a t}{m V}, $$

$$ \frac{d\theta}{dt}=\frac{mV}{ma+kVat}, $$

$$ \theta=\frac{m}{ka}\log\left[m+kVt\right]+C, $$

$$ \theta=\frac{m}{ka}\log\left[1+\frac{kVt}{m}\right]; $$

and the problem of the \htmladdnormallink{motion}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} is completely solved.

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