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<record version="1" id="291">
 <title>example of generalized coordinates for constrained motion on a horizontal circle</title>
 <name>ExampleOfGeneralizedCoordinatesForConstrainedMotionOnAHorizontalCircle</name>
 <created>2008-07-21 00:51:29</created>
 <modified>2008-07-21 00:51:29</modified>
 <type>Example</type>
<parent id="290">generalized coordinates for constrained motion</parent>
 <creator id="1" name="bloftin"/>
 <modifier id="1" name="bloftin"/>
 <comment>kinetic energy equation had a typo, forgot the dot on theta</comment>
 <author id="1" name="bloftin"/>
 <classification>
	<category scheme="msc" code="45.20.Jj"/>
 </classification>
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 <content>Let a particle of mass $m$, constrained to move on a smooth horizontal circle of radius $a$, be given an initial velocity $V$, and let it be resisted by the air with a force proportional to the square 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 kinetic energy

$$
T=\frac{m}{2}a^{2}\theta^{2},
$$

and we have 

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

\quad Our differential equation 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 motion is completely solved.</content>
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