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<record version="1" id="287">
 <title>polar coordinate motion example of generalized coordinates</title>
 <name>PolarCoordinateMotionExampleOfGeneralizedCoordinates</name>
 <created>2008-07-18 20:11:07</created>
 <modified>2008-07-18 20:11:07</modified>
 <type>Example</type>
<parent id="286">generalized coordinates  for free motion</parent>
 <creator id="1" name="bloftin"/>
 <modifier id="1" name="bloftin"/>
 <author id="1" name="bloftin"/>
 <classification>
	<category scheme="msc" code="45.20.-d"/>
 </classification>
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 <content>As an example let us get the equations in polar coordinates for motion in a plane

Here
$$ x=r\cos\phi,\,\,\,\,\,\,\,\,\,  y=r\sin\phi$$

$$ \dot{x}^{2}+\dot{y}^{2}=v^{2}=\dot{r}^{2}+r^{2}\dot{\phi}^{2} $$

and 

$$ T = \frac{m}{2}\left [ \dot{r}^{2}+r^{2}\dot{\phi}^{2} \right ] $$

$$
\frac{\partial T}{\partial \dot{r} } = m \dot{r}
$$

$$
\frac{\partial T}{\partial r}=m r \dot{\phi}^{2}.
$$
$$
\delta_{r}W=m[\ddot{r}-r\dot{\phi}^{2}]\delta r=R\delta r
$$
if $R$ is the impressed force resolved along the radius vector.
$$
\frac{\partial T}{\partial\dot{\phi}}=m r^{2}\dot{\phi},
$$

$$
\frac{\partial T}{\partial \phi}=0.
$$

$$ \delta_{\phi}W=m\frac{d}{dt}(r^{2}\dot{\phi})\delta\phi=\Phi r\delta\phi$$

if $\Phi$ is the impressed force resolved perpendicular to the radius vector.

In a more familiar form

$$
m \left [\frac{d^{2}r}{dt^{2}}-r \left ( \frac{d\phi}{dt} \right)^2 \right ]=R,
$$
$$
\frac{m}{r}\frac{d}{dt} \left ( r^{2}\frac{d\phi}{dt} \right )=\Phi.
$$</content>
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