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{\bf Question}

A particle moves in a plane under the action of the force with
potential $U(r)$.  Write the Lagrangain in terms of the polar
coordinates $(r,\phi)$ and derive the equations of motion of the
particle.

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{\bf Answer}

$\ds {\bf v} = \dot r {\bf e}_r + r \dot \phi {\bf e}_\phi$
Therefore $K.E. = \frac{1}{2}m(\dot r^2 + r^2 \dot \phi^2)$

The Lagrangian $L = K.E. - P.E. = \frac{1}{2}m(\dot r^2 + r^2 \dot
\phi^2) - U(r)$

Equations of motion:

$$\frac{\pl L}{\pl r} = mr\dot \phi^2 - U'; \hspace{.1in}
\frac{\pl L}{\pl \dot r} = m \dot r; \hspace{.1in} \frac{\pl
L}{\pl \phi} = 0;
  \hspace{.1in}\frac{\pl L}{\pl \dot \phi} = mr^2 \dot \phi.$$

Therefore

$\ds \frac{d}{dt} \left(\frac{\pl L}{\pl \dot r}\right) -
\frac{\pl L}{\pl r} = 0 \Rightarrow m[\ddot r - r \dot \phi^2] =
-U'$

$\ds \frac{d}{dt} \left(\frac{\pl L}{\pl \dot \phi}\right) -
\frac{\pl L}{\pl \phi} = 0 \Rightarrow \frac{d}{dt}(mr^2 \dot
\phi) = 0 \Rightarrow r^2 \dot \phi =$ constant

Note that these are just the radial and tangential components of
Newton's 2nd law in polar coordinates.



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