<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\begin{document}
\parindent=0pt

{\bf Question}

A massless spring of natural length $b$ and spring constant $k$
connects two particles of masses $m_1$ and $m_2$.  The system
rests on a smooth table and may oscillate and rotate.  Find the
Lagrangian of the system in terms of ($x_1, \, y_1$), the
Cartesian coordinates of $m_1$ and ($r, \, \phi$) the polar
coordinates of the relative position of $m_2$ to $m_1$.  Show that
\begin{description}
\item[(a)] $\ds  m_1 \dot x_1 + m_2 \dot x_2 =$ constant and
$\ds m_1 \dot y_1 + m_2 \dot y_2 =$ constant.
\item[(b)] $\ds \ddot r - r \dot \phi^2 + \ddot x_1 \cos \phi +
\ddot y \sin \phi + \frac{k}{m_2}(l - b) = 0.$
\item[(c)] Find the equation of motion for $\phi$.
\end{description}

\vspace{.25in}

{\bf Answer}

$$$$
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$\ds x_2 = x_1 + r \cos \phi$

$\ds y_2 = y_1 + r \sin \phi$

$\dot x_2 = \dot x_1 + \dot r \cos \phi - r \dot \phi \sin \phi
\hspace{.5in} (*)$

$\dot y_2 = \dot y_1 + \dot r \sin \phi + r \dot \phi \cos \phi$

\begin{eqnarray*} L &amp; = &amp; K.E. - P.E. \\ &amp; = &amp; \frac{1}{2}(m_1 +
m_2)(\dot x_1^2 + \dot y_1^2) + \frac{1}{2}m_2(\dot x_2^2 + \dot
y_2^2) - \frac{k}{2}(r - b)^2 \\ &amp; = &amp; \frac{1}{2}m_1(\dot x_1^2 +
\dot y_1^2) + \frac{1}{2}m_2(\dot r^2 + r^2 \dot \phi^2) + m_2
\dot r(\dot x_1 \cos \phi + \dot y_1 \sin \phi) \\ &amp; &amp; + m_2 r
\dot \phi (\dot y_1 \cos \phi - \dot x_1 \sin \phi) -
\frac{1}{2}k(r - b)^2 \end{eqnarray*}

Equations of motion
\begin{description}
\item[(a)]
$x_1$:

$\ds \frac{d}{dt}[(m_1 +m_2) \dot x_1 + m_2 \dot r \cos \phi - m_2
r \dot \phi \sin \phi] = 0$

$\ds \Rightarrow m_1 \dot x_1 + m_2 \dot x_2 =$ constant,
(using(*))

$y_2$:

$y_2:$ Similarly gives $\ds m_1 \dot y_1 + m_2 \dot y_2 =$
constant, (using(*))


\item[(b)]
$r:$

$\ds \frac{d}{dt} \left(\frac{\pl L}{\pl \dot r}\right) -
\frac{\pl L}{\pl r} = 0$ gives the required answer after some
algebra.


\item[(c)]
$\phi$:
\begin{eqnarray*}0 &amp; = &amp; \frac{d}{dt}(m_2r^2 \dot \phi + m_2 r(\dot y_1 \cos \phi - \dot
x_1 \sin \phi))  \\ &amp; &amp; - (m_2 \dot r(-\sin \phi \dot x_1 + \cos
\phi \dot y_1) + m_2 r \dot \phi (-\dot y_1 \sin \phi - \dot x_1
\cos \phi)) \end{eqnarray*}

Therefore
\begin{eqnarray*} 0 &amp; = &amp; 2m_2 \dot \phi \dot r r + m_2 r(\ddot y_1 \cos \phi - \ddot
x_1 \sin \phi)  + m_2r^2 \ddot \phi + \\ &amp; &amp; m_2 \dot r(\dot y_1
\cos \phi - \dot x_1 \sin \phi)  + m_2 r \dot \phi(-\dot y_1 \sin
\phi - \dot x_1 \cos \phi) + \\ &amp; &amp; m_2 \dot r(\dot x_1 \sin \phi
- \dot y_1 \cos \phi)  + m_2 r \dot \phi(\dot y_1 \sin \phi + \dot
x_1 \cos \phi) \end{eqnarray*}

Therefore $$\ddot \phi + 2\frac{\dot \phi \dot r}{r} + \frac{\cos
\phi}{r} \ddot y_1 - \frac{\sin \phi}{r} \ddot x_1=0$$


\end{description}
\end{document}
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