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{\bf Question}

A particle of mass $m$ slides under gravity on a smooth wire in
the shape of the cycloid $x = a(\theta - \sin \theta), \, y = a(1
+ \sin \theta), \, 0 \leq \theta \leq 2\pi$ as shown below

PICTURE

\begin{description}
\item[(a)] Show that the kinetic energy of the bead is $ma^2(1 -
\cos \theta) \dot \theta^2$.

\item[(b)] Find the Lagrangian and deduce the equation of motion
$$ 2 \sin \frac{\theta}{2} \ddot \theta + \cos \frac{\theta}{2}
\dot \theta ^2 - \frac{g}{a} \cos \frac{\theta}{2} = 0.$$

\item[(c)] Rewrite the above equation in terms of $u = \cos
\frac{\theta}{2}$ and deduce that, irrespective of the starting
position, the period of oscillation is the same as that of a plane
pendulum of length $4a$ undergoing small oscillations.

\item[(d)] Suppose two identical beads are released from rest with
$\theta = 0$ and $\theta = \frac{\pi}{2}$ respectively.  Where do
they collide?  At what time do they collide?  What happens
subsequently? (Assume that the coefficient of restitution, $e$, is
unity.)
\end{description}

\vspace{.25in}

{\bf Answer}

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$\ds x = a(\theta - \sin \theta) \hspace{.2in} y = a(1 + \cos
\theta)$

$\ds \dot x = a \dot \theta(1 - \cos \theta) \hspace{.2in} \dot y
= -a \dot \theta \sin \theta$

\begin{eqnarray*}  K.E. & = & \frac{1}{2} m (\dot x^2 + \dot y^2)
\\ & = & \frac{1}{2} m a^2 \dot \theta^2 [1 - 2\cos \theta +
\cos^2 \theta + \sin^2 \theta] \\ & = & ma^2 \dot \theta^2(1 -
\cos \theta) \\ \\ P.E. & = & mgy  \\ &  = & mga(1+ \cos \theta)
\\ \\ L & = & K.E. - P.E. \\ & = & ma^2 \dot \theta^2(1 - \cos
\theta) - mga(1+ \cos \theta) \\ \\ \frac{\pl L}{\pl\dot\theta} &
= & 2ma^2(1 - \cos \theta) \dot \theta \\ & = & 4ma^2\dot \theta
\sin^2 \frac{\theta}{2} \\ \\ \frac{\pl L}{\pl \theta} & = & ma^2
\dot \theta^2 \sin \theta + mga\sin\theta
\end{eqnarray*}

Euler-Lagrange equation:
\begin{eqnarray*} \frac{d}{dt}\left(4ma^2\dot \theta
\sin^2 \frac{\theta}{2}\right) -  ma^2 \dot \theta^2 \sin \theta -
mga\sin\theta & = & 0 \\ 4ma^2\left[ \ddot \theta \sin^2
\frac{\theta}{2} + \dot \theta^2 \sin \frac{\theta}{2}\cos
\frac{\theta}{2}\right] - ma(a\dot \theta^2 + g).2 \sin
\frac{\theta}{2}\cos \frac{\theta}{2} & = & 0 \\ 2\sin
\frac{\theta}{2} + (\cos \frac{\theta}{2}) \dot \theta^2 -
\frac{g}{a} \cos \frac{\theta}{2} & = & 0  \hspace{.1in} (*)
\end{eqnarray*}

(cancelling $a\sin\frac{\theta}{2}\ne 0$ for general motion.)

$\ds u = \cos \frac{\theta}{2} \hspace{.1in} \Rightarrow
\hspace{.1in} \dot u = -\frac{1}{2} \sin \frac{\theta}{2} \dot
\theta \hspace{.1in} \Rightarrow \hspace{.1in} \ddot u =
-\frac{1}{2} \left[ \frac{1}{2} \cos \frac{\theta}{2} \dot
\theta^2 + \sin \frac{\theta}{2} \ddot \theta\right]$

Hence (*) can be written as: $$\ddot u + \frac{g}{4a}u = 0,$$
which implies Simple Harmonic Motion with frequency $\ds
\sqrt{\frac{g}{4a}}$ i.e. a pendulum with length 4a.

The period is independent of the starting position so the beads
will

collide at the bottom of the cardioid $(\pi a, 0)$ after a quarter
of a period

$\ds \left(2\pi \frac{\sqrt{\frac{4a}{g}} }{4} =
\pi\sqrt{\frac{a}{g}}\right)$.


The collisions are perfectly elastic at the bottom.

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\put(0,3){Before:} \put(0,1){After:} \put(3,2){\circle{1}}
\put(5,2){\circle{1}} \put(2.5,3){\vector(1,0){1}}
\put(4.5,3){\vector(1,0){1}} \put(2.5,1){\vector(1,0){1}}
\put(4.5,1){\vector(1,0){1}} \put(2.9,1.9){$m$} \put(4.9,1.9){$m$}
\put(2.9,3.1){$u_1$} \put(4.9,3.1){$u_1$} \put(2.9,0.8){$v_1$}
\put(4.9,0.8){$v_1$}

\end{picture}

linear momentum: $\left.\begin{array}{rcl} mu_1 + mu_2 & = & mv_1
+ mv_2 \\ v_2-v_1 & = & -(u_2 - u_1) \end{array} \right\}
\Rightarrow \begin{array}{ccc} v_2 & = & u_1 \\ v_1 & =&  u_2
\end{array}$

i.e. the beads exchange their speeds and then repeat the motion
(and collisions at the bottom) indefinitely.



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