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A light rod of length $2a$ has a particle of mass $m$ attached to each end and it moves in  a vertical plane. The midpoint of the rod has coordinates $(x,y)$, where the $x$-axis is horizontal (within the plane of motion) and $y$ is the height above a horizontal table. Initially, the rod is vertical, and at time $t$ later it is inclined at an angle $\theta$ to the vertical.

Show that the velocity of one particle can be written in the form
\[
\begin{pmatrix}
\dot x + a \dot\theta \cos\theta  \\
 \dot y - a \dot\theta \sin\theta
\end{pmatrix}
\]
and that 
\[
m\begin{pmatrix}
 \ddot x + a\ddot\theta \cos\theta - a \dot\theta^2 \sin\theta 
\\
 \ddot y- a\ddot\theta \sin\theta - a \dot\theta^2 \cos\theta  
\end{pmatrix}
=-T\begin{pmatrix}

\sin\theta
\\
\cos\theta
\end{pmatrix}
-mg
\begin{pmatrix}
0 \\ 1
\end{pmatrix}
\]
where the dots denote differentiation with respect to time $t$ and $T$ is the tension in the rod.  Obtain the corresponding equations for the other particle.
Deduce that $\ddot x =0$, $\ddot y = -g$ and $\ddot\theta =0$.
Initially, the  midpoint of the rod is a height $h$ above the table, the velocity   of the higher particle is $\Big(\begin{matrix} \, u \, \\ v \end{matrix}\Big)$, and the velocity of the lower particle is $\Big(\begin{matrix}\, 0  \, \\ v\end{matrix}\Big)$. Given that the two  particles hit the table for the first time simultaneously, when the rod has rotated by $\frac12\pi$, show that
\[
2hu^2 = \pi^2a^2 g - 2\pi uva \,.
\]

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