Problem source

A long, light, inextensible string  passes through a small, smooth ring
fixed at  the point $O$. One end of the string 
is attached to a particle $P$ of mass $m$ 
which hangs freely below $O$. The other end is attached
to a bead, $B$, also of mass $m$, which is threaded on a smooth rigid wire
fixed in the same vertical plane as $O$.
The distance $OB$ is $r$, the distance $OH$ is $h$ 
and the  height of the bead above the horizontal plane through~$O$ is $y$,
as shown
in the diagram.


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\rput[tl](0.46,0.58){$\theta$}
\rput[tl](0.06,-0.27){$O$}
\rput[tl](0.15,-1.35){$P$}
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\rput[tl](0.77,1.7){$r$}
\rput[tl](1.84,2.68){$B$}
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The shape of the wire
is such that the system can be in static equilibrium for all positions
of the bead.
By considering potential energy, show that the 
equation of the wire is $y+r =2h$.
The bead is initially at $H$. It is then projected along the wire with
initial speed $V$. Show that, in the subsequent motion, 
\[
\dot \theta = -\frac {h \dot r }{r \sqrt{rh -h^2}}\,
\] 
where $\theta$ is given by
$\theta = \arcsin(y/r)$.
Hence show that the speed of the particle $P$ is 
$V \Big(\dfrac{r-h}{2r-h}\Big)^{\!\frac12}\,$.
\noindent[{\it Note that $\arcsin \theta$ is another notation for $\sin^{-1}\theta$.}]