Problem source

$\,$
\begin{center}
\psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.3pt,arrowsize=3pt 2,arrowinset=0.25}
\begin{pspicture*}(-2.8,-2.34)(3.26,4.36)
\pspolygon[linewidth=0.4pt](0.22,-0.18)(0.4,0.04)(0.18,0.22)(0,0)
\pscircle(0,0){2}
\psline(2,0)(2,4)
\rput[tl](2.24,2.64){$\pi b$}
\rput[tl](2.22,0.02){$C$}
\rput[tl](2.24,4.3){$A$}
\rput[tl](1.32,2.1){$Q$}
\rput[tl](-0.42,0.08){$O$}
\rput[tl](1.6,-1.38){$P$}
\psline[linestyle=dashed,dash=3pt 3pt](1.27,1.54)(0,0)
\psline[linestyle=dashed,dash=3pt 3pt](0,0)(1.54,-1.27)
\rput[tl](0.06,1.1){$2b$}
\begin{scriptsize}
\psdots[dotsize=2pt 0,dotstyle=*](2,4)
\end{scriptsize}
\end{pspicture*}
\end{center}
A uniform circular disc of radius $2b,$ mass $m$ and centre $O$
is free to turn about a fixed horizontal axis through $O$ perpendicular
to the plane of the disc. A light elastic string of modulus $kmg$,
where $k>4/\pi,$ has one end attached to a fixed point $A$ and the
other end to the rim of the disc at $P$. The string is in contact
with the rim of the disc along the arc $PC,$ and $OC$ is horizontal.
The natural length of the string and the length of the line $AC$
are each $\pi b$ and $AC$ is vertical. A particle $Q$ of mass $m$
is attached to the rim of the disc and $\angle POQ=90^{\circ}$ as
shown in the diagram. The system is released from rest with $OP$
vertical and $P$ below $O$. Show that $P$ reaches $C$ and that
then the upward vertical component of the reaction on the axis is
$mg(10-\pi k)/3$.