Problem source

$\,$
\begin{center}
\begin{tikzpicture}
    % Draw horizontal line
    \coordinate (O) at (0,0);
    \coordinate (P) at ({2*cos(300)},{2*sin(300)});
    \coordinate (X) at (2, 0);
    \coordinate (A) at ({2*cos(315)},{2*sin(315)});
    \coordinate (C) at ({3*cos(315)},{3*sin(315)});
    \coordinate (B) at ({2*cos(225)},{2*sin(225)});
    \coordinate (D) at ({3*cos(225)},{3*sin(225)});
    \draw (-2.75, -2.25) -- (2.75,-2.25);
    \draw[dashed] (O) -- (P);
    \draw[dashed] (O) -- (X);
    \draw (A) -- (C) -- (D) -- (B);
    \draw[domain = 180:360, samples=180, variable = \x]  plot ({2*cos(\x)},{2*sin(\x)}); 
    \filldraw  (O) circle (1pt);
    \filldraw  (P) circle (1pt);
    \node at (O) [left] {$O$};
    \node at (P) [right, below] {$P$};
    \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = P--O--X};
\end{tikzpicture}
\end{center}
A smooth hemispherical bowl of mass $2m$ is rigidly mounted on a
light carriage which slides freely on a horizontal table as shown
in the diagram. The rim of the bowl is horizontal and has centre $O$.
A particle $P$ of mass $m$ is free to slide on the inner surface
of the bowl. Initially, $P$ is in contact with the rim of the bowl
and the system is at rest. The system is released and when $OP$ makes
an angle $\theta$ with the horizontal the velocity of the bowl is
$v$? Show that 
\[3v=a\dot{\theta}\sin\theta
\]
and that 
\[
v^{2}=\frac{2ga\sin^{3}\theta}{3(3-\sin^{2}\theta)},
\]
where $a$ is the interior radius of the bowl. 
Find, in terms of $m,g$ and $\theta,$ the reaction between the bowl
and the particle.