Problem source

The points $O,A,B$ and $C$ are the vertices of a uniform square lamina of mass $M.$ The lamina can turn freely under gravity about a horizontal axis perpendicular to the plane of the lamina through $O$. The sides of the lamina are of length $2a.$ When the lamina is haning at rest with the diagonal $OB$ vertically downwards it is struck at the midpoint of $OC$ by a particle of mass $6M$ moving horizontally in the plane of the lamina with speed $V$. The particle adheres to the lamina. Find, in terms of $a,M$ and $g$, the value which $V^{2}$ must exceed for the lamina and particle to make complete revolutions about the axis.

Solution source

\begin{center}
        \begin{tikzpicture}[scale = 2]
        \draw[dashed] (-3,0) -- (3, 0);

        \coordinate (O) at (0,0);
        \coordinate (A) at (1, -1);
        \coordinate (B) at (0, -2);
        \coordinate (C) at (-1, -1);

        \coordinate (Ar) at ({sqrt(2)*cos(30)},{sqrt(2)*sin(30)});
        \coordinate (Br) at ({2*cos(30+45},{2*sin(30+45)});
        \coordinate (Cr) at ({sqrt(2)*cos(30+90)},{sqrt(2)*sin(30+90)});

        \draw (O) -- (A) -- (B) -- (C) -- cycle;
        \draw[dashed] (O) -- (Ar) -- (Br) -- (Cr) -- cycle;

        \node[above] at (O) {$O$};
        \node[right] at (A) {$A$};
        \node[below] at (B) {$B$};
        \node[left] at (C) {$C$};
        \node[right] at (Ar) {$C'$};
        \node[above] at (Br) {$B'$};
        \node[left] at (Cr) {$A'$};

        \filldraw (-2, -0.5) circle (1pt); 
        \filldraw ($(O)!0.5!(Ar)$) circle (1pt); 

        \draw[-latex, red, ultra thick] (-2.2, -0.7) -- (-1.8, -0.7) node[right] {$V$}; 
        
    \end{tikzpicture}
\end{center}

Consider the moment of inertia of the lamina. The MoI about the centre of mass is $\frac1{12}M((2a)^2 + (2a)^2) = \frac23Ma^2$.

//el axis theorem, tells us the moment of inertia about $O$ is $I_O = I_G + Md^2_{OG} = \frac23Ma^2 + M2a^2 = \frac83Ma^2$

Moment of inertia of particle is $6Ma^2$

Total moment of inertial is: $\frac{26}{3}Ma^2$.

Conservation of angular momentum states that $6M \frac{\sqrt{2}}2Va = \frac{26}{3}Ma^2 \omega \Rightarrow \omega = \frac{9\sqrt{2}V}{26a}$

Consider the centre of mass (in the frame drawn)

\begin{array}{c|c|c}
\text{Shape} & \text{Mass} & \text{COM} \\ \hline
\text{Square} & M & (0,-\sqrt{2}a) \\
\text{Particle} & 6M & (-\frac{\sqrt{2}}2a, -\frac{\sqrt{2}}{2}a) \\
\text{combined} & 7M & \left ( \frac{-3\sqrt{2}}{7} a, -\frac{4\sqrt{2}}{7}a \right)
\end{array}


The lamina/particle system will complete full circles if it still has positive angular velocity at the peak, ie:

\begin{align*}
&& \underbrace{\frac12 I \omega^2}_{\text{initial rotational energy}} + mgh_{start} &\geq mgh_{top} \\
&& \frac 12  \frac{26}{3} Ma^2 \frac{9^2 \cdot 2 V^2}{26^2 a^2} - (7M)g\frac{4\sqrt{2}}{7}a &\geq (7M)g\frac{5\sqrt{2}}{7}a \\
\Rightarrow && \frac{V^2 \cdot 27}{26} &\geq 9\sqrt{2}ga \\
\Rightarrow && V^2 & \geq \frac{26\sqrt{2}}{3}ga
\end{align*}