Problem source

A particle is attached to one end of a light inextensible string of length $b$. The other end of the string is attached to a fixed point $O$.
Initially the particle hangs vertically below $O$.
The particle then receives a horizontal impulse.
The particle moves  in a circular arc with the string taut until the acute angle between the string and the upward vertical is $\alpha$, at which time it becomes slack.
Express $V$, the speed of the particle when the string becomes slack, in terms of $ b$, $g$ and $\alpha$.
  
Show that the string becomes taut again a time $T$ later, where
\[
gT = 4V \sin\alpha     
\,,\]
and that just before this time the trajectory of the  particle makes an angle $\beta $ with the horizontal where $\tan\beta = 3\tan \alpha \,$.
When the string becomes taut, the momentum of the particle in the direction of the string is destroyed. Show that the particle comes instantaneously to rest at this time if and only if 
\[
\sin^2\alpha = \dfrac {1+\sqrt3}4 \,.
\]

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \coordinate (O) at (0,0);
        \coordinate (X) at (0,2);
        \coordinate (P) at ({2*cos(30)},{2*sin(30)});
        \draw[dashed] (O) circle (2);

        \filldraw (O) circle (1pt);
        \filldraw (P) circle (1pt);
        \draw (O) -- (P);

        \draw[dashed] (X) -- (O);

        \pic [draw, angle radius=0.8cm, "$\alpha$"] {angle = P--O--X};

        \draw[-latex, blue, ultra thick] (P) -- ($(O)!0.5!(P)$) node[left] {$T$};
        \draw[-latex, blue, ultra thick] (P) -- ++(0, -1) node[below] {$mg$};
        
        
    \end{tikzpicture}
\end{center}

\begin{align*}
\text{N2}(\swarrow): &&T +mg \cos \alpha &= m \frac{V^2}{b} \\
\end{align*}
So the string goes slack when $bg\cos \alpha  = V^2 \Rightarrow V = \sqrt{bg \cos \alpha}$.

Once the string goes slack, the particle moves as a projectile. It's initial speed is $V\binom{-\cos \alpha}{\sin \alpha}$ and it's position is $\binom{b\sin \alpha}{b\cos \alpha}$:

\begin{align*}
&& \mathbf{s} &= \binom{b\sin \alpha}{b\cos \alpha}+Vt \binom{-\cos \alpha}{\sin \alpha} + \frac12 gt^2 \binom{0}{-1} \\
&&&= \binom{b\sin \alpha - Vt \cos \alpha}{b\cos \alpha + Vt \sin \alpha - \frac12 gt^2} \\
|\mathbf{s}|^2 = b^2 \Rightarrow && b^2 &= \left ( \binom{b\sin \alpha}{b\cos \alpha}+Vt \binom{-\cos \alpha}{\sin \alpha} + \frac12 gt^2 \binom{0}{-1} \right)^2  \\
&&&= b^2 + V^2t^2+\frac14 g^2 t^4 -gb\cos \alpha t^2-V\sin \alpha gt^3 \\
\Rightarrow && 0 &= V^2t^2 + \frac14 g^2 t^4 - V^2 t^2- V \sin \alpha g t^3 \\
&&&= \frac14 g^2 t^4 - V \sin \alpha gt^3 \\ 
\Rightarrow &&  gT &= 4V \sin \alpha
\end{align*}

The particle will have velocity $\displaystyle \binom{-V \cos \alpha}{V \sin \alpha - 4V \sin \alpha} = \binom{-V \cos \alpha}{-3V \sin \alpha}$ so the angle $\beta$ will satisfy $\tan \beta = 3 \tan \alpha$.

The particle will come to an instantaneous rest if all the momentum is destroyed, ie if the particle is travelling parallel to the string.

\begin{align*}
&& 3 \tan \alpha &= \frac{b\cos \alpha + Vt \sin \alpha - \frac12 gt^2}{b\sin \alpha - Vt \cos \alpha}  \\
&&&= \frac{\frac{V^2}{g}+\frac{4V^2\sin^2\alpha}{g} - \frac{8V^2\sin^2 \alpha}{g}}{\frac{V^2\sin \alpha}{g \cos \alpha} - \frac{4V^2 \sin \alpha \cos \alpha}{g}} \\
&&&= \frac{1 -4\sin^2 \alpha}{\tan \alpha(1 - 4\cos^2 \alpha)} \\
\Leftrightarrow&& 3 \frac{\sin^2 \alpha}{1-\sin^2 \alpha} &= \frac{1- 4 \sin^2 \alpha}{-3+4\sin^2 \alpha} \\
\Leftrightarrow && -9 \sin^2 \alpha + 12 \sin^4 \alpha &= 1 - 5 \sin^2 \alpha + 4 \sin^4 \alpha \\
\Leftrightarrow && 0 &= 1+4 \sin^2 \alpha - 8\sin^4 \alpha \\
\Leftrightarrow && \sin^2 \alpha &= \frac{1 + \sqrt{3}}4
\end{align*}
(taking the only positive root)