Problem source

A heavy particle lies on a smooth horizontal table, and is attached to one end of a light inextensible string of length $L$. The other end of the string is attached to a point $P$ on the circumference of the base of a vertical post which is fixed into the table. The base of the post is a circle of radius $a$ with its centre at a point $O$ on the table. Initially, at time $t=0$, the string is taut and perpendicular to the line $OP.$ The particle is then struck in such a way that the string starts winding round the post and remains taut.
At a later time $t$, a length $a\theta(t)\ (< L)$ of the string is in contact with the post. Using cartesian axes with origin $O$, find the position and velocity vectors of the particle at time $t$ in terms of $a,L,\theta$ and $\dot{\theta},$ and hence show that the speed of the particle is $(L-a\theta)\dot{\theta}.$ 
If the initial speed of the particle is $v$, show that the particle hits the post at a time $L^{2}/(2av).$

Solution source

\begin{center}
    \begin{tikzpicture}[scale=1]
        \draw[->, dashed] (-3, 0) -- (3,0);
        \draw[->, dashed] (0, -3) -- (0,3);
        \draw (0,0) circle (1);
        \draw[ultra thick, red] (1,0) -- (1,3);
        \draw[domain = 0:45, samples=50, variable = \x, red, ultra thick]  plot ({cos(\x)},{sin(\x)});
        \draw[ultra thick, red] ({1/sqrt(2)},{1/sqrt(2)}) -- ({(-3+pi/4+1)/sqrt(2)},{(3-pi/4+1)/sqrt(2)});

        \coordinate (O) at (0,0);
        \coordinate (P) at (1,0);
        \coordinate (Q) at ({1/sqrt(2)},{1/sqrt(2)});

        \filldraw (1,3) circle (1.5pt);
        \filldraw ({(-3+pi/4+1)/sqrt(2)},{(3-pi/4+1)/sqrt(2)}) circle (1.5pt);
        
        
        \draw[dashed] (O) -- (Q);
        \pic [draw, angle radius=1cm, "$\theta$"] {angle = P--O--Q};
    \end{tikzpicture}
\end{center}

As the string wraps around, the total length in contact will be $a \theta$. The end contact point will be at $(a\cos \theta, a\sin \theta)$ and the string will be tangential to that. The tangent (unit) vector will be $\binom{-\sin \theta}{\cos \theta}$, and so the particle will be at $\binom{a\cos \theta - (L-a\theta) \sin \theta}{a \sin \theta + (L-a \theta) \cos \theta}$.

The velocity will be:

\begin{align*}
\frac{\d}{\d t} \binom{a\cos \theta - (L-a\theta) \sin \theta}{a \sin \theta + (L-a \theta) \cos \theta} &= \binom{-a \sin \theta \cdot \dot{\theta} -(L-a \theta) \cos \theta \cdot \dot{\theta} + a \sin \theta \cdot \dot{\theta} }{a \cos \theta \cdot \dot{\theta} + (L-a \theta) \sin \theta \cdot \dot{\theta} - a \cos \theta \cdot \dot{\theta}} \\
&= \binom{-(L-a \theta) \cos \theta \cdot \dot{\theta}  }{ (L-a \theta) \sin \theta \cdot \dot{\theta}} \\
\end{align*}

Therefore the speed is $(L-a\theta) \dot{\theta}$.

By conservation of energy, we must have that speed is constant, ie:

\begin{align*}
&& (L - a \theta)\dot{\theta} &= v \\
\Rightarrow && \int_0^{L/a} (L - a \theta)\d \theta &= \int_0^T v \d t \\
\Rightarrow && vT &= \frac{L^2}{a} - a\frac{L^2}{2a^2} \\
&&&= \frac{L^2}{2a} \\
\Rightarrow && T &= \frac{L^2}{2av}
\end{align*}

as requried