Problem source

The plot of `Rhode Island Red and the Henhouse of Doom' calls for the heroine to cling on to the circumference of a fairground wheel of radius $a$ rotating with constant angular velocity $\omega$ about its horizontal axis and then let go. Let $\omega_{0}$ be the largest value of $\omega$ for which it is not possible for her subsequent path to carry her higher than the top of the wheel. Find $\omega_{0}$ in terms of $a$ and $g$. If $\omega>\omega_{0}$ show that the greatest height above the top of the wheel to which she can rise is
\[\frac{a}{2}\left(\frac{\omega}{\omega_{0}}
-\frac{\omega_{0}}{\omega}\right)^{\!\!2}.\]

Solution source

\begin{center}
    \begin{tikzpicture}

        \coordinate (O) at (0,0);
        \coordinate (A) at (0,2);
        \coordinate (P) at ({2*cos(140)},{2*sin(140)});

        \draw (O) circle (2);

        \filldraw (O) circle (1.5pt) node[above right] {$O$};
        \filldraw (P) circle (1.5pt) node[above left] {$P$};

        \draw[dashed] (A) -- (O);
        \draw (O) -- (P);

        \pic [draw, angle radius=.7cm, angle eccentricity=1.3, "$\theta$"] {angle = A--O--P};

        \draw[-latex, red, thick] (P) -- ++({sin(140)}, {-cos(140)});
    
    
    \end{tikzpicture}
\end{center}

\begin{align*}
    \uparrow: && v &= u + at \\
    \Rightarrow && T &= \frac{a \omega \sin \theta}{g} \\
    && s &= ut + \frac12 gt^2 \\
    \Rightarrow && s &=  a\omega \sin \theta \cdot \frac{a \omega \sin \theta}{g} - \frac12 g \left ( \frac{a \omega \sin \theta}{g} \right) ^2 \\
    &&&= \frac1{2g} a^2 \omega^2 \sin^2 \theta \\
    s < \text{distance to top}: && \frac1{2g} a^2 \omega^2 \sin^2 \theta &< a(1- \cos \theta) \\
    \Rightarrow && \omega^2 &< \frac{2g}{a} \frac{1-\cos \theta}{\sin^2 \theta} \\
    &&&= \frac{2g}{a} \frac{2 \sin^2 \tfrac12 \theta}{4 \sin^2 \tfrac12 \theta \cos^2 \tfrac12 \theta} \\
    &&&= \frac{g}{a} \sec^2 \tfrac12 \theta \\
    &&&\leq \frac{g}{a} \tag{since it holds for all $\theta$ it holds for min $\theta$}\\
    \Rightarrow && \omega_0 &= \sqrt{\frac{g}{a}} \\
    \\
    && \text{max height} &= \frac1{2g} a^2 \omega^2 \sin^2 \theta - a(1-\cos \theta) \\
    &&&= \frac1{2g} a^2 \omega^2 (1-\cos^2 \theta) - a(1-\cos \theta) \\
    &&&= \frac{a}{2} \left (- \frac{\omega^2}{\omega_0^2} \cos^2 \theta + 2 \cos \theta + \frac{\omega^2}{\omega_0^2}-2 \right) \\
    &&&= \frac{a}{2} \left (-\left (\frac{\omega_0}{\omega}- \frac{\omega}{\omega_0} \cos \theta  \right)^2 + \frac{\omega^2}{\omega_0^2}-2+\frac{\omega_0^2}{\omega^2} \right) \\
    &&&= \frac{a}{2} \left ( \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} \right)^2 - \frac{a}{2} \left (\frac{\omega_0}{\omega}- \frac{\omega}{\omega_0} \cos \theta  \right)^2
\end{align*}

If $\omega > \omega_0$ we can find a $\theta$ such that the second bracket is $0$, hence the maximium height is as desired.