Problem source

In this question, $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors and $\mathbf{j}$ is vertically upwards.
A smooth hemisphere of mass $M$ and radius $a$ rests on a smooth horizontal table with its plane face in contact with the table. The point $A$ is at the top of the hemisphere and the point $O$ is at the centre of its plane face.
Initially, a particle $P$ of mass $m$ rests at $A$. It is then given a small displacement in the positive $\mathbf{i}$ direction. At a later time $t$, when the particle is still in contact with the hemisphere, the hemisphere has been displaced by $-s\mathbf{i}$ and $\angle AOP = \theta$.
\begin{questionparts}
\item Let $\mathbf{r}$ be the position vector of the particle at time $t$ with respect to the initial position of $O$. Write down an expression for $\mathbf{r}$ in terms of $a$, $\theta$ and $s$ and show that
$$\dot{\mathbf{r}} = (a\dot{\theta} \cos \theta - \dot{s})\mathbf{i} - a\dot{\theta} \sin \theta \mathbf{j}.$$
Show also that
$$\dot{s} = (1 - k)a\dot{\theta} \cos \theta,$$
where $k = \frac{M}{m + M}$, and deduce that
$$\dot{\mathbf{r}} = a\dot{\theta}(k \cos \theta \mathbf{i} - \sin \theta \mathbf{j}).$$
\item Show that
$$a\dot{\theta}^2 \left(k \cos^2 \theta + \sin^2 \theta\right) = 2g(1 - \cos \theta).$$
\item At time $T$, when $\theta = \alpha$, the particle leaves the hemisphere. By considering the component of $\ddot{\mathbf{r}}$ parallel to the vector $\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}$, or otherwise, show that at time $T$
$$a\dot{\theta}^2 = g \cos \alpha.$$
Find a cubic equation for $\cos \alpha$ and deduce that $\cos \alpha > \frac{2}{3}$.
\end{questionparts}

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \def\a{2};
        \coordinate (P) at ({\a*cos(70)}, {\a*sin(70)});
        \coordinate (O) at (0,0);
        \coordinate (A) at ({0}, {\a});
        
            
        \filldraw (P) circle (1.5pt) node[above right] {$P$};
        \filldraw (A) circle (0.5pt) node[above] {$A$};
        \filldraw (O) circle (0.5pt) node[below] {$O$};
        
        \draw[domain = 0:180, samples=100]
            plot({\a*cos(\x)}, {\a*sin(\x)});

        \draw[dashed] (P) -- (O) -- (A);
        \pic [draw, angle radius=1.2cm, angle eccentricity=1.5, "$\theta$"] {angle = P--O--A};
    \end{tikzpicture}
\end{center}


\begin{questionparts}
    \item $\mathbf{r} = (a \sin \theta - s) \mathbf{i}+a\cos \theta\mathbf{j}$, so
    \begin{align*}
        && \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j}\\
        \\
        \text{COM}(\rightarrow): && 0 &= M(-\dot{s}) + m(a \dot{\theta} \cos \theta - \dot{s}) \\
        \Rightarrow && \dot{s} &= \frac{ma \dot{\theta} \cos \theta}{m+M} \\
        &&&= \left ( 1- \frac{M}{m+M} \right) a\dot{\theta} \cos \theta \\
        &&&= (1 - k)  a\dot{\theta} \cos \theta \\
        \\
        \Rightarrow &&  \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\
        &&&= (a \dot{\theta} \cos \theta - (1 - k)  a\dot{\theta} \cos \theta) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\
        &&&= a\dot{\theta} \left ( k \cos \theta \mathbf{i} - \sin \theta \mathbf{j} \right)
    \end{align*}

\item $\,$ \begin{align*}
    COE: &&\underbrace{0}_{\text{k.e.}}+ \underbrace{mga}_{\text{GPE}} &= \underbrace{\frac12 m \mathbf{\dot{r}}\cdot\mathbf{\dot{r}}}_{\text{k.e. }P} + \underbrace{mg a\cos \theta}_{\text{GPE}} + \underbrace{\frac12 M \dot{s}^2}_{\text{k.e. hemisphere}} \\
    \Rightarrow && 2amg(1-\cos \theta) &= a^2m \dot{\theta}^2(k^2 \cos^2 \theta + \sin^2 \theta)+ M(1 - k)^2  a^2\dot{\theta}^2 \cos^2 \theta \\
    \Rightarrow && 2mg(1-\cos \theta) &= a  \dot{\theta}^2 \left (m\sin^2 \theta + (mk^2 + M(1-k)^2)\cos^2 \theta \right) \\
    &&&= a  \dot{\theta}^2 \left (m\sin^2 \theta + mk\cos^2 \theta \right) \\
    \Rightarrow && 2g(1-\cos \theta) &= a  \dot{\theta}^2 \left (\sin^2 \theta + k\cos^2 \theta \right) \\
\end{align*}

\item The equation of motion is $m \ddot{\mathbf{r}} = \mathbf{R} - mg\mathbf{j}$ and the particle will leave the surface when $\mathbf{R} = 0$. If we take the component in the directions suggested:
\begin{align*}
    && \ddot{\mathbf{r}} &= a\ddot{\theta}(k \cos \theta \mathbf{i}- \sin \theta \mathbf{j}) + a \dot{\theta}(-k\dot{\theta} \sin \theta \mathbf{i}- \dot{\theta} \cos \theta \mathbf{j}) \\
    &&&= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \mathbf{i} -a(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta) \mathbf{j} \\
    \Rightarrow && \mathbf{\ddot{r}} \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \sin \theta -ak(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta)\cos \theta \\
    &&&= - ak \dot{\theta}^2 \\
    && (-g\mathbf{j}) \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= -gk \cos \theta \\
    \mathbf{R} = 0: && gk \cos \theta &= ak \dot{\theta}^2 \\
    \Rightarrow && g \cos \theta &= a \dot{\theta}^2
\end{align*}



\item $\,$
\begin{align*}
     && 2g(1-\cos \theta) &= a  \dot{\theta}^2(k \cos^2 \theta + \sin^2 \theta) \\
     && a  \dot{\theta}^2 &= g \cos \alpha \\
     \Rightarrow &&  2g(1-\cos \alpha) &=  g \cos \alpha(k \cos^2 \alpha + (1-\cos^2 \alpha)) \\
     \Rightarrow && 0 &= g(k-1)c^3+3gc-2g \\
     \Rightarrow && 0 &= (k-1)c^3+3c - 2
\end{align*}

When $c =1, f(c) = k > 0$ when $c = \frac23, f(c) = k-1 < 0$. Therefore there is a root with $\cos \alpha > \frac23$
    
\end{questionparts}