Problem source

A uniform rod $PQ$ of mass $m$ and length $3a$ is freely  hinged at $P$. The rod is held horizontally and a particle of mass $m$ is placed on top of the rod at a distance~$\ell$ from $P$, where $\ell <2a$. The coefficient of friction between the rod and the particle is $\mu$. The rod is then released. Show that, while the particle does not  slip along the rod,
\[
(3a^2+\ell^2)\dot \theta^2 = g(3a+2\ell)\sin\theta \,,
\]
where $\theta$ is the angle through which the rod has turned, and the dot denotes the time derivative.
Hence, or otherwise, find an expression for $\ddot \theta$ and show that the normal reaction of the rod on the particle is non-zero when~$\theta$ is acute. 
Show further that, when the particle is on the point of slipping,
\[
\tan\theta = \frac{\mu a (2a-\ell)}{2(\ell^2 + a\ell +a^2)} \,.
\]  
What happens at the moment the rod is released  if, instead,  $\ell>2a$?

Solution source

\begin{tikzpicture}[scale=2]
    % Define points
    \coordinate (P) at (0,0);
    \coordinate (Q) at (3*12/13,-3*5/13);
    \coordinate (M) at (1.5*12/13,-1.5*5/13);
    \coordinate (A) at (12/13, -5/13);
    \coordinate (H) at (2*12/13, 0);
    
    % Draw lines
    \draw[thick] (P) -- (Q);
    \draw[dashed] (P) -- (H);
    
    % Add points labels
    \node[above] at (P) {$P$};
    \node[above] at (Q) {$Q$};
    \node at (A)[circle,fill,inner sep=1.5pt]{};

    % % Add angle φ at T
    \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = A--P--H};
    
    % % Add an arrow for weight if needed
    \draw[-latex, blue, ultra thick] (M) -- ++(0,-0.8) node[below] {$mg$};
    \draw[-latex, blue, ultra thick] (A) -- ++(0,-0.8) node[below] {$mg$};
    \draw[-latex, blue, ultra thick] (A) -- ++(-0.4*12/13,0.4*5/13) node[left] {$\scriptstyle F \leq \mu R$};
    \draw[-latex, blue, ultra thick] (A) -- ++(0.8*5/13,0.8*12/13) node[above] {$R$};

    \draw[-latex, red, ultra thick] (2.4*12/13, 0) -- (2.1*12/13, 0) node[above] {$0 $ GPE};

    \draw [decorate,decoration={brace,amplitude=10pt},xshift=-0.5cm,yshift=0pt]
    (P) -- (A) node [black,midway,yshift=0.5cm,xshift=0.2cm]
    {$\ell$};

    \draw [decorate,decoration={brace,amplitude=20pt},xshift=-0.5cm,yshift=0pt]
    (P) -- (Q) node [black,midway,yshift=0.8cm,xshift=0.2cm]
    {$3a$};

\end{tikzpicture}


By energy considerations, the initial energy is $0$.

\begin{center}
\begin{tabular}{l|c|c}
     & Inital & \@ $\theta$ \\ \hline
    Rotational KE of rod & $0$ & $\frac{1}{2}I\dot{\theta}^2 = \frac{1}{2} \frac{1}{3} m (3a)^2 \dot{\theta}^2 = \frac32 m a^2 \dot{\theta}^2$\\
    KE of particle & $0$ & $\frac12 m \ell^2\dot{\theta}^2$\\
    GPE of rod & $0$ & $-\frac{3}{2}mga \sin \theta$\\
    GPE of particle & $0$ & $-mg \ell \sin \theta$ \\ \hline
    Total & $0$ & $\frac12m \l \l 3a^2 + \ell^2\r \dot{\theta}^2 - \l 3a + 2\ell  \r g \sin \theta \r$
\end{tabular}
\end{center}

Therefore:
\begin{align*}
&& \l 3a^2 + \ell^2\r \dot{\theta}^2 &= \l 3a + 2\ell  \r g \sin \theta \\
\Rightarrow && \l 3a^2 + \ell^2\r 2\dot{\theta} \ddot{\theta} &= \l 3a + 2\ell  \r g \cos\theta \dot{\theta} \tag{$\frac{\d}{\d t}$} \\
\Rightarrow && 2\l 3a^2 + \ell^2\r  \ddot{\theta} &= \l 3a + 2\ell  \r g \cos\theta  \\
\Rightarrow &&  \ddot{\theta} &= \boxed{\frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta}  \\
\end{align*}

\begin{align*}
    \text{N}2(\perp PQ): && mg \cos \theta - R &= m \ell \ddot{\theta} \\
    && R &= mg \cos \theta - m \ell \l \frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta \r \\
    && &= mg\cos \theta \l 1 - \ell \frac{3a + 2\ell }{2(3a^2 + \ell^2)} \r \\
    && &= mg \cos \theta \l \frac{6a^2 + 2\ell^2 - 3a\ell - 2\ell^2}{2(3a^2 + \ell^2)} \r \\
    && &= mg \cos \theta \l \frac{3a(2a  - \ell)}{2(3a^2 + \ell^2)} \r > 0 \tag{since $2a > \ell$}
\end{align*}

At limiting equilibrium, $F = \mu R$.

\begin{align*}
    \text{N}2(\parallel PQ): && \mu R - mg \sin \theta &= m \ell \dot{\theta}^2 \\ 
    \Rightarrow && \mu mg \cos \theta \l \frac{3a(2a  - \ell)}{2(3a^2 + \ell^2)} \r - mg \sin \theta &= m \ell \frac{(3a+2\ell)}{(3a^2+\ell^2)} g \sin \theta \\
    \Rightarrow && \mu  \l 3a(2a  - \ell) \r -  \l 2(3a^2 + \ell^2) \r \tan \theta &= 2\ell (3a+2\ell) \tan \theta \\
    \Rightarrow && \mu  \l 3a(2a  - \ell) \r &= \l 6a\ell + 6a^2 + 6\ell^2 \r \tan \theta \\
    \Rightarrow && \tan \theta &= \boxed{\frac{\mu a(2a-\ell)}{2(a^2 + a\ell + \ell^2)}}
\end{align*}

If $\ell > 2a$, then the initial reaction force will be $0$, ie the particle will have no contact with the rod. In other words, the rod will rotate faster than the particle will free-fall and the particle immediately loses contact with the rod.