Problem source

A uniform rod, of mass $3m$ and length $2a,$ is freely hinged at one end and held by the other end in a horizontal position. A rough particle, of mass $m$, is placed on the rod at its mid-point. If the free end is then released, prove that, until the particle begins to slide on the rod, the inclination $\theta$ of the rod to the horizontal satisfies the equation 
\[
5a\dot{\theta}^{2}=8g\sin\theta.
\]
The coefficient of friction between the particle and the rod is $\frac{1}{2}.$ 
Show that, when the particle begins to slide, $\tan\theta=\frac{1}{26}.$

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \def\a{1};
        \def\t{-20};
        \coordinate (O) at (0,0);
        \coordinate (R) at ({2*\a*cos(\t)},{2*\a*sin(\t)});
        \coordinate (A) at ({\a*cos(\t)},{\a*sin(\t)+0.05});
        \coordinate (Z) at (5, 0);

        \draw (O) -- (R);

        \filldraw (A) circle (1pt);

        \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = R--O--Z};

        \draw[-latex, blue, ultra thick] (A) -- ++ (0,-0.25) node[below] {$4mg$};

    \end{tikzpicture}
\end{center}

While the particle is not sliding, we can consider the whole system.

Considering the moment of inertia about the end, we have:
\begin{align*}
I &= \frac13 \cdot 3m \cdot (2a)^2  + m a^2 \\
&= 5ma^2 
\end{align*}
Taking the level of the pivot as the $0$ GPE level, the initial energy is $0$.

The energy once it has rotated through an angle $\theta$ is: \begin{align*} && 0 &= \text{rotational ke} + \text{gpe} \\
&&&= \frac12 I \dot{\theta}^2 - 4mg \sin \theta \\
&&&= \frac12 5am \dot{\theta}^2 -4mg \sin \theta \\
\Rightarrow && 5a\dot{\theta}^2 &= 8g \sin \theta
\end{align*}
as required.

We also have $5a \ddot{\theta} = 4g \cos \theta$

The acceleration towards the pivot required to maintain circular motion is $m \frac{v^2}{r} = m a \dot{\theta}^2$.

When we are on the point of sliding:

\begin{center}
    \begin{tikzpicture}[scale=2]
        \def\a{1};
        \def\t{-40};
        \coordinate (O) at (0,0);
        \coordinate (R) at ({2*\a*cos(\t)},{2*\a*sin(\t)});
        \coordinate (A) at ({\a*cos(\t)},{\a*sin(\t)+0.05});
        \coordinate (Z) at (5, 0);

        \draw (O) -- (R);

        \filldraw (A) circle (1pt);

        \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = R--O--Z};

        \draw[-latex, blue, ultra thick] (A) -- ++ (0,-0.25) node[below] {$4mg$};
        \draw[-latex, blue, ultra thick] (A) -- ($(A)!0.3!(O)$) node[left] {$\mu R$};
        \draw[-latex, blue, ultra thick] (A) -- ++({-0.3*sin(\t)},{0.3*cos(\t)}) node[above] {$R$};

    \end{tikzpicture}
\end{center}


\begin{align*}
\text{N2}(\nearrow): && R - mg\cos \theta &= -ma \ddot{\theta} \\
\Rightarrow && R &= mg \cos \theta - ma \frac{4mg \cos \theta}{5a} \\
&&&= \frac15mg \cos \theta
\end{align*}

Therefore we must have:

\begin{align*}
\text{N2}(\nwarrow):&&\mu R - mg \sin \theta &= ma \dot{\theta}^2 \\
&& \frac12 \cdot \frac 15 mg \cos \theta &= m \frac{13}5 g \sin \theta \\
\Rightarrow && \tan \theta &= \frac{1}{26}
\end{align*}