Problem source

A hollow circular cylinder of internal radius $r$ is held fixed with its axis horizontal. A uniform rod of length $2a$ (where $a  < r$) rests in equilibrium inside the cylinder inclined at an angle of $\theta$ to the horizontal, where $\theta\ne0$. The vertical plane containing the rod is perpendicular to the axis of the cylinder. The coefficient of friction between the cylinder and each end of the rod is $\mu$, where $\mu > 0$.  Show that, if the rod is on the point of slipping, then the normal reactions $R_1$ and $R_2$ of the lower and higher ends of the rod, respectively, on the cylinder are related by
\[
\mu(R_1+R_2) = (R_1-R_2)\tan\phi
\]
where $\phi$ is the angle between the rod and the radius to an end of the rod.
Show further that
\[
\tan\theta = \frac {\mu r^2}{r^2 -a^2(1+\mu^2)}\,.
\]
Deduce that $\lambda <\phi $, where $\tan\lambda =\mu$.

Solution source

\begin{center}
    \begin{tikzpicture}
        \def\a{2};
        \coordinate (O) at (0,0);
        \coordinate (A) at ({\a*cos(10)},{\a*sin(10)});
        \coordinate (B) at ({\a*cos(240)},{\a*sin(240)});
        \coordinate (M) at ($(A)!0.5!(B)$);
        \coordinate (X) at ($(M) + (0.5, 0)$);
        
        % 1. Draw the circle and name it
        \draw (O) circle (\a);

        \node[right] at (A) {$B$};
        \node[below] at (B) {$A$};
        \node[above] at (O) {$O$};

        \draw[thick] (A) -- (B);

        \draw[dashed] (A) -- (O) -- (B);
        \draw[dashed] (M) -- (X);
        \draw[dashed] (O) -- (M) node[right, pos=0.25] {$d$};

        \draw[ultra thick, blue, -latex] (M) -- ++ (0,-0.6) node[below] {$W$};

        \draw[ultra thick, blue, -latex] (A) -- ($(A)!0.2!(O)$) node[above left] {$R_2$};
        \draw[ultra thick, blue, -latex] (B) -- ($(B)!0.4!(O)$) node[above] {$R_1$};
        \draw[ultra thick, blue, -latex] (A) -- ++({-0.4*sin(10)},{0.4*cos(10)}) node[above] {$\mu R_2$};
        \draw[ultra thick, blue, -latex] (B) -- ++({-0.8*sin(240)},{0.8*cos(240)}) node[above] {$\mu R_1$};

        \pic [draw, angle radius=.4cm, angle eccentricity=1.5, "$\theta$"] {angle = X--M--A};
        \pic [draw, angle radius=.5cm, angle eccentricity=1.5, "$\phi$"] {angle = O--A--B};

    \end{tikzpicture}      
\end{center}

Let $M$ be the midpoint of $AB$, then

\begin{align*}
\overset{\curvearrowright}{M}: && R_1 \sin \phi-\mu R_1 \cos \phi &= R_2 \sin \phi+\mu R_2 \cos \phi \\
\Rightarrow && (R_1-R_2) \tan \phi &= \mu(R_1+R_2)
\end{align*}

As required.

\begin{align*}
&& \cos \phi = \frac{a}{r} &,\,\, \sin \phi = \frac{\sqrt{r^2-a^2}}{r} \\

\text{N2}(\rightarrow): && R_1\cos(\phi + \theta)+\mu R_1 \sin(\phi + \theta) &= R_2 \cos(\theta - \phi) + \mu R_2 \sin(\theta - \phi) \\
\Rightarrow && R_1(\cos \theta \cos \phi - \sin \theta \sin \phi)+ \mu R_1 (\sin \theta \cos \phi + \cos \theta \sin \phi) &= R_2 (\cos\theta \cos \phi + \sin \theta \sin \phi)+ \mu R_2 (\sin \theta \cos \phi - \cos \theta \sin \phi) \\
&& R_1 (1 - \tan \theta \tan \phi)+\mu R_1 (\tan \theta + \tan \phi) &= R_2(1 + \tan \theta \tan \phi) +\mu R_2 (\tan \theta - \tan \phi) \\
&& 0 &= (R_1-R_2)(1+\mu \tan \theta)+(R_1+R_2)(-\tan \theta \tan\phi+\mu \tan \phi) \\
\Rightarrow && \frac{R_1+R_2}{R_1-R_2} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\
\Rightarrow && \frac{\tan \phi}{\mu} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\
\Rightarrow && \tan^2 \phi &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\
\Rightarrow && \frac{r^2-a^2}{a^2} &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\
\Rightarrow && \tan \theta (r^2-a^2-a^2\mu^2) &= \mu a^2+\mu(r^2-a^2) \\
\Rightarrow && \tan \theta &= \frac{\mu r^2}{r^2-(1+\mu^2)a^2}
\end{align*}

Since $\mu r^2 > 0$ we must also have $r^2 > a^2(1+\mu^2)$ ie $\\sec^2 \phi > 1 + \mu^2 = \sec^2 \lambda$ and the result follows.