Problem source

The point $A$ is vertically above 
the point $B$. A light inextensible
string, with a smooth ring $P$ 
of mass $m$ threaded onto it, has its ends
attached at $A$ and $B$. The plane $APB$ rotates
about $AB$ with constant angular velocity $\omega$
so that $P$ describes a horizontal circle of radius $r$
and the string is taut. The angle $BAP$ has value
$\theta$ and the angle $ABP$ has value $\phi$.
Show that
\[\tan\frac{\phi-\theta}{2}=\frac{g}{r\omega^{2}}.\]
Find the tension in the string
in terms of $m$, $g$, $r$, $\omega$
and $\sin\frac{1}{2}(\theta+\phi)$.
Deduce from your results that if $r\omega^2$ is small compared with $g$, then the tension is approximately $\frac{mg}{2}$

Solution source

None
\begin{multicols}{2}
\begin{center}
    
\begin{tikzpicture}[scale=1]
    % Define points
    \coordinate (A) at (0,4);
    \coordinate (B) at (0,0);
    % \coordinate (T) at (3,3);
    \coordinate (P) at (2,1.5);
    
    % Draw lines
    \draw[thick] (A) -- (P);
    \draw[thick] (B) -- (P);
    \draw[dashed] (A) -- (B);
    % \draw[thick] (T) -- (P);
    
    % Add points labels
    \node[left] at (A) {$A$};
    \node[left] at (B) {$B$};
    % \node[right] at (T) {$T$};
    \node[right] at (P) {$P$};
    
    % Add angle θ at B
    \pic [draw, angle radius=0.8cm, "$\phi$"] {angle = P--B--A};
    
    % Add angle φ at T
    \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = B--A--P};
    
    % Add weight label
    \node[right] at ($(P)+(0,-0.5)$) {$mg$};
    
    % Add an arrow for weight if needed
    \draw[-latex, blue, ultra thick] (P) -- ++(0,-0.8) node[right] {};
    \draw[-latex, blue, ultra thick] (P) -- ($(P)!0.25!(A)$) node[right] {$T$};
    \draw[-latex, blue, ultra thick] (P) -- ($(P)!0.35!(B)$) node [above] {$T$};
    % \draw[-latex, thick] (P) -- ({P+(A-P)/2}) node[right] {};
\end{tikzpicture}
\end{center}
\columnbreak
\begin{align*}
N2(\uparrow): && T \cos \theta - T \cos \phi - mg &= 0 \\
N2(\rightarrow): &&  T \sin \theta + T \sin \phi &= m r \omega^2
\\
\\
&&  T \cos \theta - T \cos \phi &= mg \tag{$*$}\\
&&  T \sin \theta + T \sin \phi &= m r \omega^2 \tag{{$**$}}
\end{align*}
\end{multicols}

Dividing $(*)$ by $(**)$ we obtain:

\begin{align*}
    \frac{g}{r\omega^2} &= \frac{\cos \theta - \cos \phi}{\sin \theta + \sin \phi} \\
    &= \frac{2 \sin \left ( \frac{\theta + \phi}2 \right )\sin \left (\frac{\phi - \theta}2 \right )}{2 \sin \left ( \frac{\theta + \phi}2 \right )\cos \left (\frac{\phi - \theta}2 \right )} \\
    &= \tan \left ( \frac{\phi - \theta}2 \right )
\end{align*}

as required.

Squaring and adding $(*)$ and $(**)$ we obtain:

\begin{align*}
    && m^2(g^2 + r^2 \omega^4) &= T^2(2 + \sin \theta \sin \phi - \cos \theta \cos \phi) \\
    && &= T^2(2 - 2\cos (\theta + \phi)) \\
    && &= T^2(2 - 2(1 - 2 \sin^2 \left ( \frac{\theta + \phi}2 \right ) )) \\
    && &= T^2(4 \sin^2 \left ( \frac{\theta + \phi}2 \right )) \\
    \Rightarrow && T &= \frac{m\sqrt{g^2 + r^2 \omega^4}}{2 \sin \left ( \frac{\theta + \phi}2 \right )} \\
    \Rightarrow && T &= \frac{mg\sqrt{1 + \frac{r^2 \omega^4}{g^2}}}{2 \sin \left ( \frac{\theta + \phi}2 \right )}
\end{align*}

If $r \omega^2 \ll g$ then $\tan \l \frac{\phi - \theta}2 \r$ is very large, so $\phi - \theta \approx \pi$ and so $\phi + \theta \approx \pi$. We can then say that

\[ T \approx \frac{mg}{2}\]