Problem source

A particle $P$ of mass $m$ is connected by two light inextensible strings to two fixed points $A$ and $B$, with $A$ vertically above $B$. The string $AP$ has length $x$.  The particle is rotating about the vertical through $A$ and $B$ with angular velocity $\omega$, and both strings are taut.  Angles $PAB$ and $PBA$ are $\alpha$ and $\beta$, respectively.
  Find the tensions $T_A$ and $T_B$ in the strings $AP$ and $BP$ (respectively), and hence show that $\omega^2 x\cos\alpha \ge g$. Consider now the case that $\omega^2 x\cos\alpha = g$. Given that $AB=h$ and $BP=d$, where $h>d$,  show that $h\cos\alpha \ge \sqrt{h^2-d^2}$. Show further that 
  \[
  mg < T_A \le \frac{mgh}{\sqrt{h^2-d^2}\,}\,.
  \]
  Describe the geometry of the strings when $T_A$ attains its upper bound.

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]

        \coordinate (A) at (0, 4);
        \coordinate (B) at (0, -2);
        \coordinate (P) at (2, 0);
        \coordinate (O) at (0, 0);

        \filldraw (P) circle (1pt);

        \draw (A) -- (P) -- (B);

        \draw[dashed] (A) -- (B);
        \draw[dashed] (P) -- (O);

        % \node[right] at ($(A)!0.5!(P)$) {$x$};
        \node[above] at ($(O)!0.5!(P)$) {$x \sin \alpha$};
        % \node[left] at ($(A)!0.5!(B)$) {$h$};
        % \node[right] at ($(P)!0.5!(B)$) {$d$};

        \node[left] at (A) {$A$};
        \node[left] at (B) {$B$};
        \node[right] at (P) {$P$};

        % \draw[dashed] (B) circle ({2*sqrt(2)});
    

        \draw[-latex, ultra thick, blue] (P) -- ($(P)!0.2!(A)$) node[above] {$T_A$};
        \draw[-latex, ultra thick, blue] (P) -- ($(P)!0.2!(B)$) node[below] {$T_B$};
        \draw[-latex, ultra thick, blue] (P) -- ++ (0, -0.5) node[below] {$mg$};
        
        \pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\alpha$"] {angle = B--A--P};
        \pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\beta$"] {angle = P--B--A};
        
    \end{tikzpicture}
\end{center}



\begin{align*}
\text{N2}(\uparrow): && T_A \cos \alpha - T_B \cos\alpha - mg &= 0 \\
\Rightarrow && T_A \cos \alpha - T_B \cos\beta &= mg \\
\text{N2}(\leftarrow, \text{radially}): && T_A \sin \alpha + T_B \sin \beta &= m x \sin \alpha \omega^2 \\
\Rightarrow && T_A(\cos \alpha \sin \beta+\sin \alpha \cos \beta) &= mg \sin \beta + mx \sin \alpha \omega^2 \cos \beta  \\
\Rightarrow && T_A &=\frac{mg\sin \beta + m x \sin \alpha  \omega^2 \cos \beta }{\sin(\alpha + \beta)} \\
\Rightarrow && T_B(\sin \beta \cos \alpha- \cos \beta \sin \alpha)&= mx \sin \alpha  \omega^2 \cos \alpha -mg \sin \alpha \\
\Rightarrow && T_B &= \frac{m x \sin \alpha \omega^2 \cos \alpha - mg \sin \alpha}{\sin(\beta - \alpha)} \\
&&&= \frac{m \sin \alpha(\omega^2  \cos\alpha - g)}{\sin (\beta - \alpha)}
\end{align*}

Since $T_B \geq 0 \Rightarrow \omega^2  \cos\alpha - g \geq 0$ as required.

\begin{center}
    \begin{tikzpicture}[scale=2]

        \coordinate (A) at (0, 4);
        \coordinate (B) at (0, -2);
        \coordinate (P) at (2, 0);
        \coordinate (O) at (0, 0);

        \coordinate (X) at ({2*sqrt(2)*cos(atan(1/2)}, {2*sqrt(2)*sin(atan(1/2))-2});

        \filldraw (X) circle (1pt);

        \filldraw (P) circle (1pt);

        \draw (A) -- (P) -- (B);

        \draw[dashed] (A) -- (B);
        \draw[dashed] (P) -- (O);

        \node[right] at ($(A)!0.5!(P)$) {$x$};
        \node[above] at ($(O)!0.5!(P)$) {$x \sin \alpha$};
        \node[left] at ($(A)!0.5!(B)$) {$h$};
        \node[right] at ($(P)!0.5!(B)$) {$d$};

        \node[left] at (A) {$A$};
        \node[left] at (B) {$B$};
        \node[right] at (P) {$P$};

        \draw[dashed] (B) circle ({2*sqrt(2)});

        \draw[dashed] (B) -- (X) -- (A);

        \draw[-latex, ultra thick, blue] (P) -- ($(P)!0.2!(A)$) node[above] {$T_A$};
        \draw[-latex, ultra thick, blue] (P) -- ($(P)!0.2!(B)$) node[below] {$T_B$};
        \draw[-latex, ultra thick, blue] (P) -- ++ (0, -0.5) node[below] {$mg$};
        
        \pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\alpha$"] {angle = B--A--P};
        \pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\beta$"] {angle = P--B--A};
        
    \end{tikzpicture}
\end{center}

$\sqrt{h^2-d^2}$ is the length of the final side on the dashed right angle triangle with hypotenuse $AB$. $h \cos \alpha$ will be clearly longer as the angle $\alpha$ will be smaller and so $\cos \alpha$ will be larger.

When $\omega^2 x \cos \alpha = g$ we must have $T_B = 0$. $T_A\cos \alpha = mg \Rightarrow T_A > mg$ since $\alpha \neq 0$.
$T_A = \frac{mg}{\cos \alpha} \leq \frac{mgh}{\sqrt{h^2-d^2}}$

$T_A$ will attain it's upper bound when $\angle APB$ is a right angle.