Problem source

The base of a non-uniform solid hemisphere, of mass $M,$ has radius $r.$ The distance of the centre of gravity, $G$, of the hemisphere from the base is $p$ and from the centre of the base is $\sqrt{p^2 + q^2} \,$.  The hemisphere rests in equilibrium with  its curved surface on a horizontal plane. 
A particle of mass $m,\,$ where $m$ is small, is attached to $A\,$, the lowest point of the circumference of the base.  In the new position of equilibrium, find the angle, $\alpha$, that the base makes with the horizontal.
The particle is removed and attached to the point $B$ of the base which is at the other end of the diameter through $A\,$. In the new position of equilibrium the base  makes an angle ${\beta}$ with the horizontal.
Show that
$$\tan(\alpha-\beta)= \frac{2mMrp}  {M^2\left(p^2+q^2\right)-m^2r^2}\;.$$

Solution source

\begin{tikzpicture}[scale=1.5]
    \draw[domain = 10:190, samples=180, variable = \x]  plot ({2*cos(\x)},{2*(1-sin(\x))}); 

    \draw ({2*cos(10)}, {2*(1-sin(10))}) --  ({2*cos(190)}, {2*(1-sin(190))});

    \draw (-2, 0) -- (2, 0);
    \draw[dashed] ({cos(10) + cos(190)}, {2 - sin(10) - sin(190)}) -- (0, 1.3);
    \draw[dashed] ({0.525*2*cos(10) + 0.475*2*cos(190)}, 
                   {0.525*2*(1 - sin(10))+ 0.475*2*(1 - sin(190))}) -- (0, 1.3);
    
    % \node[circle,fill=black,inner sep=0pt,minimum size=1pt] at (0, 1.1) {$G$};
    \node[below] at (0, 1.3) {$G$};
    \node[above] at (0.1, 2) {$p$};
    \node[right] at (0.1, 1.7) {$q$};
    \node[left] at (-0.05, 1.7) {$\sqrt{p^2+q^2}$};
    \filldraw (0, 1.3) circle (0.5pt); 

    \node[above] at ({2*cos(10)}, {2*(1-sin(10))}) {$A$};
    \node[above] at ({2*cos(190)}, {2*(1-sin(190))}) {$B$};

    
\end{tikzpicture}

In the coordinate system where $(0,0)$ is the centre base of the hemisphere, $G$ is at $(p, q)$.

Once the mass is attached at $A$, the new centre of mass will satisfy: $M \begin{pmatrix} p \\ q \end{pmatrix} + m \begin{pmatrix} r \\ 0 \end{pmatrix} = (M+m)\bar{x} \Rightarrow \bar{x} = \frac{1}{M+m} \begin{pmatrix} Mp+mr \\ Mq \end{pmatrix}$

The angle between the horizontal and $AB$, $\alpha$ will satisfy:

$$\tan \alpha = \frac{Mp + mr}{Mq}$$

Similarly, when the mass is attached at $B$, the new centre of mass will satisfy: $M \begin{pmatrix} p \\ q \end{pmatrix} + m \begin{pmatrix} -r \\ 0 \end{pmatrix} = (M+m)\bar{x} \Rightarrow \bar{x} = \frac{1}{M+m} \begin{pmatrix} Mp-mr \\ Mq \end{pmatrix}$
The angle between the horizontal and $AB$, $\beta$ will satisfy:
$$\tan \beta = \frac{Mp - mr}{Mq}$$

We are trying to find:
\begin{align*}
    \tan \l \alpha - \beta \r &= \frac{\tan \alpha - \tan \beta}{1+ \tan \alpha \tan \beta} \\
    &= \frac{\frac{Mp + mr}{Mq} - \frac{Mp - mr}{Mq}}{1 + \frac{Mp + mr}{Mq} \frac{Mp - mr}{Mq}} \\
    &= \frac{(Mp + mr)Mq - (Mp - mr)Mq}{M^2q^2 + (Mp + mr)(Mp - mr)} \\
    &= \frac{2Mmrp}{M^2(q^2+p^2) -m^2r^2} \\
\end{align*}