Problem source

In this question, $n \geq 2$.
\begin{questionparts}
\item A solid, of uniform density, is formed by rotating through $360°$ about the $y$-axis the region bounded by the part of the curve $r^{n-1}y = r^n - x^n$ with $0 \leq x \leq r$, and the $x$- and $y$-axes.
Show that the $y$-coordinate of the centre of mass of this solid is $\frac{nr}{2(n+1)}$.
\item Show that the normal to the curve $r^{n-1}y = r^n - x^n$ at the point $(rp, r(1-p^n))$, where $0 < p < 1$, meets the $y$-axis at $(0, Y)$, where $Y = r\left(1 - p^n - \frac{1}{np^{n-2}}\right)$.
In the case $n = 4$, show that the greatest value of $Y$ is $\frac{1}{4}r$.
\item A solid is formed by rotating through $360°$ about the $y$-axis the region bounded by the curves $r^3y = r^4 - x^4$ and $ry = -(r^2 - x^2)$, both for $0 \leq x \leq r$.
$A$ and $B$ are the points $(0, -r)$ and $(0, r)$, respectively, on the surface of the solid.
Show that the solid can rest in equilibrium on a horizontal surface with the vector $\overrightarrow{AB}$ at three different, non-zero, angles to the upward vertical. You should not attempt to find these angles.
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{center}
    \begin{tikzpicture}[scale=2]
    \draw[->] (0, -0.5) -- (0, 1.5);
    \draw[->] (-2, 0) -- (2, 0);

    \draw[domain = 0:1, variable = \x]  plot ({\x},{1-(\x)^3}); 
    \draw[domain = 0:1, variable = \x]  plot ({-\x},{1-(\x)^3}); 

    
    % \draw (2, 0) -- (2, 4.2426);
    \draw[domain = 3.14:6.81, variable = \x]  plot ({(1-(0.6)^3)*cos(deg(\x))}, {0.5+sin(deg(\x))/10}); 
    \draw[domain = 0:6.81, variable = \x]  plot ({(1-(0.6)^3)*cos(deg(\x))}, {0.55+sin(deg(\x))/10}); 
    
    
    % \node at (2.05, 1.6) {\tiny $\delta x$};
    % \node at (2.5, 1.4) {\tiny $(x,y)$};
    
    \node[below] at (1, 0)  {\small $r$};
    \node[below] at (-1, 0)  {\small $-r$};
    \node[above left] at (0, 1)  {\small $r$};
    
\end{tikzpicture}
\end{center}

By symmetry, the centre of mass will lie on the $y$ axis. 

Notice that a single slice (when revolved around the $y$-axis) has volume $y \cdot \pi \cdot ((x+ \delta x)^2 - x^2) = 2 \pi x y \delta x$, and COM at height $\frac12 y$ so we can conclude:

\[ \overline{y} \sum_{\delta x} 2 \pi x y \delta x = \sum_{\delta x} \pi xy^2  \delta x\]

\begin{align*}
&& \overline{y} \int_0^r 2xy  \d x &= \int_0^r  y^2 x \d x \\
\Rightarrow && \overline{y} 2\int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)x \d x &= \int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)^2 x \d x \\
\Rightarrow && \overline{y} \left [r \frac{x^2}{2} - \frac{1}{r^{n-1}} \frac{x^{n+2}}{n+2}  \right]_0^r &= \left [r^2 \frac{x^2}{2} - \frac{2}{r^{n-2}} \frac{x^{n+2}}{n+2} + \frac{1}{r^{2n-2}} \frac{x^{2n+2}}{2n+2} \right]_0^r \\
\Rightarrow && 2\overline{y} \left (\frac{r^3}{2} - \frac{r^3}{n+2} \right) &= \left (\frac12 r^4 - \frac{2}{n+2}r^4 + \frac{1}{2n+2}r^4 \right) \\
\Rightarrow && \overline{y}r^3 \frac{n}{(n+2)} &= r^4\frac{(n+1)(n+2)-2\cdot2\cdot(n+1)+(n+2)}{2(n+1)(n+2)} \\
\Rightarrow && \overline{y} \frac{n}{(n+2)} &= r \left ( \frac{n^2}{2(n+1)(n+2)} \right) \\
\Rightarrow && \overline{y} &= \frac{nr}{2(n+1)} \\
&&&= r \left (1 -p^n  \right)
\end{align*}

as required.

\item \begin{align*}
&& r^{n-1}y &= r^n - x^n \\
\frac{\d}{\d x}: && r^{n-1} \frac{\d y}{\d x} &= -n x^{n-1} \\
&& \frac{\d y}{\d x} &= -np^{n-1}
\end{align*}

Therefor the normal has the equation:

\begin{align*}
&& \frac{y-r(1-p^n)}{x-rp} &= \frac{1}{np^{n-1}} \\
\Rightarrow && Y &= \frac{-rp}{np^{n-1}} + r(1-p^n) \\
&&&= r \left (1 - p^n - \frac{1}{np^{n-2}} \right)
\end{align*}

If $n = 4$ then 
\begin{align*}
&& Y &= r\left (1 - p^4 - \frac{1}{4p^{2}} \right) \\
\Rightarrow && \frac{\d Y}{\d p} &= r \left (-4p^3 + \frac{1}{2p^3} \right)
\end{align*}

Therefore there is a stationary point if $p^6 = \frac18 \Rightarrow p =2^{-1/2}$. Clearly this will be a maximum (sketch or second derivative) therefore, $Y = r(1 - \frac14 - \frac{2}{4}) = \frac14 r$

\item The centre of mass of this shape can be found using this table:

\begin{array}{|c|c|c|} \hline
\text{} & \overline{y} & \text{mass} \\ \hline
r^3y = r^4 - x^4 & \frac{2r}{5} & \frac{4\pi r^3}{6} = \frac23 \pi r^3\\
ry = -(r^2 - x^2) & -\frac{r}{3}& \frac{2 \pi r^3}{4}=\frac12\pi r^3 \\ 
\text{combined} & \frac{(\frac25 \cdot \frac23-\frac13 \cdot \frac12)r^4}{\frac76 r^3} = \frac3{35}r & \frac76 \pi r^3\\
\hline 
\end{array}

Normals to the surface through points on the upper surface will meet the $y$-axis between $(-\infty, \frac14 r)$, and since $p = 0 \to -\infty$ and $p = 1 \to -\frac14 r$, so normals will pass through $(0, \frac3{35}r)$ from two different points.

Normals to the surface through points on the lower surface will go through $-r(1 - p^2 - \frac12) =- r(\frac12 -p^2)$ which ranges monotonically from $\frac12 r \to -\frac12 r$ so there will be one point where the normal goes through $\frac3{35}r$. Therefore there are three angles where the vector $\overrightarrow{AB}$ is not vertical but the normal to the surfaces runs through the centre of mass (ie the the solid can rest in equilibrium)

\end{questionparts}