An example that a 2D shape with its centre of mass on its boundary [closed]

The object has constant density. Could any body suggest one for me?

Solution 1:

Take the unit disk, drill a hole of radius $r$ centered at $(1-r,0)$. The boundary of the hole will be touching the boundary of the unit disk. The resulting shape will be sort of like a crescent.

Let $(x,0)$ be the CM of this shape. If we combine it with the disk of radius $r$ that get removed, we known the combined CM is the origin. This means

$$\pi(1-r^2) x + \pi r^2 (1-r) = 0 \quad\implies\quad x = -\frac{r^2}{r+1}$$

If $x = 1-2r$, the CM of the "crescent" shape will be lying on its boundary.

Solving $\displaystyle\;1 - 2r = -\frac{r^2}{r+1}\;$ gives us $\displaystyle\;r = \frac{\sqrt{5}-1}{2} = \frac{1}{\phi}$, the inverse of the golden ratio!

The final figure looks like this:

$\hspace1in$

Solution 2:

The Moon.

When it is full, the center of mass is at the center of the circle.

When it is very thin, the center of mass is in the concave side of the crescent.

In between, there is a moment when the center of mass is on the boundary.

Addendum:

Neglecting the finiteness of the Moon/Sun and Moon/Earth distances, a crescent is made of a half circle and a half ellipse, say of vertical long semi-axis $1$ and short semi-axis $s$.

The horizontal position of the center of mass is given by the ratio

$$\frac{\int_{-1}^1(1-s)\sqrt{1-y^2}\frac12(1+s)\sqrt{1-y^2}dy}{\int_{-1}^1(1-s)\sqrt{1-y^2}dy}=\frac{4(1+s)}{3\pi}.$$

It varies between $0$ (full Moon) to $\dfrac8{3\pi}\approx85\%$ of the radius (vanishing).

Coincidence occurs for $s=\dfrac4{3\pi-4}\approx74\%$ of the radius.