An example that a 2D shape with its centre of mass on its boundary The object has constant density. Could any body suggest one for me?
 A: 
An  '' odd rose'' , i.e. the plane surface limited by the boundary of equation $r=a\cos (n \theta)$ with $n$ odd, has the center if mass at $(0,0)$ that is a point of the boundary.
And from this you can imagine many other similar solutions.
A: It's hard to improve on @YvesDaoust 's answer, but this athletic feat suggests another: in a well executed high jump, the jumper's center of gravity stays well under the bar, so is outside his (or her) body. At some point in the jump it's on his boundary.
Pictures here, including one that is a direct answer to the question:
http://nrich.maths.org/2742
A: 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:
$\hspace1in$ 
A: The shape must be concave, in the sense that there exists a pair of points inside the shape, such that the segment joining them is not entirely inside the shape.
One of the shapes can be an L shaped thing.
A: 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.
A: Here is one with the CM at $M$

A: Take a $2 \times 2$ square.  Take another $4 \times 4$ square around it. The area in between them is a $2-d$ shape. Center of mass is exactly at the center of the square. Now cut it in half. You will get the center of mass on boundary.
A: Consider an L-shaped region obtained as follows:


*

*take the square with corners at $(0,0), (0,1), (1,0), (1,1)$;

*remove the square with corners at $(0,0), (0,t), (t,0), (t,t)$ from this square.
Since this is diagonally symmetric, it has its center of mass at $(f(t),f(t))$ for some function $f$ that we need to work out.
Now, we can view the L-shape as made of two rectangles:


*

*rectangle $R_1$, with corners at $(0,t), (t,t), (0,1), (t,1)$.  This has area $t(1-t)$ and the $x$-coordinate of its center of mass is $t/2$.

*rectangle $R_2$, with corners at $(t,0), (1,0), (t,1), (1,1)$.  This has area $(1-t)$ and the $x$-coordinate of its center of mass is $(1+t)/2$.
So the $x$-coordinate of the center of mass of the whole L-shaped region $R_1 \cup R_2$ is the average of these two $x$-coordinates weighted according to the area.  That is,
$$ f(t) = {t(1-t) \times (t/2) + (1-t) \times (1+t)/2 \over t(1-t) + (1-t)} $$
Canceling out the $1-t$ gives
$$ f(t) = {t(t/2) + (1+t)/2 \over 1+t} = {t^2 + t + 1 \over 2(t+1)} $$
If $f(t) = t$ exactly, then $(t, t)$ will be on the boundary of the L-shaped region.  Solving for $f(t) = t$ we get
$$ t = {t^2 + t + 1 \over 2(t+1)} $$
or, after multiplying through, $2t^2 + 2t = t^2 + t + 1$.  This simplifies to $t^2 - t - 1$ and so, by the quadratic formula, setting $t = (\sqrt{5}-1)/2$ does the trick.
(This is inspired by Yves Daoust's answer about the Moon, but with shapes made up of rectangles to make the calculations easier.)
A: Visualize a triangle -- $(0\mid 0),(90\mid 0),(0\mid 90)$
and a square -- $(0\mid 0),(A\mid 0),(A\mid A),(0\mid A)$.
The square is cut out of the triangle.    
When $A=0$ the CM of the entire figure is $(30\mid 30)$ which is inside the entire figure.
When $A=45$ the CM of the entire figure is $(60\mid 60)$ which is outside the entire figure.
At some point between $A=0$ and $A=45$ the boundary of the entire figure has been crossed. I would complete it myself, except that the margins of my notebook are too small to contain it. Besides, my dogs are dancing around with their leashes in their mouths and holding their crotches ...
