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

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closed as off-topic by Harish Chandra Rajpoot, Jon Mark Perry, kamil09875, wythagoras, Chris Godsil Feb 23 at 20:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Harish Chandra Rajpoot, kamil09875, wythagoras, Chris Godsil
If this question can be reworded to fit the rules in the help center, please edit the question.

Concentric semi-circle laminas radii a,b<a can be determined by calculus by varying b. – Narasimham Feb 23 at 15:28
Please make the body of your Question reasonably self-contained, not relying on the title alone to express the problem to be solved. – hardmath Feb 23 at 16:03
up vote 21 down vote accepted

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 crescent in golden ratio

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This is very clever! It's basically the oldest definition of the golden ratio – You're In My Eye Feb 23 at 16:33
Despite being a non-mathematician, the math around Φ is continually fascinating to me. – Stonetip Feb 23 at 16:48

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.

enter image description here


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


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.

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@tim: I mean the center of mass of the visible part. – Yves Daoust Feb 23 at 15:13
@Tim No, this is a perfectly serious answer. The point at which the moving boundary of the moon cuts the horizontal axis varies continuously with time, as does the horizontal coordinate of the centre of mass. Then, by the intermediate value theorem, there must be some point at which these values coincide. – Donkey_2009 Feb 23 at 15:15
New moon is wrong, it has no visible surface, i.e. no mass in this representation. But still, great answer – You're In My Eye Feb 23 at 15:16
@tim: center of mass or gravity center, the distinction doesn't matter. The property holds for both, as a consequence of continuity. – Yves Daoust Feb 23 at 15:30
+1 Suggested clarification: the answer to the OPs question is the plane image of the appropriate crescent moon, not the moon. (That's implicitly clear.) – Ethan Bolker Feb 23 at 15:34

Here is one with the CM at $M$ enter image description here

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Can you tell how you arrived at this? – N.S.JOHN Feb 23 at 15:04
@N.S.JOHN: I wanted the area distribution above and below the $x$ axis to be the same. Essentially I took a square, cut the bottom half apart, and slid the pieces sideways. The position of the CM doesn't change, but now the boundary goes through it. – Ross Millikan Feb 23 at 15:07

enter image description here

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.

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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:

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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.

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This is not correct, but you are thinking in the right direction. You need some more area on one side. – Ross Millikan Feb 23 at 14:59

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.

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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.)

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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 ...

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Are dogs physically capable of holding their crotches? Do dogs even have crotches? – immibis Feb 24 at 5:07
@immibis \ Well, somehow they make their wants known to me. Good question, though. – Senex Ægypti Parvi Feb 24 at 7:54

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