# How to find the center of mass of several objects in a 2d plane?

We have the following scenario in a 2d plane: a big rectangle with a lot of smaller similar uniform rectangles in it, all of them weight differently.

Where is the center of mass of this object found by a formula?

To generalise it (In a 2d plane):

We have a shape, composed of different shapes, which might be similar or not. We know all the small shapes' centers of mass.

Where is the center of mass of the larger shape?

For example one, here is an image:

• How were you taught to calculate the center of mass? Did you use vectors? Try to apply what you have learned about that calculation to this case. – CuriousOne Dec 20 '14 at 12:03
• No, we mainly learned in school how to do it with two weights. It is a personal question. – MikhailTal Dec 20 '14 at 12:13
• I see. Sofia down there gave you the formula. Are you comfortable with the sum notation and is it obvious that her $\vec{R}$ and $\vec{r_i}$ quantities denote vectors and not just scalar numbers? – CuriousOne Dec 20 '14 at 12:21
• Yes, because I am a math olympiadist, I know the notations. I was just thinking of this today by coincidence. – MikhailTal Dec 20 '14 at 12:22
• @sofia \vec I think – MikhailTal Dec 20 '14 at 12:32

Let bottomleft point be $(0,0)$ and assuming each small segment is a uniform square of side 1 unit. The $y$ coordinate of the center of mass will be

$y_{cm}=\dfrac{28\cdot4.5+35\cdot3.5+27\cdot2.5+27\cdot1.5+26\cdot.5}{143}\approx2.58$

Similarly the x coordinate will be,

$x_{cm}=\dfrac{23\cdot4.5+31\cdot3.5+21\cdot2.5+28\cdot1.5+40\cdot.5}{143}\approx2.28$

So, if the bottom left point of the body is considered as the origin the center of mass will be located at $(2.58,2.28)$.

Note:- This method is also applicable if small segments are uniform rectangles, provided their lengths and breadths.

• is the diagonal center a reasonable center of mass for a uniform rectangle – Gowtham Dec 20 '14 at 13:37
• why not?${}{}{}{}{}$ – Dheeraj Kumar Dec 20 '14 at 13:38
• if the body is uniform, then geometrical center can also be considered as center of mass. – Dheeraj Kumar Dec 20 '14 at 13:39
• okay , i am not saying it isnt , just curious if there are exceptions ,thats all :P – Gowtham Dec 20 '14 at 13:41
• I gave it as a picture example rather than to be solved. – MikhailTal Dec 21 '14 at 18:45

The center of mass (or gravity) is given by the formula (see Wikipedia, http://en.wikipedia.org/wiki/Center_of_mass)

$$(1) \ \Sigma_{i=0}^N \ m_i (\vec r_i - \vec R) = 0$$

If someone has a problem with the uniformity of the shapes, then we can replace the sum by an integral, and inside each shape consider the mass as a function of $\vec r$, i.e. $m(\vec r)$

$$(1') \ \vec i\int dx \frac {∂m}{∂x} \ (x - X) + \ \vec j\int dy \frac {∂m}{∂y} \ (y - Y) + \ \vec k\int dz \frac {∂m}{∂z} \ (z - Z) = 0$$

But it is more convenient to work with sums, and assume that each point $m_i$ is sufficiently small to have inside itself a homogeneous structure.

From the formula (1) we can elaborate and find $\vec R$ as a function of the centers of mass of the different shapes that you are given.

The procedure below is easy to generalize to 3D.

Let's separate the index $i$ into two indexes, $j$ and $k_j$, where $j$ runs over the shapes, and $k_j$ over each particular shape, if we divide each particular shape into very small points carrying mass, as said above.

For instance, for the shape no. $j$, we have

$$\Sigma_{k_j = 0}^{N_j} \ m_{k_j}[ (\vec r_{k_j} - \vec R_j) - (\vec R_j - \vec R)]$$

$$= \Sigma_{k_j = 0}^{N_j} \ m_{k_j} (\vec r_{k_j} - \vec R_j) - \Sigma_{k_j = 0}^{N_j}(\vec R_j - \vec R)$$

$$= \Sigma_{k_j = 0}^{N_j} \ m_{k_j} (\vec r_{k_j} - \vec R_j) - \Sigma_{k_j = 0}^{N_j} \ m_{k_j}(\vec R_j - \vec R)$$

Now, let me denote by $M_j$ the mass of the shape no. $j$.

Therefore

$$\Sigma_{k_j = 0}^{N_j} \ m_{k_j}[(\vec r_{k_j} - \vec R_j) - (\vec R_j - \vec R)]$$

$$= \Sigma_{k_j = 0}^{N_j} \ m_{k_j} (\vec r_{k_j} - \vec R_j) - M_j(\vec R_j - \vec R)$$

Now, I require for the object no. j that the position $\vec R_j$ be its center-of-mass, i.e. in line with the general formula (1),

$$\Sigma_{k_j = 0}^{N_j} \ m_{k_j} (\vec r_{k_j} - \vec R_j) = 0,$$

s.t. we have

$$(2) \ \vec R_j = \frac {\Sigma_{k_j = 0}^{N_j} \ m_{k_j} \vec r_{k_j}}{M_j}$$

Now we can return safely to the situation of all our block of shapes. I will denote by $J$ the number of shapes

$$(3) \ \Sigma_{i=0}^N \ m_i (\vec r_i - \vec R) = \Sigma_{j=0}^J \ M_j(\vec R_j - \vec R) = 0.$$

That's it! We have

$$(4) \ \vec R = \frac {\Sigma_{j=0}^J \ M_j \vec R_j} {\Sigma_{j=0}^J \ M_j} .$$

where each $R_j$ is given, as you say.