Animated Solution for Physics - System of Particles: The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass $4\text{ kg}$. (The coordinates of the same are shown in figure) are

The concept of the center of mass is one of the most elegant and powerful tools in physics. It allows us to take a complex, extended object and treat it as if all its mass were concentrated at a single point. In this problem, we are given a uniform flag-shaped lamina and asked to find its center of mass. At first glance, the shape might look a bit tricky, but the beauty of physics lies in breaking down complex problems into simpler, manageable pieces.

Imagine you are holding this flag-shaped thin flat plate. It has a total mass of $4\text{ kg}$, and its coordinates are given on a 2D plane. The shape is not a standard geometric figure like a simple circle or a square, so we cannot just guess its center of mass. However, we can observe that this complex shape is actually composed of two very simple shapes: rectangles!

We can divide the flag into two parts: 1. Part I (The Pole): A vertical rectangle that starts from the origin $(0,0)$ and goes up to $(1,3)$. 2. Part II (The Flag): A horizontal rectangle attached to the top right of the pole, starting from $(1,2)$ and ending at $(2,3)$.

By breaking the object down like this, we have transformed a difficult problem into a very simple one.

For a system of discrete particles, the coordinates of the center of mass are given by the mass-weighted average of their positions:

But wait, we are dealing with a continuous lamina, not discrete particles! Here is the crucial insight: because the lamina is uniform, its mass is evenly distributed across its area. This means the mass of any part of the lamina is directly proportional to its area ($m\propto A$). Therefore, we can replace the masses in our formula with the areas of the respective parts:

Notice how the total mass of $4\text{ kg}$ given in the problem is actually redundant! We don't even need it to find the coordinates.

Now, let's find the area and the center of mass for each of our two rectangles. The center of mass of a uniform rectangle lies exactly at its geometric center.

For Part I (The Pole): The width is $1\text{ m}$ (from $x=0$ to $x=1$). The height is $3\text{ m}$ (from $y=0$ to $y=3$). The area is $A_1=1\times 3=3{\text{ m}}^2$. The center of mass $(x_1,y_1)$ is at the midpoint of its width and height: $x_1=0.5\text{ m}$ and $y_1=1.5\text{ m}$.

For Part II (The Flag): The width is $1\text{ m}$ (from $x=1$ to $x=2$). The height is $1\text{ m}$ (from $y=2$ to $y=3$). The area is $A_2=1\times 1=1{\text{ m}}^2$. The center of mass $(x_2,y_2)$ is at the midpoint of its width and height: $x_2=1+0.5=1.5\text{ m}$ and $y_2=2+0.5=2.5\text{ m}$.

We have all the pieces of the puzzle. Let's plug them into our master equations!

Calculating the X-coordinate:

Calculating the Y-coordinate:

And there we have it! The center of mass of the entire flag-shaped lamina is located at the coordinates $(0.75\text{ m},1.75\text{ m})$.

This problem perfectly illustrates how breaking a complex system into simpler, symmetric components can make finding the center of mass a breeze. Always look for those hidden symmetries!