Animated Solution for Physics - System of Particles: An $L$-shaped object made of thin rods of uniform mass density is suspended with a string as shown in figure. If $AB=BC$ and the angle is made by $AB$ with downward vertical is $\theta$, then

A Center of Mass Masterclass

Have you ever tried balancing a ruler on your finger? It balances perfectly when your finger is exactly under its center of mass. Now, what happens if the object isn't a simple straight ruler, but an L-shaped rod? Let's dive into this fascinating equilibrium problem!

We have an L-shaped object made of two identical uniform rods, AB and BC. It's suspended from point A by a string. The rod AB makes an angle $\theta$ with the downward vertical. Our goal is to find this angle $\theta$ when the system is in perfect equilibrium.

The Golden Rule of Equilibrium: For any suspended object to be in rotational equilibrium, the net torque about the point of suspension must be zero. This happens only when the center of mass (CM) of the entire system lies exactly on the vertical line passing through the suspension point.

Let's set up a coordinate system. Imagine the suspension point A is at the origin $(0,0)$. The downward vertical is our y-axis, and the horizontal is our x-axis. Since the rods are uniform, their individual centers of mass lie exactly at their midpoints. Let's call them $G_1$ for rod AB and $G_2$ for rod BC.

Let the length of each rod be $L$. Rod AB makes an angle $\theta$ with the y-axis. Using basic trigonometry, the coordinates of its midpoint $G_1$ are:

Now for rod BC. First, the coordinates of point B are $(-L\sin \theta ,L\cos \theta )$. Since rod BC is perpendicular to AB, it makes an angle of $90^{\circ }-\theta$ with the vertical, which means it makes an angle $\theta$ with the horizontal. Moving from B to the midpoint $G_2$, we add the horizontal and vertical components of this half-length. The coordinates of $G_2$ become:

For the entire system to balance, the x-coordinate of the combined center of mass must be zero. Since both rods have the same mass $m$, the x-coordinate of the system's CM is simply the average of the x-coordinates of $G_1$ and $G_2$.

Setting the average x-coordinate to zero, we get:

Simplifying this, we have:

And there we have it! The angle $\theta$ for perfect equilibrium is ${\tan }^{-1}(1/3)$. The beauty of physics lies in how complex shapes can be broken down into simple, manageable parts.