Animated Solution for Physics - Rotational Motion: A massless equilateral triangle $EFG$ of side $a$ (as shown in figure) has three particles of mass $m$ situated at its vertices. The moment of inertia of the system about the line $EX$ perpendicular to $EG$ in the plane of $EFG$ is $\frac{N}{20}m{a}^2$, where $N$ is an integer. The value of $N$ is ............ .

Imagine a rigid, massless equilateral triangle $EFG$ with side length $a$. At each of its three corners, a point mass $m$ is firmly attached. We are asked to find the moment of inertia of this entire system about a very specific axis: the line $EX$.

This line $EX$ passes through the vertex $E$ and is perpendicular to the base $EG$. To visualize this, you can imagine placing the triangle on an $xy$-coordinate system. Let the vertex $E$ be at the origin $(0,0)$, and the base $EG$ lie along the positive $x$-axis. The line $EX$, being perpendicular to $EG$ at $E$, perfectly aligns with the $y$-axis!

The moment of inertia $I$ for a system of discrete point masses is a measure of how difficult it is to change the system's rotational motion. It depends not just on the masses, but crucially on how far each mass is distributed from the axis of rotation. The fundamental formula is:

Here, $m_i$ is the mass of the $i$-th particle, and $r_i$ is its perpendicular distance from the axis of rotation. Our task is simply to find this perpendicular distance for all three masses relative to the $y$-axis (line $EX$).

Let's break it down particle by particle:

1. The Mass at Vertex E: This mass is sitting right at the origin $(0,0)$. Since it lies exactly on the axis of rotation $EX$, its perpendicular distance $r_E$ is zero.

2. The Mass at Vertex G: Vertex $G$ lies on the $x$-axis at a distance $a$ from the origin. The perpendicular distance from the $y$-axis (line $EX$) to this point is simply the length of the segment $EG$. Thus, $r_G=a$.

3. The Mass at Vertex F: This is the only tricky part. Vertex $F$ is the top peak of the equilateral triangle. We need its perpendicular distance to the $y$-axis. If we drop a perpendicular from $F$ down to the base $EG$, it bisects the base because the triangle is equilateral. This means the $x$-coordinate of $F$ is exactly half of the side length, $a/2$. This $x$-coordinate is precisely the perpendicular distance from the $y$-axis! So, $r_F=a/2$.

Now, we simply sum up the individual moments of inertia to find the total moment of inertia of the system:

The problem states that the moment of inertia is given in the form $\frac{N}{20}m{a}^2$. To find the integer $N$, we need to express our result with a denominator of 20. We can do this by multiplying the numerator and denominator by 5:

Comparing this with the given expression $\frac{N}{20}m{a}^2$, it is crystal clear that:

And there we have it! By carefully identifying the axis of rotation and using the geometric properties of an equilateral triangle, we've successfully unraveled the rotational inertia of the system.