Animated Solution for Physics - Rotational Motion: Four equal masses, $m$ each are placed at the corners of a square of length ($l$) as shown in the figure. The moment of inertia of the system about an axis passing through $A$ and parallel to $DB$ would be

The problem asks us to find the moment of inertia of a system of four identical masses placed at the corners of a square. The axis of rotation is a specific line: it passes through corner $A$ and is parallel to the diagonal $DB$.

Imagine a square $ABCD$ with side length $l$. At each corner, there is a point mass $m$.

The axis of rotation, let's call it $X{X}^{\prime }$, is a straight line passing through $A$. We are given that this axis is parallel to the diagonal $DB$.

To find the moment of inertia of a system of discrete point masses, we use the fundamental formula: where $m_i$ is the mass of the $i$-th particle and $r_i$ is its perpendicular distance from the axis of rotation.

Let's calculate the perpendicular distance for each mass from the axis $X{X}^{\prime }$:

1. Mass at A: Since the axis $X{X}^{\prime }$ passes directly through corner $A$, the perpendicular distance is zero ($r_A=0$).

2. Masses at B and D: The axis $X{X}^{\prime }$ is parallel to the diagonal $DB$. The distance between the parallel lines $X{X}^{\prime }$ and $DB$ is exactly half the length of the other diagonal $AC$. The length of the diagonal of a square of side $l$ is $\sqrt{2}l$. Therefore, the perpendicular distance from $B$ and $D$ to the axis $X{X}^{\prime }$ is half of this diagonal: The moment of inertia for these two masses will be:

3. Mass at C: The corner $C$ lies on the diagonal $AC$. The perpendicular distance from $C$ to the axis $X{X}^{\prime }$ (which passes through $A$ and is perpendicular to $AC$) is the full length of the diagonal $AC$. The moment of inertia for the mass at $C$ is:

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

Adding the terms together:

The Parallel Axis Theorem

Physics often rewards us with multiple paths to the truth. Let's verify our result using the Parallel Axis Theorem.

First, let's find the moment of inertia of the system about the diagonal $BD$. This axis passes through the center of mass of the system. - The masses at $B$ and $D$ lie exactly on this axis, so their distance is zero. - The masses at $A$ and $C$ are at a perpendicular distance of half the diagonal, which is $\frac{l}{\sqrt{2}}$.

The moment of inertia about the center of mass axis ($I_{CM}$) is:

Now, we want to shift our axis from the diagonal $BD$ to the parallel axis $X{X}^{\prime }$ passing through $A$. The perpendicular distance $d$ between these two parallel axes is half the diagonal, $d=\frac{l}{\sqrt{2}}$. The total mass of the system is $M=4m$.

According to the Parallel Axis Theorem:

Both methods yield the exact same beautiful result!