Four particles each of mass m are kept at the four corners of a square of edge a. Find the moment of inertia of the system about a line perpendicular to the plane of the square and passing through the centre of the square.

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Solution

The perpendicular distance of every particle from the given line is a / $\sqrt{2}$
The moment of inertia of one particle is, therefore, m $(a / \sqrt{2})^{2} = \frac{1}{2} m a^{2}$ . The moment of inertia of the system is, therefore,
$4 \times \frac{1}{2} m a^{2} = 2 m a^{2}$

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Four particles each of mass m are kept at the four corners of a square of edge a. Find the moment of inertia of the system about a line perpendicular to the plane of the square and passing through the centre of the square.

The perpendicular distance of every particle from the given line is a /√2The moment of inertia of one particle is, therefore, m (a/√2)2=12ma2. The moment of inertia of the system is, therefore, 4×12ma2=2ma2

The perpendicular distance of every particle from the given line is a / $\sqrt{2}$
The moment of inertia of one particle is, therefore, m $(a / \sqrt{2})^{2} = \frac{1}{2} m a^{2}$ . The moment of inertia of the system is, therefore,
$4 \times \frac{1}{2} m a^{2} = 2 m a^{2}$