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 ineria 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/√2. The moment of inertia of one particle is, therefore, m(a/√2)2=12ma2. The moment of ineria of the system is, therefore4×12ma2=2ma2

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