Four point masses m, 2m, 3m and 4m are kept at the corners A, B, C and D respectively of a square ABCD of side ' b '. The moment of inertia of the system about an axis perpendicular to the p

Question

Four point masses m, 2m, 3m and 4m are kept at the corners A, B, C and D respectively of a square ABCD of side ' b '. The moment of inertia of the system about an axis perpendicular to the plane of the square and passing through the point D is

TestHub · Accepted Answer

The moment of inertia of a system of point masses is given by $I = \sum m_{i} r_{i}^{2}$ , where $r_{i}$ is the perpendicular distance of the $i$ -th mass from the axis of rotation.

The axis of rotation is perpendicular to the plane of the square and passes through point D. The distances of the masses from the axis are:

For mass at A ( $m_{A} = m$ ): $r_{A} = b$

For mass at B ( $m_{B} = 2 m$ ): $r_{B} = b 2$ (diagonal of the square)

For mass at C ( $m_{C} = 3 m$ ): $r_{C} = b$

For mass at D ( $m_{D} = 4 m$ ): $r_{D} = 0$

Substituting these values into the formula:

$I = m (b)^{2} + 2 m (b 2)^{2} + 3 m (b)^{2} + 4 m (0)^{2}$

$I = m b^{2} + 2 m (2 b^{2}) + 3 m b^{2} + 0$

$I = m b^{2} + 4 m b^{2} + 3 m b^{2} = 8 m b^{2}$

Answer: $8 m b^{2}$

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