Moment of Inertia of Three Point Masses – Rankers Physics
Topic: Rotational Motion
Subtopic: Moment of Inertia

Moment of Inertia of Three Point Masses

Three point masses \(m\), \(2m\) and \(3m\) are located at the vertices of an equilateral triangle of side length \(L\). The moment of inertia of the system about an axis passing through mid-point of the side (connecting \(m\) and \(2m\)) and perpendicular to the plane of the triangle, is
\(mL^2\)
\(2mL^2\)
\(3mL^2\)
\(4mL^2\)

Solution:

The distance of masses \(m\) and \(2m\) from the midpoint of their side is \(L/2\). The third mass \(3m\) lies at a distance of \(h = \frac{\sqrt{3}}{2}L\) (the height of the triangle). The total moment of inertia is \(I = m\left(\frac{L}{2}\right)^2 + 2m\left(\frac{L}{2}\right)^2 + 3m\left(\frac{\sqrt{3}}{2}L\right)^2 = \frac{mL^2}{4} + \frac{2mL^2}{4} + \frac{9mL^2}{4} = 3mL^2\).

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