JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Rotational Mechanics

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Three point masses of 2 kg, 4 kg and 6 kg are placed at the corners of an equilateral triangle of side 2 m. The moment of inertia of the system about an axis passing through the centroid and perpendicular to the plane of the triangle is ______

kgm^{2}

Answer: 4

Solution

r = \frac{a}{2 3} = \frac{1}{3} m

I = m_{1} r^{2} + m_{2} r^{2} + m_{3} r^{2}

l = r^{2} (m_{1} + m_{2} + m_{3})

I = (\frac{1}{3})^{2} (2 + 4 + 6)

I = 4 kgm^{2}

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Integer answer

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