Question:

Three balls of masses 2kg2 \, \text{kg}, 4kg4 \, \text{kg}, and 6kg6 \, \text{kg} respectively are arranged at the centre of the edges of an equilateral triangle of side 2m2 \, \text{m}.The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of the triangle, will be ____ kgm2\text{kg} \, \text{m}^2

Updated On: Mar 22, 2025
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Correct Answer: 4

Solution and Explanation

For an equilateral triangle, the distance r r from the center of each edge to the centroid C C is:
r=13r = \frac{1}{\sqrt{3}}
The moment of inertia I I about point C C and perpendicular to the plane is given by:
I=r2[2+4+6]I = r^2 \left[ 2 + 4 + 6 \right]
Substitute r=13 r = \frac{1}{\sqrt{3}} :
I=(13)2×12I = \left( \frac{1}{\sqrt{3}} \right)^2 \times 12
I=13×12=4kgm2I = \frac{1}{3} \times 12 = 4 \, \text{kg} \cdot \text{m}^2

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