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Three point masses, each of mass m, are placed at the corner of an equilateral triangle of side.Then the moment of inertia of this system about an axis along one side of the triangle is?
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Three point masses, each of mass m, are placed at the corner of an equ...
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Calculation of Moment of Inertia of Point Masses
- Given: Three point masses (m) at the corners of an equilateral triangle of side (s).
- Moment of inertia (I) can be calculated using the formula: I = Σmiri^2, where mi is the mass of the point mass and ri is the distance of the point mass from the axis of rotation.
- For an equilateral triangle, the distance of each point mass from the axis along one side is s/2.
- Let's consider the point masses A, B, and C placed at the corners of the equilateral triangle.
- The moment of inertia of mass A about the axis passing through side BC is: IA = m(s/2)^2 = ms^2/4.
- Similarly, the moment of inertia of masses B and C about the same axis will also be ms^2/4.
- The total moment of inertia of the system is the sum of the individual moments of inertia: I_total = IA + IB + IC = 3ms^2/4 = 3/4 * m * s^2.

Conclusion
- The moment of inertia of the system of three point masses about an axis along one side of the equilateral triangle is 3/4 * m * s^2.
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Three point masses, each of mass m, are placed at the corner of an equ...
3/4 ma^2
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Three point masses, each of mass m, are placed at the corner of an equilateral triangle of side.Then the moment of inertia of this system about an axis along one side of the triangle is?
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