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MHT CET 2020 (19 Oct, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2020 (19 Oct, Shift 2)

Q141MHT CET 2020 · 19 OCT · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

Three point masses, each of mass mm are placed at the corners of an equilateral triangle of side ℓ\ell. The moment of inertia of the system about an axis passing through one of the vertices and parallel to the side joining other two vertices, will be

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Answer: (c)

  1. The mass at the vertex is on the axis and contributes zero.
  2. The other two masses are at height 32ℓ\frac{\sqrt{3}}{2}\ell from the axis.
  3. I=2m⋅3ℓ24=32mℓ2I = 2m \cdot \frac{3\ell^2}{4} = \frac{3}{2}m\ell^2
  4. So the answer is (c) 32mℓ2\frac{3}{2}m\ell^2.
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