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NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice

A system of two identical, uniform discs with identical circular cavities, is shown in the figure. Different relevant coordinates are given in the figure. The coordinates of the centre of mass of the system are

Options

  1. A3 R 2 , 5 R 4
  2. B1 9 R 6 , 5 R 2
  3. CR 2 , R 4
  4. D2 0 R 6 , 5 R 2

Correct answer

B. 1 9 R 6 , 5 R 2

Step-by-step solution

Mass of the removed portion = M π R 2 × π R 2 2 = M 4 x cm = M · 0 - M 4 × R + M × 7 R - M 4 × 8 R M - M 4 + M - M 4 = 1 9 R 6 y cm = M · 0 - M 4 0 + M 5 R - M 4 5 R M - M 4 + M - M 4 = 5 R 2

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