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NEET2015PhysicsCenter of Mass, Momentum and CollisionActual

Point masses m 1 and m 2 are placed at the opposite ends of rigid rod of length L , and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point L on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω 0 is minimum, is given by:

Options

  1. Ax = m 1 m 2 L
  2. Bx = m 2 m 1 L
  3. Cx = m 2 L m 1 + m 2
  4. Dx = m 1 L m ! + m 2

Correct answer

C. x = m 2 L m 1 + m 2

Step-by-step solution

K . E . = 1 2 I ω 2 I is min. about the centre of mass So, m 1 x = m 2 L - x x = m 2 L m 1 + m 2

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