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AP EAMCET202125 Aug 2021Morning ShiftPhysicsRotational MotionActual

The moment of inertia of a thin rod of mass M and length L about an axis passing through a point at a distance L 4 from its centre and perpendicular to its length is

Options

  1. AM L^3 48
  2. BM L^2 48
  3. CM L^2 12
  4. D7 M L^2 48

Correct answer

D. 7 M L^2 48

Step-by-step solution

Let, M be the mass of the thin rod and L be the length of thin rod. Moment of inertia of thin rod about its centre of mass, I_ COM = 1 12 M L^2 By parallel axis theorem, Moment of inertia about an axis = Moment of inertia about centre of mass + Mass (distance from axis )^2 aligned I & =I_ coM +M(d)^2 & = 1 12 M L^2+M ( L 4 )^2 ( Given, d= L 4 ) & = ( 4+3 48 ) M L^2= 7 48 M L^2 aligned

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