Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is x/12ML² kg m². The value of x is ____ .

Solution

Physics sir · 4 steps

Take each rod separately about P, and remember the bottom rod's centre sits a full L away from P.

1. Vertical rod. P is at the top end of the vertical rod, and the axis is perpendicular to the plane. For a rod about one end, I₁ = 1/3ML² = 4/12ML².
2. Horizontal rod. About its own centre, the horizontal rod has 1/12ML². Its centre is where the vertical rod meets it, a distance L below P. The parallel axis theorem adds ML².

   I₂ = 1/12ML² + ML² = 13/12ML²

3. Add them. Moments of inertia about the same axis add: I = 4/12ML² + 13/12ML² = 17/12ML².
4. So. Comparing with x/12ML² gives x = 17.

The common slip: 5. Leaving out the parallel axis term ML² for the horizontal rod gives 4/12 + 1/12, so x = 5 instead of 17.

The idea to keep. For a combined body, find each part's moment of inertia about the same axis and add. Use the parallel axis theorem with the distance from the axis to that part's own centre of mass.

Answer from the NTA final answer key. Solution by PrepWiser.