A rod whose ends are A & B of length 25 cm is hanged in vertical plane. When hanged from point A and point B the time periods calculated are 3 s & 4 s respectively. Given the moment of inertia of rod about axis perpendicular to the rod is in ratio 9 : 4 at points A and B. Find the distance of the centre of mass from point A.
Physics · Simple Harmonic Motion
JEE Mains Advanced

Question

A rod whose ends are A & B of length 25 cm is hanged in vertical plane. When hanged from point A and point B the time periods calculated are 3 s & 4 s respectively. Given the moment of inertia of rod about axis perpendicular to the rod is in ratio 9 : 4 at points A and B. Find the distance of the centre of mass from point A.

Physics · Simple Harmonic Motion

G1 Physics · Accepted Answer

For a physical pendulum T²∝I/h. Thus 3²/4²=(Iₐ/hₐ)/(Iᵦ/hᵦ). Given Iₐ/Iᵦ=9/4, this gives hₐ/hᵦ=4. Since hₐ+hᵦ=25 cm, hₐ=20 cm and hᵦ=5 cm. Required distance from A: 20 cm.