Question

Particles of masses $1g,2g,3g,.....,100g$ are kept at the marks $1cm,2cm,3cm,.......,100cm$ respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale.

Answer 1

Perpendicular bisection will be at the mark $50$cm on the scale.

$1,2,3,4,\dots .49$ masses are placed at one side.

$50$ masses $(51,52\dots ..100)$

Are placed at other side and ${50}^{th}$ mass is at middle.

So moment of inertia,

$I=2(m\times 1^2+m\times 2^2+m\times 3^2+...m\times {49}^2)+m\times {50}^2=2m[49\times (49+1)(98+1)/6]+m\times 2500=83350mg.{cm}^2$