Question

Particles of masses $1g,2g,3g...,100g$ are kept at the marks $1cm,2cm.3cm,...,100cm$ respectively on a metre scale. The moment of inertia of the system of particles about a perpendicular bisector of the metre scale is about $x\times {10}^{6}gm-c{m}^2$. Find $x$.

Answer 1

$I=50(0{)}^2+(49+51\left)\right(1{)}^2+(48+52)(2{)}^2+........+(1+99\left)\right(49{)}^2$
$=100\left[\right(1{)}^2+(2{)}^2+.....+(49{)}^2]$ where we have used $1^2+2^2+3^2+...............+n^2=\frac{n(n+1)(2n+1)}{6}$
$4.3\times {10}^{6}gm-c{m}^2=0.402kg-m^2$