Question

Particles of mass $1g,2g,3g.....100g$ are kept at the marks $1cm,2cm,3cm,.....100cm$ respectively on a metre scale. The MOI of the system of particles about perpendicular bisector of the metre scale is given as $\frac{40+x}{100}kg{m}^2$. Find $x$.

Answer 1

$MOI=MO{I}_1+MO{I}_2+....+MO{I}_{100}=1\times {49}^2+2\times {48}^2+.....+49\times 1^2+51\times 1^2+52\times 2^2+100\times {50}^2=100\times {50}^2+(1+99){49}^2+....+(49+51)\times 1^2=100({50}^2+{49}^2+...+1^2)=100\times \frac{50\times 51\times 101}{6}$

Sum of square $=\frac{n(n+1)(2n+1)}{6}$

$=4292500g{cm}^2=0.42925kg{m}^2=\frac{40+x}{100}kg{m}^{2}x=2.925$