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 metre scale.

Answer 1

As we know,

Masses of $1gm,2gm....100gm$ are kept at the marks $1cm,2cm,....1000cm$ on the $x$ axis respectively. A perpendicular axis is passed at the ${50}^{th}$ particle.

Therefore on the L.H.S. side of the axis, there will be $49$ particles and on the R.H.S. side, there are $50$ particles.

Consider the two particles at the position $49cm$ and $51cm$.

Moment inertia due to these two-particle will be $=49\times 1^2+51+1^2=100gm-c{m}^2$

Similarly if we consider ${48}^{th}$ and ${52}^{nd}$ tern will get

$100\times 2^{2}gm-c{m}^2$

Therefore we will get $49$ such set and one lone particle at $100cm$.

Therefore total moment of inertia $=100\{1^2+2^2+3^2+...+{49}^2\}+100(50{)}^2$.

$=100\times (50\times 51\times 101)/6=4292500gm-c{m}^2$

$=0.429kg-m^2=0.43kg-m^2$.