Question

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

• A. $0.43kg-m^2$
• B. $0.56kg-m^2$
• C. $4.3\times {10}^{6}kg-m^2$
• D. $5.6\times {10}^{6}kg-m^2$

Answer 1

The correct option is A

$0.43kg-m^2$

Masses of 1 g, 2 g ...... 100 gm are kept at the marks 1 cm, 2 cm, .....100 cm on the x axis respectively. A perpendicular axis is passing through the 50th 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 49 cm and 51 cm.

Moment of inertia due to these two particle will be =

$49\times 1^2+51\times 1^2$ = $100gm-c{m}^2$

Similarly if we consider ${48}^{th}$ and ${52}^{nd}$ term we will get $100\times 2^{2}gm-c{m}^2$

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

Therefore total moment of inertia =

$100(1^2+2^2+3^2+...+{49}^2)+100(50{)}^2$.

= $\frac{100\times (50\times 51\times 101)}{6}$ = $4292500gm-c{m}^2(1^2+2^2+.......+n^2$ = $\frac{n(n+1)(2n+1)}{6})$

= $0.43kg-m^2$