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. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.

Answer 1

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

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

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

Moment of inertia due to these two particles will be:

= 49 $\times$ (1)${}^2$+51 $\times$ (1)${}^2$

= 100 $\times$ 1 = 100 gm - cm${}^2$

Similarly, if we consider 48${}^{th}$ and 52${}^{nd}$ term we will get 100 $\times$ $(2{)}^2$ gm -cm${}^2$.

Therefore, we will get 49 such set and one lone particle at 100 cm. Therefore total moment of inertia =

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

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

=$100\times \frac{(50\times 51\times 101)}{6}$

= $100\times 25\times 17\times 101$

= $100\times 101\times 425$

= $4292500{\text{gm-cm}}^2$

= $429{\text{kg-m}}^2\simeq 0.43\text{kg}-{\text{m}}^2$