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 meter scale, Find the moment of Inertia of the system of particles about a perpendicular bisector of a meter scale?

Answer 1

Given in the question :- 

Perpendicular axis is passing through 50th particle.
So the L.H.S will be 49th particle & on the other side there would be 50th particles.

Now consider 49 th particle and 50th particle
 
 Moment of Inertia of these two particle,
I = (49 × 1² + 51 × 1²)
= 100 gm-cm²

Similarly , consider 48 and 52 th particle. we get
= (100 × 2²) gm-cm²


From above condition we understood that we get 49 sets and one lone pair at 100 cm.

$100 * (1^2 +2^2 + 3^2 + ......+ 49^2 ) + 100(50^2)$ 
$100* \frac{(50*51*101)}{6}$ 
= 4292500 gm.cm²
= 0.43 kg.m².


Hope it Helps :-)