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

It is given that the perpendicular bisector of the metre scale is passed through the 50^thparticle.

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

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

Moment of inertia due to these two particles = 49 × (1)² + 51 × (1)²

I₁ = 100 × 1 = 100 gm-cm²

Similarly, if we consider particles positioned at 48 cm and 52 cm, we get

I₂  = 100 × (2)² gm-cm²

Thus, we will get 49 such sets and one particle at 100 cm. Therefore, total moment of inertia,

\[I = \left( I_1 + I_2 + I_3 . . . . . + I_{49} \right) + I'\]

Here, I' is the moment of inertia of particle at 100 cm.

\[So, I = 100 \left( 1^2 + 2^2 + 3^2 + . . . + {49}^2 \right) + 100 \left( 50 \right)^2 \]

\[ = 100 \left( 1^2 + 2^2 + 3^2 + . . . + {50}^2 \right)\]

\[ = 100 \times \frac{\left( 50 \times 51 \times 101 \right)}{6}\]

\[ = 100 \times 25 \times 17 \times 101 = 4292599 gm - {cm}^2 \]

Or, I = 0.429 kg-m² ≃ 0.43 kg-m²