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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.

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Solution

It is given that the perpendicular bisector of the metre scale is passed through the 50th particle.
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=I1+I2+I3.....+I49 +I'
Here, I' is the moment of inertia of particle at 100 cm.
So, I=100 12+22+32+...+492+100 502 =100 12+22+32+...+502 =100×50×51×1016 =100×25×17×101=4292599 gm-cm2
Or, I = 0.429 kg-m2 ≃ 0.43 kg-m2


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