Question

Particles of masses 1g, 2g, 3g â¦.100g are kept at the marks 1cm, 2cm, 3cm â¦., 100 cm respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular besector of the metre scale.

Answer 1

The moment of inertia of the system of particles is 4292500 gm cm²

• Moment if inertia for a discrete system is given by ∑mr².
• So for the above system moment of inertia about perpendicular bisector of the meter scale is given by ∑Mi*(Ri-50)², where Mi and Ri varies from 1 to 100.
• As Mi and Ri are numerically same for each particle, hence replacing them with n(n varies from 1 to 100)

• By expanding we get ∑ n(n²-100n+2500) = ∑(n³-100n²+2500n)
• By using ∑n³ = n²(n+1)²/4 , ∑n² = n(n+1)(2n+1)/6 , ∑n = n(n+1)/2 , we can solve the above equation.

Answer 2

Thus the moment of inertia of the system is I = 83350 mg.cm^2

Explanation:

Perpendicular bisection will be at the mark 50 cm on the scale.

1,2,3,4,….49 masses are placed at one side.

50 masses (51,52…..100)

Are placed at other side and 50  the  mass is at middle.

So moment of inertia

I=2(m×1^2  + m×2^ 2  + m × 3^ 2  + ... m × 49^ 2  )   + m × 50^ 2

= 2 m [49 × ( 49 + 1 ) ( 98 + 1 ) / 6 ] + m × 2500

I = 83350 mg.cm^2

Thus the moment of inertia of the system is I = 83350 mg.cm^2