Particles having masses of 1 g , 2 g , 3 g , ..., 100 g are placed at the marks 1 cm , 2 cm , 3 cm , ..., 100 cm respectively on a metre scale. Determine the moment of inertia of this system of particles about the perpendicular bisector of the metre scale.

Step-by-step solution

The perpendicular bisector of a metre scale passes through its midpoint at 50 cm . The moment of inertia of the system about this axis is given by: I = _ i=1 ¹⁰⁰ m_i r_i^2 where m_i = i g and the distance from the axis is r_i = |i - 50| cm . I = _ i=1 ¹⁰⁰ i (i - 50)^2 Let j = i - 50 . As i varies from 1 to 100 , j varies from -49 to 50 . I = _ j=-49 ⁵⁰ (j + 50) j^2 = _ j=-49 ⁵⁰ j^3 + 50 _ j=-49 ⁵⁰ j^2 Since j^3 is an odd power, the sum of symmetric terms is zero, giving _ j=-49 ⁴⁹ j^3 = 0 . _ j=-49 ⁵⁰ j^3 = 50^3 =