Particles of mass $1 g, 2 g, 3 g . . . . .100 g$ are kept at the marks $1 c m, 2 c m, 3 c m, . . . . .100 c m$ respectively on a metre scale. The MOI of the system of particles about perpendicular bisector of the metre scale is given as $\frac{40 + x}{100} k g m^{2}$ . Find $x$ .

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Solution

$M O I = M O I_{1} + M O I_{2} + . . . . + M O I_{100} = 1 \times 49^{2} + 2 \times 48^{2} + . . . . . + 49 \times 1^{2} + 51 \times 1^{2} + 52 \times 2^{2} + 100 \times 50^{2} = 100 \times 50^{2} + (1 + 99) 49^{2} + . . . . + (49 + 51) \times 1^{2} = 100 (50^{2} + 49^{2} + . . . + 1^{2}) = 100 \times \frac{50 \times 51 \times 101}{6}$
Sum of square $= \frac{n (n + 1) (2 n + 1)}{6}$

$= 4292500 g {c m}^{2} = 0.42925 k g m^{2} = \frac{40 + x}{100} k g m^{2} x = 2.925$

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