Particles of masses 1g, 28, 3g ..., 100g are kept at the marks 1 cm, 2 cm, 3 cm ... , 100 cm respectively on a meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale.

Explanation

Given Masses of 1g, 28, 3g ....... 100 g are 1, 2, 3 ...... 100 cm on a scale.

i) Sum of masses `Sigma m= underset(i=1)overset(n) "ni"`
Sum of n natural numbers S
`=(n(n+1))/(2) = (100xx101)/(2) = 5051`
`therefore` Total mass M = 5051 gr = 5.051 kg ` to (1)`
ii) Centre of mass of all these masses is given by
`X_(CM) = (m_1x_1 +m_2x_2 + .....)/(Sigma m)`
`(1xx 1+2xx 2 +3xx3 + ........+ 100 xx 100)/(M)`
= `(1^2 + 2^2 + 3^2 + ......+ 100^2)/(M)`
But sum of squares of 1st n natural numbers is `S = (n(n+1)(2n+1))/(6)`
`therefore CM = (n(n+1)(2n+1))/(6xxn(n+1)) "use " M = (n(n+1))/(2)`
`therefore CM = 201/2 = 67 CM `to (2)`
iii) M.O.I = m_1 r_1^2 + m_2r_2^2 + m_3r_3^2 + ......+ m_100 r_100^2`
`therefoer 1 = 1xx 1xx 1+ 2 xx 2^2 + 3xx 3^2 + ....... + 100 xx 100^2`
`= 1+2^3 + 3^3 + ......+100^3 xx 10^(-7) kgm^2`
`(because 1g = 10^(-3) kg & 1.CM =10^(-2) m)`
Sum of cubes of 1st n natural numbers is
`S = (n^2(n+1)^2)/(4)`
`therefore I = (100^2 xx (101)^2)/(4) g.cm^2`
`= 2.550 xx 10^7 g.cm^2 = 2.550 kg.m^2` (From one end of scale)
M.O.I about C.M = `I_G - I - MR^2`
`= 2.550 - 5.05xx 0.67xx 0.67`
` = 2.550 - 2.267 = 0.238 kgm^2`
iv) Perpendicular bisector is at 50 CM.
so shift M.O.I from centre of mass to `x_1` = 50cm point from x = 67 CM
`therefoer` Distance between the axis
`R= 67- 50 = 17cm = 0.17 M`
M.O.I about this axis `I = I_G + MR^2`
` = 0.283 + 5.05xx 0.17 xx 0.17`
`= 0.283 + 0.146 = 0.429 kgm^2`
`therefore` M.O.I about perpendicular bisector of scale = `0.429 kg -m^2`