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InterviewSolution
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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. |
Answer» Solution :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` |
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