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

13 years ago

we know that moment of inertia of the particle of mass m and radius r  is mr^2 add up monent of inertia to all the particles u get a sequence,solve sequence to get ur answer like m(1+4+9+..............................

mn(n+1)/6 as u know sum of square of natural no is n(n+1)/6 Badiuddin askIITians.ismu Expert
148 Points
13 years ago

Dear niharika

perpendicular bisector will be at the mark 50 cm on the scale. 49 masses is pleced at one side of the scale at (1,2,3,4 ....49)

and 50 masses is placed at the other side of the scale at ( 51,52,53,54,......100) ,and one mass is placed at the 50th mark.

so moment of inertia I = 2* [m*12 + m*22 + m* 32 + .....m*492  ] + m* 502

=2m[49(49+1)(98+1)/6  + m *2500

=83350m    g cm2

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