Question #c4ad9

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Jane
Feb 4, 2018

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The moment of inertia of a rod of mass m and length (l)about an axis passing through its centre and making an angle of theta w.r.t it is given as (ml^2)/12 sin^2 theta

here,as the axis passes through the bisector of the angle between the rods,so theta =45

So,w.r.t the mentioned axis, the rods have a total moment of inertia of 2*(ml^2)/12* sin ^2 45=(ml^2)/12 Kg.m^2

Now,from perpendicular axis theorem, the moment of inertia of a ring of radius l/2 and mass m w.r.t an axis passing through its centre and parallel to its plane is (m(l/2)^2)/2=(ml^2)/8 Kg.m^2

So,w.r.t the given axis,the moment of inertia of the whole system is (ml^2)/12 + (ml^2)/8 =(5(ml^2))/24Kgm^2

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