Question

Question #c4ad9

Answer 1

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`