Question

A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid thermocol spheres of mass 25 g and radius 10 cm each are at the two ends of the rod. Calculate moment of inertia of the dumbbell when rotated about an axis passing through its centre and perpendicular to the length.

Answer 1

Data : M_sph = 50 g, R_sph = 10 cm, M_rod = 60 g, L_rod = 20 cm

The MI of a solid sphere about its diameter is I_sph,CM = 2/5 M_sphR_sph

The distance of the rotation axis (transverse symmetry axis of the dumbbell) from the centre of sphere, h = 30 cm.

The MI of a solid sphere about the rotation axis,  I_{sph =}  I_sph, CM ₊ M_sphh²

For the rod, the rotation axis is its transverse symmetry axis through CM.

The MI of a rod about this axis,

I_rod = 1/12 M_rodL²_rod

Since there are two solid spheres, the MI of the dumbbell about the rotation axis is