Given :
Mass of sphere M_{s}=50g,
Radius of sphere R_{s} = 10 cm,
Mass of rod M_{r} = 60g,
Length of rod L_{r} = 20 cm
Solution :
The MI of a solid sphere about its diameter is I_{s-CM} = \(2\over 5\) M_{s}R_{s}^{2 }
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_{s} = I_{s-CM} + M_{s }h^{2}
For the rod, the rotation axis is its transverse symmetry axis through CM.
The MI of a rod about this axis,
Ir =\(1\over12\) M_{r}L_{r}^{2}
Since there are two solid spheres, the MI of the dumbbell about the rotation axis is
I= 2I_{s} + I_{r}
\(= 2M_s ( \frac{2}{5}R_s^2 + h^2) + \frac{1}{12}M_rL_r^2\)
\(= 2(50) (\frac{ 2}{5}(10)^2 + (30)^2 ) + \frac{1}{12}(60)(20)^2\)
\(= 100 (40 + 900) + 5(400) = 94000 + 2000\)
= 96000 g cm^{2}