Question

A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid spheres of mass 50 g and radius 10 cm each are at the two ends of the rod. Calculate moment of inertia of the dumb-bell when rotated about an axis passing through its centre and perpendicular to the length

Answer 1

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_sR_s² 

 

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 h²

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_rL_r²

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 cm²