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 = $\frac{2}{5}$ 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 =$\frac{1}{12}$ M_rL_r²

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

I= 2I_s + I_r

$=2M_s(\frac{2}{5}{R}_s^2+h^2)+\frac{1}{12}{M}_r{L}_r^2$ 

$=2(50)(\frac{2}{5}(10{)}^2+(30{)}^2)+\frac{1}{12}(60)(20{)}^2$ 

$=100(40+900)+5(400)=94000+2000$ 

= 96000 g cm²