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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
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Given :

Mass of sphere Ms=50g,

Radius of sphere Rs = 10 cm,

Mass of rod Mr = 60g,

Length of rod Lr = 20 cm

Solution :

The MI of a solid sphere about its diameter is Is-CM = $$2\over 5$$ MsRs

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, Is  = Is-CM + Mh2

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

Ir =$$1\over12$$ MrLr2

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

I= 2Is + Ir

$$= 2M_s ( \frac{2}{5}R_s^2 + h^2) + \frac{1}{12}M_rL_r^2$$

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

$$= 100 (40 + 900) + 5(400) = 94000 + 2000$$

= 96000 g cm2

by Expert (11.5k points)