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Asked by Prashant DIGHE | 26 Jan, 2020, 09:36: PM
If point O is centroid of triangle ABC, the distance OA = (2/3) of height of triangle from base BC

Hence OA = (2/3) ×(√3/2) × (2R) = (2/3)(√3R)

where R is radius of sphere

By parallel-axis theorem, Moment of inertia I of each sphere about an axis passing through O is given by,

I= (2/5)MR2 + (4/3)MR2

where M is mass of sphere.

Hence Moment of inertia Io of all three sphers about an axis passing through O is given by,

Io = 3 × [ (2/5)MR2 + (4/3)MR2 ] = (26/5) MR2 ......................(1)

If Axis of rotation is passing through A, then moment of inertia IA of whole system is given by

IA = (2/5)MR2 + [ (2/5)MR2 + 4MR2 ] + [ (2/5)MR2 + 4MR2 ] = (46/5) MR2 .......................(2)

Using eqn.(1) and eqn.(2),   ( Io / IA ) = 26/46 = 13/23
Answered by Thiyagarajan K | 27 Jan, 2020, 09:52: AM

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