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Question
A metallic ring of mass 1 kg has a moment of inertia 1 kg m2 when rotating about one of its diameters. It is molten and remolded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own axis.
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Solution
Given:
mass of ring and disc is M =1 kg
Moment of inertia of ring at diameter (Ir)d = 1 kg m2
Rr = Rd
To find:
Moment of inertia of disc about own axis = Id =?
Solution:
Using theorem of perpendicular axes, for a ring M.I about its axis passing through C.M and perpendicular to its plane is twice the M.I about its any diameter, which is given by,
(Ir)c = 2 (Ir)d
= 2 × 1
MRr2 = 2 kg m2
Rr2 = Rd2 = 2 meter
Hence,
Moment of inertia of disc about own axis is given by,
Id =`1/2` MRd2
= `1/2` × 1 × 2
Id = 1 kg m2
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