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A metallic ring of mass 1 kg has moment of inertia 1 kg m2 when rotating about one of its diameters. It is molten and remoulded 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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The MI of the thin ring about its diameter,

\(Ir = \frac{1}{2}MR^2 = 1 kg\ m^2\) 

Since the ring is melted and recast into a thin disc of same radius R, the mass of the disc equals the mass of the ring = M.  The MI of the thin disc about its own axis (i.e. transverse symmetry axis) is

\(I_d= \frac{1}{2}MR^2 = I_r\) 

\(∴ I_d = 1 kg\ m^2\) 

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