Question

A metal ring has a moment of inertia 2 kg·m² about a transverse axis through its centre. It is melted and recast into a thin uniform disc of the same radius. What will be the disc's moment of inertia about its diameter?

Answer 1

The MI of the thin ring about its transverse symmetry axis through its centre,

I_ring = MR² = 2 kg∙m²

Since the ring is melted and recast into a thin disc, the mass of the disc equals the mass of the ring = M. Also, the disc has the same radius as the ring. Then, the MI of the thin disc about its diameter is

`I_(disc) = 1/4 MR^2`

∴ `I_(disc) = 1/4` I_ring

= `1/4 xx 2`

 0.5 kg.m²