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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.

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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.

Sum
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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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Chapter 1: Rotational Dynamics - Exercises [Page 25]

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Balbharati Physics [English] Standard 12 Maharashtra State Board
Chapter 1 Rotational Dynamics
Exercises | Q 19 | Page 25

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