Question

metal ring has moment of inertia 2 kg m about a transverse axis through its centre. It is

melted and recast into a this uniform disc of the same radius. What will be the disc's moment

of inertia about its diameter?​

Answer 1

Let the moment of inertia of the ring about any of the two perpendicular axes passing through the diametral plane be I. Using perpendicular axis theorem we get 2I=MR

2

or I=

2

MR

2

. Now using parallel axis theorem we get the moment of inertia about tangential axis will be

2

MR

2

+MR

2

=

2

3

MR

2

. Now substituting the values we get

3=

2

3

(2R

2

)

or

R=1m