Question

three identical rings each of mass M and radius R are placed in the same plane touching each other such that their centres form the vertices of an equilateral triangle. the MI of the systems about an axis passing through the centre of one of the rings and perpendicular to its plane is
MR2/2
MR2
11MR2

Answer 1

The moment of Inertia of the system = The moment of Inertia of Ring rotating about an axis perpendicular to its plane + The moment of Inertia of Ring rotating about an axis in its own plane.


The moment of Inertia of Ring rotating about an axis perpendicular to its plane= MR2


The moment of Inertia of Ring rotating about an axis in its own plane.= MR2/2


So the moment of inertia of the system is (3/2)MR2