# two rings of same radius and same mass are placed such that their centers are at a common point and their planes are perpendicular to each other. THE MOMENT OFINERTIA ABOUT THE CENTER PASSING THROUGH CENTER AND PERPENDICULAR TO PLANE OF ONE OF THE RINGS ?

Swapnil Saxena
102 Points
12 years ago

The moment of Inertia of such a system would be (3/2)MR2 .

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

Raunaq Mehta
39 Points
7 years ago
The moment of Inertia of such a system would be (3/2)MR2.
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
5 years ago