Question

The moment of inertia of a circular ring (radius R, mass M) about an axis which passes through tangentially and perpendicular to its plane will be?
(A) $ \dfrac{{M{R^2}}}{2} $
(B) $ M{R^2} $
(C) $ \dfrac{{3M{R^2}}}{2} $
(D) $ 2\,M{R^2} $

Answer 1

Hint
We will first find the relation between the moment of inertias of the axes of the circular ring. Then using the parallel axis theorem to calculate the moment of inertia of the axis which passes through tangentially and perpendicular to its plane.

Complete step by step answer
We have to calculate the moment of inertia of a circular ring about an axis which passes through tangentially and perpendicular to its plane. Let this axis K as shown.
                 

Now axis Y passes through the origin, let its moment of inertia be $ {I_Y} $ .
Axis K is tangential to the ring and lies in its pane and also axis K is parallel to axis Y.
We know that moment of inertia of a ring if the axis passes through its plane is $ {I_Y} = M{R^2} $ where R is radius and M is mass.
Now we know by parallel axis theorem i.e. $ {I_K} = {I_Y} + M{R^2} $ .
Now we put the value of $ {I_Y} $ into this equation, we get:
  $ {I_K} = M{R^2} + M{R^2} = 2M{R^2} $ .
So, the moment of inertia about an axis which passes through tangentially and perpendicular to its plane will be $ 2\,M{R^2} $ .
So, the correct option is D.

Note
Moment of inertia of a part of a rigid body is the same as the moment of inertia of the whole body. Theorem of parallel axis is applicable for any type of rigid body whether it is two dimensional or three dimensions, while the theorem of perpendicular axis is applicable for two dimensional bodies only.