Question

The moment of inertia of a circular ring is $I$ about an axis perpendicular to its plane and passing through its centre. About an axis passing through tangent of ring in its plane, its moment of inertia is :

• A. $\frac{I}{2}$
• B. $\frac{2I}{3}$
• C. $\frac{3I}{2}$
• D. $\frac{I}{3}$

Answer 1

The correct option is C $\frac{3I}{2}$

Moment of Inertia of ring about an axis perpendicular to its plane is $I=M{R}^2$
By perpendicular axis theorem , moment of Inertia about any diameter $I^{\prime }=\frac{I}{2}=\frac{M{R}^2}{2}$
Moment of Inertia about any tangent is
$I_1=I^{\prime }+M{R}^2=\frac{M{R}^2}{2}+M{R}^2=\frac{3M{R}^2}{2}=\frac{3I}{2}$