Question

The moment of inertia of a thin ring of mass M and radius R about its diameter is

• A. $\frac{1}{2}M{R}^2$
• B. $\frac{3}{2}M{R}^2$
• C. $\frac{M{R}^2}{2}$
• D. $2M{R}^2$

Answer 1

The correct option is C $\frac{M{R}^2}{2}$

The moment of inertia about O is given by =$M{R}^2$

By perpendicular axis theorem,

$I_x+I_y=I_z$

Since both $I_x$ and $I_y$ are along the diameter,

$\begin{array}{c}{I}_x=I_y\\ 2I_x=I_z=M{R}^2\\ I_{diameter}=\frac{M{R}^2}{2}\end{array}$