Question

Moment of inertia of a hoop suspended from a peg about the peg is:

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

Answer 1

Answer: (C)

2MR$^2$

Answer 2

The moment of inertia of a hoop about an axis through its center and perpendicular to its plane is $MR^2$. When suspended from a peg, the axis of rotation passes through a point on its circumference and is parallel to the central axis. Using the parallel axis theorem, $I = I_{CM} + Md^2$, where $I_{CM} = MR^2$ and $d=R$. Thus, $I = MR^2 + MR^2 = 2MR^2$.