alleryStart learning →By **parallel-axis theorem**, the moment of inertia of a thin ring of mass $M$, radius $R$, about a tangent in its plane is:
- $MR^2$
- $\tfrac{1}{2}MR^2$
- $\tfrac{3}{2}MR^2$
- $2MR^2$
Show answer and solution
Correct answer: C. $\tfrac{3}{2}MR^2$
About diameter (in plane), $I_d = MR^2/2$ (perpendicular-axis theorem: $I_z = 2I_d = MR^2$ gives $I_d = MR^2/2$).
**Tangent in plane:** $I = MR^2/2 + MR^2 = 3MR^2/2$.