By **parallel-axis theorem**, the moment of inertia of a thin ring of …

Question

By **parallel-axis theorem**, the moment of inertia of a thin ring of mass $M$, radius $R$, about a tangent in its plane is:

Accepted Answer

C. $	frac{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:** $I = MR^2/2 + MR^2 = 3MR^2/2$.

Answer

$MR^2$

Answer

$	frac{1}{2}MR^2$

Answer

$2MR^2$

By parallel-axis theorem, the moment of inertia of a thin ring of mass $M$, radius $R$, about a tangent in its plane is:

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