Second moment of area of a thin circular ring: For a thin ring with external diameter D and internal diameter d, what is the area moment of inertia about the centroidal axis perpendicular to the plane of the ring?

Introduction / Context:
The second moment of area (area moment of inertia) quantifies a cross-section’s resistance to bending. For circular shapes and annuli, the dependence on the fourth power of diameter is particularly strong, which is important in comparing stiffness of tubes versus solid rods for the same material and weight.

Given Data / Assumptions:

• Thin circular ring (annulus) with outer diameter D and inner diameter d.
• Centroidal axis perpendicular to the plane (i.e., polar axis through center).
• Standard thin-walled/annulus formulae are applicable.

Concept / Approach:
For a solid circle about its centroidal axis perpendicular to the plane, the second moment of area about x or y is I = (π/64) D^4; the polar moment J = I_x + I_y = (π/32) D^4. For an annulus, subtract the inner hole contribution of corresponding diameter d. Thus I = (π/64) (D^4 − d^4) about a centroidal in-plane axis, and J = (π/32) (D^4 − d^4). Therefore the dependence is on the difference of fourth powers, not their sum or product.

Step-by-Step Solution:

Verification / Alternative check:
If d → 0, the expression reduces to solid circle; if d → D, I → 0, consistent with vanishing area.

Why Other Options Are Wrong:

• (D^4 + d^4) or its half: incorrect dependence; would not vanish as d → D.
• (D^4 * d^4): dimensionally inconsistent without normalization.
• Zero: only true in the trivial limit of zero thickness.

Common Pitfalls:
Confusing polar moment with planar second moment; mixing up constants (π/64 vs π/32).

Final Answer:
Proportional to (D^4 - d^4)