Question

Polar moment of inertia of a circle (\(I_{\text{0}}\)) is given by

• A. \(\frac{\pi}{64} d^4\)
• B. \(\frac{\pi}{32} d^4\)
• C. \(\frac{\pi}{16} d^4\)
• D. \(\frac{\pi}{32} d^3\)

Hint:

Remember the distinction:
- Area Moment of Inertia (bending resistance): \(I = \frac{\pi d^4}{64}\)
- Polar Moment of Inertia (torsion resistance): \(J = 2I = \frac{\pi d^4}{32}\)
This factor of 2 is a useful shortcut for circular cross-sections.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The moment of inertia of a planar area represents its resistance to bending.
The polar moment of inertia (\(I_{\text{p}}\) or \(I_{\text{0}}\)) represents its resistance to torsional deformation (twisting) about an axis perpendicular to the plane of the area.
Key Formula or Approach:
According to the Perpendicular Axis Theorem, the polar moment of inertia about the centroidal z-axis is the sum of the moments of inertia about the x and y axes:
\[ I_{\text{0}} = I_{\text{xx}} + I_{\text{yy}} \]

Step 2: Detailed Explanation:
Let us apply this theorem to a circular area of diameter \(d\):
- The area is perfectly symmetrical about its centroidal axes, so:
\[ I_{\text{xx}} = I_{\text{yy}} \]
- The area moment of inertia of a circle about its diameter is:
\[ I_{\text{xx}} = \frac{\pi d^4}{64} \]
- Applying the perpendicular axis theorem to find the polar moment of inertia:
\[ I_{\text{0}} = I_{\text{xx}} + I_{\text{yy}} \]
\[ I_{\text{0}} = \frac{\pi d^4}{64} + \frac{\pi d^4}{64} \]
\[ I_{\text{0}} = \frac{2 \pi d^4}{64} \]
\[ I_{\text{0}} = \frac{\pi}{32} d^4 \]

Step 3: Final Answer:
The polar moment of inertia of a circle is \(\frac{\pi}{32} d^4\). Hence, the correct option is (B).