Question
Download Solution PDFPolar moment of inertia about central axis of a semi-circular lamina with radius R is given as ______.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
The polar moment of inertia (J) about the central axis for a semi-circular lamina is calculated from the formula for a full circle and then taking half of it since the semi-circle is half of a full circle.
For a full circular lamina (disk) of radius (R), the polar moment of inertia (J) about the central axis is given by the formula :\( J_{\text{full circle}} = \frac{1}{2} \pi R^4 \)
For a semi-circular lamina, we take half of this value: \( J_{\text{semi-circle}} = \frac{1}{2} \times \frac{1}{2} \pi R^4 = \frac{\pi R^4}{4} \)
Additional Information
The formula of the moment of inertia for various other figures is given below.
Moment of inertia of different section:
|
Shape of cross-section |
INA |
Ymax |
Z |
|
Rectangle |
\(I = \frac{{b{d^3}}}{{12}}\) |
\({Y_{max}} = \frac{d}{2}\) |
\(Z = \frac{{b{d^2}}}{6}\) |
|
Circular |
\(I = \frac{π }{{64}}{D^4}\) |
\({Y_{max}} = \frac{d}{2}\) |
\(Z = \frac{π }{{32}}{D^3}\) |
|
Triangular |
\(I = \frac{{B{h^3}}}{{36}}\) |
\({Y_{max}} = \frac{{2h}}{3}\) |
\(Z = \frac{{B{h^2}}}{{24}}\) |
Last updated on May 28, 2025
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