Tags: math
Derivation of the Moment of Inertia of a Rectangle About Its Base
We calculate the moment of inertia of a rectangle about its base, where the base is along the \(x\)-axis and the height extends from \(y = 0\) to \(y = h\).
1. Moment of Inertia Formula
The formula for the moment of inertia is:
\[I = \int y^2 \, dA\]
Here: - \(y\): distance from the axis of rotation (the base of the rectangle in this case), - \(dA\): the infinitesimal area element.
2. Setup for the Rectangle
- The rectangle has a width of \(b\) and height \(h\).
- The height extends from \(y = 0\) to \(y = h\).
- The infinitesimal area element is \(dA = b \, dy\).
The integral becomes:
\[I_{\text{base}} = \int_{0}^{h} y^2 \, b \, dy\]
3. Solve the Integral
Factor \(b\) (constant width) outside of the integral:
\[I_{\text{base}} = b \int_{0}^{h} y^2 \, dy\]
The integral of \(y^2\) is:
\[\int y^2 \, dy = \frac{y^3}{3}\]
Apply the limits \(y = 0\) to \(y = h\):
\[I_{\text{base}} = b \left[ \frac{y^3}{3} \right]_{0}^{h}\]
Substitute the limits:
\[I_{\text{base}} = b \left( \frac{h^3}{3} - \frac{0^3}{3} \right)\]
Simplify:
\[I_{\text{base}} = b \cdot \frac{h^3}{3}\]
4. Final Result
The moment of inertia of a rectangle about its base is:
\[I_{\text{base}} = \frac{b h^3}{3}\]
Comparison with the Centroidal Moment of Inertia
The moment of inertia about the base is four times greater than that about the centroidal \(x\)-axis:
\[I_{\text{base}} = 4 \cdot I_{\text{centroidal}}\]
where:
\[I_{\text{centroidal}} = \frac{b h^3}{12}\]