Question D.3: Determine the moments of inertia Ix and Iy for the parabolic......
Determine the moments of inertia I_x and I_y for the parabolic semisegment OAB shown in Fig. D-12. The equation of the parabolic boundary is
\quad\quad\quad\quad y = f(x) = h{\Bigg\lgroup}1-{\frac{x^{2}}{b^{2}}}{\Bigg\rgroup}\quad\quad (a)
(This same area was considered previously in Example D-1.)
To determine the moments of inertia by integration, use Eqs. (D-9a and b).
I_x = \int y^2dA \quad I_y = \int x^2 dA \qquad (D-9a,b)The differential element of area dA is selected as a vertical strip of width dx and height y, as shown in Fig. D-12. The area of this element is
\quad\quad\quad\quad dA = y dx = h{{\Bigg\lgroup}1-{\frac{x^{2}}{b^{2}}}{\Bigg\rgroup}}\,d x\quad\quad (b)
Since every point in this element is at the same distance from the y axis, the moment of inertia of the element with respect to the y axis is x^2 dA. Therefore, the moment of inertia of the entire area with respect to the y axis is obtained as
\quad\quad\quad\quad I_{y}\,=\,\int x^{2}d A=\,\int_{0}^{b}x^{2}h{\Bigg\lgroup}1-{\frac{x^{2}}{b^{2}}}{\Bigg\rgroup}d x={\frac{2h b^{3}}{15}}\quad\quad (c)
To obtain the moment of inertia with respect to the x axis, note that the differential element of area dA has a moment of inertia dI_x with respect to the x axis equal to
\quad\quad\quad\quad d I_{x}={\frac{1}{3}}(d x)y^{3}={\frac{y^{3}}{3}}d{{x}}
as obtained from Eq. (D-12).
Hence, the moment of inertia of the entire area with respect to the x axis is
\quad\quad\quad\quad I_{x}\,=\,\int_{0}^{b}{\frac{y^{3}}{3}}d x\,=\,\int_{0}^{b}{\frac{h^{3}}{3}}{\Bigg\lgroup}1-{\frac{x^{2}}{b^{2}}}{\Bigg\rgroup}^{3}\,d x={\frac{16 b h^{3}}{105}}\quad\quad (d)
These same results for I_x ~ and ~ I_y can be obtained by using an element in the form of a horizontal strip of area dA = xdy or by using a rectangular element of area dA = dxdy and performing a double integration. Also, note that the preceding formulas for I_x ~and ~ I_y agree with those given in Case 17 of Appendix E.