Question 12.4: The parabolic semisegment OAB shown in Fig. 12-15 has base b......
The parabolic semisegment OAB shown in Fig. 12-15 has base b and height h. Using the parallel-axis theorem, determine the moments of inertia I_{x_c} and I_{y_c} with respect to the centroidal axes x_c and y_c.
We can use the parallel-axis theorem (rather than integration) to find the centroidal moments of inertia because we already know the area A. the centroidal coordinates \bar{x} and \bar{y}. and the moments of inertia I_x and I_y with respect to the x and y axes. These quantities were obtained earlier in Examples 12-1 and 12-3. They also are listed in Case 17 of Appendix D and are repeated here:
A=\frac{2bh}{3}\quad \bar{x}=\frac{3b}{8}\quad \bar{y}=\frac{2h}{5}\quad I_x=\frac{16bh^3}{105}\quad I_y=\frac{2hb^3}{15}
To obtain the moment of inertia with respect to the x_c axis, we write the parallel-axis theorem as follows:
I_x=I_{x_c}+A\bar{y}^2
Therefore, the moment of inertia I_{x_c} is
I_{x_c}=I_x\ -\ A\bar{y}^2=\frac{16bh^3}{105}\ -\ \frac{2bh}{3}\left(\frac{2h}{5}\right)^2=\frac{8bh^3}{175} (12-13a)
In a similar manner, we obtain the moment of inertia with respect to the y_c axis:
I_{x_c}=I_y\ -\ A\bar{x}^2=\frac{2hb^3}{15}\ -\ \frac{2bh}{3}\left(\frac{3b}{8}\right)^2=\frac{19bhb^3}{480} (12-13b)
Thus, we have found the centroidal moments of inertia of the semisegment.