Moments of Inertia
Area and mass moments of inertia, polar moment, parallel-axis theorems, and radius of gyration.
Area and mass moments of inertia, polar moment, parallel-axis theorems, and radius of gyration.
Find the centroid of the region in the first quadrant bounded by $y=9-x^2$, the $x$-axis, and the $y$-axis.
The area is
$$A=\int_0^3(9-x^2)\,dx=18.$$
For the $x$-coordinate:
$$\bar{x}=\frac{1}{18}\int_0^3x(9-x^2)\,dx=\frac{1}{18}\left[\frac{9x^2}{2}-\frac{x^4}{4}\right]_0^3=\frac{9}{8}$$
For the $y$-coordinate:
$$\bar{y}=\frac{1}{18}\cdot\frac{1}{2}\int_0^3(9-x^2)^2\,dx=\frac{18}{5}$$
$$\boxed{\left(\frac{9}{8},\frac{18}{5}\right)}$$
Find $I_y$ of the region bounded by $x^2=8y$, $y=2$, and $x=0$.
At $y=2$, $x=4$. Use vertical strips from $x=0$ to $x=4$. The strip height is $2-\frac{x^2}{8}$.
$$I_y=\int_0^4x^2\left(2-\frac{x^2}{8}\right)dx$$
$$=\left[\frac{2x^3}{3}-\frac{x^5}{40}\right]_0^4=\boxed{17.07}$$
Find $I_x$ of the first-quadrant region bounded by $y^2=4x$, $y=2$, and the $y$-axis.
Use horizontal strips. Since $x=\frac{y^2}{4}$, the strip length is $\frac{y^2}{4}$.
$$I_x=\int_0^2y^2\left(\frac{y^2}{4}\right)dy=\frac{1}{4}\int_0^2y^4\,dy$$
$$I_x=\frac{1}{4}\left[\frac{y^5}{5}\right]_0^2=\boxed{1.6}$$
For the same region bounded by $y^2=4x$, $y=2$, and the $y$-axis, find the radius of gyration about the $x$-axis.
The area is
$$A=\int_0^2\frac{y^2}{4}\,dy=\frac{2}{3}.$$
From the previous result, $I_x=1.6$.
$$k_x=\sqrt{\frac{I_x}{A}}=\sqrt{\frac{1.6}{2/3}}=\boxed{1.55}$$