This applet illustrates how the area moments of inertia and the product of inertia vary under rotation. The area shown is a circle that is rotated about the origin. The circle has a radius r[sub]1[/sub] and it rotates at a radius r[sub]2[/sub].[br]The moments are all about the axes or origin.[br][br]Values shown include the moment with respect to the x-axis,[math]I_x[/math], the moment of inertia with respect to the y axis,[math]I_y[/math] the product of inertia,[math]I_{xy}[/math] and the polar moment of inertia,[math]J_O[/math], about the origin. The calculations are shown below.[br][br]The right graph shows the relation between the moment of inertia and the product of inertia. The moments of inertia are shown on the abscissa ( x-axis) vs the product of inertia on the ordinate (y-axis). Points are added as you rotate by changing the angle [math]\theta[/math]. Note the scale changes as the radii are changed. Checking the box will plot the points (Iyy,-Ixy) instead of (Iyy,Ixy) and add a title the resulting figure. To clear the right graph change either radii on the left graph.

Rotate the area by moving the [math]\theta[/math] slider or selecting the play button. What shape appears in the right side graph?[br][br]How does the shape change when you change the radii on the left graph?[br][br]How does the axis scale vary when you change the radii on the left graph?[br][br]How does the angle on the right change compared to the angle on the left?[br][br]How would the rotation angle change if you rotated the axes instead of the circle?[br][br]What happens when r[sub]2[/sub] is set to zero?[br][br]How does the polar moment of inertia vary under a rotation?[br][br]Do you notice a relation between the [math]\left\langle I_{xx},I_{xy}\right\rangle[/math] point and the [math]\left(I_{yy},I_{xy}\right)[/math] point?[br][br]Check the (I[sub]yy[/sub], -I[sub]xy[/sub]) box and notice what happens to the relation between the[math]\left(I_{xx},I_{xy}\right)[/math] point and the [math]\left(I_{yy},-I_{xy}\right)[/math] point?[br][br]How would you calculate the radius r[sub]3[/sub] given the moments of inertia and the product of inertia at any rotation angle?[br][br]How would you calculate the rotation angle, [math]\theta[/math], where I[sub]xy[/sub] is zero given I[sub]xx[/sub], I[sub]yy[/sub] and I[sub]xy[/sub] for a shape?[br][br]Note: The observed behavior under rotation is the same for any shape.

[size=150][b]Moments of Inertia[br][/b][/size][br]Moments of inertia about the circle centroid from tables:[br][math]I_x_{Center}=I_y_{Center}=\frac{1}{4}\pi\left(r_1\right)^4[/math][br][br]Apply the parallel axis theorem: [math]I_x=I_{x'}+d^2A[/math][br][math]I_x=I_{xCenter}+\left(r_2\sin\theta\right)^2\left(\pi\left(r_1\right)^2\right)=\pi\left(r_1\right)^2\left(\frac{\left(r_1\right)^2}{4}+\left(r_2\right)^2\sin^2\theta\right)[/math][br][math]I_y=I_{yCenter}+\left(r_2\cos\theta\right)^2\left(\pi\left(r_1\right)^2\right)=\pi\left(r_1\right)^2\left(\frac{\left(r_1\right)^2}{4}+\left(r_2\right)^2\cos^2\theta\right)[/math][br][br]Some trigonometric identities to note to help explain Mohr's circle results:[br][math]\sin^2\theta=\frac{1-\cos\left(2\theta\right)}{2}[/math] and [math]\cos^2\theta=\frac{\cos\left(2\theta\right)+1}{2}[/math][br][size=150][b][br]Product of Inertia[br][/b][size=100]For a circle the product of inertia about the center is always zero due to symmetry.[br][math]I_{xyCenter}=0[/math][br][/size][/size][br]Apply parallel axis theorem for product of inertia: [math]I_{xy}=I_{x'y'}+d_xd_yA[/math][br][math]I_{xy}=I_{xyCenter}+\left(r_2\sin\theta\right)\left(r_2\cos\theta\right)\left(\pi\left(r_1\right)^2\right)=\pi\left(r_1\right)^2\left(r_2\right)^2\sin\theta\cdot\cos\theta[/math][br][br]Another notable trigonometric identity: [math]\sin\theta\cdot\cos\theta=\frac{\sin\left(2\theta\right)}{2}[/math][br][br][size=150][b]Polar Moment of Inertia[/b][/size][br][br]From Pythagorean theorem and polar moment definition[br][math]J_O=\int r^2dA=\int\left(x^2+y^2\right)dA=\int x^2dA+\int y^2dA=I_x+I_y[/math][br][br]Combining the above formulas for moments of inertia[br][math]\begin{matrix}J_O=I_x+I_y\end{matrix}[/math][br][math]=\pi\left(r_1\right)^2\left(\frac{\left(r_1\right)^2}{4}+\left(r_2\right)^2\sin^2\theta\right)[/math][br][math]+\pi\left(r_1\right)^2\left(\frac{\left(r_1\right)^2}{4}+\left(r_2\right)^2\cos^2\theta\right)[/math] and noting the trigonometric identity [math]\sin^2\theta+\cos^2\theta=1[/math] gives[br][math]J_O=\pi\left(r_1\right)^2\left(\frac{\left(r_1\right)^2}{2}+\left(r_2\right)^2\right)[/math][br][br]