Q 51. - Calculus

Question

Let $C=\left\{\left(x,y\right)|x^2+y^2\le 1\right\}$

If the density at each point in C is proportional to the point’s distance from the origin, find the moments of inertia about the x-axis, the y-axis, and the origin. Use these answers to find the radii of gyration of C about the x-axis,

the y-axis, and the origin.

Step-by-Step Solution

Answer

$\begin{array}{rcl}{I}_x& =& \frac{1}{5}\mathrm{\pi k}\\ I_y& =& \frac{1}{5}\mathrm{\pi k}\\ I_0& =& \frac{2}{5}k\mathrm{\pi }\\ R_x& =& \frac{\sqrt{30}}{10}\\ R_y& =& \frac{\sqrt{30}}{10}\\ R_0& =& \frac{\sqrt{15}}{5}\end{array}$

Step 1. Given information

$C=\left\{\left(x,y\right)|x^2+y^2\le 1\right\}$

Step 2. Explanation

$C=\left\{\left(x,y\right)|x^2+y^2\le 1\right\}$

$\begin{array}{rcl}{I}_y& =& 4{\int }_0^{\pi /2}{\int }_0^{1}k·r·r^2{\cos }^2\theta rdrd\theta \\ & =& 4k{\int }_0^{\pi /2}{\int }_0^1{r}^4{\cos }^2\theta drd\theta \\ I_y& =& 4k{\int }_0^{\pi /2}{\left[\frac{r^5}{5}\right]}_0^1{\cos }^2\theta d\theta \\ & =& \frac{4}{10}k{\int }_0^{\pi /2}(1+\cos 2\theta )d\theta \left(\text{ Use }{\cos }^2\theta =\frac{1+\cos 2\theta }{2}\right)\\ I_y& =& \frac{4}{10}k{\left[\theta +\frac{1}{2}\sin 2\theta \right]}_0^{\pi /2}\\ & =& \frac{1}{5}k\pi \end{array}$

$\begin{array}{rcl}{I}_x& =& {\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}{y}^2\rho (x,y)dydx \left(\text{ use }\rho (x,y)=k\sqrt{x^2+y^2}\right)\\ & =& {\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}k{y}^2\sqrt{x^2+y^2}dydx\\ I_x& =& 4{\int }_0^{\pi /2}{\int }_0^{1}kr·r^2{\sin }^2\theta rdrd\theta \\ & =& 4k{\int }_0^{\pi /2}{\int }_0^1{r}^4{\sin }^2\theta drd\theta \\ I_x& =& 4k{\int }_0^{\pi /2}{\left[\frac{r^5}{5}\right]}_0^1{\sin }^2\theta d\theta \\ & =& \frac{4k}{5}{\int }_0^{\pi /2}{\sin }^2\theta d\theta \\ & =& \frac{4k}{10}{\int }_0^{\pi /2}(1-\cos 2\theta )d\theta \left(\text{ Use }{\sin }^2\theta =\frac{(1-\cos 2\theta )}{2}\right)\\ I_x& =& \frac{4k}{10}{\left[\theta -\frac{1}{2}\sin 2\theta \right]}_0^{\pi /2}\\ & =& \frac{1}{5}k\pi \end{array}$

$\begin{array}{rcl}{I}_x& =& \frac{1}{5}\mathrm{\pi k}\\ I_y& =& \frac{1}{5}\mathrm{\pi k}\\ I_0& =& \frac{2}{5}k\mathrm{\pi }\\ R_x& =& \frac{\sqrt{30}}{10}\\ R_y& =& \frac{\sqrt{30}}{10}\\ R_0& =& \frac{\sqrt{15}}{5}\end{array}$