Q 21. - Calculus

Question

Find the moments of inertia about the $x$- and $y$-axes for the semicircular lamina described in Example $2$. Assume that the density at every point is proportional to the distance of the point from the origin.

Step-by-Step Solution

Answer

The moments are:

$I_x=\frac{k}{5}\left[\frac{821}{210\sqrt{2}}+\frac{1}{8}\ln (3+2\sqrt{2})\right]$

$I_y=\frac{k}{5}\left[\frac{673\sqrt{2}}{35}-2{\tanh }^{-1}\left(\tan \left(\frac{\pi }{8}\right)\right)\right]$

Step 1: Given Information

We need to find the moment of inertia about $x \mathrm{and} y$ axis of semicircular lamina.

Density is proportional to distance of point from origin.

Step 2: Simplification

Density is given by $\rho (x,y)=k\sqrt{x^2+y^2}$

Moment of inertia about $x$ axis

$I_x={\iint }_{\Omega }{y}^2\rho (x,y)dA={\iint }_{\Omega }{y}^{2}k\sqrt{x^2+y^2}dydx$

Use $x=r\cos \theta ,y=r\sin \theta \text{ and }dxdy=rdrd\theta$

Step 3: Calculation of Inertia about x axis

We know equation of circle is

$r=2\cos \theta$

Equation of line $x=1$ in polar form is $r=\sec \theta$

$I_x={\int }_{-\pi /4}^{\pi /4}{\int }_{\sec \theta }^{2\cos \theta }k{r}^4{\sin }^2\theta drd\theta$

Integration inner integral first

$I_x=k{\int }_{-\pi /4}^{\pi /4}{\left[\frac{r^5}{5}\right]}_{\sec \theta }^{2\cos \theta }{\sin }^2\theta d\theta$

$I_x=\frac{k}{5}{\int }_{-\pi /4}^{\pi /4}\left[32{\sin }^2\theta {\cos }^5\theta -{\sin }^2\theta {\sec }^5\theta \right]d\theta$

$I_x=\frac{k}{5}\left[\frac{821}{210\sqrt{2}}+\frac{1}{8}\ln (3+2\sqrt{2})\right]$

Step 4: Calculation of Moment of Inertia about y axis

It is given by $I_y={\iint }_{\Omega }{x}^2\rho (x,y)dA$

$I_y={\iint }_{\Omega }{x}^{2}k\sqrt{x^2+y^2}dydx$

Use $x=r\cos \theta ,y=r\sin \theta \text{ and }dxdy=rdrd\theta$

$r=2\cos \theta$ is equation of circle

$r=\sec \theta$ is equation of line $x=1$ in polar form

$I_y={\int }_{-\pi /4}^{\pi /4}{\int }_{\sec \theta }^{2\cos \theta }k{r}^4{\cos }^2\theta drd\theta$

$I_y=k{\int }_{-\pi /4}^{\pi /4}{\left[\frac{r^5}{5}\right]}_{\sec \theta }^{2\cos \theta }{\cos }^2\theta d\theta$

Using limits 

$I_y=\frac{k}{5}{\int }_{-\pi /4}^{\pi /4}\left[32{\cos }^2\theta {\cos }^5\theta -{\cos }^2\theta {\sec }^5\theta \right]d\theta$

$I_y=\frac{k}{5}{\int }_{-\pi /4}^{\pi /4}\left[32{\cos }^7\theta -{\sec }^3\theta \right]d\theta$

$I_y=\frac{k}{5}\left[\frac{673\sqrt{2}}{35}-2{\tanh }^{-1}\left(\tan \left(\frac{\pi }{8}\right)\right)\right]$