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 moment of inertia about the x and y axes are

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

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

part (a) step 1: Given information

The objective of this problem is to find the moments of inertia about x - axis and y- axis for the semicircular lamina . Density of region is proportional to the distance of the point from origin.

part (a) step 1: calculation

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

Moment of inertia about x - axis

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

Substitute x=r \cos \theta, y=r \sin \theta$ and d x d y=r d r d \theta

Equation of circle

$r=2\cos \theta$

Equation of line$x=1inpolarformr=\sec \theta$

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

Integrate the inner integral with respect to r first.

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

Substitute the limits

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

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

Moment of inertia about y - axis

$$\begin{gathered} I_y={\iint }_{\Omega }{x}^2\rho (x,y)dA \\ {I}_y={\iint }_{\Omega }{x}^{2}k\sqrt{x^2+y^2}dydx \end{gathered}$$

Substitute$x=r \backslash \cos \backslash theta, y=r \backslash \sin \backslash theta\$ and d x d y=r d r d \backslash theta$

Equation of circle

$x=r\cos \theta$, $y=r\sin \theta anddxdy=rdrd\theta$ 

Equation of line $x=1inpolarformr=\sec \theta$

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

Integrate the inner integral with respect to r first.

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

Substitute the limits

$$\begin{gathered} I_y=\frac{k}{5}{\int }_{-\pi /4}^{\pi /4}\left[32{\cos }^2\theta {\cos }^5\theta -{\cos }^2\theta {\sec }^5\theta \right]d\theta \\ {I}_y=\frac{k}{5}{\int }_{-\pi /4}^{\pi /4}\left[32{\cos }^7\theta -{\sec }^3\theta \right]d\theta \\ {I}_y=\frac{k}{5}\left[\frac{673\sqrt{2}}{35}-2{\tanh }^{-1}\left(\tan \left(\frac{\pi }{8}\right)\right)\right] \end{gathered}$$