Question

lamina with density ρ(x,y) = 3 √{x2+y2} occupies region D, enclosed by the curve r = 1−sin(θ). Which of the following statements is the best description of the center of mass of the lamina?

Find the moments of intertia about the x-axis, the y-axis, and the origin for the lamina. Yes, the integrals can be done by hand, but why put yourself through that? You may round your answers to the nearest 0.01.

Answer 1

First thing to note is that the figure is symmetric about the y-axis because

$r(\theta)=r(\pi-\theta)$

and also the density, this means the x coordinate of COM is 0. We need to find the y coordinate.

$y_{com}=\frac{\iint r\sin\theta\rho rdrd\theta}{\iint\rho rdrd\theta}$

where

$\rho=3r$

$y_{com}=\frac{\int_0^{2\pi}\int_0^{1-\sin\theta} 3r^3\sin\theta drd\theta}{\int_0^{2\pi}\int_0^{1-\sin\theta} 3r^2drd\theta}$

$y_{com}=-\frac{21}{20}$

MOI about x-axis

$I_{x}=\iint (r\sin\theta)^2\rho rdrd\theta$

$I_{x}=\int_0^{2\pi}\int^{1-\sin\theta}_0 3r^4\sin^2\theta drd\theta$

$I_{x}=\frac{279\pi}{40}$

Similarly about y-axis,

$I_{x}=\iint (r\cos\theta)^2\rho rdrd\theta$

$I_{x}=\int_0^{2\pi}\int^{1-\sin\theta}_0 3r^4\cos^2\theta drd\theta$

$I_{x}=\frac{99\pi}{40}$

By perpendicular axis theorem

$I_{x}+I_{y}=I_{o}$

$I_{o}=\frac{189\pi}{20}$