Respuesta :
Answer:
Required center of mass [tex](\bar{x},\bar{y})=(\frac{2}{\pi},0)[/tex]
Step-by-step explanation:
Given semcircles are,
[tex]y=\sqrt{1-x^2}, y=\sqrt{16-x^2}[/tex] whose radious are 1 and 4 respectively.
To find center of mass, [tex](\bar{x},\bar{y})[/tex], let density at any point is [tex]\rho[/tex] and distance from the origin is r be such that,
[tex]\rho=\frac{k}{r}[/tex] where k is a constant.
Mass of the lamina=m=[tex]\int\int_{D}\rho dA[/tex] where A is the total region and D is curves.
then,
[tex]m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k[/tex]
- Now, x-coordinate of center of mass is [tex]\bar{y}=\frac{M_x}{m}[/tex]. in polar coordinate [tex]y=r\sin\theta[/tex]
[tex]\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA[/tex]
[tex]=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta[/tex]
[tex]=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta[/tex]
[tex]=3k\int_{0}^{\pi}\sin\theta d\theta[/tex]
[tex]=3k\big[-\cos\theta\big]_{0}^{\pi}[/tex]
[tex]=3k\big[-\cos\pi+\cos 0\big][/tex]
[tex]=6k[/tex]
Then, [tex]\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}[/tex]
- y-coordinate of center of mass is [tex]\bar{x}=\frac{M_y}{m}[/tex]. in polar coordinate [tex]x=r\cos\theta[/tex]
[tex]\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA[/tex]
[tex]=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta[/tex]
[tex]=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta[/tex]
[tex]=3k\int_{0}^{\pi}\cos\theta d\theta[/tex]
[tex]=3k\big[\sin\theta\big]_{0}^{\pi}[/tex]
[tex]=3k\big[\sin\pi-\sin 0\big][/tex]
[tex]=0[/tex]
Then, [tex]\bar{x}=\frac{M_y}{m}=0[/tex]
Hence center of mass [tex](\bar{x},\bar{y})=(\frac{2}{\pi},0)[/tex]
