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Viktor [21]
3 years ago
15

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

in them. Find the center of mass of the lamina if the density at any point is inversely proportional to its distance from the origin.
Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

Required center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

Step-by-step explanation:

Given semcircles are,

y=\sqrt{1-x^2}, y=\sqrt{16-x^2} whose radious are 1 and 4 respectively.

To find center of mass, (\bar{x},\bar{y}), let density at any point is \rho and distance from the origin is r be such that,

\rho=\frac{k}{r} where k is a constant.

Mass of the lamina=m=\int\int_{D}\rho dA where A is the total region and D is curves.

then,

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

  • Now, x-coordinate of center of mass is \bar{y}=\frac{M_x}{m}. in polar coordinate y=r\sin\theta

\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta

=3k\int_{0}^{\pi}\sin\theta d\theta

=3k\big[-\cos\theta\big]_{0}^{\pi}

=3k\big[-\cos\pi+\cos 0\big]

=6k

Then, \bar{y}=\frac{M_x}{m}=\frac{2}{\pi}

  • y-coordinate of center of mass is \bar{x}=\frac{M_y}{m}. in polar coordinate x=r\cos\theta

\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta

=3k\int_{0}^{\pi}\cos\theta d\theta

=3k\big[\sin\theta\big]_{0}^{\pi}

=3k\big[\sin\pi-\sin 0\big]

=0

Then, \bar{x}=\frac{M_y}{m}=0

Hence center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

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