Respuesta :
Answer:
The center mass of the lamina = [tex]{\begin {pmatrix} 0, \dfrac{1755}{146 \pi } \end {pmatrix}[/tex]
Step-by-step explanation:
Let take a look at the boundary of the lamina that comprises of the semicircles [tex]y = \sqrt{1-x^2}[/tex] and [tex]y = \sqrt{64-x^2}[/tex]
We are also informed that the density at any point is proportional to its distance from the origin.
Thus, the main task required is to find the center mass of the lamina.
In as much as the density is proportional to the distance from the origin;
Then;
[tex]\rho (x,y) = k \sqrt{x^2 + y^2} = kr[/tex]
Thus; the mass of the lamina is determined as:
[tex]m = \iint \limits _ D \rho (x,y) \ dA= \int ^{\pi}_{0}\int ^{8}_{1} kr.r dr d \theta[/tex]
[tex]m = \int ^{\pi}_{0}\int ^{8}_{1} kr^2 dr d \theta[/tex]
[tex]m = k \int ^{\pi}_{0} \begin {bmatrix} \int \limits ^8_1 r^2 dr \end {bmatrix} d \theta[/tex]
[tex]m = k \int ^{\pi}_{0} \begin {bmatrix} \int \limits\dfrac{r^3}{3} \end {bmatrix} ^8_1d \theta[/tex]
[tex]m = k \int ^{\pi}_{0} \begin {bmatrix} \dfrac{512}{3}- \dfrac{1}{3} \end {bmatrix} d \theta[/tex]
[tex]m = \dfrac{511 \ k}{3} \int \limits ^{\pi}_{0} \ d \theta[/tex]
[tex]m = \dfrac{511 \ k}{3} \bigg (\theta \bigg ) ^{\pi}_{0}[/tex]
[tex]m = \dfrac{511 \ k \pi}{3}[/tex]
Now, about the y-axis, the moment of the entire lamina can be computed as:
[tex]M_y = \iint \limits _D x\rho (x,y) \ dA = \int \limits^{\pi}_{0} \int \limits ^{8}_{1} r cos \theta kr .r \ dr \ d \theta[/tex]
[tex]M_y = k \int \limits ^{\pi}_{0} \int ^{8}_{1} \ cos \theta . r^3 \ dr \ d \theta[/tex]
[tex]M_y = k \begin {bmatrix} \int \limits ^{\pi}_{0} cos \theta d \theta \end {bmatrix} \begin {bmatrix} \int \limits ^8_1 r^3 dr \end {bmatrix}[/tex]
[tex]M_y = k (sin \ \theta )^{\pi}_{0} \begin {bmatrix} \dfrac{r^4}{4} \end {bmatrix}^8_1[/tex]
[tex]M_y = k(sin \pi - sin 0 ) \bigg ( \dfrac{8^4}{4}- \dfrac{1^4}{4} \bigg)[/tex]
[tex]M_y = 0[/tex]
Similarly; about the x-axis, the moment of the entire lamina can be computed as:
[tex]M_x = \iint \limits _D y \rho (x,y) \ dA = \int \limits^{\pi}_{0} \int \limits ^{8}_{1} r sin \theta kr .r \ dr \ d \theta[/tex]
[tex]M_x = k \int \limits ^{\pi}_{0} \int ^{8}_{1} \ sin \theta . r^3 \ dr \ d \theta[/tex]
[tex]M_x = k \begin {bmatrix} \int \limits ^{\pi}_{0} sin \theta d \theta \end {bmatrix} \begin {bmatrix} \int \limits ^8_1 r^3 dr \end {bmatrix}[/tex]
[tex]M_x= k (- cos \ \theta )^{\pi}_{0} \begin {bmatrix} \dfrac{r^4}{4} \end {bmatrix}^8_1[/tex]
[tex]M_x = k(-cos \pi +cos 0 ) \bigg ( \dfrac{8^4}{4}- \dfrac{1^4}{4} \bigg)[/tex]
[tex]M_x = \dfrac{4095}{2}k[/tex]
Hence, The center mass of the lamina is:
[tex]( \overline x, \overline y) = \bigg ( \dfrac{M_y}{m}, \dfrac{M_x}{m} \bigg )[/tex]
[tex]( \overline x, \overline y) = \begin {pmatrix} \dfrac{0}{\dfrac{511 \ k \pi}{3}}, \dfrac{\dfrac{4095 \ k}{2}}{\dfrac{511 \ k \pi}{3}} \end {pmatrix}[/tex]
[tex]( \overline x, \overline y) = \begin {pmatrix} 0, \dfrac{1755}{146 \pi } \end {pmatrix}[/tex]
