Answer:
CM = (0, 0)
Explanation:
We can apply
me = σ(x, y)*Ae
where Ae = π*(7)² = 49*π
then
me = (σ₀*x² + y²)*49*π
cm_e = (cm_ex, cm_ey) = (0, 0)
mi = σ(x, y)*Ai
where Ai = π*(2)² = 4*π
then
mi = (σ₀*x² + y²)*4*π
cm_i = (cm_ix, cm_iy) = (0, 0)
We can apply the equation
mt = me -mi
where is the total mass of the region
then
mt = me - mi = (σ₀*x² + y²)*49*π - (σ₀*x² + y²)*4*π
⇒ mt = 45*π*(σ₀*x² + y²)
then we apply the equation
x_cm = (me*cm_ex - mi*cm_ix) / mt
x_cm = ((σ₀*x² + y²)*49*π*(0) - (σ₀*x² + y²)*4*π*(0)) / (45*π*(σ₀*x² + y²))
x_cm = 0
y_cm = (me*cm_ey - mi*cm_iy) / mt
y_cm = ((σ₀*x² + y²)*49*π*(0) - (σ₀*x² + y²)*4*π*(0)) / (45*π*(σ₀*x² + y²))
y_cm = 0
finally, we get
CM = (x_cm, y_cm) = (0, 0)
A plane lamina occupies the region in the xy-plane between the two circles x2 + y2 = 4 and x2 + y2 = 49 and above the x-axis. The center of
mass of the lamina if its mass density is [tex]\mathbf{\sigma (x, y) = \dfrac{\sigma _o}{\sqrt{x^2 + y^2 }} \ kg/m^2}[/tex] is:
[tex]\mathbf{\bar x = \dfrac{0}{M} = 0 \ \ ; \ \ \bar y = \dfrac{45 \sigma_0}{5 \pi \sigma_o} = \dfrac{9}{\pi}}[/tex]
The center of the mass of the lamina can be computed by using the formula:
[tex]\mathbf{\bar x = \dfrac{\int \int \sigma xdA }{M} , \ \bar y = \dfrac{\int \int \sigma y dA }{M}}[/tex]
The mass of the region R can be estimated by using the formula:
[tex]\mathbf{A = \int \int _R \sigma \ dA}[/tex]
To determine the area by using polar coordinates, we have:
[tex]\mathbf{M = \int^{\pi}_{0} \int ^7_{2} \sigma rdr d\theta}[/tex]
[tex]\mathbf{\implies \int^{\pi}_{0} \int ^7_{2} \dfrac{\sigma_0}{\sqrt{r^2 cos^2 \theta + r^2 sin^2 \theta }} \ rdr d\theta}}[/tex]
[tex]\mathbf{\implies \sigma_0\int^{\pi}_{0} \Big [ r \Big] ^7_2 \ d \theta}[/tex]
i.e.
[tex]\mathbf{M = \sigma_o \int^{\pi}_{0}\Big[ 7-2\Big] d \theta = 5 \pi \sigma _o}[/tex]
Now, we will also need to find the numerator;
i.e.
[tex]\mathbf{\implies \int \int_{R} \sigma xdA = \int ^{\pi}_{0} \int ^7_{2} \dfrac{\sigma_0}{\sqrt{r^2 cos^2 \theta + r^2 sin^2 \theta }} \ rcos \theta r \ dr \ d\theta}}[/tex]
[tex]\mathbf{=\sigma _o \int ^{\pi}_{0} \int ^7_2 \ rcos \theta \ dr d \theta}[/tex]
[tex]\mathbf{=\sigma _o \int ^{\pi}_{0} \Big [ \dfrac{r^2}{2}\Big]^7_2 \ rcos \theta \ d \theta = \sigma_o \Big[ sin \theta \Big]^{\pi}_{0} ( \dfrac{7^2-2^2}{2} )=0}[/tex]
[tex]\mathbf{\implies \int \int_{R} \sigma ydA = \int ^{\pi}_{0} \int ^7_{2} \dfrac{\sigma_0}{\sqrt{r^2 cos^2 \theta + r^2 sin^2 \theta }} \ rsin \theta r \ dr \ d\theta}}[/tex]
[tex]\mathbf{=\sigma _o \int ^{\pi}_{0} \int ^7_2 \ rsin \theta \ dr d \theta}[/tex]
[tex]\mathbf{=\sigma _o \int ^{\pi}_{0} \Big [ \dfrac{r^2}{2}\Big]^7_2 \ rsin \theta \ d \theta = \sigma_o \Big[ -cos \theta \Big]^{\pi}_{0} ( \dfrac{7^2-2^2}{2} )=45 \sigma _o}}[/tex]
Therefore, the center of the mass (m) is:
[tex]\mathbf{\bar x = \dfrac{0}{M} = 0 \ \ ; \ \ \bar y = \dfrac{45 \sigma_0}{5 \pi \sigma_o} = \dfrac{9}{\pi}}[/tex]
Learn more about the lamina of a plane here:
https://brainly.com/question/15556258?referrer=searchResults
