Find M_x,M_y , and ($\bar{x}$ ,$\bar {y}$) for the lamina of uniform density $\rho$ text bounded by the graphs A) $\begin{array} { l } M _ { x } = \frac { 1,024 } { 5 } , M _ { y } = 0 \ \bar { x } = \frac { 112 } { 13 } , \bar { y } = \frac { 7 } { 3 } \end{array}$ B) $\begin{array} { l } M _ { x } = \frac { 128 } { 5 } \rho\ , M _ { y } = \frac { 8,192 } { 11 } \rho \ \bar { x } = \frac { 112 } { 13 } , \bar { y } = \frac { 14 } { 5 } \end{array}$ C) $\begin{array} { l } M _ { x } = \frac { 1,024 } { 5 } \rho , M _ { y } = \frac { 8,192 } { 11 } \rho , \ \bar { x } = \frac { 112 } { 11 } , \bar { y } = \frac { 14 } { 5 } \end{array}$ D) $\begin{array} { l } M _ { x } = \frac { 512 } { 5 } \rho\ , M _ { y } = \frac { 8,192 } { 11 } \rho \ \bar { x } = \frac { 112 } { 11 } , \bar { y } = \frac { 7 } { 3 } \end{array}$ E) $ M_{x}=0, M_{y}=\frac{8,192}{11} \rho $ $ \bar{x}=\frac{112}{9}, \bar{y}=\frac{56}{3} $

Question

Quizplus · Accepted Answer

We know that the moments are given by the double integrals:
\begin{align*}
M_{x} &= \iint x \, dA \
M_{y} &= \iint y \, dA
\end{align*}
And the coordinates of the center of mass are given by
\begin{align*}
\bar{x} &= \frac{1}{M}\iint x \, dA\
\bar{y} &= \frac{1}{M}\iint y \, dA
\end{align*}
where $M$ is the mass of the lamina, given by
\begin{align*}
M &= \iint ho \, dA
\end{align*}
where $ho$ is the density of the lamina.

A) We have $M_{x}=\frac{1,024}{5}$ and $M_{y}=0$, which means that the centroid of the lamina lies on the $x$-axis. However, since the $y$ moment is zero, the lamina must be symmetric about the $x$-axis. But this is not possible since the $x$ moment is nonzero. Therefore, this choice is not possible.

B) We have $M_{x}=\frac{128}{5}ho$ and $M_{y}=\frac{8192}{11}ho$, which means that the lamina is not symmetric about either the $x$-axis or the $y$-axis. The centroid lies at $\left(\frac{112}{13},\frac{14}{5}ight)$, but we do not know the value of $ho$. Therefore, this choice is not possible.

C) We have $M_{x}=\frac{1,024}{5}ho$ and $M_{y}=\frac{8192}{11}ho$. The coordinates of the centroid are given as $\left(\frac{112}{11},\frac{14}{5}ight)$. Substituting the given values in the equation $M=\iint ho \, dA$, we get
\begin{align*}
\frac{1,024}{5}ho &= \iint x \, ho \, dA = ho \iint x \, dA = ho \cdot M_{x}\
\frac{8,192}{11}ho &= \iint y \, ho \, dA = ho \iint y \, dA = ho \cdot M_{y}
\end{align*}
Solving for $ho$ in both equations and setting them equal, we get
\begin{align*}
\frac{1,024}{5}\cdot \frac{11}{8,192} = \frac{112}{11} \cdot \frac{14}{5} \
\implies ho = \frac{55}{104}
\end{align*}
Therefore, the answer is choice C.

D) We have $M_{x}=\frac{512}{5}ho$ and $M_{y}=\frac{8192}{11}ho$. The coordinates of the centroid are given as $\left(\frac{112}{11},\frac{7}{3}ight)$. Substituting the given values in the equation $M=\iint ho \, dA$, we get
\begin{align*}
\frac{512}{5}ho &= \iint x \, ho \, dA = ho \iint x \, dA = ho \cdot M_{x}\
\frac{8,192}{11}ho &= \iint y \, ho \, dA = ho \iint y \, dA = ho \cdot M_{y}
\end{align*}
Solving for $ho$ in both equations and setting them equal, we get
\begin{align*}
\frac{512}{5}\cdot \frac{11}{8,192} = \frac{112}{11} \cdot \frac73 \
\implies ho = \frac{165}{2048}
\end{align*}
Therefore, this choice is not possible.

E) We have $M_{x}=0$ and $M_{y}=\frac{8192}{11}ho$, which means that the centroid of the lamina lies on the $x$-axis. However, the $x$ moment is zero, which means that the centroid lies on the $y$-axis. This is a possible scenario, but the coordinates of the centroid are given as $\left(\frac{112}{9},\frac{56}{3}ight)$, which is not consistent with the moment equation $M_{y}=\frac{8192}{11}ho$. Therefore, this choice is not possible.

Find Mx,My , and xˉ\bar{x}xˉ yˉ\bar {y}yˉ​ ) for the Lamina of Uniform Density

Find M_x,M_y , and $\bar{x}$ $\bar {y}$ ) for the Lamina of Uniform Density