Question

determine the coordinates of the centroid of the plane area with figure with reference to the axis as shown​

Answer 1

The coordinates of center of mass of,

• square plane of side 2a is (a, a), where origin is at one of its corners and axes are along its sides.

• semicircular disc of radius r is (0, 4r/3π), where origin is at center of its curvature and axes are along its base diameter.

• quarter circular disc of radius r is (4r/3π, 4r/3π) where origin is at the vertex and axes are along its sides.

Let M be the mass of the whole rectangular plane without any voids, whose position of center of mass is $\sf{\left<7,\ 6\right>.}$

Note that $\sf{\left<x,\ y\right>=x\,\hat i+y\,\hat j.}$

Consider the square void of mass $\sf{m_1}$ whose position of center of mass is $\sf{\left<0,\ 8\right>+\left<2,\ 2\right>=\left<2,\ 10\right>.}$

As areal density is same,

$\longrightarrow\sf{\dfrac{m_1}{4\times4}=\sf{\dfrac{M}{14\times12}}$

$\longrightarrow\sf{m_1=\sf{\dfrac{4M}{42}}$

Consider the semicircular void of mass $\sf{m_2}$ whose position of center of mass is $\sf{\left<6,\ 0\right>+\left<0,\ \dfrac{16}{3\pi}\right>=\left<6,\ \dfrac{16}{3\pi}\right>}.$

As areal density is same,

$\longrightarrow\sf{\dfrac{m_2}{8\pi}=\sf{\dfrac{M}{14\times12}}$

$\longrightarrow\sf{m_2=\sf{\dfrac{2M\pi}{42}}$

Consider the quarter circular void of mass $\sf{m_3}$ whose position of center of mass is $\sf{\left<14,\ 12\right>+\left<-\dfrac{16}{3\pi},\ -\dfrac{16}{3\pi}\right>=\left<14-\dfrac{16}{3\pi},\ 12-\dfrac{16}{3\pi}\right>.}$

As areal density is same,

$\sf{\longrightarrow \dfrac{m_3}{4\pi}=\dfrac{M}{14\times12}}$

$\sf{\longrightarrow m_3=\dfrac{M\pi}{42}}$

Now the position of center of mass of our plane area is,

$\sf{\longrightarrow\bar r=\dfrac{M\left<7,\ 6\right>-m_1\left<2,\ 10\right>-m_2\left<6,\ \dfrac{16}{3\pi}\right>-m_3\left<14-\dfrac{16}{3\pi},\ 12-\dfrac{16}{3\pi}\right>}{M-m_1-m_2-m_3}}$

$\small\textsf{ \sf{\longrightarrow\bar r=\dfrac{M\left<7,\ 6\right>-\dfrac{2M}{21}\left<2,\ 10\right>-\dfrac{M\pi}{21}\left<6,\ \dfrac{16}{3\pi}\right>-\dfrac{M\pi}{42}\left<14-\dfrac{16}{3\pi},\ 12-\dfrac{16}{3\pi}\right>}{M-\dfrac{2M}{21}-\dfrac{M\pi}{21}-\dfrac{M\pi}{42}}} }$

$\small\textsf{ \sf{\longrightarrow\bar r=\dfrac{\left<7,\ 6\right>-\dfrac{2}{21}\left<2,\ 10\right>-\dfrac{\pi}{21}\left<6,\ \dfrac{16}{3\pi}\right>-\dfrac{\pi}{42}\left<14-\dfrac{16}{3\pi},\ 12-\dfrac{16}{3\pi}\right>}{1-\dfrac{3\pi+4}{42}}} }$

$\small\textsf{ \sf{\longrightarrow\bar r=\dfrac{\left<7,\ 6\right>-\left<\dfrac{4}{21},\ \dfrac{20}{21}\right>-\left<\dfrac{2\pi}{7},\ \dfrac{16}{63}\right>-\left<\dfrac{\pi}{3}-\dfrac{8}{63},\ \dfrac{2\pi}{7}-\dfrac{8}{63}\right>}{\dfrac{38-3\pi}{42}}} }$

$\sf{\longrightarrow\bar r=\dfrac{\left<7-\dfrac{4}{21}-\dfrac{2\pi}{7}-\dfrac{\pi}{3}+\dfrac{8}{63},\ 6-\dfrac{20}{21}-\dfrac{16}{63}-\dfrac{2\pi}{7}+\dfrac{8}{63}\right>}{\dfrac{38-3\pi}{42}}}$

$\sf{\longrightarrow\underline{\underline{\bar r=\dfrac{2}{114-9\pi}\big<437-13\pi,\ 310-18\pi\big>}}}$

This is the coordinates of the centroid or center of mass of our plane area.