Center of Mass — Question 9

Question 9

A lamina is bounded by $y=\sqrt{x}$, $y=0$, $x=4$. A circular hole of radius (1/2), centered at ((3,1/2)), is removed. Find the new centroid using subtraction.

Question 9 – Solution

Step 1: Find the area of the original region. $$A_0=\int_0^4\sqrt{x}\,dx =\left[\frac23x^{3/2}\right]_0^4=\frac{16}{3}.$$

Step 2: Find the original moments. $$\begin{align*} M_{y,0}&=\int_0^4x\sqrt{x}\,dx =\left[\frac25x^{5/2}\right]_0^4=\frac{64}{5},\\ M_{x,0}&=\frac12\int_0^4(\sqrt{x})^2\,dx =\frac12\int_0^4x\,dx=4. \end{align*}$$ Hence the original centroid is $$(\bar x_0,\bar y_0) =\left(\frac{M_{y,0}}{A_0},\frac{M_{x,0}}{A_0}\right) =\left(\frac{12}{5},\frac34\right).$$

Step 3: Record the hole’s area and moments. The hole has radius $1/2$ and centroid $(3,1/2)$, so $$A_h=\pi\left(\frac12\right)^2=\frac\pi4,\quad M_{y,h}=3\left(\frac\pi4\right)=\frac{3\pi}{4},\quad M_{x,h}=\frac12\left(\frac\pi4\right)=\frac\pi8.$$

Step 4: Subtract the hole. $$A=\frac{16}{3}-\frac\pi4,\quad M_y=\frac{64}{5}-\frac{3\pi}{4},\quad M_x=4-\frac\pi8.$$

Step 5: Divide the remaining moments by the remaining area. $$\boxed{\bar x=\frac{\frac{64}{5}-\frac{3\pi}{4}}{\frac{16}{3}-\frac\pi4},\qquad \bar y=\frac{4-\frac\pi8}{\frac{16}{3}-\frac\pi4}}$$