Center of Mass — Question 4

Question 4 – Solution

Step 1: Use symmetry. The lamina is symmetric about the $y$-axis, so $$\bar x=0.$$

Step 2: Find the area. The region is one-half of a disk of radius $R$: $$A=\frac12\pi R^2.$$

Step 3: Write the upper boundary. For $-R\le x\le R$, $$f(x)=\sqrt{R^2-x^2}.$$ A vertical strip of height $f(x)$ has centroid height $f(x)/2$.

Step 4: Find the moment about the $x$-axis. $$\begin{align*} M_x&=\frac12\int_{-R}^{R}f(x)^2\,dx =\frac12\int_{-R}^{R}(R^2-x^2)\,dx\\ &=\frac12\left[R^2x-\frac{x^3}{3}\right]_{-R}^{R} =\frac{2R^3}{3}. \end{align*}$$

Step 5: Divide the moment by the area. $$\bar y=\frac{M_x}{A} =\frac{\frac{2R^3}{3}}{\frac12\pi R^2} =\frac{4R}{3\pi}.$$ $$\boxed{(\bar x,\bar y)=\left(0,\frac{4R}{3\pi}\right)}$$