calculusintegration
Problem:
Derive the Center of Mass of a semi-circular disk of mass $M$ and radius $R$.
My attempt:
$$Y_{CM}=\int ydm$$
Now, $$dm=\sigma dA$$ where $\sigma$ is mass per unit area.
Converting into Cylindrical Coordinates,$$dA=rdrd\theta$$. Also, $$y=r\sin\theta$$
Hence the integral can be rewrittenn as
$$\int_0^R\int_0^{\pi}r^2\sin\theta d\theta dr$$
However this Integral gives me the wrong value of the Y coordinate of the Center of Mass.
I would be truly grateful for any help with this problem.
To integrate this region in polar coordinates, it is advisable to break up the integral into two parts, as shown in the figures below:
The two parts of the integral are divided by the diagonal line through the upper right corner of the rectangle. Since the sides of the rectangle are $a$ and $b$, this diagonal line is at the angle $\arctan \frac ba.$
For $0 \leq \theta \leq \arctan \frac ba,$ you would integrate over $0 \leq r \leq a \sec\theta,$ and for $\arctan \frac ba \leq \theta \leq \frac\pi2,$ you would integrate over $0 \leq r \leq b \csc\theta.$
If you actually try this, I think you'll find that it is no easier than doing the integration in rectangular coordinates. It may even be worse.
An alternative approach, rather than combining $x^2+y^2$ into $r^2$, is to integrate the terms separately:
$$\begin{eqnarray} m &=& \int_0^a\int_0^b (1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b dy\,dx +\int_0^a\int_0^b x^2 \,dy\,dx +\int_0^a\int_0^b y^2 \,dy\,dx \end{eqnarray}$$ $$\begin{eqnarray} m\bar{x} &=& \int_0^a\int_0^b x(1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b x \,dy\,dx +\int_0^a\int_0^b x^3 \,dy\,dx +\int_0^a\int_0^b xy^2 \,dy\,dx \end{eqnarray}$$ $$\begin{eqnarray} m\bar{y} &=& \int_0^a\int_0^b y(1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b y \,dy\,dx +\int_0^a\int_0^b x^2y \,dy\,dx +\int_0^a\int_0^b y^3 \,dy\,dx \end{eqnarray}$$
Now you have nine integrals to solve, but they're all quite simple.
$$\bar{\cos{\theta}} = \int dr \, r^2 f(r) \, \int d\phi \, \sin{\phi} \, \int d\theta \, \cos{\theta}$$ $$\bar{\sin{\phi}} = \int dr \, r^3 f(r) \, \int d\phi \, \sin^2{\phi} \, \int d\theta \, $$ $$\bar{r} = \int dr \, r^3 f(r) \, \int d\phi \, \sin{\phi} \, \int d\theta \, $$
Then
$$\begin{align}\bar{r} \,\bar{\sin{\phi}}\,\bar{\cos{\theta}} &= \frac{\int dr \, r^3 f(r)\left (\int dr \, r^3 f(r) \right )^2 \int d\phi \, \sin^2{\phi} \left ( \int d\phi \, \sin{\phi} \right )^2\int d\theta \, \cos{\theta} \left ( \int d\theta \right )^2}{\left (\int dr \, r^2 f(r) \right )^3 \left ( \int d\phi \, \sin{\phi} \right )^3 \left ( \int d\theta \right )^3}\\ &= \frac{\int dr \, r^3 f(r) \, \int d\phi \, \sin^2{\phi} \,\, \int d\theta \, \cos{\theta} }{\int dr \, r^2 f(r)\, \int d\phi \, \sin{\phi}\, \int d\theta }\\ &= \bar{x}\end{align}$$
But I do not think you can work with the bare $\phi$ and $\theta$ because the trig functions introduce a nonlinearity that belies the linearity of the integrals.
Best Answer
The center of mass of a uniform half-disk obviously lies on the perpendicular bisector of the base diameter, at distance $d$ from the centre of the disk. By the Pappus centroid theorem,
$$ 2\pi d \cdot \frac{\pi}{2}R^2 = \frac{4\pi}{3}R^3, $$ hence $d=\color{red}{\large\frac{4R}{3\pi}}$.