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.


Notice, your formula $Y_{CM} = \displaystyle{\int y\ dm}$ is not correct.

the center of mass is given as $$Y_{CM}=\frac{\displaystyle{\int y\ dm}}{\displaystyle{\int dm}}$$ Now, substituting the values $y=r\sin \theta$ & $dm=\sigma rdrd\theta$, we get

$$Y_{CM}=\frac{\displaystyle{\int_{0}^{R}\int_{0}^{\pi} \sigma r^2\sin\theta d\theta\ dr}}{\displaystyle{\int_{0}^{R}\int_{0}^{\pi}\sigma r d\theta\ dr}}$$

$$=\frac{\displaystyle{\int_{0}^{R}\left(\int_{0}^{\pi}\sin\theta d\theta\right)r^2\ dr}}{\displaystyle{\int_{0}^{R}\left(\int_{0}^{\pi}d\theta\right)r\ dr}}$$

$$ = \frac{\displaystyle{\int_{0}^{R}\left(2\right)r^2\ dr}}{\displaystyle{\int_{0}^{R}\left(\pi\right)r\ dr}} = \frac{\displaystyle{2\int_{0}^{R}r^2\ dr}}{\displaystyle{\pi\int_{0}^{R}r\ dr}}$$

$$ = \frac{2\left[\frac{r^3}{3}\right]_{0}^{R}}{\pi\left[\frac{r^2}{2}\right]_{0}^{R}}$$

$$ = \frac{4R^3}{3\pi R^2} = \frac{4R}{3\pi}$$

$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{Y_{CM}=\frac{4R}{3\pi}}}$$

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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 $d=\color{red}{\large\frac{4R}{3\pi}}$.

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  • $\begingroup$ What is the reason behind the downvote? It looks like a pretty straightforward solution to me. $\endgroup$ – Jack D'Aurizio Oct 18 '16 at 19:16
  • 6
    $\begingroup$ Please remember that answers are not designed for the OP only: they serve a community. So, if you do not understand a correct answer by your own limitations, or think it is "too advanced" by your personal standards, please do not punish through a downvote the good will to provide a contribute to MSE, but just ignore it. $\endgroup$ – Jack D'Aurizio Oct 18 '16 at 19:22
  • $\begingroup$ Can u please explain how u wrote 2\pi d \cdot \frac{\pi}{2}R^2 = \frac{4\pi}{3}R^3 $\endgroup$ – Riya Verma Nov 16 '19 at 11:56
  • $\begingroup$ @RiyaVerma: have you looked at the statement of Pappus' theorem? $\endgroup$ – Jack D'Aurizio Nov 16 '19 at 13:55
  • $\begingroup$ Correct me if Im wrong,are you trying to say semi disks rotated by 2πd create volume of sphere. If it is so, why we need d(com) in 2πd we can take 2πr. $\endgroup$ – Riya Verma Nov 16 '19 at 15:18

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