Find the center of mass of a uniform semicircular plate of radius R

[annamal]

Homework Statement

Find the center of mass of a uniform this semicircular plate of radius R. Let the origin be at the center of the semicircle, the plate arc from the +x axis to the -x axis, and the z axis be perpendicular to the plate.

Relevant Equations

$rcm = \frac 1 M*\int(r*dm)$

$$rcm = \frac{1}{M}\int_0^\pi(rdm)$$
$$dm = \sigma{dA}$$
$$dA = (\pi*R^2)*\frac{d\theta}{2\pi}$$
$$\sigma = \frac{M}{\frac{\pi*R^2}{2}}$$
$$dm = M*\frac{d\theta}{\pi}$$
$$r = R(cos(\theta)\vec i + sin(\theta)\vec j)$$
$$rcm = \int_0^\pi{\frac{R}{\pi}(cos(\theta)\vec i + sin(\theta)\vec j)} = \frac{2*R}{\pi}\vec j$$

Where did I go wrong?

 

[haruspex]

Relevant Equations:: $rcm = \frac 1 M*\int(r*dm)$
Where did I go wrong?

Right there. It is not $\int r.dm$.

 

[annamal]

Right there. It is not $\int r.dm$.

I don't understand what you're saying. That formula is correct. (Search integral)
Source: https://scientificsentence.net/Equations/Mechanics/collisions/index.php?key=yes&Integer=center_mass

 

[haruspex]

I don't understand what you're saying. That formula is correct. (Search integral)
Source: https://scientificsentence.net/Equations/Mechanics/collisions/index.php?key=yes&Integer=center_mass

Read the text there carefully. The r in that formula is a position vector, (x, y, z). So the equation becomes $(x_{cm}, y_{cm}, z_{cm})=\frac 1M\int(x,y,z).dm$.
That works because all the x are parallel. The r in your usage are not all parallel.

 

[annamal]

Read the text there carefully. The r in that formula is a position vector, (x, y, z). So the equation becomes $(x_{cm}, y_{cm}, z_{cm})=\frac 1M\int(x,y,z).dm$.
That works because all the x are parallel. The r in your usage are not all parallel.

Yes, so $\vec r = R(cos(\theta)\vec i + sin(\theta)\vec j)$
I meant $\vec r_{cm} = \frac 1 M*\int(\vec r*dm)$
What's the problem?

 

[haruspex]

Yes, so $\vec r = R(cos(\theta)\vec i + sin(\theta)\vec j)$

That's only true of a point on the circumference, so you have found the mass centre of a semicircular arc.

 

[annamal]

That's only true of a point on the circumference, so you have found the mass centre of a semicircular arc.

Ah, I see. This doesn't work either

$$\sigma = \frac{2M}{\pi*R^2}$$
$$dM = \sigma dA$$
$$dA = \pi ydy$$
$$y_{cm} = \frac{1}{M}\int ydM = \frac{\sigma*\pi}{M}\int{y^2}{dy} = 2MR/3$$