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Center of Massby smeagol
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#1
Dec1703, 09:14 PM

P: 6

Show that the center of mass of a uniform semicircular disk of radius R is at a point (4/(3*Pi))R from the center of the circle.
well I know I am suppose to find this by integration. By this equation [tex]M\vec{r_{cm}}=\int\vec{rdm}[/tex] But, I am not sure how to find dm in this case... do I divide mass by area? or by circumference? and since it's a disk, and it was 2 variables, would I have to integrate for 2 variables? (x and y)? 


#2
Dec1703, 09:27 PM

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In general, CM calculations end up being double or triple integrals, but this one is pretty straightforward since the disk has constant density.
I'd be inclined to cut the halfdisk into parralel strips and integrate along the CM of each strip. Since the density is uniform, you might as well assume that it's one. 


#3
Dec1703, 09:45 PM

P: 6

so I divide M by [tex]2R\sin{\theta}[/tex]?



#4
Dec1803, 05:30 AM

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Center of Mass
Since you are talking about a semicircular disk, I would be inclined to use polar (actually cylindrical coordinates).
The differential of area in polar coordinates is r dr dθ so the differential of volume in cylindrical coordinates is r dr dθ dz. Taking the density to be ρ(r,θ,z), dm= ρ(r,θz) dr dθ dz. The mass, for semicircular disk of radius R and thickness h would be M= ∫(z=0 to h)∫(θ=0 to π)∫(r= 0 to R) ρ(r,θ,z)r dr dθ dz. If ρ is a constant, this is just ρπr^{2}h. Since x= r cosθ, the formula for the x_{cm} would be Mx_{cm}= ∫(z=0 to h)∫(&theta= 0 to π)∫(r= 0 to R)(r cos θ)(ρ(r,θ,z)r dr dθ dz)= ∫(z=0 to h)∫(&theta= 0 to π)∫(r= 0 to R)(ρ(r,θ,z)r^{2}cosθ dz dθ dz. Since y= r sinθ, the formula for y_{cm} would be My_{cm}= ∫(z=0 to h)∫(θ= 0 to π)∫(r= 0 to R)(&rho(r,θ,z)r^{2}sinθ dz dθ dz. The formula for z_{cm} would be Mz_{cm}= ∫(z=0 to h)∫(θ= 0 to π)∫(r= 0 to R)(&rho(r,θ,z)zr dr dθ dz. Of course, if ρ is a constant, from symmetry (cosine is an even function) x_{cm}= 0 and z_{cm}= h/2. 


#5
Dec2003, 02:47 PM

P: 6

Ok, I got that one.
Moving on, the next problem is confusing me as well. A baseball bat of length L has a peculiar linear density (mass per unit length) given by [tex]\lambda=\lambda_0(1+x^2/L^2)[/tex] so what I've done is [tex]\int_{0}^{L}x\lambda_0(1+x^2/L^2)dx[/tex] which gives [tex]\frac{3\lambda_0L^2}{4}[/tex] so I know that M * x_cm = that but how do I find M so I can find x_cm? 


#6
Dec2003, 08:57 PM

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[tex]M=\int_0^L\lambda dx[/tex] 


#7
Dec2103, 12:28 PM

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krab: Watch it now! You're starting to sound like me!



#8
Dec2103, 12:30 PM

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By the way, why were these posted under the "classical physics" forum? The look like fairly standard calculus homework problems.



#9
Dec2103, 03:36 PM

P: 6

well I got it from a "physics for scientists and engineers" from tipler. So I figured it would fit better under physics.



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