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Center of Mass

by smeagol
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smeagol
#1
Dec17-03, 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)?
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NateTG
#2
Dec17-03, 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 half-disk 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.
smeagol
#3
Dec17-03, 09:45 PM
P: 6
so I divide M by [tex]2R\sin{\theta}[/tex]?

HallsofIvy
#4
Dec18-03, 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 ρπr2h.

Since x= r cosθ, the formula for the xcm would be Mxcm= ∫(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)r2cosθ dz dθ dz.

Since y= r sinθ, the formula for ycm would be Mycm= ∫(z=0 to h)∫(θ= 0 to π)∫(r= 0 to R)(&rho(r,θ,z)r2sinθ dz dθ dz.

The formula for zcm would be Mzcm= ∫(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) xcm= 0 and zcm= h/2.
smeagol
#5
Dec20-03, 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?
krab
#6
Dec20-03, 08:57 PM
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Originally posted by smeagol
...but how do I find M so I can find x_cm?
C'mon. You know the mass per unit length. How can you not know the mass?

[tex]M=\int_0^L\lambda dx[/tex]
HallsofIvy
#7
Dec21-03, 12:28 PM
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krab: Watch it now! You're starting to sound like me!
HallsofIvy
#8
Dec21-03, 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.
smeagol
#9
Dec21-03, 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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