Finding the Center of Mass of a Uniform Semicircular Disk Using Integration

[smeagol]

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)?

 

[NateTG]

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]

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

 

[HallsofIvy]

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 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)r²cosθ 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)r²sinθ 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.

 

[smeagol]

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]

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]

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

 

[HallsofIvy]

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

 

[smeagol]

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