- #1
Emspak
- 243
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This might be more of a notational issue, I am working on this problem and just trying to figure out what my prof/ TA did in the solution sheet -- but also to make sure I understand what I am doing.
Given a spherical shell of radius R and surface density σ, and a particle at position a from the center where 0<a<R. Assume the position a is along the z (vertical) axis.
What is the force on the particle from the smaller hemisphere, described by the part that extends from the top (at z=R) down to where the particle is (at z=a) and where is the center of mass of that hemisphere?
I am using θ as the angle between the origin and the point on the shell that intersects a line drawn through a that is perpendicular to the z axis. (I wish I knew how to post a diagram here, that would make it easier I think, but I hope people get the idea).
Repeat for the larger part of the shell.
Now, I know intuitively (and from the symmetry principle) that the rings, each with a mass of dM = 2πRsinθRdθ will exert a net zero force in the x-y plane as we are assuming symmetry here.
So, I only need the force in the z-direction. (By the way I do understand that with a shell you get zero net force and I know how to do that one, this is a bit different).
Anyhow, I want to know the total mass of the smaller part of the shell. That's going to be an integral from 0 to θ. Since the radius of the strip is R cosθ, and the force vector is along the z-axis, I should get this:
[tex]
\int dF_z = \int_0^{\theta} \frac{Gm2\pi R \sin \theta R d \theta}{r^2}\cos\theta [/tex]
where r is the distance from my particle at a to the piece of the shell I want. So far so good.
Now, the distance r is going to be determined by the law of cosines, if I want to get it in terms of θ. So I can say [itex]r^2 = a^2 + R^2 - 2aR \cos \theta[/itex].
I'm also multiplying stuff by the cosθ which in tems of R and a is [itex]\frac{R^2+a^2-r^2}{2aR}[/itex]
If I differentiate [itex]r^2[/itex] -- because I am setting up an integral here -- I get [itex]2rdr = 2aR \sin \theta d \theta[/itex]. A little algebra gets me [itex] \sin \theta d \theta = \frac{2rdr}{2aR} = \frac{rdr}{aR}[/itex] and I should be able to plug that back into the integral and get:
[tex]
\int dF_z = \sigma Gm2\pi R^2 \int_0^{\theta} \frac{rdr}{aR} \frac{1}{r^2} \frac{R^2+a^2-r^2}{2aR} [/tex]
and we have to change the limits, which here would be arccos(a/R) so I have:
[tex]
\int dF_z = \sigma Gm2\pi R^2 \int_0^{\cos^{-1} \frac{a}{R}} \frac{rdr}{aR} \frac{1}{r^2} \frac{R^2+a^2-r^2}{2aR} = \sigma Gm\pi \int_0^{\cos^{-1} \frac{a}{R}} \frac{rdr}{a^2} \frac{1}{r^2} (R^2+a^2-r^2)[/tex]
[tex]= \sigma Gm\pi \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{a^2} \frac{1}{r} dr = \frac{\sigma Gm\pi}{a^2} \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{r} dr[/tex] and because [itex] \sigma = \frac{M}{4 \pi R^2}[/itex] the integral looks like:
[tex]= \frac{M Gm}{4R^2a^2} \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{r} dr[/tex]
So my question is, how am I doing so far? I feel like there is something I am missing -- because the answer sheet says that I should get [tex]\frac{M Gm}{4Ra^2}2(\sqrt{R^2-a^2}-2R[/tex]
and I am not entirely sure if I got the final integral right. I don't think I did...
thanks in advance.
Homework Statement
Given a spherical shell of radius R and surface density σ, and a particle at position a from the center where 0<a<R. Assume the position a is along the z (vertical) axis.
What is the force on the particle from the smaller hemisphere, described by the part that extends from the top (at z=R) down to where the particle is (at z=a) and where is the center of mass of that hemisphere?
I am using θ as the angle between the origin and the point on the shell that intersects a line drawn through a that is perpendicular to the z axis. (I wish I knew how to post a diagram here, that would make it easier I think, but I hope people get the idea).
Repeat for the larger part of the shell.
The Attempt at a Solution
Now, I know intuitively (and from the symmetry principle) that the rings, each with a mass of dM = 2πRsinθRdθ will exert a net zero force in the x-y plane as we are assuming symmetry here.
So, I only need the force in the z-direction. (By the way I do understand that with a shell you get zero net force and I know how to do that one, this is a bit different).
Anyhow, I want to know the total mass of the smaller part of the shell. That's going to be an integral from 0 to θ. Since the radius of the strip is R cosθ, and the force vector is along the z-axis, I should get this:
[tex]
\int dF_z = \int_0^{\theta} \frac{Gm2\pi R \sin \theta R d \theta}{r^2}\cos\theta [/tex]
where r is the distance from my particle at a to the piece of the shell I want. So far so good.
Now, the distance r is going to be determined by the law of cosines, if I want to get it in terms of θ. So I can say [itex]r^2 = a^2 + R^2 - 2aR \cos \theta[/itex].
I'm also multiplying stuff by the cosθ which in tems of R and a is [itex]\frac{R^2+a^2-r^2}{2aR}[/itex]
If I differentiate [itex]r^2[/itex] -- because I am setting up an integral here -- I get [itex]2rdr = 2aR \sin \theta d \theta[/itex]. A little algebra gets me [itex] \sin \theta d \theta = \frac{2rdr}{2aR} = \frac{rdr}{aR}[/itex] and I should be able to plug that back into the integral and get:
[tex]
\int dF_z = \sigma Gm2\pi R^2 \int_0^{\theta} \frac{rdr}{aR} \frac{1}{r^2} \frac{R^2+a^2-r^2}{2aR} [/tex]
and we have to change the limits, which here would be arccos(a/R) so I have:
[tex]
\int dF_z = \sigma Gm2\pi R^2 \int_0^{\cos^{-1} \frac{a}{R}} \frac{rdr}{aR} \frac{1}{r^2} \frac{R^2+a^2-r^2}{2aR} = \sigma Gm\pi \int_0^{\cos^{-1} \frac{a}{R}} \frac{rdr}{a^2} \frac{1}{r^2} (R^2+a^2-r^2)[/tex]
[tex]= \sigma Gm\pi \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{a^2} \frac{1}{r} dr = \frac{\sigma Gm\pi}{a^2} \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{r} dr[/tex] and because [itex] \sigma = \frac{M}{4 \pi R^2}[/itex] the integral looks like:
[tex]= \frac{M Gm}{4R^2a^2} \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{r} dr[/tex]
So my question is, how am I doing so far? I feel like there is something I am missing -- because the answer sheet says that I should get [tex]\frac{M Gm}{4Ra^2}2(\sqrt{R^2-a^2}-2R[/tex]
and I am not entirely sure if I got the final integral right. I don't think I did...
thanks in advance.