Force from a Hemispherical shell

[Emspak]

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.

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:

<br /> \int dF_z = \int_0^{\theta} \frac{Gm2\pi R \sin \theta R d \theta}{r^2}\cos\theta

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 r^2 = a^2 + R^2 - 2aR \cos \theta.

I'm also multiplying stuff by the cosθ which in tems of R and a is \frac{R^2+a^2-r^2}{2aR}

If I differentiate r^2 -- because I am setting up an integral here -- I get 2rdr = 2aR \sin \theta d \theta. A little algebra gets me \sin \theta d \theta = \frac{2rdr}{2aR} = \frac{rdr}{aR} and I should be able to plug that back into the integral and get:

<br /> \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}

and we have to change the limits, which here would be arccos(a/R) so I have:

<br /> \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)
= \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 and because \sigma = \frac{M}{4 \pi R^2} the integral looks like:

= \frac{M Gm}{4R^2a^2} \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{r} dr

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 \frac{M Gm}{4Ra^2}2(\sqrt{R^2-a^2}-2R

and I am not entirely sure if I got the final integral right. I don't think I did...

thanks in advance.

 

[haruspex]

The error is a confusion over the angle theta.
Let the centre of the full sphere be O and the particle be at P. Consider a point Q on the band of the sphere distance r from P. According to the equation $r^2=R^2+a^2-2aR\cos(\theta)$, theta is the angle POQ. The force from the element at Q exerted at P is in the direction PQ, not in the direct OQ. Thus, to get the component along the line OP we need to multiply by the cosine of pi - angle QPO, not the cosine of angle theta. That is, the cos(theta) factor needs to be changed to $\frac{R\cos(\theta)-a}r$.

There are also some errors in the limits. In the first integral, theta is both the variable of integration and the upper limit. That's a common 'pun' in indefinite integrals, but here we want a definite integral. The upper limit should be $\arccos(a/R)$. Later that limit is shown, but now the variable of integration is r, so the limits should change to reflect that.