Force on Particle due to Sphere with Cylindrical Hole

• Leyic
In summary, the conversation discusses a problem involving determining the force felt by a test particle placed on the axis of a cylindrical hole in a homogeneous sphere of radius R_s made of a material that exerts a force obeying the inverse r-squared law. The conversation also includes a discussion of the solution attempt using cylindrical shells and integrals, as well as a suggestion for a simpler approach using the superposition principle.
Leyic
Note: This is my first time using LaTeX. Any formatting advice would be appreciated.

Homework Statement

Consider a homogeneous sphere of radius R_s made of a material that exerts a force obeying the inverse r-squared law (i.e. a particle of this material exerts a force on a test particle $$\vec{F}=\frac{K}{r^2}\hat{r}$$, where the vector r points from the particle to the test particle and K is a real number such that positive values are repulsive and negative values attractive). Next, consider a cylindrical hole of radius R_d that is removed from the sphere such that the axis of the cylinder passes through the center of the sphere. The hole passes completely through the sphere. Finally, consider a test particle arbitrarily placed on the axis of the cylinder a distance R from the center of the sphere. Determine the force felt by the test particle due to this configuration.

The Attempt at a Solution

Due to the cylindrical hole, I decided to construct the sphere by integrating cylindrical shells. I defined the x-axis to be the axis of the cylinder and the y-axis to be orthogonal to the x-axis. Considering two particles equidistant from the test particle and the x-axis, I found the force on the test particle to be $$\vec{F} = \frac{2 K r_{x}}{r^3}\hat{x}$$, with $$r=\sqrt{r_{x}^2+r_{y}^2}$$ and r_x and r_y being the x- and y-components of the distance from the particles to the test particle, respectively.

Next, I revolved this about the x-axis to produce a ring of material of radius r_y. Now, $$\vec{F}=\frac{2 K r_{x}}{r^3}\hat{x} \pi r_{y} n$$, where $$n$$ is a number density term. I then note that the distance from the center of the sphere to the center of the ring is $$r_{d}=\sqrt{R_{s}^2+r_{y}^2}$$ and that r_x = R - r_d. I can then rewrite the force caused by the ring as $$\vec{F}=\frac{2 \pi K n (R-r_{d})\sqrt{R_{s}^2-r_{d}^2}}{(R^2+R_{s}^2-2 r_{d} R)^{3/2}}\hat{x}$$.

Next, I integrate over the length of a cylinder such that rings at the ends coincide with the surface of the sphere. Noting that r_d is the distance from the center of the sphere to a ring, I can show that the length of a cylinder of a given radius will be 2*r_d. The force is now $$\vec{F}=\int_{-r_{d}}^{r_{d}}\frac{2 \pi K n (R-r_{d}\prime)\sqrt{R_{s}^2-r_{d}^2}}{(R^2+R_{s}^2-2 r_{d}\prime R)^{3/2}}\hat{x}dr_{d}\prime$$, noting that the r_d in $$\sqrt{R_{s}^2-r_{d}^2}$$ should not be primed as this represents the radius and is constant for a given cylinder. Also, n is adjusted accordingly. Putting this into MATLAB, I get the solution $$\vec{F}=2 \pi K n \hat{x} \frac{\sqrt{R_{s}^2-r_{d}^2}}{R^2}(\frac{r_{d} R-R_{s}^2}{\sqrt{R^2+R_{s}^2-2 r_{d} R}}+\frac{r_{d} R+R_{s}^2}{\sqrt{R^2+R_{s}^2+2 r_{d} R}})$$.

Finally, I integrate over all cylinders in the sphere by integrating the cylinder half-length r_d from 0 to $$\sqrt{R_{s}^2-R_{d}^2}$$. This results in the following equation:
$$\vec{F}=2 \pi K n \hat{x} R^{-2}(R \int_{0}^{\sqrt{R_{s}^2-R_{d}^2}} \frac{\sqrt{R_{s}^2-r_{d}^2} r_{d} dr_{d}}{\sqrt{R^2+R_{s}^2-2 r_{d} R}} - R_{s}^2 \int_{0}^{\sqrt{R_{s}^2-R_{d}^2}} \frac{\sqrt{R_{s}^2-r_{d}^2} dr_{d}}{\sqrt{R^2+R_{s}^2-2 r_{d} R}} + R \int_{0}^{\sqrt{R_{s}^2-R_{d}^2}} \frac{\sqrt{R_{s}^2-r_{d}^2} r_{d} dr_{d}}{\sqrt{R^2+R_{s}^2+2 r_{d} R}} + R_{s}^2 \int_{0}^{\sqrt{R_{s}^2-R_{d}^2}} \frac{\sqrt{R_{s}^2-r_{d}^2} dr_{d}}{\sqrt{R^2+R_{s}^2+2 r_{d} R}})$$.

To simplify this equation, I multiply the integrands by $$\sqrt{\frac{1/2R}{1/2R}}$$, and then let r_d = x, R_s = a, and (R^2+R_s^2)/2R = b. I then find that the integrals are of the following forms:
$$1. \int_{A}^{B} \frac{\sqrt{a^2-x^2} x dx}{\sqrt{b-x}} \ \ \ 2. \int_{A}^{B} \frac{\sqrt{a^2-x^2} dx}{\sqrt{b-x}} \ \ \ 3. \int_{A}^{B} \frac{\sqrt{a^2-x^2} x dx}{\sqrt{b+x}} \ \ \ 4. \int_{A}^{B} \frac{\sqrt{a^2-x^2} dx}{\sqrt{b+x}} \ \ \$$

I have not found any solutions to these integrals either by using MATLAB or by looking up integral tables. Nor can I see any other way of constructing the sphere that would properly account for the hole without further complicating the problem. Any advice on how I might move forward would be appreciated.

Last edited:
Leyic said:
Due to the cylindrical hole, I decided to construct the sphere by integrating cylindrical shells. I defined the x-axis to be the axis of the cylinder and the y-axis to be orthogonal to the x-axis. Considering two particles equidistant from the test particle and the x-axis, I found the force on the test particle to be $$\vec{F} = \frac{2 K r_{x}}{r^3}\hat{x}$$, with $$r=\sqrt{r_{x}^2+r_{y}^2}$$ and r_x and r_y being the x- and y-components of the distance from the particles to the test particle, respectively.

Next, I revolved this about the x-axis to produce a ring of material of radius r_y. Now, $$\vec{F}=\frac{2 K r_{x}}{r^3}\hat{x} \pi r_{y} n$$, where $$n$$ is a number density term.

Yikes!

There is a much easier way to approach this problem. Any force that obeys an inverse square law, also obeys the superposition principle. That means that is you add the forces due to a solid charged sphere of number density $n$ and a cylinder of number density $-n$ that lies inside the sphere exactly like the hole you had before, you get the same thing as the force due to the sphere with a hole in it.

gabbagabbahey said:
Any force that obeys an inverse square law, also obeys the superposition principle.
I don't think that is true. Can you explain?

gabbagabbahey said:
There is a much easier way to approach this problem. Any force that obeys an inverse square law, also obeys the superposition principle. That means that is you add the forces due to a solid charged sphere of number density $n$ and a cylinder of number density $-n$ that lies inside the sphere exactly like the hole you had before, you get the same thing as the force due to the sphere with a hole in it.

Assuming superposition works*, the cylinder would need spherical end caps to properly coincide with the sphere, otherwise there would be "left over" material and I would find an approximate solution. I suppose I could construct the cylinder in a piecewise manner using integration by disks, remembering to vary the radius of the disks appropriately for the end caps. However, on an earlier attempt at this problem I had tried to construct the complete sphere and hole using integration by disks and found the final integration to be more troublesome than what I found integrating over cylindrical shells, so I'm somewhat reluctant to try integration by disks again. Nevertheless, I'll make the attempt and update accordingly.

*Superposition should work since two particles at the same location and with equal but opposite charges should cancel each other out. The inverse r-squared law shouldn't even have anything to do with it. The downside is I would have to form the sphere and the cylinder separately, meaning there would be more integrals to solve.

Update: I've taken the earlier stated equation for a ring and integrated over radius to create a disk, which I then integrated over the diameter of the sphere to create the solid sphere. Specifically, I find the force on a particle due to a solid sphere is
$$\vec{F}=\int_{-R_{s}}^{R_{s}}2 \pi K n \hat{x} (1-\frac{R-r_{d}}{\sqrt{R^2+R_{s}^2-2 r_{d} R}}) dr_{d} = \frac{-2 \pi K n \hat{x}}{3 R^2}(2 R^3-3 R_{s} R^2-R_{s}^3-(2 R^3-3 R_{s} R^2+R_{s}^3)csgn(R-R_{s}))$$
where csgn is the complex sign function. Hopefully the integration over the cylinder won't present any extra difficulties.

Last edited:

1. What is the force on a particle due to a sphere with a cylindrical hole?

The force on a particle due to a sphere with a cylindrical hole is the sum of the forces exerted by the sphere and the cylinder. This force can be calculated using the formula F = GmM/r2, where G is the gravitational constant, m and M are the masses of the particle and the sphere, and r is the distance between them.

2. How does the size of the hole in the sphere affect the force on the particle?

The size of the hole in the sphere does not have a significant effect on the force on the particle, as long as the hole is small compared to the diameter of the sphere. This is because the majority of the mass of the sphere is still concentrated in the solid parts, so the force will be primarily determined by the mass of the sphere and the distance between the particle and the sphere.

3. Does the position of the hole in the sphere affect the force on the particle?

No, the position of the hole in the sphere does not affect the force on the particle. As long as the particle is at the same distance from the center of the sphere, the force will be the same regardless of the position of the hole.

4. Is the force on the particle always attractive when interacting with a sphere with a cylindrical hole?

Yes, the force on the particle will always be attractive when interacting with a sphere with a cylindrical hole. This is because the force of gravity is always attractive between two masses, and the presence of a hole does not change this fundamental nature of gravity.

5. How does the force on the particle change if the particle is inside the cylindrical hole?

If the particle is inside the cylindrical hole, the force on the particle will be reduced compared to if it was outside the hole. This is because the mass of the sphere is partially blocked by the hole, resulting in a decrease in the force of gravity between the particle and the sphere. The exact change in force will depend on the size and position of the hole and the distance between the particle and the sphere.

Replies
11
Views
1K
Replies
1
Views
1K
Replies
26
Views
3K
Replies
5
Views
2K
Replies
4
Views
1K
Replies
3
Views
2K
Replies
19
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
421
• Special and General Relativity
Replies
4
Views
519
• Linear and Abstract Algebra
Replies
1
Views
1K