A narrow uniform rod has length 2 a. The linear mass density...

A narrow uniform rod has length $2 a$. The linear mass density of the rod is $\rho,$ so the mass $m$ of a length $l$ of the rod is $\rho l$. (a) A point mass is located a perpendicular distance $r$ from the center of the rod. Calculate the magnitude and direction of the force that the rod exerts on the point mass. (Hint: Let the rod be along the $y$ -axis with the center of the rod at the origin, and divide the rod into infinitesimal segments that have length $d y$ and that are located at coordinate $y$. The mass of the segment is $d m=\rho d y$. Write expressions for the $x$ - and $y$ -components of the force on the point mass, and integrate from $-a$ to $+a$ to find the components of the total force. Use the integrals in Appendix B.) (b) What does your result become for $a \gg r ?$ (Hint: Use the power series for $(1+x)^{n}$ given in Appendix B.) (c) For $a \gg r,$ what is the gravitational field $g=\boldsymbol{F}_{g} / m$ at a distance $r$ from the rod? (d) Consider a cylinder of radius $r$ and length $L$ whose axis is along the rod. As in part (c), let the length of the rod be much greater than both the radius and length of the cylinder. Then the gravitational ficld is constant on the curved side of the cylinder and perpendicular to it, so the gravitational flux $\Phi_{g}$ through this surface is cqual to $g A$, where $A=2 \pi r L$ is the area of the curved side of the cylinder (see Problem 13.59 ). Calculate this flux. Write your result in terms of the mass $M$ of the portion of the rod that is inside the cylindrical surface. How does your result depend on the radius of the cylindrical surface?

Key Concepts

Gravitational Flux and Gauss's Law for Gravity

Gravitational flux involves the amount of gravitational field passing through a given surface, and Gauss's law for gravity relates this flux to the total mass enclosed by the surface. Utilizing this law helps to simplify the calculation of gravitational fields, especially when symmetry allows the flux to be easily integrated over geometric surfaces such as cylinders.

Gravitational Field

The gravitational field represents the force per unit mass at a given point in space, providing a description of the gravitational influence produced by a mass distribution. It is a vector field that encapsulates both the strength and direction of gravity and is closely related to the gravitational force experienced by objects.

Series Expansion Approximation

Series expansion is used to simplify complex expressions when parameters are in limiting cases, such as when one length scale is much larger than another. In gravitational problems, using a power series approximation can reduce complicated integrals to simpler forms that reveal the dominant behavior of the gravitational force or field in a specific limit.

Linear Mass Density

Linear mass density is a measure of how mass is distributed along a one-dimensional object, such as a rod. It is defined as the mass per unit length and is crucial for analyzing problems where the object’s mass is continuously distributed along its length, aiding in the determination of mass elements for integration.

Gravitational Force from Continuous Mass Distributions

This concept refers to computing the net gravitational force produced by an extended object by considering it as a collection of infinitesimal mass elements. Each mass element contributes a small gravitational force, and by integrating these contributions over the entire mass distribution, one can determine the overall force exerted on a test mass.

Integration over Infinitesimal Mass Elements

This technique involves breaking an extended object into small segments where each segment can be treated as having a small mass. By integrating the force contributions from these infinitesimal segments, one can calculate the net effect of the entire mass distribution on another object. This method is key in deriving expressions for forces and fields in continuous systems.

Transcript

00:01 In this problem, we're going to determine the gravitational force on a small mass that's approaching a long, thin rod of material.

00:13 We're going to see whether the gravitational force goes as one over r squared.

00:19 And you may guess that the length of the rod is going to change the dependence of the gravitational force with distance.

00:28 We're also going to relate the gravitational field that that little mass experiences to the field you would expect using a very beautiful mathematical relationship that's often used in electricity as well, and that is galses law.

00:50 But let's first of all figure out the gravitational force on that little mass.

00:58 And see how it behaves.

01:01 So the way to start is to imagine that the rod is broken up into little chunks of mass.

01:12 And ordinarily, the force of gravity involves two masses that are interacting, a big mass and a little tiny mass.

01:22 And it's the big mass that we're breaking up into tinier chunks.

01:26 And each of those little chunks are going to contribute a little tiny amount to the total force, which we can write down with the universal law of gravity.

01:43 And notice i am going to write this down in its vector form.

01:49 Okay, so that's still a one over r square dependence, but we now have a vector which separates are two particles.

02:02 So what r is, is it is a vector that points from the source to the observer.

02:11 And i'll just show that r in our picture.

02:15 And it probably should set up a coordinate system, so the usual x, y, in a plane.

02:26 Okay, now we can see that r has two coordinates to it.

02:34 It has a variable x component that is somewhere along the line, and our observer is up there at position b.

02:53 Okay, so those are the x and y coordinates.

02:58 And our little particle of mass inside of that rod is equal to the density of the rod, a linear density times how long that little chunk is.

03:13 So we're going to call that dx, little differential hunk of material.

03:20 And so we're ready to write down our integral for force.

03:28 It is just going to take that little differential force and integrate it.

03:34 And it's going to have two components, one in the x direction, and one in the y direction.

03:43 And i'll write those out with some of the constants taken out of the integration.

03:51 So in the x direction, both of these integrals, by the way, are going to happen along the length of the rod, adding up the contributions from each of the tiny masses.

04:05 And i just realized i got the wrong mass in there.

04:07 This is a little m that's up at the observation point.

04:15 So it may be a spacecraft approaching a large rod -shaped space station.

04:22 And the dm is inside the rod, and r is the separation that connects one of these little points in the rod to the spacecraft up at the axis, waiting to try to get.

04:44 To the rod.

04:45 Okay, but both integrals for the x and the y components of the force are over the length of the rod.

04:53 So we've set that up with the center of the rod in the middle.

04:59 And let's see, we've got row x, dx, and r, by the way, you should point this out, that we have a right triangle so that we can write down the magnitude of r as the square root of x squared plus b squared.

05:30 And we will need that in the one over r cubed dependence.

05:42 We'll show up with a three -aft power because it's arc cubed.

05:48 And i know that from symmetry, it may seem very clear that the rod will exert no force on little m in the x direction.

06:04 And we can see that through the integral because it is an odd function of x, where we're kind of finding the area of an odd function, which is just as much negative as positive.

06:22 And so i really don't need to integrate that.

06:25 And i'll just kind of show a little sketch there that an odd function of x, we'll just call it f of x in general, but i'll show it's perfectly anti -symmetric.

06:39 So it would have as much area in the negative part of the region as it does in the positive region.

06:50 And those two regions give you an area of zero.

06:55 And the real definition of an odd function means that if you reflect x through the origin, the function simply becomes negative.

07:14 The y component of force, on the other hand, is going to turn out to be an even function.

07:20 And i will have to wind up integrating this from x as minus a to plus a.

07:38 And i haven't quite got all my constants out of the integral, but we'll do that in the next step.

07:46 The only thing i'm integrating over is x.

07:55 So what i'll need, let me just kind of delineate the part that i'm going to integrate, is what i'll need is to look up that integral.

08:05 And that's perfectly a valid thing to do, is to look up an integral in an integral table, this is an even function, which means that f of minus x reflected through the origin is positive f of x.

08:28 And so that will result in a non -zero integral...

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