The resultant forces at the bottom of a pendulum are not zero, but they are not perpendicular to the velocity either. I think you should include in the “and so forth” the tension of the pendulum arm. So, the only unresolved force is air friction, which opposes the velocity.
ras2000, why would you disagree with the book?
And, the “earth pendulum” would stop for the same reason that normal pendulums stop. Now, we can argue whether normal pendulums stop…
Sir Isaac Newton provided us with the formula that showed that the mass attraction (i.e., gravity) between two bodies is proportional to the product of their respective masses and inversely proportional to the square of the distance between them.
To analyze this problem, it is useful to remember that mass attraction exists between EVERY PAIR of particles of matter in the universe. So, between a person and a planet, the mass attraction is the vector sum of the attractions between every particle of the person and every partical of the planet, the result being that the composite vector acts between the centers of mass of the two bodies.
In the process of falling through this imaginary tunnel, the person will be accelerated toward the center by a force inversely proportional to the square of his (her?) distance from the center. His/her velocity will be the integral of that acceleration minus the force of wind resistance (which is proportional to the frontal area of the person and the square of his/her velocity). The person will achieve a terminal velocity where these acelerations are equal.
As a result, I believe that there will still be velocity at the center. Once the center had been passed, the gravitational acceleration will start to build up in the opposite direction (i.e., deceleration), while the air resistance will still be a decelerating force. I picture the person eventually coming to rest at the center after a damped oscillation of several cycles. (And I’m ignoring the increase in density of the air column as it increases in depth toward the center!)
A side comment to the folks arguing about the hollow sphere: if the sphere is made of anything but imagination, every particle that comprises it will have mass. As such, these particles will attract the person in the same way as the tunneled solid sphere, but with much less overall acceleration. But he/she will STILL be accelerated toward the center of mass of the body!
Omnivar, except when we decide to get more complex the velocity should still be at a maximum at the centre (of the first pass), this is as the oscialltor will be isochronous as the dampening will be proportional to the velocity. Heavy dampeing (which I do not think will be the case) will result in the displacement exponentially decresing to zero with no osciallations.
There will be NO accelartion towards the centre inside a hollow sphere: it’s a well known property of Newtonian gravity that a hollow sphere exerts zero mass inside it’s radius, which means that you only have to consider the mass of the sphere that’s radius is the distance from the object to the centre. You’ll just have to take my word for it as the proof is not partcularly simple
jimpeel, is this an accepted model that I haven’t heard of? If I remember correctly, gravitational pull increases by a square of the distance of the mass as you get closer to the mass. So, on the assumption that this has some mass, if you were 1 meter from one wall in a 100 meter diameter sphere, wouldn’t you fall towards the near wall, due to it being much closer? Or does the fact that most of the mass of the shell is “behind” you (away from the near wall) cancel this out?
i.e. force of small, nearby mass = force of large, far-away mass?
This is probably bad math, but based on this equation, a=(W-D)/m. a is acceleration, W is weight, D is drag, m is mass (which will remain constant). Assuming constant density of the earth, as I fall, I will before long reach terminal velocity (weight and drag balance). As weight decreases the closer we get to the center (where weight will be briefly 0), we will end up with a negative value on the right side of the equation, therefore a negative acceleration. The negative acceleration will result in a lower speed, which results in lower drag.
I don’t know if I’m clear here, but it seems to me that the drag decreases as a result of an already decreased velocity. After terminal velocity is reached, deceleration due to drag at any point in time will be higher than acceleration due to gravity. At some final point in time, this effect will cause me to stop at the earth’s centerpoint.
I’m tempted to try to computer-model this, but it may be way beyond my meager skilz.
Not quite sure what you’re saying, RM. Let me clarify: At the bottom of it’s swing, the resultant of the forces on the pendulum bob, not including aerodynamic drag, are perpendicular to the velocity of the bob.
If you’re pointing out that the total resultant force on the bob, including aerodynamic drag, isn’t perpendicular to the velocity, then yes, I agree, and my wording from a couple posts above is sloppy.
If you’re saying that forces on the bob other than drag are canceled out (as I infer from the comment that “the only unresolved force is air friction”), then that is incorrect. The tension in the arm is greater than the pendulum weight by a factor of mv[sup]2[/sup]/r (I refer you to this analysis), and so there is an unresolved force that accelerates the weight upward in addition to the unresolved drag force that slows the swing down.
ras2000, I see your points. However, I would argue a couple things. First, agreed that classical aerodynamics would predict a speed decrease that asymptotes to zero. However, I doubt that classical aerodynamic theory is applicable at the low velocities (near zero) that would occur. Second, whatever the correct theory to apply at that scale, I would argue that absolute zero velocity is essentially meaningless, as eventually Brownian motion would dominate movement.
Additionally, I understand your point about the pendulum. However, “terminal velocity” is simply the wrong term to use for this situation. A ball at the top of it’s arc does not have a terminal velocity of zero because there is an additional force (gravity) on the ball that will accelerate it to a higher velocity. Thus, zero velocity is in no sense the end, or “terminal” velocity.
Believe it or not, everyone is making this problem too simple… Too simple because no one has even thought about the gravitational pull of all of the mass around the earth, unless of course all mass rotates around the earth… but of course it doesn’t therefore as the ball, person, or whatever falls towards the center of the earth the pull of all that other mass would cause the ball to drift towards the center of all mass and away from the center of the earth, so the problem is solved…your welcome
Perhaps the pull of the sun or the moon will cause some drift (that is to say I can’t be bothered to work it out, a = GM/r[sup]2[/sup] would be that’s needed), but that would be only a minor factor and you needn’t consider the mass of anything else as the effects would be nominal. Also there is no such thing as “the centre of all mass”.
Nope, just sloppy. ras2000’s point, which I was responding to, was that the tension and gravitational forces on a pendulum at the bottom of its swing are perpendicular to the direction of motion (and thus the drag force). The resultant of tension and gravitational force is still perpendicular to the drag. There’s no requirement that a “resultant force” include all forces on a body. Clearly, though, the “resultant force” mentioned in the dictionary definition of “terminal velocity” is intended to be the resultant of all forces on the body…which is why my usage above was sloppy.
I would think that that resultant is zero. But you said “Resultant forces on a pendulum are not zero, as you point out. The fact that the resultant force is perpendicular to the velocity is irrelevent.”
The resultant forces are not zero, if you include the drag–but then it is not perpendicular to the velocity, contrary to what you said. If you don’t include the drag, then they are zero–also contradictory to what you said.
Ah, but the tension in the arm is greater than the weight of the pendulum. Take a look at my earlier post. The tension in the arm is greater than the pendulum weight by a factor of mv[sup]2[/sup]/r (I refer you to this analysis), and so the resultant of the tension and gravitational forces is simply mv[sup]2[/sup]/r; most certainly not zero. This force accelerates the weight upward, and is wholly seperate from the drag force that slows the swing down.
There are a bunch of things that haven’t been taken into account, for example the difference in air pressure between the two sides, magnetism, the effect of distance suns, and so forth. So for the record, let’s just look at the frictionless tube, where the acceleration decreases/increases at some weird function from one end (g) to the other (-g), and the drag would change at some other strange function, and maybe throw the rotation of the earth in there for good measure.
If we look at a simplified version where we have g at one side of the center, and -g at the other side, shifting abrubtly between the two, and the air density is the same all the way through, I think the body would reach terminal velocity after a couple of seconds, and stay at this speed untill it passes the center. This speed would then determine how far the body would pass the center by, and that would be the last time terminal velocity would be reached. Whether or not the body would ever stop bobbing up and down is still open. There is a small difference between this pendulum, and the usual one, in that there is a friction in the string, the arm, and at the hinge, which behaves differently than fluids, there is a minimum.
In the “real” example, I’m not really sure what speed the body would have when it passes the center, but I still feel that the body would never stop moving, untill someone can tell me what the minimum amount of force required to make a body move is, when the only opposing force is wind drag, and what the wind drag would be at the center of the earth.
Actually, I’m not really sure that terminal velocity would ever be reached.
If you look at a body falling, and imagine that g is constant, then the resulting force (g*m - f_wind) would near 0, but the closer it comes, the smaller the resulting force. Wouldn’t the speed come really close to terminal velocity, but never reach it, as well?
Yeah, that’s true. With a constant g, velocity of a falling object asymptotes to terminal velocity, so, mathematically, terminal velocity is never reached. Of course, the question then is: how close to terminal velocity do you need to get before the difference is physically meaningless? Certainly you would rapidly get to the point where the difference between actual velocity and terminal velocity is so small that this difference is dwarfed by the uncertainty introduced by the effects you’ve ignored (like, say, the molecular make-up of the air, interaction of the air with the wall of the tunnel, minor and major changes in air density and gravity, etc).
I know I’m coming back at this after a time, so I may be repeating something, but wouldn’t that sort of reasoning also apply to a pendulum, for instance? Why wouldn’t you also say that a pendulum would never stop? Or do you think they never do, in the same sense? Is that the sense you mean?
Hey RM Mentock
I wrote a couple of notes up:
There is a small difference between this pendulum, and the usual one, in that there is a friction in the string, the arm, and at the hinge, which behaves differently than fluids, there is a minimum.
The post you quoted was to me, not from me. I do, however, understand the concept which is apparently accepted Physics 101 fare.
If the mass, as a whole, is acting upon the subject body then the distance from one side cancels out the gravity generated to the other side. It would, for all intents and purposes, be the umltimate fun ride as you would be in 0g wherever you went inside the sphere.