Bart's Second Order Nonlinear Differential Problem?

Question

See The cat's question:

http://answers.yahoo.com/question/index;_ylt=ApZo01JFGON920IYeGRPJUHsy6IX;_ylv=3?qid=20071213163311AAyMl2Z&show=7#profile-info-AA12298210

Assuming that Earth is 8,000 miles in diameter and rotates 24 hours a day, and disregarding gravity, if Bart skates frictionlessly from the north pole to the equation via a straight tunnel, how fast will he be going by the time he exits the tunnel? Acceleration will be due to centrifugal force of the spinning Earth, and you may simplify the problem by also disregarding coriolis forces outside of the plane of the tunnel and center of the earth.

Of course, assume that Bart kick starts with some nominal velocity so that he can get off the north pole.

Gee, Alexander, that explains why seawater isn't being flung out into space! I have a new theory about Noah's Flood---I say that after it was over, God cancelled gravity for a day (he stopped the sun for a day too at one time, didn't he?), and thus most of the flood waters go flung out into space at the equator, and so that's why we have the missing mass of water today that we can't account for.

Unless someone else is brave enough to go up against your answer with one that disregards gravity, it looks like you'll be getting your 10 points by default. But thanks for the very erudite answer, I did enjoy it.

ksoileau, I did say, "disregard gravity". This problem involves only centrifugal force, so that the equation has the form x''(t) = k (x(t))².

Alexander, probably this is a badly worded question. I'm asking for the velocity vector in line with the tunnel, not tangent to the surface of the earth at the exit. Let's imagine that a factory machine has a spinning tube set at 45², and marbles are dropped into it at the "north pole". The exit velocity is going to have both radial and tangential velocities, so that the final velocity will be a vector sum. It's really about solving the differential equation x''(t) = k (x(t))², which I'm finding out to be really hard to do.

ksoileau, ditto, see above.

Steve H, I'll wait for your answer.

I see that I should have asked what will be Bart's velocity just prior to exiting the tunnel, relative to the tunnel length.

ksoileau, I've decided to extend this question so that you have a chance to respond. I don't understand why you feel that the total energy of the system should be conserved, while leaving out the rotational energy of the earth. If you leave out earth's energy, then the total energy of the remaining system, namely Bart, isn't conserved, because earth is imparting energy to Bart. We can imagine a playground merry-go-round with a radial track for marbles. It seems to me that putting marbles on the track anywhere but the center, would have them flying out of the perimeter.

Answer 1

Bart's initial energy=1/2*m*v0^2
Bart's final energy=1/2*m*v1^2+1/2*m*ve^2
where
ve=rotational speed at the equator
v0=kickoff speed at north pole
v1=speed of emergence out of tunnel at equator
ve=2*pi*R/T
R=radius of Earth at equator
T=period of Earth's rotation
By conservation of energy,
1/2*m*v0^2=1/2*m*v1^2
+1/2*m*ve^2
1/2*m*v0^2=1/2*m*v1^2
+1/2*m*(2*pi*R/T)^2
v0^2=v1^2+(2*pi*R/T)^2
v1^2=v0^2-(2*pi*R/T)^2

v1=sqrt(v0^2-(2*pi*R/T)^2)
which is approximately
sqrt(v0^2-1082436.933) mph.

(I'm using actual values for the equatorial radius of the Earth and rotation rate, not 4000 miles and 24 hours.)

Note that Bart must kick off with speed at least 2*pi*R/T~1040.402294 mph to make it all the way to the end of the tunnel.

Note: Bart actually feels himself gradually slowing down (with respect to the tunnel floor) as he travels down the tunnel, a sort of "polarpetal" force (acceleration toward the north pole).

PS I don't think gravity matters much at all here, since at the point of kick off and the point of emergence Bart has nearly, but not exactly the same gravitational potential energy.

@scythian: Yes, I understand that the problem doesn't involve gravity, and that's what my solution reflects. I was simply remarking that gravity (perhaps surprisingly) doesn't really much affect the kickoff speed required.

@Steve H: I understand your feeling about the deceleration effect. I look at it this way: At the pole, before he kicks off, Bart has no energy, potential or kinetic. When he emerges at the equator, he is going over 1000 mph in revolution around the Earth's center. He makes an initial energy deposit by kicking off, then as he travels down the tunnel, and therefore farther from the Earth's axis, more and more of that deposit becomes rotational energy. This leaves less energy in the budget to account for down-tunnel speed, so his down-tunnel speed must gradually decrease. This is the very definition of deceleration: decrease in speed.

@scythian: "ksoileau, ditto, see above."
I did, and didn't see any refutation of what I wrote. If you can find a flaw in my analysis, please do. If not, I'd like 10 points for my answer.

Yes, I see your point, my calculations show that the downtunnel acceleration is 4*pi^2/T^2/sqrt(2)*s miles/hour^2 where T is rotation period in hours (roughly 24) and s is the amount of downtunnel travel in miles. The resulting differential equation can be solved, yielding ds/dt=sqrt(4*pi^2/T^2/sqrt(2)*s^2+v0^2), where v0 is Bart's kickoff speed at the pole. At emergence s=sqrt(2)*R where R is the equatorial radius of the Earth (roughly 4000 miles). Using these values, emergence speed for positive v0 will be sqrt(1550858.711+v0^2). For tiny values of v0, this emergence speed will be approximately
1245 mph,
and is an increasing function of v0.

Answer 2

This strikes me as an interesting problem. I can see that working it through in a rotating frame of reference is probably the neat way to do it, but I am trying to tackle it from a 'stationary' one.

First I was going to try to work out the easier problem of a tunnel straight from the center of the earth out to the equator.

My instinct bristles at the claim Bart will slow down as he slides from the pole to the equator in the given problem -- that just doesn't sound right. I'll try to come back with more later ...

----

Okay, I've been putzing around a little with the simpler problem of a particle just getting whipped out form the cener of the earth by its rotation.

I'm getting a curious radial motion that goes like the hyperbolic cosine of the angle rotated through. It seems to agree with the force always being perpendicular to the tunnel, but I'm not at all sure if I haven't slipped up somewhere. I'll keep playing ...

By the way, the above arguments about energy conservation seem to be ignoring the energy of the rotating earth -- isn't that where Bart's energy is coming from?

Answer 3

In frame of refernece rotating with the Earth there are forces acting on any body:

gravity mg(x) = -m dP(r)/dr
centrifugal mΩ²R = -m d[-1/2 Ω²R²]
coriolis m 2[Ω x v]

r denotes 3d coordinate, and R is the distance to the axis

The first two forces are potential.
The third one is not potential, but perpendicular to v, and thus cannot do any mechanical work.

Consequently we have conservation of energy:
m [P(r) - 1/2 Ω²R² + 1/2 v²] = const

Because elements of water at ocean surface are in equilibrium and have zero velocity, the shape of the planet satisfies equation
TotalPotential(r) = P(r) - 1/2 Ω²R² = const_at_surface

For Bart emerging everywhere on surface out of tunnel of arbitrary shape
1/2 Vo² + const_at_surface = 1/2 V1² + const_at_surface

Because Vo is zero, V1 is zero too.

Note:
gravitation potential is not supposed to be harmonic, it can be arbitrary. The only thing that matters is that g(r) is potential field, which is consequence of superpositon of elemetary 1/r² fields.


***************
All right, all right.
Disregarding gravity you throw away the P(r) part and end up with
v² = Ω²R²
v = ΩR

This is velocity in rotating frame of reference, in stationary frame of reference you need to add up perpendicular velocity of the equator for a total of √2 ΩR = √2/π 8000/24 mph


*********************
The exit velocity relative to the equator is ΩR along the tunnel. It follows from conservation of energy:
m [-1/2 Ω²R² + 1/2 v²] = const
m [-1/2 Ω²0² + 1/2 Vo²] = 0 = m [-1/2 Ω²R² + 1/2 V1²]
v = ΩR



I have a strange feeling that the asker wants us to actually integrarate the dynamic equation
m d²x/dt² - F(x) = 0.

As long as F(x) is potential force, that is
F(x) = -d/dx P(x)
no physicist in his right mind will bother to actaully integrate this equation, because the result is know in advance (namely mv²/2 + P = const).

Let me do the integration once again:
m d²x/dt² - F(x) = 0
multiply both sides by dx/dt repalcing forces with powers:
m dx/dt d²x/dt² + dx/dt dP(x)/dx = 0
m v dv/dt + dx/dt dP(x)/dx = 0
m d[v²/2]/dt + dP(x)/dt = 0
d[mv²/2 + P(x)]/dt = 0
mv²/2 + P(x) = const

Finally, the centrifugal force is not proportional to r², it is proprtional to r (namely F = mΩ²R).