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        riddles >> easy >> Off the Edge of the Earth.
        
(Message started by: rloginunix on Oct 11^th, 2014, 10:16pm)

Title: Off the Edge of the Earth.
Post by rloginunix on Oct 11^th, 2014, 10:16pm

Off the Edge of the Earth.

A material point is sliding off the top of a stationary sphere of radius R without friction and without initial velocity.

At what height above the ground level will the point separate from the surface of the sphere?

Title: Re: Off the Edge of the Earth.
Post by dudiobugtron on Oct 11^th, 2014, 11:23pm

How is it sliding off without initial velocity? :P

-----
I think I could try to solve this using maths and physics equations, but I'll look for a way to solve it more intuitively first, since it's in the easy forum. :)

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 12^th, 2014, 10:46am

on 10/11/14 at 23:23:52, dudiobugtron wrote:

How is it sliding off without initial velocity? :P

I had the same question when I first read the problem. According to Newton "a body at rest remains at rest unless acted upon by an external force".

A sphere has a constant non-zero curvature at any point of its surface so the equilibrium of a material point dead on the sphere's North Pole is inherently unstable. Combined with the absence of friction I made peace with this issue in two ways: either the external force (initial velocity) or the deviation off the North Pole is infinitesimally small - if you conduct a series of experiments with these values progressively diminishing without bound then the answer that you get should converge to the one obtained with the above condition standing. As such either of the parameters can be neglected as are other things: [hide]resistance to air and sound waves production[/hide], for example. However, there is no harm in assuming that V₀ is known.

Separately. Once you know the height you can calculate the angle of departure. Once the angle of departure is known you can calculate the length of the arc traveled and, hence, the travel time. But these are extras for experts.

Title: Re: Off the Edge of the Earth.
Post by rmsgrey on Oct 13^th, 2014, 4:09am

on 10/11/14 at 23:23:52, dudiobugtron wrote:

How is it sliding off without initial velocity? :P

Incredibly slowly...

Title: Re: Off the Edge of the Earth.
Post by towr on Oct 13^th, 2014, 11:57am

I have revised and replaced my previous answer (in case anyone saw it), I'll now go with [hide]3/2 R[/hide] until such time I discovered why I'm wrong this time.

[hide]If we take the component of gravity directed at the center of the sphere g*cos(alpha), that gives an escape velocity via v_e²/2 = g*R*cos(alpha) [edit](mistakenly wrote down sin here before)[/edit]
We can combine this with the released potential energy from dropping a certain height to get v²/2 = g*R*(1-cos(alpha)).
At which point (if this time I didn't screw up where to use which trigonometric function), it's quite obvious they have to meet right in the middle between 2R and R.[/hide]

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 13^th, 2014, 8:14pm

Good going, towr. I did not see your original answer, may be it was right, :). Numerically the current one I am seeing, [hide]3/2[/hide], is just a bit off.

Your [hide]conservation of energy[/hide] observation and equation are both correct.

While the point is [hide]on the sphere it is acted upon by mg and N - the normal force of reaction of the sphere, so in the vector form we have ma = mg + N. At the point of escape the contact is lost: N = 0 and projecting the vectors on the line that runs into the sphere's center you get mgcos(alpha) = ma_n = mw²_eR = mv²_e/R where w is angular speed and v is linear speed at the point of escape[/hide]. So I think you have a typo in your first equation: [hide]v²_e/2 - it should be over R and instead of sin it should be cos. So now you have 3 unknowns (h, cos, v) and 3 equations[/hide]...

This drawing, that I think follows towr's train of thought, may help but, for the rest of the audience, it is a spoiler (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1396709569;start=100#118). Sub index e stands for Escape and the angle is the actual one.

Title: Re: Off the Edge of the Earth.
Post by towr on Oct 13^th, 2014, 10:43pm

on 10/13/14 at 20:14:49, rloginunix wrote:

[hide] it should be over R and instead of sin it should be cos.[/hide]

It is over R, but R is on the other side of the equation I used; specifically I used [hide]v_e= sqrt(2gr) from [/hide][hide]http://en.wikipedia.org/wiki/Escape_velocity#Calculating_an_escape_velocity[/hide] (http://en.wikipedia.org/wiki/Escape_velocity#Calculating_an_escape_velocity) and moved some terms around for convenience. And I did use cosine doing the calculations, so I don't know why I wrote sine.
I don't see any issues, or where I'd need a third equation.

[edit]Maybe [hide]the point doesn't need to escape out into the universe, and reaching orbit is enough :P So perhaps I should use the orbital velocity equation instead of escape velocity.[/hide] In that case [hide]cos(alpha) = 2/3, so take-off is at 5/3 R[/hide][/edit]

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 14^th, 2014, 8:57am

Yes, your edited answer of [hide]5/3[/hide] is correct.

By three equations I meant:

1). [hide]mgh = mgR(1 - cos(alpha)), potential energy change[/hide].

2). [hide]mgR(1 - cos(alpha)) = mv²_e/2, conservation of energy[/hide].

3). [hide]mgcos(alpha) = mv²_e/R, projection of F=ma[/hide].

I must say your [hide]energy conservation[/hide] approach computationally is a bit simpler than mine. I took it more literally: [hide]2mgR = mv²_e/2 + mgH_e[/hide]. The [hide]projection of F=ma is the same, mgcos(alpha)=mv²_e/R[/hide] and [hide]cos(alpha) = (H_e - R)/R[/hide]. From here [hide]H_e = (5/3)*R[/hide].

Anyone interested in seeing what the answer will be if V₀ is given?

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 15^th, 2014, 9:43am

In my calculations the initial velocity of V₀ adds a [hide]V²₀/3g[/hide] term to H_e and [hide]V²₀/3gR[/hide] to [hide]arccos[/hide] of alpha:

[hide]H_e = (5/3)*R + V²₀/3g
alpha_e = arccos(2/3 + V²₀/3gR)[/hide]

If R = 1 meter and V₀ = 1 meter per second, or 3.6 kilometers per hour, (g = 9.81 meters per second squared) the numerical addition is 0.0339789330

If V₀ = 0.1 meters per second the addition is 0.000339789330, etc.

Title: Re: Off the Edge of the Earth.
Post by SWF on Oct 16^th, 2014, 10:12pm

If started out with zero intial velocity, find the equation for the time taken to go from any height to the separation point. I am asking for the zero initial velocity case, because that allows for an answer in terms of the elementary functions.

For the issue of starting from a point of unstable equilibrium, could also think of launching the particle from the ground so it flies through the air, then comes in tangent to the sphere with just the right velocity to slide along the surface and stop at the top.

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 21^st, 2014, 9:34am

So far I have found three ways to arrive at the same integral. Here is one based on the Lagrangian.

Our point can execute an independent movement in only one direction characterized by the angle of deviation from the sphere's rotational axis passing through the sphere's center perpendicular to the plane the sphere is resting on (Z axis). In spherical coordinates it is the polar angle theta http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif. As such the point has one degree of freedom and we choose http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif as our generalized coordinate.

The Lagrangian L = T - U, where T is kinetic and U is potential energy, in this case will be (the prime, single tick character, denotes a regular first derivative over time):

T = mV²/2 = mhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif²R²/2 = mR²http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'²/2, where http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif= http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'

U = mgR + mgRCos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif)

L = mR²http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'²/2 - mgR - mgRCos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif)

The Euler-Lagrangian equations "split" into just one equation:

d( http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifL/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif')/dt = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifL/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifL/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif' = mR²http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'

d( http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifL/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif')/dt = mR²http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif''

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifL/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif= mgRSin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif)

and hence:

mR²http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'' - mgRSin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif) = 0

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'' - (g/R)Sin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif) = 0

For convenience let us put g/R = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀² and integrate the above equation once:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'²/2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀²Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif) + C = 0

To solve for C we observe that initially the point is at rest: http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'(0) = 0 and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif= 0:

0²/2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀²Cos(0) + C = 0

C = -http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀²

and the integrated once equation becomes (after multiplying both sides by 2):

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'² + 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀²Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif) - 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀² = 0

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif'² = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀² - 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀²Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif)

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/vert.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif' http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/vert.gif= http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/largetimes.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif))

Here we show the differentials:

dhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/dt = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/largetimes.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif))

and rewrite the equation in preparation for integration:

dt = (1/(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2)) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/largetimes.gifd http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif))

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 21^st, 2014, 9:36am

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdt = (1/(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2)) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif))

From the trigonometric identity (1 - Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif)) = 2Sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/2) for the integral on the right we have:

(1/(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2)) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(2) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/largetimes.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/vert.gifSin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/2) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/vert.gif)

(1/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀)) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/vert.gifSin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/2) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/vert.gif

The only easy substitution I see here is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/2 = x. And dhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif= 2dx:

(1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdx/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/vert.gifSin(x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/vert.gif

At this point I am not really sure which substitution will work, it's been a while. Does anyone know how to proceed here?

Assuming we have calculated the above integral if we take the boundaries of integration from 0 to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_e we will get the total travel time till escape/departure (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_e is known).

If we integrate from 0 to an arbitrary angle http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/leqslant.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_e we will get the travel time from start to this arbitrary angle.

To do what SWF as asking for we would integrate from an arbitrary initial angle http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₀ to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_e. And since an arbitrary initial height H₀ is:

H₀ = R + RCos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₀)

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₀ = arccos((H₀ - R)/R)

we will get the required travel time as a function of H₀.

Now, any ideas about the integral?

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 22^nd, 2014, 11:19am

In order to calculate this integral we employ the idea of "reduce the current problem to an already solved one".

We know that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdx/x = ln|x| + C. This allows us to easily calculate http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdx/(1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifx) which allows us to easily calculate http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifSin(x)dx/(1 - Cos(x)) with the following substitution:

z = 1 - Cos(x)

dz = d(1 - Cos(x)) = Sin(x)dx which is the expression in the numerator of the integral. Hence:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifSin(x)dx/(1 - Cos(x)) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdz/z = ln|z| + C = ln|1 - Cos(x)| + C

In our particular case let z = Cos(x), dz = -Sin(x)dx and then:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdx/Sin(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifSin(x)dx/Sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifSin(x)dx/(1 - Coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif(x)) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifSin(x)dx/((1 - Cos(x))(1 + Cos(x))) = -http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdz/((1 - z)(1 + z))

Now we can use the method of Partial Fractions:

x/(1 - z) + y/(1 + z) = -dz/((1 - z)(1 + z))

x + xz + y -yz = -dz

1). x + y = -dz
2). (x - y)z = 0

or

3). x + y = 0
4). (x - y)z = -dz

3) and 4) are no good - putting their solutions back yields the original integral. We choose 1) and 2) solving which yields x = y = -dz/2. Hence our integral becomes:

-dz/((1 - u)(1 + u)) = -dz/(2*(1 - z) - dz/(2*(1 + z))

-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdz/((1 - z)(1 + z)) = (1/2)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdz/(1 - z) + (1/2)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdz/(1 + z) = (1/2)ln|1 - z| - (1/2)ln|1 + z| + C =

= (1/2)ln|1 - Cos(x)| - (1/2)ln|1 + Cos(x)| + C = ln|http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif((1 - Cos(x))/(1 + Cos(x))) + C = ln|Tan(x/2)| + C

where I have used the tangent of a half-angle formula.

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 22^nd, 2014, 11:22am

Now we perform the reverse substitution of x = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/2:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdt = (1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀)ln|Tan(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/4)|

integrated for the boundaries discussed above. Tan(0) = 0, ln(0) is not defined and consequently the integral is not determinate. Conclusion - must start with a very small initial angle http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₀.

It is interesting that my initial settlement with zero initial velocity issue shows up here - a small deviation off the North Pole.

In any case,

T_e - T₀ = (1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀)(ln|Tan(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_e/4)| - ln|Tan(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₀/4)|) = (1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀)ln|Tan(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_e/4)/Tan(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₀/4)|

where the expression for http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₀ is known, see above.

We can now invert the above finding and write a formula for "angle as a function of time":

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(t) = 4ArcTan(e^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀t} Tan(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₀/4))

(Who the hell put this in the easy section?)

Title: Re: Off the Edge of the Earth.
Post by SWF on Oct 22^nd, 2014, 7:34pm

That looks good, rloginunix. The integral of 1/sin(x) is ln(1/sin(x)-cos(x)/sin(x)). It is one of the more obscure of the basic integrals (commonly printed inside cover of calculus book). The way I have seen to solve it is multiply 1/sin(x) by f(x)/f(x) and note that the result is 1/f(x) * d(f(x)), where f(x)= 1/sin(x)-cos(x)/sin(x). This approach never seemed that satisfying, since it is almost like already knowing the answer.

The way I found the differential equation was conservation of energy:
mv²+mgR*cos(phi)=mgR*cos(phi0).
Replace v with v=R*d(phi)/dt, and solve for d(phi)/dt.

If the inital velocity is non-zero, the solution involves elliptic integrals.

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 23^rd, 2014, 1:17pm

That's right. Energy was the shortest path I could find. Then comes projecting the Newtonian vectors. Then the Lagrangian. There should be yet another way - via the angular momentum but I do not remember the details well enough to post them here.

Yes, if we consider V₀ then the denominator of the integral over dhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif becomes http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif((V₀/R)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif+ (4g/R)Sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif/2)) which, after the usual manipulations, we can reduce to a generic form of:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdx/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 + khttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gifSinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif(x))

If, however, we ask ourselves: "what V₀ should be given to a point so that it flies off the sphere without ever touching it?" then from the earlier Newtonian projections for the normal force of reaction N we have:

N = m(gCos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif) - Vhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif/R)

From the conservation of energy we find V:

Vhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif= Vhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif₀ + 2gR(1 - Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif))

and hence:

N = m(gCos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif) - Vhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif₀/R - 2g + 2gCos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif)) = m(3gCos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif) - Vhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif₀/R - 2g)

For the point to fly off N and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif must be 0:

0 = 3g - Vhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif₀/R - 2g

V₀ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifgR

Title: Re: Off the Edge of the Earth.
Post by SWF on Oct 23^rd, 2014, 8:54pm

Projecting force in the tangent direction and setting equal to acceleration in tangent direction is pretty easy to get a differential equation too:
mR*d²phi/d t²= mg*sin(phi)
The tougher part is integrating, but use the same method as when deriving energy from F=ma: multiply by d phi/dt and integrate. Using initial condition of zero velocity at phi=0 to evaluate constant of integration, results in same equation as before.

By the way, another way to integate dx/sin(x):
= dx/( 2* sin(x/2)*cos(x/2))
= (cos(x/2)/sin(x/2)) *dx/(2*cos²(x/2)))
=(1/tan(x/2))*d(tan(x/2))
=d ln(tan(x/2))

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 25^th, 2014, 11:02am

on 10/16/14 at 22:12:58, SWF wrote:

For the issue of starting from a point of unstable equilibrium, could also think of launching the particle from the ground so it flies through the air, then comes in tangent to the sphere with just the right velocity to slide along the surface and stop at the top.

I was thinking. If we finish solving the original problem in "forward movement" then we can use the obtained parameters to solve the above problem if we "reverse" them.

When the point separates from the sphere it begins a parabola-following free flight. The initial parameters are known at this point: velocity V_e, angle http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_e, height H_e. From these we can find the landing velocity V_L - its scalar value and the angle it forms with the ground http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_L. We can also find the free flight time T_FF and the horizontal landing distance from the sphere D_L.

We now flip the V_L vector 180 degrees and keep the rest of the parameters the same.

In the drawing below the parabola is not real - I just invented it for demonstration purposes to keep it somewhat realistic:

http://www.ocf.berkeley.edu/~wwu/YaBBAttachments/rlu_oteote2.png

I will just highlight the solution: the horizontal component V_ex will remain the same throughout the free flight:

V_ex = V_e*Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_e) = V_Lx

We can find the scalar value V_L from the conservation of energy. And hence:

Cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_L) = V_ex / V_L

and hence we know the vertical component V_Ly = V_L*Sin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_L)

The horizontal landing distance:

D_L = LE + ET = V_ex*T_FF + ET

where T_FF can be found by solving either a quadratic equation for the vertical coordinate y or a linear equation for a vertical velocity component:

V_Ly = V_e*Sin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_e) + gT_FF

etc.

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 25^th, 2014, 11:03am

Lastly, after looking at the above drawing you can't help but formulate a similar but different problem (inspired by SWF's idea):

A material point is launched off the ground over a stationary sphere of radius R. What must the initial velocity - angle and a scalar value - and distance from sphere be so that the point barely makes it over the top of the sphere and lands on the other side?

In mathematical terms optimize or find the smallest V₀, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₀, D₀.

Title: Re: Off the Edge of the Earth.
Post by SWF on Oct 28^th, 2014, 8:18pm

The peak of the parabola in that last figure should be below the top of circle, and is not on same vertical line as the center of the circle. That the peak is below top of circle should be obvious from conservation of energy. The peak is at a height of (50/27)*R.

If instead of a sliding particle, it is a ball rolling off the sphere, the angle that it separates is given by a similar result: cos(angle)= A/B. (I will leave that as a problem to find the integers A and B). The result does not depend on the radius of either sphere.

Title: Re: Off the Edge of the Earth.
Post by rloginunix on Oct 29^th, 2014, 7:28pm

I got lazy (and sloppy) when I shouldn't have but I did the math now. The y(x) equation for the above horns-down parabola is:

y(x) = (5/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2)x - (27/(16R))xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif

It peaks at (50/27)R, as you said, at X_p = (20http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(2)/27)R if nothing stands in its way - compare that height to (50/30)R escape height.

on 10/28/14 at 20:18:20, SWF wrote:

If instead of a sliding particle, it is a ball rolling off the sphere, the angle that it separates is given by a similar result: cos(angle)= A/B. (I will leave that as a problem to find the integers A and B). The result does not depend on the radius of either sphere.

Assuming no slippage and a uniform density of a solid ball I get A = [hide]10[/hide] and B = [hide]17[/hide]. For a small hollow sphere I get A = [hide]6[/hide] and B = [hide]11[/hide].

And since we went that far. Making these guys behave in 3D is very difficult so it is admittedly a rather outlandish scenario but say we somehow managed to do it (or simply switched to 2D) and are rolling a discus (a solid circle) and a ring (a perimeter or a circumference) of uniform density without slippage with all the above conditions standing.

For a discus I get A = [hide]4[/hide] and B = [hide]7[/hide]. And for a ring I get A = [hide]1[/hide] and B = [hide]2[/hide].

The ring will jump off the lowest - all of its mass is far away from the axis of rotation, expensive to get it rolling. Then comes the hollow sphere - a bit of its mass gets closer to the axis of rotation. Then comes the discus - some of its mass is close to the axis of rotation. The solid sphere will jump off the highest - a good amount of its mass is close to the axis of rotation.