[Robert A Cook PE]

Congrats!

You won't ever regret it .... I think. 8<)

(Thinking of Monday night at 12:30 working with Jean on her Physics 3302 problems: many mechanical, rotation inertia; sliding objects problems ...)

One funny problem I need to consult Eala, Maxie and Cyber need to think about: (Xsex can probably do it in her sleep ... so I won't ask her.)

Imagine an eskimo sitting on a prefectly round frictionaless(hemispherical) igloo at the north pole. A very tiny force starts moving him sideways: at what point does his bottom leave the igloo's surface?

Way I figgured it: ya gotta calculate the x,y,z vectors of his weight (forces down, out, and sideways) as a sin/cos function of his angular position starting from the top and going down all the to the extreme side point.

This gives you a vector of sideways acceleration AND downwards acceleration that changes with his radial position: so the "perfect" trajectory laws of motion (free fall in vacuum where there is NO horizontal acceleration and a constant downwards accelration don't apply)!

Assume there is no igloo in the way, and calc the trajectory for his path using the changing horizontal acceleration to get (x,t) and the changing (vertical) acceleration to get (y,t): plot the curve.

Plot a second curve of a quarter circle.

find the intersection of the two - this is where his modified" parabolic trajectory leaves the surface of the igloo.

Sound right? (Jean doesn't think so...)

[xsmommy]

oh that's so easy.....

[Conspiracy Guy]

I hate igloos.

[camle]

if he's sitting atop an igloo, he's probably stick to it...

[Eala]

Robert, I think you're making it too difficult.

I'm going to add a few assumptions, such as no air, pefectly round igloo, and I am assuming the point you're seeking is referenced to the igloo-earth contact point in line with the direction in which he started.

So... it gets easier if you change your frame of reference. The first step is to calculate how long it takes him to reach the ground moving in a single plane; that's the hard part (but computationally easier than figuring the 3-D trajectory). But once you have that, all you need to do is to figure out how far the Earth has rotated in that time, and that's your contact point.