| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1752.1 | | AUSSIE::GARSON | nouveau pauvre | Sun May 23 1993 00:40 | 3 |
| re .0
Sounds like you need some PHYSICS to answer this.
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| 1752.2 | Differential equation | IOSG::CARLIN | Dick Carlin IOSG, Reading, England | Sun May 23 1993 04:00 | 10 |
| Derek, you're right, it is physics. But I soon got to the point where
the angle swept by the body satisfies the differential equation (wrt
time):
y'' = c * sin(y) y(0) = 0, y'(0) = 0
which I suppose is still physics though. Differential equations are so
messy, give me number theory any day.
Dick
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| 1752.3 | 2nd try | IOSG::CARLIN | Dick Carlin IOSG, Reading, England | Mon May 24 1993 05:19 | 42 |
| Well, I think honour may be restored. .2 in isolation was really a dead
end.
If penguin has traversed angle y
Resolving along tangent:
m*g*sin(y) = m*r*y'' (I) ( ' = differential wrt time)
and along normal:
m*g*cos(y) - N = m*r*(y')� (II) ( N = normal force, not constant)
Differentiate (II):
-m*g*sin(y)*y' - N' = 2*m*r*y'*y''
Eliminate y'' using (I):
N' = -3*m*g*sin(y)*y'
Integrate:
N = 3*m*g*cos(y) + C (that came out suspiciously conveniently,
implying that this is an over-complicated
way of deriving it)
Initial condition N = m*g when y = 0, so C = -2*m*g:
N = 3*m*g*cos(y) - 2*m*g
Penguin becomes airborne when N = 0, ie:
cos(y) = 2/3 (just over 1/2 way round)
------------
So it seems that trying to solve for y in terms of t was
(a) unnecessary
(b) difficult
Dick
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| 1752.4 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Mon May 24 1993 15:22 | 9 |
| Computing y(t) is neat. If the object starts at the top of
the hemisphere with 0 kinetic energy, then it stays there
forever. If instead it starts at cos(y) = 2/3 with the exact
amount of kinetic energy directed upwards to get back to the
top with 0 kinetic energy left, then it takes forever to
actually get to the top, although it does approach arbitrarily
closely (going more and more slowly) as t -> oo.
Dan
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