Quote:
That could happen, but that's not what the equations you set up represent. That would be more like:
Ball: y = 10 - v*t - (1/2)g*t^2
Platform: y = 0 + (1/2)a*t^2
|
Ohh, so it can happen because if y = 10, then
0 = -v*t - (1/2)g*t^2
v*t = - (1/2)g*t^2
v = - (1/2)gt
-v = (1/2)gt
and the velocity will be a number so time can pass! And the negative is probably because gravity and the velocity upward are not going in the same direction?
So if y = d
Ball: 0 = -v*t - (1/2)g*t^2
Platform: d = d' + (1/2)a*t^2
at' = v where t' is how much time the platform has been accelerating in total
And then to plug that into 0 = -v*t - (1/2)g*t^2
at't = - (1/2)gt^2
a = (-(1/2)gt^2)/(t't)
So... given the time the platform has been moving and given the time it took for the ball to hit the platform, it's possible to find an acceleration for which the ball falls lands on the platform where it started? It looks a lot less convenient than I'd imagined in the beginning!
Uh, granted I have no clue if this is messed up or not, or if it's a quadratic equation. It doesn't really look like one, I think.
Quote:
|
This isn't quite representing the same system -- the previous system is a special case of this one where r = 0, but this system of equations can represent more possibilities.
|
At this point I think it's starting to go in over my head X'D. I don't really understand how equations relate to each other, or what systems are.
Quote:
|
You can add the force vectors together (this is a two-dimensional operation) to determine the net force that the person would feel. You could theoretically put a diagonal platform inside that box perpendicular to that net force vector, and the person could stand on it.
|
Ohh that's really cool! How much of a stretch would it be to say that in the moment a person is being pushed or pulled, their idea of gravity is changed and so they lose their balance?
As for the centrifuge, I sadly haven't learned a thing about rotating systems yet, or calculus '~'.