a thing @empty

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|dw:1444251095743:dw|
for the rollycoaster to not fall off, h=f(r)=?
so far I have for conservation of energy: 1/2 mv^2=mgh -> v^2=2gh

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|dw:1444251172019:dw| is N up or down? I think it's down but I can't tell.
is the forces equation -N-mg=mv^2/r? or N-mg=mv^2/r
Energy is scalar, forces are vectors, so things are usually easier to do with just looking at energy considerations. |dw:1444251302235:dw| In order to get through the loop you have to have some energy, since energy is conserved (not lost through friction) then you just need \[U_i > mg 2r\] at the top of the hill and that will get you through the loop.
I wrote that thing on there but that would describe your minimum velocity at the bottom of the hill, ignore that.
so 2 r is enuf?
ya the diameter lol |dw:1444251872307:dw|
it's not taking it x.x
With no friction, the ball rolls back and forth up the height and back down again. If you want forces, \[F=-\frac{d U}{dx}\] Huh what do you mean it's not taking it what kind of answer are you plugging in.
tried r, 2*r, 4*r, none were accepted x.x
I think the problem is that it has enough energy to get to that point, but the velocity is still too low cause it needs moar centripetal acceleration
Oh so at the top of the hill it's not enough to have 0 velocity it needs what a=v^2/r or something?
I was thinking it was on a track then, where not having 2r height if it started from rest would mean it wouldn't get through the loop
it would stop and roll backwards so ok simple fix then, since you want kinetic energy at the height to be \[K_f = \frac{1}{2} m v^2\] with \[ar=v^2\] so let me work through this to make sure.
uhh ok
does this not make sense?
|dw:1444252751446:dw|
huh
idk if that works, cause it's non uniform circular motion... v at the base is different from v at the top of the loop
good question so we have: \[mgh=\frac{1}{2}mv^2+mg2r\] and \[\frac{v^2}{r}=a\] at that point, and the acceleration at that point is gravity, a=g: \[mgh=\frac{1}{2} m gr+mg2r\] Oh look it's the christmas miracle \[h=\frac{5}{2} r\] I think that should work try it.
LOL lemme try
YES
ok so no forces, forces are whack energy is where it's at.
wanna help with a nastier (imo) one?
sure I'll try no guarantees that I'm actually doing this right though.

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