inkyvoyd
  • inkyvoyd
a thing @empty
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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inkyvoyd
  • inkyvoyd
|dw:1444251095743:dw|
inkyvoyd
  • inkyvoyd
for the rollycoaster to not fall off, h=f(r)=?
inkyvoyd
  • inkyvoyd
so far I have for conservation of energy: 1/2 mv^2=mgh -> v^2=2gh

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inkyvoyd
  • inkyvoyd
|dw:1444251172019:dw| is N up or down? I think it's down but I can't tell.
inkyvoyd
  • inkyvoyd
is the forces equation -N-mg=mv^2/r? or N-mg=mv^2/r
Empty
  • Empty
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.
Empty
  • Empty
I wrote that thing on there but that would describe your minimum velocity at the bottom of the hill, ignore that.
inkyvoyd
  • inkyvoyd
so 2 r is enuf?
Empty
  • Empty
ya the diameter lol |dw:1444251872307:dw|
inkyvoyd
  • inkyvoyd
it's not taking it x.x
Empty
  • Empty
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.
inkyvoyd
  • inkyvoyd
tried r, 2*r, 4*r, none were accepted x.x
inkyvoyd
  • inkyvoyd
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
Empty
  • Empty
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?
Empty
  • Empty
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
Empty
  • Empty
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.
inkyvoyd
  • inkyvoyd
uhh ok
Empty
  • Empty
does this not make sense?
Empty
  • Empty
|dw:1444252751446:dw|
inkyvoyd
  • inkyvoyd
huh
inkyvoyd
  • inkyvoyd
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
Empty
  • Empty
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.
inkyvoyd
  • inkyvoyd
LOL lemme try
inkyvoyd
  • inkyvoyd
YES
Empty
  • Empty
ok so no forces, forces are whack energy is where it's at.
inkyvoyd
  • inkyvoyd
wanna help with a nastier (imo) one?
Empty
  • Empty
sure I'll try no guarantees that I'm actually doing this right though.

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