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inkyvoyd

  • one year ago

a thing @empty

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  1. inkyvoyd
    • one year ago
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    |dw:1444251095743:dw|

  2. inkyvoyd
    • one year ago
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    for the rollycoaster to not fall off, h=f(r)=?

  3. inkyvoyd
    • one year ago
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    so far I have for conservation of energy: 1/2 mv^2=mgh -> v^2=2gh

  4. inkyvoyd
    • one year ago
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    |dw:1444251172019:dw| is N up or down? I think it's down but I can't tell.

  5. inkyvoyd
    • one year ago
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    is the forces equation -N-mg=mv^2/r? or N-mg=mv^2/r

  6. Empty
    • one year ago
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    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.

  7. Empty
    • one year ago
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    I wrote that thing on there but that would describe your minimum velocity at the bottom of the hill, ignore that.

  8. inkyvoyd
    • one year ago
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    so 2 r is enuf?

  9. Empty
    • one year ago
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    ya the diameter lol |dw:1444251872307:dw|

  10. inkyvoyd
    • one year ago
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    it's not taking it x.x

  11. Empty
    • one year ago
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    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.

  12. inkyvoyd
    • one year ago
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    tried r, 2*r, 4*r, none were accepted x.x

  13. inkyvoyd
    • one year ago
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    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

  14. Empty
    • one year ago
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    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?

  15. Empty
    • one year ago
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    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

  16. Empty
    • one year ago
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    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.

  17. inkyvoyd
    • one year ago
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    uhh ok

  18. Empty
    • one year ago
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    does this not make sense?

  19. Empty
    • one year ago
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    |dw:1444252751446:dw|

  20. inkyvoyd
    • one year ago
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    huh

  21. inkyvoyd
    • one year ago
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    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

  22. Empty
    • one year ago
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    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.

  23. inkyvoyd
    • one year ago
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    LOL lemme try

  24. inkyvoyd
    • one year ago
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    YES

  25. Empty
    • one year ago
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    ok so no forces, forces are whack energy is where it's at.

  26. inkyvoyd
    • one year ago
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    wanna help with a nastier (imo) one?

  27. Empty
    • one year ago
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    sure I'll try no guarantees that I'm actually doing this right though.

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