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inkyvoyd
 one year ago
a thing @empty
inkyvoyd
 one year ago
a thing @empty

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inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444251095743:dw

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1for the rollycoaster to not fall off, h=f(r)=?

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1so far I have for conservation of energy: 1/2 mv^2=mgh > v^2=2gh

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444251172019:dw is N up or down? I think it's down but I can't tell.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1is the forces equation Nmg=mv^2/r? or Nmg=mv^2/r

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Energy 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
 one year ago
Best ResponseYou've already chosen the best response.1I wrote that thing on there but that would describe your minimum velocity at the bottom of the hill, ignore that.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1ya the diameter lol dw:1444251872307:dw

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1it's not taking it x.x

Empty
 one year ago
Best ResponseYou've already chosen the best response.1With 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
 one year ago
Best ResponseYou've already chosen the best response.1tried r, 2*r, 4*r, none were accepted x.x

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1Oh 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
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1it 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1does this not make sense?

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1idk 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
 one year ago
Best ResponseYou've already chosen the best response.1good 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1ok so no forces, forces are whack energy is where it's at.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1wanna help with a nastier (imo) one?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1sure I'll try no guarantees that I'm actually doing this right though.
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