## inkyvoyd one year ago a thing @empty

1. inkyvoyd

|dw:1444251095743:dw|

2. inkyvoyd

for the rollycoaster to not fall off, h=f(r)=?

3. inkyvoyd

so far I have for conservation of energy: 1/2 mv^2=mgh -> v^2=2gh

4. inkyvoyd

|dw:1444251172019:dw| is N up or down? I think it's down but I can't tell.

5. inkyvoyd

is the forces equation -N-mg=mv^2/r? or N-mg=mv^2/r

6. 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.

7. Empty

I wrote that thing on there but that would describe your minimum velocity at the bottom of the hill, ignore that.

8. inkyvoyd

so 2 r is enuf?

9. Empty

ya the diameter lol |dw:1444251872307:dw|

10. inkyvoyd

it's not taking it x.x

11. 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.

12. inkyvoyd

tried r, 2*r, 4*r, none were accepted x.x

13. 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

14. 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?

15. 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

16. 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.

17. inkyvoyd

uhh ok

18. Empty

does this not make sense?

19. Empty

|dw:1444252751446:dw|

20. inkyvoyd

huh

21. 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

22. 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.

23. inkyvoyd

LOL lemme try

24. inkyvoyd

YES

25. Empty

ok so no forces, forces are whack energy is where it's at.

26. inkyvoyd

wanna help with a nastier (imo) one?

27. Empty

sure I'll try no guarantees that I'm actually doing this right though.