Loop-the-loop in roller coaster

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  • #1
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Quick question: In a loop-the-loop, is the speed at which the coaster enters the loop equal to the speed at which it leaves?
 

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  • #2
berkeman
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Not quite. Quiz question -- why not?
 
  • #3
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Not quite. Quiz question -- why not?
Without friction: YES

WITH friction (real life): NO
 
  • #4
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Not quite. Quiz question -- why not?
Ummm...No?

As noted. Friction.

Also:
1) Bearing losses in wheels.
2) Air braking. Think when people are more likely to have thier arms in the air.
3) Lost purses, junk etc. ejected at top of loop? (Fanciful I know.)
4) Geometry of the loop itself.
 
  • #5
rcgldr
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Most roller coasters are very massive and have relatively little frontal area compared to the amount of mass (they's heavy and long), so the losses due to aerodynamic drag aren't that much more than the losses due to friction. As the grease in the coaster wheels warms up, the coasters will go faster, unless there are speed controls (usually spinning tires under the track to regulate coaster speed).
 
  • #6
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Of course the coaster would leave the loop with less speed than when it entered it - friction - but how do you calculate the speed of the coaster when it is leaving the loop?
 
  • #7
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If we assume the loop was level going in and out, then you'd have losses. If downhill, then you have gravity which could make it faster going out.
 
  • #8
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Lets assume its level coming out, how would go you go about finding its velocity?
 
  • #9
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You would need more information, such as the loss of kinetic energy in the coaster, or the average force of friction.
 
  • #10
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so there isnt an equation dealing with the degree of a loop and the velocity, mass, of coaster and pull of gravity to determine the end velocity? if friction and air resistance were negligible.
 
  • #11
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If friction and air resistance are negligible, it's a whole 'nother question you're dealing with. The change in velocity would depend entirely in the change in graviational potential energy (the vertical work done by gravity), otherwise energy is conserved and Ek is still the same. If Ek is still the same, velocity is still the same.

It's similar to how if you have a ball, and you throw it up in the air. No matter what angle you throw it at (as long as it's up), it'll land back in your hand with the same magnitude of velocity (assuming your hand stays at the same height). Regarding the rollercoaster, if we neglect friction, it's just as if it were a ball flying through the air.

So I'm guessing delta h and v1 are your only dependent variables. You could write an equation for v2 using conservation of energy, kinetic energy, and gravitational potential energy.

Keep in mind I'm basing this off what I've learned in Grade 12 Physics. There may be circumstantial information our teacher has withheld from us. But in the meantime, I hope this helps.
 
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  • #12
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Hmm i guess thats true. I think i am maybe trying to over analyze my roller coaster problem. Thank you for your time and effort, much appriciated.
 

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