How Does a Roller Coaster's Loop Affect Your Apparent Weight?

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SUMMARY

The discussion centers on the physics of roller coasters, specifically the apparent weight experienced by riders at the top and bottom of a circular vertical loop. The calculations demonstrate that the difference in apparent weight is 6 g's, derived from the equations of motion and conservation of mechanical energy. The apparent weight at the bottom is calculated as FNorm. Bot. = (2mgh/R) + mg, while at the top it is FNorm. Top. = (2mg(h-2R))/R - mg. The conclusion emphasizes that as long as the speed exceeds the minimum required, the results are independent of the loop's size or the speed through it.

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  • Understanding of circular motion dynamics
  • Familiarity with conservation of mechanical energy principles
  • Knowledge of forces acting on objects in motion
  • Basic algebra and equation manipulation skills
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  • Study the principles of circular motion in physics
  • Learn about the conservation of energy in mechanical systems
  • Explore the effects of friction on roller coaster dynamics
  • Investigate the relationship between speed, radius, and apparent weight in vertical loops
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ctpengage
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Show that a roller coaster with a circular vertical loop. The difference in your apparent weight at the top of the circular loop and the bottom of the circular loop is 6 g's-that is, six times your weight. Ignore friction. Show also that as long as your speed is above the minimum needed, this answer doesn't depend on the size of the loop or how fast your go through it.

My working for the first half of the problem, the 6g's part is as follows

Radius if loop is R
Height from which it is released is h

The speed at bottom of the loop is determined by the conservation of mechanical energy
1/2 mvbottom2=2mgh

Apparent weight at the bottom of the loop is obtained by the below:

mvBot2= FNorm. Bot.-mg
Therefore apparent weight at bottom is
FNorm. Bot.=mvbot2/R+mg
FNorm. Bot.=2mgh/R+mg (using result obtained via conservation of energy)

To find speed at top of the loop we have from Conservation of Energy
1/2 mvtop2+mg(2R)=mgh
mvtop2=2mg(h-2R)
Therefore using the above the apparent weight at the top of the loop is

mvTop2/R = FNorm. Top.+mg
Therefore Apparent weight is :
FNorm. Top. = (2mg(h-2R))/R - mg

Hence
FNorm. Bot. - FNorm. Top. =
2mgh/R + mg - [((2mg(h-2R))/R - mg)]=
2mgh/R + mg - 2mgh/R + 4mg + mg=
6mg

That's how I proved the first part of the problem. Can anyone please tell me how to complete the second part of the problem; namely proving that as long as your speed is above the minimum needed, the answer doesn't depend on the size of the loop or how fast your go through it. This part of the problem is really bugging me and I've tried heaps of ways but can't come up with the answer.
 
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ctpengage said:
The speed at bottom of the loop is determined by the conservation of mechanical energy
1/2 mvbottom2=2mgh

To find speed at top of the loop we have from Conservation of Energy
1/2 mvtop2+mg(2R)=mgh
mvtop2=2mg(h-2R)

Hi ctpengage! :smile:

I'm a little confused …

you seem to have got the right result with the wrong equations. :confused:

You have two equations for conservation of energy, one for the top and one for the bottom …

but conservation means that the energy for top and bottom should be in the same equation, doesn't it? :smile:
 

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