What Calculations Determine Roller Coaster Dynamics on a Vertical Track?

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SUMMARY

The discussion focuses on the calculations necessary to determine the dynamics of a roller coaster traversing a vertical circular track. Key equations include Newton's second law (EF=ma) and energy conservation principles (Etop=1/2mv²(top)+mgh). The speed required at the top of the loop is derived as v=sqrt(rg), where r is the radius and g is the acceleration due to gravity. The normal force on passengers at both the top and bottom of the loop is also analyzed, emphasizing the importance of applying Newton's laws to solve for these forces.

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  • Understanding of Newton's laws of motion
  • Basic principles of circular motion
  • Familiarity with energy conservation in physics
  • Knowledge of gravitational acceleration (g = 9.81 m/s²)
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  • Explore the derivation of centripetal force in circular motion
  • Learn about energy conservation in mechanical systems
  • Investigate the effects of varying radius on roller coaster dynamics
  • Study the impact of speed on normal force experienced by passengers
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Homework Statement



roller coaster traverses a vertical circular track.
a)what speed must car have to make it over the top w/o leaving the track? assume it is not attached to track.
b)what will speed be at bottom?
c)what will normal force on passenger be at the bottom of loop if radius is 14.5m?
d)at the top?

Homework Equations


a) EF=ma
n-mg=ma2
n=0 at top
b)Etop=1/2mv2(top)+mgh

The Attempt at a Solution


a) v=square root of (rg)
b)1/2mrg+mg(2*3.14) =2.5mgr
c)?
d)?
 
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Hi dr2112,

dr2112 said:

Homework Statement



roller coaster traverses a vertical circular track.
a)what speed must car have to make it over the top w/o leaving the track? assume it is not attached to track.
b)what will speed be at bottom?
c)what will normal force on passenger be at the bottom of loop if radius is 14.5m?
d)at the top?

Homework Equations


a) EF=ma
n-mg=ma2
n=0 at top
b)Etop=1/2mv2(top)+mgh

The Attempt at a Solution


a) v=square root of (rg)
b)1/2mrg+mg(2*3.14) =2.5mgr

The first term (1/2 mrg) looks right for the kinetic energy at the high point. What do the other two terms represent, and how did you get those numeric values?

c)?

Do the same thing for part c that you did in part a--use Newton's law for the forces.

d)?

You've already answered this in the working for part a.
 

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