Max Velocity and Height of Roller Coaster Track

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SUMMARY

The discussion focuses on calculating the maximum initial velocity (v_{o}) for a roller coaster cart to remain on track at point A and determining the height (h') required for the cart to just reach point B. The equations used include energy conservation (Ei = Ef) and centripetal force (mv^2/r = mg). The final expressions derived are v_{o} = √(g*(r-(2/3)h)) for maximum velocity and h' = r/2 + 2h/3 for the required height at point B.

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  • Understanding of energy conservation principles in physics
  • Familiarity with centripetal force equations
  • Knowledge of basic kinematics and motion in circular paths
  • Ability to manipulate algebraic equations involving square roots
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  • Study the principles of energy conservation in mechanical systems
  • Learn about centripetal acceleration and forces in circular motion
  • Explore the dynamics of roller coaster design and safety parameters
  • Investigate the effects of friction on roller coaster performance
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Physics students, mechanical engineers, and anyone interested in the dynamics of roller coaster design and safety analysis.

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Homework Statement


b7mank.jpg

A man starts riding on a frictionless roller coaster track at initial velocity v_{o}. At point A the radius of the bend is r.

a) What is the maximum v_{o} such that when the cart gets at point A it stays on the track?
b) Using you answer in a) what should h' be such that the cart only just makes it to point B?

Homework Equations



Ei = Ef
mv^2/r

The Attempt at a Solution


So this question was kind of bugging me, and I want to know if I got it right so for the a) part this is what I did firstly

mgh + (1/2)mv_{o}^{2} = mg(2/3)h+1/2mv^{2}
v = \sqrt{2gh/3 + Vo^2}

and then since at point be its a circle and its looking for the max velocity until the normal force = 0 I did
mv^2/r = mg
v = \sqrt{rg}
then
rg = 2gh/3 + v_{o}^{2}
v_{o} = \sqrt{g*(r-(2/3)h)}THEN for b) I just used Energy again.
so
v = \sqrt{rg}
Ei = (1/2)mrg + mg(2/3)h
Ef = mgh'
h' = r/2 + 2h/3
 
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Looks good to me. :thumbs:
 
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Likes   Reactions: 1 person
Oh thank you so much!This was one of two questions I was unsure of so I just went with my gut on it.
 

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