Maximum Height for a Car to Roll Over a Hill on a Frictionless Track

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SUMMARY

The maximum height \( h \) from which a car can start on a frictionless track without flying off the hilltop is determined by the relationship between gravitational potential energy and centripetal force. The car must start from a height of at least \( 2R \) to maintain contact with the track at the top of the hill, where \( R \) is the radius of the circular hill segment. The calculations involve equating potential energy \( mgh \) to the necessary centripetal force \( \frac{mv^2}{R} \), leading to the conclusion that \( h = 2R \) is the critical height for safe traversal.

PREREQUISITES
  • Understanding of gravitational potential energy
  • Knowledge of centripetal force and its formula
  • Familiarity with basic physics concepts such as energy conservation
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the principles of energy conservation in mechanical systems
  • Learn about centripetal motion and the forces involved
  • Explore the implications of frictionless surfaces in physics problems
  • Investigate real-world applications of circular motion in automotive engineering
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the dynamics of motion on curved tracks.

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Homework Statement


A car on the frictionless track starts from rest at height h. The tracks valley and hill
consists of a circular-shaped segments of radius R.
What is the maximum height h from which the car can start so as not to fly off the track when going over the hill?
Give your answer as a multiple of R.

The Attempt at a Solution


ok I know that its potential energy is mgh when we let it go. And we need to put it at least
the height of R to make it the top of the hill on the other side.
But I am not sure how fast it can be going and not lift off the ground.
I know that it would have to exceed its weight. I am not sure how to relate the energy into force.
 

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Think about the path of the car over the top of the hill. It's a curve of radius R so what force do objects moving on a curved path experience?
 
ok thanks.
so now I have
\frac{mv^2}{r}=mg
v=\sqrt{rg}
then
mg(h-r)=m\frac{rg}{2}
and this gets me the correct answer.
thanks for the help .
 

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