How Fast Must a Kid Go to Complete a Frictionless Loop-the-Loop?

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SUMMARY

The discussion focuses on calculating the minimum tangential velocity required for a kid weighing 56.7 kg to successfully navigate a frictionless loop-the-loop with a radius of 1.17 meters. The analysis begins with the conservation of energy principle, leading to the equation gh[top] = 1/2v^2 + gh[loop]. However, the lack of information regarding the height at the top of the ramp complicates the solution. Ultimately, the correct approach involves applying the centripetal force equation, resulting in the formula v = sqrt(g * r), where g represents gravitational acceleration.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with centripetal force equations
  • Knowledge of gravitational acceleration (g = 9.81 m/s²)
  • Ability to analyze free body diagrams
NEXT STEPS
  • Study the principles of conservation of mechanical energy in physics
  • Learn about centripetal acceleration and its applications
  • Explore free body diagram techniques for analyzing forces
  • Investigate real-world applications of loop-the-loop dynamics in amusement park rides
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of motion in circular paths, particularly in the context of amusement park physics.

mohabitar
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We're analyzing this video here:
https://www.youtube.com/watch?v=ye3rF1MY9Ho

Radius of the loop is 1.17
No friction
Kid weighs 56.7 kg.
He starts from rest at the top of the ramp

What is the minimum tangential velocity required to successfully navigate the loop?

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So I did 1/2mv^2+mgh[top]=1/2mv[f]^2+mgh[loop]
The first 1/2m^2 cancels since v is zero.
The m's cancel out on all sides, and we're left with gh[top]=1/2v^2+gh[loop]
However, we have two unknowns in this equation: the height at the top of the ramp and the final velocity, so we can't solve. And the top height is not given to us. Am I missing something here? Am I doing this wrong?
 
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Nobody? This seems like a simple one to help out with?
 
well.. I am not sure but the way that i did the first question was that I used a free body diagram for when the dude was at the top of the loop.
so the only force acting on the dude would be gravity. so you can say that the force of gravity is equal to the force of centripetal acceleration, which means that
the net force is equal to mv^2/r
so...
mg=mv^2/r
v=sqrt(g*r)
 

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