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

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To determine the minimum tangential velocity for a kid to complete a frictionless loop-the-loop with a radius of 1.17 meters and a weight of 56.7 kg, the analysis begins with energy conservation principles. The initial kinetic energy is zero since the kid starts from rest, leading to the equation gh[top] = 1/2v^2 + gh[loop]. However, the height at the top of the ramp is unknown, complicating the solution. A free body diagram approach shows that at the top of the loop, gravitational force must equal the centripetal force, resulting in the equation mg = mv^2/r. This simplifies to the velocity formula v = sqrt(g*r), providing the necessary minimum speed to successfully navigate the loop.
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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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