Child sliding down a frictionless slide

AI Thread Summary
A child sliding down a frictionless slide will lose contact with the slide at a height h that depends on the initial height H and the radius R of the slide's curvature. The relationship involves equating potential energy and kinetic energy, leading to the equation mgH = mgh + 1/2mv^2. The centripetal force equation, F = mv^2/r, is also relevant for determining the conditions under which contact is maintained. The derived solution indicates that h equals 2H/3 if R is greater than or equal to 2H/3; otherwise, h equals R. This analysis combines principles of energy conservation and centripetal motion.
jfreimer
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Homework Statement



A child starts from rest and slides down a frictionless slide. In terms of R and H,
at what height h will he lose contact with the section of radius R?

(child starts at the top with height H, at the bottom the slide flattens out and then forms a quarter circle to the ground with radius R)


Homework Equations





The Attempt at a Solution


I tried figuring it out using potential energy and kinetic energy since that is the chapter. I think it has something to do with mgH=mgh+1/2mv^2 and then solving v^2 for some value using centripetal motion but I wasn't sure how to set that step up.
 
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I had you attempt at a solution in mind as well. Equating Potential Energies and solving. Now, what is the equation for Centripetal Force?

F=\frac{mv^2}{r}

Just work from there.
 
My attempt for a solution has given the answer h=2H/3 if R >=2H/3 , else h=R
Is it possible?
 
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