## Halliday and Resnick - Boy sliding down ice mound

1. The problem statement, all variables and given/known data

This is from Fundamentals of Physics 8th ed, by Halliday and Resnick:
A boy is initially seated on the top of a hemispherical ice mound of radius R = 13.8m. He begins to slide down the ice, with a negligible initial speed. Approximate the ice as being frictionless. At what height does the boy lose contact with the ice?

2. Relevant equations

This is classified under Conservation of Mechanical Energy
There's a picture of a semicircle and a boy sliding down the circumference of the semicircle.

3. The attempt at a solution

I don't understand what conditions would make the boy go off the ice rather than stay on the ice. I thought it had to do with centripetal forces at first (cause he's going in a circle ... for a bit) but I couldn't make that idea work.
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 Quote by Farnak I thought it had to do with centripetal forces at first (cause he's going in a circle ... for a bit) but I couldn't make that idea work.
Try it again--that's the right track. Apply Newton's 2nd law. And conservation of energy.
 So far what I did was PEtop= KE bottom mgh=.5mv^2 gh=.5v^2 (9.81)h=(0.5)v^2 then at the point where the boy just barely stays on the ice mound, I used the equation Fn= Fw-Fc, which would be 0 at that point 0=mg-(mv^2/r) mg=(mv^2/r) g=v^2/r gr=v^2 (9.81)(13.8)=v^2 v^2= 135.378 m/s Plug that v^2 back into the initial equation: (9.81)h= (0.5)(135.378) h= 6.9m Does that look right?

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## Halliday and Resnick - Boy sliding down ice mound

 Quote by jlee888 then at the point where the boy just barely stays on the ice mound, I used the equation Fn= Fw-Fc, which would be 0 at that point
It's true that the normal force will be zero at that point, but only a component of the weight contributes to the centripetal force. Not the full mg, which acts downward. (Find the radial component.)