Halliday and Resnick - Boy sliding down ice mound

AI Thread Summary
The discussion centers on a physics problem involving a boy sliding down a frictionless hemispherical ice mound. The key focus is on determining the height at which the boy loses contact with the ice, using principles of conservation of mechanical energy and Newton's second law. Initial calculations suggest that the boy's potential energy at the top converts to kinetic energy at the bottom, leading to a derived speed. The critical point of analysis is when the normal force equals zero, indicating loss of contact, which requires considering only the radial component of gravitational force for centripetal acceleration. The final calculated height at which the boy loses contact is approximately 6.9 meters.
Farnak
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Homework Statement



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?

Homework 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.

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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Farnak said:
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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jlee888 said:
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.)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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