Halliday and Resnick - Boy sliding down ice mound

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Homework Help Overview

The problem involves a boy sliding down a hemispherical ice mound, with the goal of determining the height at which he loses contact with the ice. The context is rooted in mechanics, particularly focusing on concepts of energy conservation and forces acting on the boy as he descends the mound.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between centripetal forces and the conditions for the boy to maintain contact with the ice. There is an attempt to apply conservation of energy principles and Newton's second law to analyze the situation.

Discussion Status

The discussion is ongoing, with some participants providing guidance on applying Newton's second law and conservation of energy. There are differing interpretations of the forces at play, particularly regarding the components of weight contributing to centripetal force.

Contextual Notes

Participants are navigating the complexities of the problem, including the assumptions of a frictionless surface and the dynamics of circular motion. There is a focus on understanding the conditions under which the boy loses contact with the ice, which remains a point of inquiry.

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.)
 

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