Boy Loses Contact with Ice Mound: Approximate Solution

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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 scenario is set in the context of classical mechanics, specifically focusing on forces, motion, and centripetal acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the boy, including gravity and centripetal force, and question the relationship between acceleration, velocity, and position at the point of losing contact. There are attempts to derive equations relating these variables, and some participants express uncertainty about their reasoning and calculations.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the forces involved and attempting to relate them mathematically. Some guidance has been offered regarding the need to consider the radial component of forces and the relationship between height and angle, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are working under the assumption that the ice is frictionless and are grappling with the implications of this assumption on the forces at play. There is also a noted confusion regarding the definitions and relationships between force, acceleration, and velocity.

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A boy is initially seated on the top of a hemispherical ice mound of radius R = 2.6 m. He begins to slide down the ice, with a negligible initial speed (Fig. 8-47). Approximate the ice as being frictionless. At what height does the boy lose contact with the ice?
 
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Ah, a classic problem!

What are your ideas to solve it? Can you say what is special about the position or velocity or acceleration at the point that contact is lost?
 
i really have no idea where to start. the velocity at that point is very small and the acceleration would be positive, the position would be a height less than 2.6 but that's all i can figure out.
 
Will the velocity really be "very small"? What would "very small" mean?

Before he leaves the ice, how is his acceleration in the radial direction related to his speed? What forces produce this acceleration? (Try drawing a free-body diagram.)

Are all those forces still present at the moment he loses contact?
 
kapopka88 said:
and the only force acting on the boy is gravity.
After he leaves the ice, yes, but before he leaves there is another force.
 
oh centripetal force is also acting on him before he looses contact
 
v=sqrt(2g(r-h))
 
kapopka88 said:
oh centripetal force is also acting on him before he looses contact
Well, I would call it the normal force of the ice on the boy.
 
kapopka88 said:
v=sqrt(2g(r-h))
Yes. (Can you explain why this equation is valid?)

So, you have one relation between v and h.

At the moment he loses contact, the normal force of the ice on the boy is zero. So, his centripetal acceleration must be produced by the force of ? And the component of that force in the radial direction is ? (Why do we want the component in the radial direction?)
 
  • #10
That's a sloppy way of saying it. I would say the centripetal acceleration is produced by the force of gravity. It's best not to confuse cause (force) with effect (acceleration).
 
  • #11
so, centripetal force =0, so gravity must produce his centripetal acceleration. how do we find the radial component with no angle?
 
  • #12
i set the centripetal force equal to gravity to solve for v, then plugged v into my first equation to get height but that was not right.
 
  • #13
kapopka88 said:
so, centripetal force =0, so gravity must produce his centripetal acceleration. how do we find the radial component with no angle?
There is an angle; the boy has moved off the top, and now a line from him to the center of the sphere makes an angle theta with respect to vertical. How is this angle related to h and R?
 
  • #14
kapopka88 said:
i set the centripetal force equal to gravity to solve for v, then plugged v into my first equation to get height but that was not right.
Show your work!
 
  • #15
sin(theta)=h/r
 
  • #16
so, so far i have F=mv^2/R, v=sqrt(2g(r-h)), and sin(x)=h/r
 
  • #17
What is F?
 
  • #18
the magnitude of the centripetal force
 
  • #19
And that should be equal to ?
 
  • #20
No, sorry. I think you need to find someone locally who can help you with some basics. (A force is not going to be equal to a length ...)
 

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