Conservation of mechanical energy for skier on sphere

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Homework Help Overview

The problem involves a skier of mass m sliding down a frictionless solid sphere of radius r, with the goal of determining the angle θ at which the skier leaves the sphere. The context centers around the conservation of mechanical energy and the forces acting on the skier as they descend.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy and kinetic energy, with one noting the condition for the skier to leave the sphere when the normal force is zero. There are attempts to set up equations based on energy conservation and force analysis, with questions about the correct expressions for height and forces involved.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to set up the problem. Some have provided hints and guidance regarding the need to express height in terms of the angle θ and to consider the vector nature of forces. There is an acknowledgment of confusion and a request for clarification on the steps involved.

Contextual Notes

Participants are grappling with the correct application of energy conservation and force equations, with specific attention to the geometry of the situation and the components of forces acting on the skier.

NathanLeduc1
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Homework Statement


A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface. At what angle θ will the skier leave the sphere?

Homework Equations


KE= 0.5mv^2
PE = mgh
Fc = (mv^2)/r


The Attempt at a Solution


I am really quite confused and don't even know how to begin. I know that the skier will fall off when the normal force is 0 but I'm not sure how to even get to that point.
I thought that PE=KE so 0.5mv^2 = mgh which simplifies to 0.5v^2 = gh
 
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NathanLeduc1 said:
I know that the skier will fall off when the normal force is 0 but I'm not sure how to even get to that point.
That's the key point. To make use of it, set up a force equation with Newton's 2nd law.
I thought that PE=KE so 0.5mv^2 = mgh which simplifies to 0.5v^2 = gh
Good. You'll need that too. (Express h in terms of θ.)
 
So rewriting the energy equations gives me
v^2 / 2 = grsin(θ).

When I write the force equations, I get
N-mg = ma
N = mg + mg
N = m(a+g)
m = N/(a+g)

If I plug that into the centripetal force equation, I get
F = (mv^2)/r
F = (Nv^2)/((r)(a+g))

Is this right? If so, where do I go from here? Sorry to ask such dumb questions, I'm just very confused on this problem.
 
NathanLeduc1 said:
So rewriting the energy equations gives me
v^2 / 2 = grsin(θ).
What you want is Δh, the drop from the original position at the top. Δh ≠ r sinθ.

When I write the force equations, I get
N-mg = ma
N = mg + mg
N = m(a+g)
m = N/(a+g)
Careful! Forces are vectors.

Hint: Consider force components perpendicular to the surface at any point.
 

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