At what height does the boy lose contact with the ice?

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The problem involves a boy sliding down a frictionless hemispherical ice mound of radius 13.8m, and the goal is to determine the height at which he loses contact with the ice. The discussion emphasizes using Newton's Second Law and the Conservation of Energy principles to analyze the forces acting on the boy, including gravity and centripetal force. Key equations involve the relationship between potential energy (PE) and kinetic energy (KE), as well as the dynamics of motion described in terms of angle theta. The approach includes calculating the forces at play and determining when the normal force becomes zero, indicating loss of contact. Understanding the geometry and forces involved is crucial for solving the problem accurately.
mikesown
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Homework Statement


"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 loose contact with the ice?"


Homework Equations


PE = mgh
KE = 1/2mv^2
PE + KE = constant

The Attempt at a Solution


I have absolutely 0 clue what to do... to start, I equated PE and KE to get:
mgh = 1/2mv^2
gh = 1/2v^2
h = 1/2v^2/g
I do not think that's correct though... Any ideas on how to approach this?
 
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You'll want to use two tools here, the first being Newton's Second Law. Using a centripetal coordinate system, how would you set up the F =ma equation? Second, friction is negligible, so you can use the Conservation of Energy to determine a height.

Think of it this way, if the boy moves a distance down the hill, he will have swept out an angle theta from his initial starting point. At this point, how would you construct Newton's Second Law? What two centripetal forces are acting on him?
 
If you look at his position at some angle theta from the top you should be able to compute the component of mg that is normal to the hemisphere. You should also be able to compute v as a function of theta and from that get the radial acceleration. If the normal component of mg drops below the value of the radial acceleration, he can't stay on the hemisphere.
 
Hmm, I'm still not completely understanding how to set up the problem. Here's what I'm understanding:
There are two forces acting on the person, the force of gravity(mg), and the centripetal force acting as he slides down.

I am still not understanding how to formalize this into mathematics. How can I describe his position as a function of theta?
 
mikesown said:
Hmm, I'm still not completely understanding how to set up the problem. Here's what I'm understanding:
There are two forces acting on the person, the force of gravity(mg), and the centripetal force acting as he slides down.

I am still not understanding how to formalize this into mathematics. How can I describe his position as a function of theta?

If you measure angle from the bottom of the hemisphere, x=R*cos(theta), y=R*sin(theta). The v can be found by considering the change in y from the top of the sphere. I'm not sure what your question is exactly.
 
The two forces acting on him is 1) a component of the weight force (this is his centripetal force) and 2) the normal force. I would suggest drawing a diagram if you haven't. If we take centripetally outward to be positive, our equation will look similar to this:

\Sigma F_{centrip} = N - \frac{cos \theta}{W} = -\frac{mv^2}{r}

Now you have two unknowns and one equation. Use the equation for the conservation of mechanical energy to get the boys speed at some angle theta, and plug it into the above equation. Then consider what force in the force equation must go to zero when the boy loses contact with the hill. This will allow you to solve for theta, which in turn will allow you to solve for the height.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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