# Hemisphere sliding down

1. Dec 13, 2007

### mikesown

1. The problem statement, all variables and given/known data
"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?"

2. Relevant equations
PE = mgh
KE = 1/2mv^2
PE + KE = constant

3. 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?

2. Dec 13, 2007

### hotcommodity

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?

3. Dec 13, 2007

### Dick

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.

4. Dec 13, 2007

### mikesown

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?

5. Dec 13, 2007

### Dick

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.

6. Dec 13, 2007

### hotcommodity

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.