A skier starts at rest on the top of Mt Circular, a strange, smooth, icy hill shaped like a hemisphere. The hill has a constant radius of R. Neglecting friction (it is icy!), show that the skier will leave the surface of the hill and become air-borne at a vertical distance of h=R/3, measured from the top of the hill.
I think the way to do it is to use forces, since at R/3 Fn=0, and combine that with conservation of energy, but everything I try just leads in one big circle and I can't get anywhere.
The normal component of weight has to less than centripetal force. Solve that .
To solve this problem, we can use the concept of conservation of energy. We know that the skier starts from rest at the top of the hill, so the initial kinetic energy is zero. We also neglect friction, so there is no dissipative force acting on the skier.
Let's consider the different points along the path of the skier. We'll take the top of the hill as our reference point (zero height) and measure the vertical distance downwards. We'll use a coordinate system where positive values are downwards.
At the top of the hill (initial position), the skier has potential energy given by U = mgh, where m is the mass of the skier and g is the acceleration due to gravity. Since the skier starts from rest, there is no initial kinetic energy.
As the skier moves down the hill, potential energy is converted into kinetic energy. At any point along the hill, we can write the total mechanical energy (sum of kinetic and potential energy) as:
E = 1/2 mv^2 + mgh
Where v is the speed of the skier at that point. We can rewrite this equation as:
E = 1/2 mv^2 + (mg)(h - y)
Where y is the vertical distance below the top of the hill.
When the skier reaches the point where y = R/3 (a vertical distance of R/3 below the top of the hill), they become airborne. At this point, the normal force acting on the skier becomes zero, i.e., the surface of the hill can no longer support the skier.
To determine the speed of the skier at this point, we can set the total mechanical energy to zero:
0 = 1/2 mv^2 + (mg)(h - R/3)
Since the skier leaves the surface and becomes airborne, the normal force equals zero. This means that the only force acting on the skier is the force of gravity, mg.
Simplifying the equation, we have:
1/2 mv^2 = (mg)(R/3)
Canceling the mass 'm' on both sides, we find:
1/2 v^2 = g(R/3)
v^2 = (2gR)/3
v = sqrt(2gR/3)
Therefore, the speed of the skier at the point where they leave the surface is v = sqrt(2gR/3).
So, using conservation of energy, we have found the speed of the skier, but we also need to show that the skier becomes airborne at a vertical distance of h = R/3.
Let's calculate the vertical distance when the skier becomes airborne:
At this point, the skier's kinetic energy is given by 1/2 mv^2, and the potential energy is zero since y = R/3. Thus, the total mechanical energy is:
E = 1/2 mv^2 + 0
Since mechanical energy is conserved, this must be equal to the initial mechanical energy:
0 = 1/2 mv^2 + mgh
Since the initial height is h (zero height) and the final height is R/3, we can rewrite this equation as:
0 = 1/2 mv^2 + mg(R/3)
Simplifying further:
0 = m(v^2 + gR/3)
Since mass 'm' is nonzero, we can conclude that v^2 + gR/3 = 0.
Substituting the value of v^2 we found earlier:
(2gR/3) + (gR/3) = 0
(3gR + gR)/3 = 0
4gR/3 = 0
This equation cannot be satisfied for any value of R, which means the skier can never become airborne at a vertical distance h = R/3 below the top of the hill.
Therefore, there might be a mistake or a misunderstanding in the problem statement. Double-check the question or seek clarification if needed.