# Skier question

1. Nov 5, 2006

### amcca064

Ok so the question is,
"A skier of mass 60kg, initially at rest, slides down from the top of a frictionless, icy, hemispherical, mountain with a radius of curvature R of 100m.

a)Draw a free body diagram and write the Newton's Equations at the moment when he/she is at some point below the top of the mountain.

b)Find the angle 'alpha' with the horizontal surface at which he/she will lose contact with the mountain surface."

Ok so for this question the free body diagram I have is set up so that the angle is measured from the centre of the hemispherical mountain to the skier. In this way, the x axis' is tangent to the semicircle, and the y axis follows the radius of the semicircle. The skier in my diagram is moving to the right along the mountain, which means that the component of the gravitational force that is acting on him/her in the x direction is
mgcos(alpha) and the component along the y direction is mgsin(alpha). The normal force is along y axis. The newton's equations are

Fx = mgcosalpha
and
Fy = n - mgsinalpha

So far I think I have that right, now the difficult part is part b. I don't know what to look for in this part! What equations should I use and how should I set them up? I think this question has to do with momentum, i.e. when does her horizontal momentum carry her off the mountain, but I don't know how to set the equations up to get there. Any help would be greatly appreciated. Thank you.
Aidan

2. Nov 5, 2006

Hint 1: didn't you forget about a force? (Circular motion!)
Hint 2: if contact is lost, which force must equal zero?

3. Nov 5, 2006

### amcca064

Even if its circular motion, that means that Fy= Fr which is still = n - mgsinalpha
And when contact is lost, the normal force will be zero, which means that Fr = -mgsinalpha

Last edited: Nov 5, 2006
4. Nov 5, 2006

Ok, I assumed the skier is skiing down 'from left to right', but this doesn't matter. You have: N = mg*sin(alpha) - m*v^2 / R. The only thing you have to find is the velocity v the skier has when the line connecting the center of curvature and the skier's position makes some angle alpha with the horizontal. You can use energy conservation.

5. Nov 5, 2006

### amcca064

ok....... so what you said is basically, mgsinalpha=m[(v^2) / r] ----> gsinalpha = (v^2)/r ----> (9.81)(100)sinalpha=v^2 so sinalpha= v^2/981
which is two unknowns and one equation. I know I need to use energy conservation to find the answer, I just don't know how.......

6. Nov 5, 2006

Select two positions of the skier so that one is at the top and the other is at some angle alpha. What is the potential energy of the skier at the top? What about the kinetic energy at the top? Further on, what is the potential energy of the skier at some point alpha? ANd what about the kinetic energy? Now use energy conservation. The sum of kinetic and potential energy of the skier at the top must equal the sum of potential and kinetic energy of the skier at some angle alpha.

7. Nov 5, 2006

### amcca064

Ki + Ui = Kf +Uf makes things a lot easier now, thanks for bringing that to my attention, it was the one thing I couldn't piece together. thanks

8. Nov 6, 2006

### mirandasatterley

I have two questions about the conservation of energy part.

First: Is the v in the final kinetic energy left as v and solved for?
Second: How can you determine the height, h, in the final potential energy, Uf?