# Homework Help: Skier on a snowball (Newtonian Mechanics)

1. Jan 28, 2013

### derravaragh

1. The problem statement, all variables and given/known data
A skier slides down a giant snowball (= sphere of radius R) with negligible friction. H starts at the top with very small velocity. Determine the angle θ_f where the skier will come off the surface. Use these principles of dynamics: (1) Energy is a constant of the motion. (2) The normal force of the contact decreases as the skier descends, and N = 0 at the point where the skier comes off the surface. (3) a = R*alpha*t(hat) - R*(w^2)*r(hat) where a is the acceleration vector, t(hat) is the time vector, and r(hat) is the position vector.

2. Relevant equations
E = KE + U where E is energy, KE is kinetic energy, and U is potential energy (I understand these may not be the standard)
KE = (1/2)mv^2

3. The attempt at a solution
For this problem, I am at a loss for what to do. I've worked it out a few times, taking into account all the factors. Using the first principle, I determined the initial kinetic energy is 0 as the initial velocity is so small, it's negligible, and the initial potential energy is merely 2Rmg. Next, I determined the final kinetic energy to be (1/2)mv^2 and final potential to be mg(R+Rcosθ_f).
Therefore, the energy equation is: 2Rmg = (1/2)mv^2 + mg(R+Rcosθ_f)

Next, I considered the fact that the centripetal force = the normal force, therefore at the point when the skier leaves the surface, the acceleration is only reliant on the tangential acceleration.

This is where I get confused. I can't seem to figure out how to make the connection between the acceleration and velocity for this problem, and be able to solve for θ_f.

Where should I go from here, assuming I'm even on the right track.

2. Jan 28, 2013

### haruspex

That's wrong. You were given, correctly, that at point of departure the normal force will be zero. What are all the forces acting on the skier at this point?

3. Jan 28, 2013

### derravaragh

The gravitational force, which I took to be just F_g = -mg at this point because there is nothing below him until he reaches the ground. The only other force I can think of would be the force of his motion.

4. Jan 28, 2013

### haruspex

Right (but motion is not a force... you may be thinking of inertia, which is momentum).
What is the skier's acceleration in the radial direction (you previously named this)? Since, a microsecond before losing contact, gravity is the only force available to provide that acceleration, what equation can you write?

5. Jan 29, 2013

### derravaragh

Ok, so the radial acceleration is what I called centripetal acceleration, so (and this may be wrong way of looking at it, but I'm going for the end result) mg = m*a_c where a_c = (v^2)/R therefore g = (v^2)/R and (v^2) = Rg which I can plug into an equation I came to before to obtain an angle of 60°.

6. Jan 29, 2013

### haruspex

Right idea, but notice that g is not in quite the right direction. What component of g is?