- #1

- 38

- 1

## Homework Statement

Consider a small frictionless puck perched at the top of a fixed sphere of radius R. If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]

## Homework Equations

F=[itex]\dot{p}[/itex]

ΔU+ΔT=0

## The Attempt at a Solution

I use y=y(t) as the height of the puck at time=t.

I managed the first part of the hint, getting down to [itex]v=\sqrt{2g(R-y)}[/itex]

The next part is what's throwing me. Intuitively, I would think that the puck leaves the sphere when the normal force is zero. This would happen when the normal force vector is perpendicular to the weight vector, making it so that the puck falls of once it gets to the equator of the sphere. That would then be my answer, but if it were correct, I wouldn't have to find the velocity of the puck. It also stands to reason that the velocity of the puck at any given time affects whether it's about to fly off the surface.

Analytically, I tried: (N for normal and W for weight)

F=N+W=[itex]\dot{p}[/itex]

N=[itex]\dot{p}[/itex]-W , setting N=0 gives

[itex]\dot{p}[/itex]=W

I considered integrating both sides with respect to dt so I could sub in my velocity equation on the left, but the right hand side would give me a function of t, so I would get something like

p=Wt

mv=m[itex]\sqrt{2g(R-y)}=Wt[/itex]

If I solve that for y I don't have y explicitly, I have function y(t). Also, I didn't put arrows above all the vectors, but consider that W points down, Wt would also point down. This would mean that p also has to point down, which shouldn't happen even after the puck leaves the sphere, which is nonsense.

The only other thing I can think of is the "centrifugal force" lessening the magnitude of N, but since it's not a real force, I don't really know how to do math with it.