A Penny Falling Off a Non-moving Sphere

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A penny released from the top of a fixed, smooth sphere slides down and eventually leaves the sphere at a specific point. To determine how far the penny falls from the point of contact with the platform, it is essential to find the angle at which it detaches. Conservation of energy should be applied to calculate the penny's speed at the detachment point, rather than using equations for constant acceleration. The discussion emphasizes the need to analyze forces acting on the penny at the moment it leaves the sphere. Understanding these principles will lead to the correct solution for the distance fallen.
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Homework Statement



A penny is released from the top of a very smooth sphere of radius 1 m. The sphere is fixed to a platform and doesn't move. The penny slides down from rest and leaves the sphere at a certain point. How far will the penny fall away from the point of contact of the sphere and the platform?

Homework Equations



(1) V = Vo + at
(2) a = (V^2)/9.8
(3) V^2 = Vo^2 + 2a(x - xo)

The Attempt at a Solution



Vo = 0 m/s
r = 1 m
a = g = 9.8 m/(s^2)
x = ? => The distance the penny fell from the top of the sphere to ground.

I used (1) in order to rearrange into

t = V/g

After doing that, I was thinking that maybe I could use (2) or (3) to rearrange the equation a little more to find the solution, but I'm sure that's not right, and I'm doing too much work. I've attached a pdf of the problem.
 

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Juan42 said:
(1) V = Vo + at
That's only valid for constant acceleration, which this is not. You need to find the point at which the penny leaves the sphere. Suppose the radius to that point makes an angle theta to the vertical. Use conservation of energy to find its speed there, then consider the forces on the penny at that point.
 
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