Particle sliding down a sphere - When does it leave the sphere?

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A particle placed on a smooth, frictionless sphere will leave the sphere when the normal force is overcome by the gravitational force. The analysis involves using the equations of motion and conservation of energy to determine the height at which the particle detaches. The calculations show that the particle will leave the sphere at a height of (2/3) R above the center. The discussion also clarifies the distinction between a "particle" sliding down and a "ball" rolling down the sphere, emphasizing the correct terminology. Overall, the solution is confirmed as accurate.
Adoniram
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Homework Statement


A particle is placed on top of a smooth (frictionless) sphere of radius R. If the particle is slightly
disturbed, at what point will it leave the sphere?

Homework Equations


Same as first question, just
F = ma = ΣF_i

The Attempt at a Solution


Similarly, we want to know when the normal force of the sphere on the particle is overcome:
F_norm = F_cent
mg CosΘ = (mv2/r)
CosΘ = y/R (where y is the height above the center of the sphere)

So:
y = v2/g

Finding v2:
Using conservation of energy, PE_initial = PE_final + KE_final
mgR = mgy + mv2/2

Solving for v2
v2 = 2g(R-y)

Placing into equation for y:
y = 2g(R-y)/g = 2(R-y)

Solving for y:
y = (2/3) R

Correct? Or am I making a horrible mistake?
 
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This is correct.
 
The title of the thread is "Ball rolling down a sphere". You solved the problem of a particle sliding down a sphere instead of a ball rolling down.
 
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You're right, if I could change the original post, I would, but there's no "Edit" button. Could be an account permissions issue...
 
I can fix it. So the title should read "particle" instead of "ball"?
 
Yes, thank you, and an even better description would be "Particle sliding down a sphere" (instead of rolling)
 
Done! And thank you to @ehild -- I had the same question when I saw the thread and posted answer. :smile:
 

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