What Angle Does a Rice Grain Leave a Frictionless Billiard Ball?

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The discussion revolves around determining the angle at which a rice grain slides off a frictionless billiard ball. Participants suggest resolving the weight of the grain into tangential and radial components to analyze the forces involved. The centripetal force required for circular motion is compared to the radial component of gravity to find the critical angle. Conservation of energy is used to relate the speed of the grain to its height on the ball's surface. The conversation concludes with a clarification on the geometry involved in resolving the forces.
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A rice grain is sitting on top of a billiard ball. It slides down. Where does it come off the ball? (You can assume that the ball is frictionless.) Find the angle with respect to the vertical diameter of the ball.

I have no idea on where to start for this problem, I was thinking finding the angle relating arclength to angle, but I'm not too sure.
 
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Here are some ideas I have for a starting point:

Resolve the weight of the grain into tangential and radial components. The radial component is what is available to provide a centripetal force. These components are both functions of theta (the angle).

Find out how the speed varies with theta. This is easy if you use conservation of energy: the speed of the grain depends only on its height above the ground.

I suspect that the grain will cease to follow the curve of the ball's surface when the centripetal force required for circular motion at speed v(θ) and radius r is greater than the centripetal force available from the radial component of gravity at that angle θ.
 
Thank you for your reply,

so I have a centripetal force that is

Fc = (mv^2)/r

and finding the relationship between v and theta
(mv^2)/2 = m g (r+rsin(theta))

v = sqrt(2*m*g*r*(1+sin(theta)))

now for the tangential part I am a little lost

EDIT:

could the tangential be
Ft = mgsintheta?
 
Pick a particular point on the surface of the ball to be the point where the grain is. Draw the vector mg (the weight of the grain) down from that point. It should be vertical. Now just resolve that vector into two components, one of which is tangent to the ball's surface at that point (ie it points in the direction the grain is going). The other component is what you have to add to the tangential component in order to have the resultant be mg. It will be perpendicular to the tangent, so it points radially inward. I don't have time to check your geometry at the moment, sorry. Good luck with it!
 
yeah I figured it out my geometry was a little messed, thank you so much.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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