A bead slides down the surface of a sphere. Ignoring the initial speed of the bead, define the normal force between the bead and the sphere with respect to the angle of displacement.

 

Given Answer: N = mg(3cosL-2) , L is the angle between the bead and the vertical taken from the centre of the sphere.

 

My answer was N = mgcosL. I don't see how the given answer could have been derived.

An object moving on a circular path must feel a centripetal force as its net force. Have you taken that into account?

 

N = mgcosL implies that there is a normal force up until the bead is at 90º. Will it stay on the sphere all the way to that point?

I didn't think of that, thanks. But I'm still a bit confused. How do I take the centripetal force into consideration? If the object leaves the surface of the sphere eventually than surely the centripetal fore, which should act towards the centre of the circle, is not the net force...

 

N = mgcosL + m(v^2)/r ?

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Bt definition, the centripetal force is the net radial force for an object moving in a circular path. That's an argument of math and geometry — it's what the acceleration vector must be. (There can also be a tangential force, which is perpendicular, causing the object to change speed.)

 

So you need to analyze this in terms of radial and tangential forces.

Ok, I'll try that