Physics
A hemisphere of radius r is placed on a horizontal plane and a small mass of "m" is placed on the top of the hemisphere.
Find the height from the ground to the mass m,at the moment it loses it's contact with the hemisphere.And the contact is frictionless.
If we take the speed of the mass m, at the top of the hemisphere, v,the forces acting on the mass in that moment are mg,R and the centripetal force.
And the moment the mass loses its contact with the hemisphere,the forces acting on it are R and mg,and the centripetal force(which is zero as the speed is zero.)(because for the mass to lose its contact with the hemisphere, the speed should be zero as R≠0)
Hope my thoughts are correct and please correct me if they are wrong!
So do we have to solve this applying the law of conservation of energy? But how do find the speed v at the top as both R and v are unknowns.We can't simply equate R and mg because of the centripetal force acting on the mass in that moment!
Any help would be appreciated! Thank you!

And in both cases R cannot be the same too.
posted by Shenaya

Try reaction=centripetal force,
mg(cos(θ))=mv^2/R
Yes, v^2 (at the time it leaves sphere) can be calculated using energy considerations.
posted by MathMate

But at the top if we apply F=ma towards the center of the hemisphere it would be,
mgR=mv^2/r right? So how do we eliminate R?posted by Shenaya

You don't have to worry about the top, just assume that it was off balance by some random motion, and acceleration will start from there where KE=0, and PE=mg(R) if we take reference elevation at the centre of the sphere. Then at at certain θ<pi/2, mg(cos(θ))=mv^2/R
PE=mgR(1cos(θ)=mv^2/2=KE
The R will cancel out when you substitute.posted by MathMate
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