A smooth semisphere of radius R is fixed to the surface of a table. A coin slide

Question

A smooth semisphere of radius R is fixed to the surface of a table. A coin slides down without friction from the surface of this semisphere. At which height above the table will the coin break contact with the semisphere?

I know the answer is (2/3)R but I am not sure how that conclusion was reached and would appreciate a step by step solution with explanations.
A smooth semisphere of radius R is fixed to the surface of a table. A coin slides down without friction from the surface of this semisphere. At which height above the table will the coin break contact with the semisphere?

I know the answer is (2/3)R but I am not sure how that conclusion was reached and would appreciate a step by step solution with explanations.

Explanation / Answer

conserving energy between the highest and the breakoff point

0.5*mv^2=mg*R*(1-cos(theta))

and since it breaks off

mv^2/R=mg*cos(theta)

==> 0.5*Rg*cos(theta)=Rg*(1-cos(theta))

==> 1.5*cos(theta)=1

==> cos(theta)=2/3

==> hight = R*cos(theta)=2R/3

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