A block of metal is placed on a top of a half-sphere made of ice. The block slid

Question

A block of metal is placed on a top of a half-sphere made of ice. The block slides on the half-shpere down to a height (h) above the ground, then it loses contact with the ice surface and falls to the floor. The half-shere has a radius of (R=1m); we can ignore the friction between the ice and the metal.

A) What is the force exerted by the half-sphere on the block at the moment in which the block loses contact with the Ice?

B) Calculate h.

I think A is zero b/c when the block loses contact with the ice the normal force is zero right?

FOR B)  PLEASE GIVE DETAILED STEPS AND ALL EQUATIONS USED AND REARRANGED!!!

Explanation / Answer

it will loose contact when the normal reaction force becomes zero that is the centripetal force matches the component of weight normal to the surface
mgcosx=mv^2/r and we can calculate 0.5mv^2 =mg(R-h)=mgR(1-cosx) where x is the angle with the verical passing from the centre to the top of the sphere=>mgcosx=2mg(1-cosx)=>cosx=2/3=>h=R(cosx)=2R/3

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