A block of mass m is placed on a smooth sphere of radius R. It slides when pushed slightly. At what vertical distance h, from the top, will it leave the sphere?

A block of mass m is placed on a smooth sphere of radius R. It slides

1.	A block of mass m is placed on a smooth sphere of radius R. It slides when pushed slightly. At what vertical distance h, from the top, will it leave the sphere?
Answer» `(R)/(4)` `(R)/(3)` `(R )/(2) ` `R` Solution : Let the block leave the sphere at point B, which is at a distance h from the top of the sphere. `THEREFORE(mv^2)/( R ) =mg COSTHETA -N` where N is the normal reaction and v is the velocity of block at point B. When the block leaves the sphere at point B, the normal reaction N becomes ZERO. `therefore(mv^2)/( R )= mgcos thetaorcos theta =(v^2)/(Rg)` From figure,` cos theta = (R-h)/( R)` ` therefore(R-h)/(R ) = (2 gh)/(Rg) "" [:.v= sqrt(2 gh)]` ` orR-h =2hor 3h=Rorh=(R )/(3 )`