1. An ice cube is placed on top of an overturned spherical bowl of radius r. (a half circle with a radius r). If the cube slides down from rest at the top of the bowl, at what angle(o) does it separate from the bowl? (hint: what happens to normal force when it leaves the bowl?) 2. Relevant equations KE=1/2mv^2 PE=mgh F=ma 3. The attempt at a solution When the normal force leaves the bowl, it doesn't exist anymore. At when the ice cube is about leave, the height is rcos(o), so PE = mgrcos(o) When the ice cue is about to hit the ground, all the PE is Ke, so: 1/2mv^2 = mgrcos(o) 1/2v^2 = grcos(o) the x distance the ice cube slides before separates is rsin(o) Um...so I'm not too sure how I'm suppose do, so far I have found random information, but I don't know what to do know.