1. The problem statement, all variables and given/known data A ball on top of a circle is acted on by a FG. The ball rolls to a certain point on the circle, and detaches. Ffr Is negligable. Given info: Diameter Of the circle, Mass of ball no ffr x = the distance from the balls posistion to the top of the big circle. x is what is needed to be found (Teal in diagram) Okay... So i think i might have it.... http://s13.postimage.org/fpzmhstnb/New_Bitmap_Image_3.png [Broken] So what i did, was assumed was that the distance from the top to the center (1/2D) and the distance from the object to the center (1/2) were the adjacent and opposite sides of an isoceles right angle triangle.... ( i don't know if this is okay..) Now I sovled the angle with TanA = 1/2D/1/2D = 1(TanInverse) = 45 degrees Now with this angle... I assumed that angle it traveled would be the angle it left at... (45 Degrees) and any corresponding angle woudl be 45 degrees anyway.. So i am left with the forces of FN which = 0 at the point it detaches and i am left with the Force of gravity... I have come to conclude that gravity is what allows centripetal accellearation... So because the angle of FN = 45 degrees, when i make my Frame of reference so that the X direction is paralell to FN, i end up with force of gravity with the angle and 2 components. So.. Fnety = mgsin45 may = mgsin45 ay = gsin45 Fnetx = mgcos45 - FN Fnetx = mgcos45 - 0 max = mgcos45 ax=gcos45 okay.. so now since i know that gravitys X component is what provides the centripetal force... R = D/2 Fc = Fnetx Mv^2/R = mgcos45 V =√[Dgcos45/2] Now.. Comparing total mechanical energy.. ET1 = EG1 = mgD Et2 = EK2 + EG2 = 1/2mv^2 + mg(d-x) Et1 = Et2 mgD = 1/2mv^2 + mg(d-x) gx = 1/2v^2 x = dgcos45/2/2g x = D (gcos45/2g) x = D (cos45/2) x = D([√2]/8) and this is my answer... x = D([√2]/8) so yea.. i'm hoping this strategy is right.. everything seems to add up mathematically.. i'm still not sure about this angle though.. i'm not sure if that is correct way to solv eit.