A piece of ice at the very top of a round silo dome slides off one side. Assuming there’s no friction between the ice and the dome, at what angle θ from the vertical would the ice lose contact (i.e. n= 0) with the silo?
The only force when it loses contact is gravity (mg) R is radius of dome Fnet = ma mg = mv^2/R g = v^2/R Rg = v^2 Now, we have to find how far down the sled goes in order to reach that velocity. h is vertical distance traveled by block I will call that Work energy theorem: Wnet = Kf - Ki mg(R-h) = .5*mv^2 mg(R-h) = .5*mRg R-h = .5R .5R = h So, the block has gone down half the radius of the dome From this we can extrapolate a right triangle that has a height of .5R and a hypotenuse of R. This can be recognized as a 30-60-90 triangle, so the answer: 60 degrees = θ
