I’m having trouble with my physics lab this week. Here is the problem which we are trying to solve: What is the minimum height at which you can release a ball in order for the ball to just make it around a loop? The ideal situation for this would be to drop it from a height at which the normal force at the top of the loop would equal 0. I’ve found an equation in which to figure out the height that would make the normal force 0. This is how I got that equation… Since PE=KE I used the formula [tex]mgh=1/2v^2[/tex] where m is the mass, g is gravity, h is the height at which to drop it from and v is the velocity at the top of the loop. I figure that v is not important so what I did was I took the centripetal acceleration formula which states that [tex]a=v^2/r[/tex] and somehow decided that [tex]a=v^2/r[/tex] is the same thing as [tex]g=v^2/r[/tex] solved that for [tex]v^2[/tex] and came up with [tex]v^2=gr[/tex]. This made my equation [tex]mgh=1/2gr[/tex] and then I solved that for my h to get the height which would be [tex]h=(1/2r)/m[/tex]. Now I don’t know if this is at all right or not that is why I am posting this. I’m also having another problem. Since that is when n=0 then it’s the absolute extreme case for which the ball to travel around the loop neglecting friction. Now I need to devise a formula or a way to come up with a more reasonable height in which to drop it from. The trick is that I have to state it physically how I came up with that height. If anyone could help me with this too that would be great. Thanks.