Critical velocity in vertical motion using energy conservation I was reading vertical motion in my physics book in which they were trying to find the critical velocity for a roller coaster cart to do the loop a loop. They did it by considering net force towards the center and finding the conditions under which it will be greater or equal to zero and they find the answer v>sqrt(g*r) where g is gravitational acceleration and r is the radius of the circle.
I want to do it another way. Using the law of conservation of energy, the kinetic energy at the bottom should at least be the potential energy at the top
so 1/2*mv^2 > mg*2*r
but I get v>2sqrt(g*r)
I don't understand what I am missing or not considering.
 A: Your assumption works if the minimum velocity at the top needed was indeed 0. Then you can equate that all kinetic energy is converted to gravitational potential energy so $v=\sqrt{2gh}$. 
But, here the roller coaster to complete circular motion, it must have some non-zero horizontal velocity at the top of the loop, or it would simply fall down. Hence that  velocity is found using the velocity for centripetal acceleration provided that the centripetal force is as low as possible which will yield the lowest velocity needed and that is the case when the downward(centripetal) force is just gravity or $mg$. This thus results that the minimum velocity is found from: $$mg=mv^2/r$$ 
If you backward calculate this you can indeed find the minimum initial kinetic energy the roller coaster must have. But there is no way to satisfy the circular motion priority with invoking just energy conservation. 
Notice that for $v=\sqrt{2gh}$. is for the velocity from the ground when considering vertical straight motion while the velocity of $v=\sqrt{gr}$ is for the top of the loop for circular motion. To find the minimum velocity needed at the bottom/ground of the loop for circular motion would be by the following setting initial kinetic energy to top’s GPE and Kinetic energy due to top’s minimum velocity:
$$1/2mu^2=2mgr+1/2(\sqrt{gr})^2$$
$$u^2=5gr$$
$$u=\sqrt{5gr}$$
So hence you see $u$ is indeed greater for circular motion.
