A loop puts two ideas together: energy gives the speed at the top, and the circle's force equation from Vertical Circles gives how hard the track pushes. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
Energy from the release height to the top of the loop ($2r$ up): $$mgh = mg(2r) + \tfrac{1}{2}mv^2.$$ At the top both the push $N$ and gravity point down, to the center: $$N + mg = \frac{mv^2}{r}.$$ Just making it round means $N = 0$, which needs $h = \tfrac{5}{2}r$.
Worked example: A block around a loop
A $0.2\ \text{kg}$ block is released from $1.4\ \text{m}$ on a smooth track with a loop of radius $0.4\ \text{m}$. Find (1) its speed at the top and (2) the track's push there.
- (1) $v = \sqrt{2(9.8)(1.4 - 0.8)} = 3.43\ \text{m/s}$.
- (2) $N = \dfrac{(0.2)(11.8)}{0.4} - (0.2)(9.8) = 3.92\ \text{N}$.
Answer: $3.43\ \text{m/s}$, $3.92\ \text{N}$
Common mistake
Writing $N - mg = mv^2/r$ at the top, as if the track pushed up. At the top of a loop the track is above the object and pushes down. Check: with $N = 0$ the equation must reduce to $mg = mv^2/r$.