QUESTION: An object of mass $m$ is released at a height $h$ above the center of a circular loop and slides down without friction. How high must the drop be so that it just barely completes the loop-the-loop?

1. Since this is a conservative force problem the potential energy (gravitational) plus the kinetic energy of the mass is a constant of the motion. Thus at the start $mgh + KE_i = \mbox{constant}$.

2. The KE+PE energy at the end point (the top of the loop) hasn't changed although any changes in kinetic energy are due to changes in the potential energy (work done by gravity).

3. The constraint is that the object must remain in circular motion throughout. If the ball is released at $h=2R$ it will rise again to a height of $2R$ but will depart from the loop when the component of the acceleration due to gravity (normal to the loop) exactly equals the centripetal acceleration.

   (This should happen when ball is at a height of 2/3 $R$ above the loop center. You should try problem this yourself! To further test yourself you might wish to locate where the ball impacts the loop.)

   This constraint implies that at the top $F_c=mv^2/R \ge mg = F_g$ or $v^2 \ge gR$.

4. The above constraint implies the KE at the TOP must be greater that $1/2~mv^2$ or $mgR/2$. Thus $mg \Delta R = mgR/2$ or $\Delta R = R/2$ to give $h= 2R + R/2 = 5R/2$.