# Is circular motion possible with zero velocity at top of vertical loop?

The key point is that the bar is rigid, so the block can reach the top of the circle with zero velocity. So the block's kinetic energy at the point where it is at the same height as the ring about which it pivots (a distance $l$ below the top of the circle) must be at least $mgl$. But the block's kinetic energy at this point is equal to the potential energy that it has lost in falling from a height $h$ above the ring, so

$mgh \ge mgl \Rightarrow h \ge l$

If the bar is replaced by a light string, then the velocity of the block at the top of the circle must be at least $\sqrt {gl}$ to keep the string taut. So the block's kinetic energy at the top of the circle must be at least $\frac 1 2 mgl$, and its kinetic energy when it is at the same height as the ring must be at least $\frac 3 2 mgl$. So we have

$mgh \ge \frac 3 2 mgl \Rightarrow h \ge \frac 3 2 l$

which is the answer that you arrived at.

Assuming the hook does not come off the ring, if you start from height L, then you can arrive back at height L with no kinetic energy. That would be a position of unstable equilibrium.