Is circular motion possible with zero velocity at top of vertical loop? The figure shows a light rod of length $l$ rigidly attached to a small heavy block at one end and a hook at the other end. The system is released from rest with the rod in a horizontal position. There is a fixed smooth ring at a depth $h$ below the initial position of the hook and the hook gets into the ring as it reaches there. What should be the minimum value of h so that the block moves in a complete circle about the ring?

So I tried to solve it like this:
$$mg(h+l)=\frac{1}{2}mv_m^2$$
$$V_m=\sqrt{5gl}$$
which is the minimum velocity needed at the bottom and got the answer $$h=\frac{3}{2}l$$
Which is wrong, the correct answer being l. So if $h=l$, doesn't it mean that the velocity at topmost point is zero therefore the centripetal acceleration should also be zero. How can the circular motion be possible?
 A: 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.
A: 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.
A: You must understand that velocity at the top is not zero but $ infinitesimally $ greater than zero and that zero here is a good approximation for calculating min velocity required.

If you get $v_{min}=5_{m/s}$ that means the velocity required for circular motion is something like 5.001or 5.0003 , anything greater than $5_{m/s}$
A: Assuming the rod has negligible mass and that l is the distance to the center of mass of the block, I would think that h=l. Since the rod is rigid, there is no reason the velocity when the block is at the peak of the circle should not be essentially zero. So the initial potential energy mgh should be equal to the potential energy when the block is at the top of the circle, mgl. mgh=mgl, so h=l.
