One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.48 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed v0, so that the rod begins to rotate upward about the pivot. What must be the value of v0, such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?

Answer 1

I'm going to take a shot and guess they're trying to equate angular KE with gravitational PE.
SO:
½ I ω^2 = mgh
Where you need to look up I for a rod spun about one end, ω = v/r, and h = r.
Pretty sure the m’s will cancel