This problem involves a uniform rod of length L with a pivot at point L/4 (so that 1/4 of the rod is behind the pivot and 3/4 is in front of it). The rod is released from a horizontal position and drops down. You need to use conservation of energy to solve for the velocity when it is at the vertical position (when the energy is all kinetic). Then, using the idea of a physical pendulum, you have to calculate the period of oscillation if it is displaced slightly.
mgy = 1/2mv^2
The Attempt at a Solution
My idea for the solution is just to ignore the portion of the rod that is to the left of the pivot point (which is equal in length to L/4) and focus only on the remaining 3L/4 portion of the rod. I don't know, however, if this is the right way to approach the problem, or if there is something else i need to take into account.