How Does Pivot Position Affect the Period of a Physical Pendulum?

[Frillth]

Homework Statement

A slender, uniform rod of mass m and length l can be pivoted on a frictionless horizontal support at any point along the rod's length. The rod then moves as a physical pendulum for small oscillations about the vertical equilibrium position. Suppose that the pivot point is at a point zl from the end of the rod, where z is a fraction between 0 and 1 Using an equation for the angular acceleration of the rod as a function of the angle θ, which measures the departure from the vertical, find the period of small-angle oscillations. Check the answer by specifying z = 0.5, where the period should become large.

Homework Equations

T = 2π*sqrt(I/(mgr))
T = period, I = moment of inertia, m = total mass, g = gravity, r = distance from pivot to center of mass

The Attempt at a Solution

I know that I for the center of mass is 1/12*m*l^2, so using the parallel axis theorem we can see that I for the point zl is:
I = 1/12*m*l^2 + m(.5 - z)^2
We can also see:
r = abs(.5 - z)*l

I now know the values for I, m, g, and r, so I should be able to use these in the equation for T. This can't be right, however, because my solution doesn't use an acceleration formula at all. How does this acceleration formula affect the period, and how do I incorporate it into my formula?

Thanks!

 

[Hootenanny]

The questions wants you to start from an expression for the acceleration of a rod and derive the formulae for the period of the pendulum which you have quoted above.