Angular Velocity of a pivoting rod

[TG3]

Homework Statement

A uniform rod 5.8m long weighing 10kg is pivoted at its center and a small weight of mass 5.15 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod. It is held at a 37 degree angle at rest, then released.
What is the angular velocity when the rod is vertical?

Homework Equations

After some work, I earlier solved that angular acceleration at it's release point is 1.6295 Don't know if that will be needed or not...
Angular Velocity = Tangential Velocity /R
VF^2 = VI^2 + 2A (X1-Xf)

The Attempt at a Solution

I dared to hope I could do this on my first try, but was disappointed. (Again.)
5.8/2 = 2.9, so radius = 2.9
sin37 2.9 = 1.745
2.9-1.745 =1.1547 m of fall

VF^2 = 0 + 2 (9.81) (1.1547)
VF^2 = 22.65
VF = 4.76

4.76 / 2.9 = 1.64
It looks so right... but is so wrong. *sigh*

 

[tiny-tim]

Hi TG3!

A uniform rod 5.8m long weighing 10kg is pivoted at its center and a small weight of mass 5.15 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod. It is held at a 37 degree angle at rest, then released.
What is the angular velocity when the rod is vertical?
…
VF^2 = VI^2 + 2A (X1-Xf) …

Sorry, not following your proof at all …

using conservation of energy is right …

but what's happened to the mass, and the moment of inertia ?