Parallel Axis Theorem | Kinetic and Potential Energy

[Lahooty]

Homework Statement

A rigid body is made of three identical rods, each with length L = 0.525 m, fastened together in the form of a letter H, as in Figure 10-56 below. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of the H is vertical?

http://www.webassign.net/hrw/10-56.gif

Homework Equations

The Attempt at a Solution

I_1 = 0

I_2 = m/L∫x^2dx from 0 to L
I_2 = (1/3)*mL^2

I_3 = mL^2

I_total = (4/3)mL^2

E_Mec,Top = E_Mec, Bot

U_T + K_T = U_B + K_B

3*m*g*L + 0 = 0 + (1/2)*Iv^2

3*m*g*L = 1.5*m*L^2*v^2

2*g = L*v^2

[(2*g)/L]^.5 = v

v = 6.11 rad/s

This is wrong I'm pretty sure that I'm calculating the Rotational Inertia incorrectly. I don't know how to use the parallel axis theorem, especially not on three continuous bodies. I know that the formula is:

I = I_com+Mh^2

When I use that to calculate the Inertia, I get:

I_com = 3/2mL^2

I = 3/2mL^2 + 3mL^2
I = 4.5mL^2

 

[tms]

U_T + K_T = U_B + K_B

3*m*g*L + 0 = 0 + (1/2)*Iv^2

Why 3mgL? And on the right hand side, the usual symbol for angular speed is \omega, not v, which is usually taken to mean translational speed.