Dynamics problem of two bars

[Pepachin]

Hello!, I am trying to solve this problem of the book beer vector mechanics for engineers statics and dynamics, I`ll be very glad if you can help me with this
http://i.imgur.com/pFkUfuv.png (Image of the problem)
Each of the two identical slender bars shown has a length L. Knowing that the system is released from rest when the bars are horizontal, calculate the equations for angular velocity(ω) of rod AB and velocity of point D for values of θ from 0 to 90 degrees.

Here are some formulas that might help:
I= 1/12 mL^2
Principle of conservation of energy T1 + V1 = T2+V2
V=mgh
T= (1/2)*Iω^2
V= ωxr

First I tried to solve as a mechanism of three bars but it didn´t work and I´m trying to solve it with moments of Inertia I= 1/12 mL^2 and kinematics but the problem doesn´t say the mass of the rods, I also tried to solve for the equations of the two positions (when the rods are horizontal and when the angle is 90) but I can not get to anywhere
Thanks for taking your time to read this.

 

[KataKoniK]

Hello there! I would be happy to help you with this problem. First, let's start by breaking down the problem into smaller parts.

1. Identify the variables and their values:
- Length of the bars (L)
- Initial angle (θ)
- Mass of the bars (m)
- Gravitational acceleration (g)
- Moment of inertia (I)

2. Write down the given formulas:
- Principle of conservation of energy: T1 + V1 = T2 + V2
- Potential energy: V = mgh
- Kinetic energy: T = (1/2)*I*ω^2
- Velocity of point D: V = ω*r

3. Determine the initial and final positions:
- Initial position: rods are horizontal (θ = 0)
- Final position: angle is 90 degrees (θ = 90)

4. Calculate the potential energy at the initial and final positions:
- Initial position: V1 = mgh = 0 (since the rods are horizontal, h = 0)
- Final position: V2 = mgh = mgL (since the angle is 90 degrees, h = L)

5. Calculate the moment of inertia for a slender rod:
- Using the formula I = 1/12 * m * L^2, we can rearrange it to solve for mass: m = 12I/L^2
- Since the rods are identical, we can assume they have the same mass: m1 = m2 = m
- Plugging in the values, we get m = 12I/L^2

6. Calculate the kinetic energy at the final position:
- Using the formula T = (1/2)*I*ω^2, we can rearrange it to solve for angular velocity: ω = √(2T/I)
- Since the rods are identical, they have the same velocity: ω1 = ω2 = ω
- Plugging in the values, we get ω = √(2mgL/I)

7. Calculate the velocity of point D at the final position:
- Using the formula V = ω*r, we can rearrange it to solve for velocity: V = ω*r = √(2mgL/I) * L = √(2mgL^2/I)

8. Now we have all the equations for angular velocity (ω) and velocity of