2D Dynamics Work-Energy Problem

[Gaps_to_fill]

Homework Statement

Homework Equations

Work done by a couple (U) = ∫M dθ
Where M is the applied bending moment and θ is angle

For a uniform, slender beam mass moment of inertia (I) through an axis perpendicular to the beam and passing through the centre of mass = $\frac{mL^{2}}{12}$
If the axis passes through either end of the bar rather than the centre:
I = $\frac{mL^{2}}{3}$
Where m is mass of the beam and L is length of the beam

ΔWork = ΔTotal Energy

Kinetic energy (T) (translational) = 0.5mv$^{2}$
T (rotational) = 0.5Iω$^{2}$
Potential energy (Vg) = mgh
Where v is velocity, ω is angular velocity, g is acceleration due to gravity and h is height

v = rω

The Attempt at a Solution

We were given a broad outline and final solution to the problem but I've been having difficulty making sense of it.

I understand how they found potential energy, treat mg as a point force acting on the bars centre of mass:
Initial Ep = mgh = mg(0.5b×cosθ)
Then because there's two beams:
Initial Ep (total) = mgb×cosθ
In the final state the bars are nearly vertical so I assumed θ is negligable:
Final Ep (total) = 2mgh = 2mg(0.5b) = mgb

I also get why U = Mθ and initial kinetic energy is 0 but I get lost when looking at their final term for kinetic energy and the final solution.

I would've thought the kinetic energy in the final state would be the rotational componant of bar OB plus the rotational and translational components of bar AB:
T = I$_{OB}$ω + I$_{AB}$ω + 0.5mv$^{2}$
= ω($\frac{mL^{2}}{3}$+$\frac{mL^{2}}{12}$) + 0.5mv$^{2}$
I can see that instead the solutions use the parallel axes theorem but I don't understand how it's used here.

If I had the right T term I'd equate the total energy (Vg + T) of the initial and final states and try to solve for the velocity from there (though I'm not quite sure how).

If anyone can help clarify what's happening in the solutions or how else I might get the velocity it would be greatly appreciated.

 

[nilesh_pat]

The solution provided is:Using the parallel axes theorem for bar AB:I_{AB} = I_{CM} + md^{2} Where d is distance from centre of mass to either end of the beamFinal T (for both bars) = 0.5(I_{OB} + I_{AB})ω^{2}= 0.5(0.5mL^{2} + 1.5mL^{2})ω^{2} = 2.5mL^{2}ω^{2}Therefore, ΔWork = ΔTotal Energy U = mgb = 2.5mL^{2}ω^{2} Therefore, ω = \sqrt{\frac{mgb}{2.5mL^{2}}} v = rω = \sqrt{\frac{r^{2}mgb}{2.5mL^{2}}}