1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 2D Dynamics Work-Energy Problem

  1. Dec 10, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant 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 = [itex]\frac{mL^{2}}{12}[/itex]
    If the axis passes through either end of the bar rather than the centre:
    I = [itex]\frac{mL^{2}}{3}[/itex]
    Where m is mass of the beam and L is length of the beam

    ΔWork = ΔTotal Energy

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

    v = rω
    3. 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 componants of bar AB:
    T = I[itex]_{OB}[/itex]ω + I[itex]_{AB}[/itex]ω + 0.5mv[itex]^{2}[/itex]
    = ω([itex]\frac{mL^{2}}{3}[/itex]+[itex]\frac{mL^{2}}{12}[/itex]) + 0.5mv[itex]^{2}[/itex]
    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.
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted