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Rotational kinetics of a rigid body

  1. Nov 7, 2009 #1
    Hi, I had a question about a dynamics problem involving the rotation of a rigid body. Here's what I've worked out so far, but I can't seem to get the correct answer.

    I'm suppose to find the magnitude of the reactionary forces at pin A. See the fbd for a drawing of the problem. I'm not sure if [tex]R_{t}[/tex] is pointing in the correct direction, but I think it shouldn't matter because the math should tell me that. Each segment of the beam weighs 10 Lb and has a length of 3.

    [tex]L = 3\ ft[/tex]
    [tex]w = 10\ Lb.[/tex]
    [tex]m = 0.3106\ slugs[/tex]
    [tex]\omega_{0} = 0\ rad/sec[/tex]
    (distance from A to G)
    [tex]r_{G} = \sqrt{\overline{x}^{2}+\overline{y}^{2}}[/tex]
    [tex]\theta = tan^{-1}(\frac{\overline{y}}{r_{G}})[/tex]
    [tex]\theta = 17.5484^{o}[/tex]
    [tex]\overline{x} = 2.25\ ft[/tex]
    [tex]\overline{y} = 0.75\ ft[/tex]

    [tex]I_{G} = 2(\frac{m*\overline{x}^{3}}{9} + \frac{m*\overline{y}^{3}}{9}+m*\overline{y}^{2})[/tex] (taking advantage of the fact that each bar has the same moment of inertia about the center of gravity)
    [tex]I_{G} = 1.1646\ slugs*ft^{2}[/tex]

    [tex]\sum m*(a_{G})_{n} = m*\omega^{2}*r_{G}[/tex]
    at time t=0 (assuming the direction of [tex]R_{n}[/tex] in the fbd is in the positive direction),
    [tex]R_{n} - 2*w*sin(\theta) = m*\omega_{0}^{2}*r_{G}[/tex]
    [tex]R_{n} = 6.030227\ N[/tex]

    [tex]\sum m*(a_{G})_{t} = m * \alpha * r_{G}[/tex]
    (assuming the direction of [tex]R_{t}[/tex] in the fbd is in the negative direction)
    [tex]-R_{t} - 2*w*cos(\theta) = m*\alpha * r_{G}[/tex]
    [tex]\sum M_{G} = I_{G}*\alpha[/tex]

    (assuming that a counter-clockwise rotation is positive)
    [tex]R_{t} * r_{G} = I_{G} * \alpha[/tex]
    [tex]-R_{t} - 2*w*cos(\theta) = \frac{m* R_{t}*r_{G}*r_{G}}{I_{G}}[/tex]
    [tex]R_{t} = -\frac{2*w*cos(\theta)}{\frac{m*r_{G}^{2}}{I_{G}} + 1}[/tex]
    [tex]R_{t} = -7.6277\ Lb.[/tex] (so my fbd was backwards, [tex]R_{t}[/tex] really points in the other direction)

    Transforming these into [tex]R_{x}[/tex] and [tex]R_{y}[/tex] (that's the way the answer is formatted),

    [tex]R_{x} = R_{n}*cos(\theta)+R_{t}*sin(\theta)[/tex]
    [tex]R_{x} = 3.45\ Lb.[/tex]

    [tex]R_{y} = R_{n} * sin(\theta) - R_{t} * cos(\theta)[/tex]
    [tex]R_{y} = 9.0909\ Lb.[/tex]

    which are both wrong...

    The correct answers are:

    [tex]R_{x} = 4.5\ Lb.[/tex]
    [tex]R_{y} = 6.5\ Lb.[/tex]

    Where did I go wrong?

    edit:

    hmm.. I seem to have put this in the wrong forum. Grr to having multiple tabs open :( Could someone move this for me?

    edit 2:

    Ok, i figured out my problem. I had a few signs backwards and somehow managed to mess up some calculations above.
     

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    Last edited: Nov 7, 2009
  2. jcsd
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