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: 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?


    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.

    Attached Files:

    • fbd.jpg
      File size:
      8.5 KB
    Last edited: Nov 7, 2009
  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