rotational kinetics of a rigid body

**helloworld922** · Nov7-09, 01:13 AM

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 $LaTeX Code: R_{t}$ 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.

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

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

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

$LaTeX Code: \\sum m*(a_{G})_{t} = m * \\alpha * r_{G}$
(assuming the direction of $LaTeX Code: R_{t}$ in the fbd is in the negative direction)
$LaTeX Code: -R_{t} - 2*w*cos(\\theta) = m*\\alpha * r_{G}$
$LaTeX Code: \\sum M_{G} = I_{G}*\\alpha$

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

Transforming these into $LaTeX Code: R_{x}$ and $LaTeX Code: R_{y}$ (that's the way the answer is formatted),

$LaTeX Code: R_{x} = R_{n}*cos(\\theta)+R_{t}*sin(\\theta)$
$LaTeX Code: R_{x} = 3.45\\ Lb.$

$LaTeX Code: R_{y} = R_{n} * sin(\\theta) - R_{t} * cos(\\theta)$
$LaTeX Code: R_{y} = 9.0909\\ Lb.$

which are both wrong...

The correct answers are:

$LaTeX Code: R_{x} = 4.5\\ Lb.$
$LaTeX Code: R_{y} = 6.5\\ Lb.$

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.