Rigid body acceleration question

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SUMMARY

The discussion focuses on calculating the centripetal acceleration (a_C) of point C in a rigid body system represented by an equilateral triangle supported by two links. Given the angular velocity (\dot{\theta} = 9 rad/s) and angular acceleration (\ddot{\theta} = 0 rad/s²), the user derived angular speeds and accelerations, ultimately calculating a_C to be 168, which is incorrect as it should be 0.202. The user seeks assistance in identifying potential errors in their calculations.

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Homework Statement


The illustrated equilateral triangle is supported by two links. d = 0.5 m. At the illustrated position,\dot{\theta}= 9 rad/s and \ddot{\theta}= 0 rad/s^2. Find the magnitude of a_C.

http://img406.imageshack.us/img406/7264/tonguech66315yx0.th.gif



The Attempt at a Solution


First I found the angle \beta. This is the angle between point B and the horizontal

\beta=30

Next I found all the angular speeds I am going to need: \omega_{DA}, \omega_{EB}, \omega_{AB}

\omega_{DA}=9 rad/s (Given)

\omega_{EB}=\omega_{DA} \frac{AD}{EB} (AD and EB are essentially given)
\omega_{EB}=18 rad/s

\omega_{AB}=-\omega_{DA}DA+ \omega_{EB}EB
\omega_{AB}=0

Next I found all angular accelerations I will need by assuming PGM:a_B=a_A+a_{B/A}

\omega_{EB}^2(EB)j-\alpha_{EB}(EB)i=\omega_{DA}^2(AD)j+\alpha_{AB}(AB)sin(90-\beta)j-\alpha_{AB}(AB)cos(90-\beta)i

I arranged the i and j components and solved finding:
\alpha_{EB}=93.53
\alpha_{AB}=93.53

Now I can find the accelerations of C:

a_Ci+acj=\omega_{EB}^2(EB)j-\alpha_{EB}(EB)i+\alpha_{AB}(BC)i

Solve resultant of a_C to equal 168 whereas it should be .202.

I'm not good at this at all and could have made some pretty big mistakes so bear with me.

Any help would be greatly appreciated!
 
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