Dec 1, 2008
Soln Video 6/8
Why cant you sum the moments about A and set that equal to I(about A) times alpha? That was my first instinct when solving this problem. It is moving but not relative to the bar, which is our FBD. Do we need to look at the whole system when solving the moment equation instead of our FBD?? I think Im a little confused.
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3 comments:
Recall that the full form of Euler's equation (using point A) looks like:
sum_moments_A = IA*alpha + m*(rG/A) x aA
Since aA is in the x-direction and rG/A is in the y-direction, rG/A x aA is NOT zero. [However, the moment equation about G does render the second term to be zero.]
If you choose point A for your moments, you need to carry along the second term on the RHS.
The absolute acceleration of A is important here, not its motion relative to the bar.
Let us know if this is not clear.
Everything you said there and everything you said in the summary at the end of the problem makes sense. But I think its simply that we dont take into account acceleration in the FBD. So when doing the moment equation it doesn't pop out as having an acceleration. Thank you for the quick response. I understand but I think im going to look over that problem again. Thanks
You make a very good point here.
My recommendation is that at Step 2 you write down the FULL form of the RHS of Euler's equation if you choose any point except the center of mass or a true fixed point. At Step 3 (kinematics) you will then need to address the question as to whether this second term on the RHS is zero or not.
If you do not want to be adventurous, then ALWAYS choose the center of mass or a fixed point for Euler's equation. You know that those two points always simplify the RHS of the equation.
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