Rod on a pivot: Linear acceleration and reaction forces

AI Thread Summary
The discussion focuses on calculating linear acceleration and reaction forces for a rod on a pivot. The equations used include arad = (L/2)(3g/L) and atan = (L/2)(3g/2L), leading to arad = 3g/2 and atan = 3g/4. A correction was made to arad, changing it to -3g/2 after considering proper signs. The initial confusion regarding reaction forces was clarified, emphasizing that R = ma is not sufficient without accounting for direction. The importance of incorporating negative signs in calculations was highlighted for accuracy.
StrawHat
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Homework Statement


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Homework Equations


r * \alpha = atan
r * \omega^2 = arad


The Attempt at a Solution


arad = (L/2)(3g/L) = 3g/2
atan = (L/2)(3g/2L) = 3g/4
 
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StrawHat said:
arad = (L/2)(3g/L) = 3g/2
How did you get this?
atan = (L/2)(3g/2L) = 3g/4
Looks OK. (But don't forget the proper signs of the components.)
 
I got arad = 3g/2 by multiplying r by ω2.

EDIT: It was actually -3g/2.

Going onto the reaction forces, I thought it was just R = ma, but apparently that's not the case...

Re-EDIT: Got it. I just needed to add the negative signs.
 
Last edited:

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