# Dynamics: rogid bodies force and acceleration

• etotheix
In summary, the conversation discusses a problem involving moments and rotational motion. The student is attempting to solve for angular acceleration using the moment about point A, but the solution uses the moment about point G. The conversation also clarifies the use of the formula for parallel axis theorem.

## Homework Statement

[PLAIN]http://img135.imageshack.us/img135/1063/1763o.jpg [Broken]

## Homework Equations

$IG = (1/12)ml^2$
$\sum Ma=(IG+m(rG)^2)$

## The Attempt at a Solution

I take the moment about A, which gives me the following:
$(4kg)(9.81m/s^2)(1m)=((1/12)(4kg)(2m)^2+(4kg)(1m)^2)\alpha$
$\alpha = 7.3575 rad/s^2$

But in the solutions they have: $\alpha = 14.7 rad/s^2$, which is 2 times more. They are taking the moment about G, using $Fa = (1/2)(4kg)(9.81m/s^2)$. Why does it yield a different answer? The way I do it must be wrong, but I don't see why.

Also they have $(aG)y = 4.905m/s^2$, but I though that $(aG)y = rG*\alpha$, and $rG = 1m$ in this case, so $(aG)y$ must be equal to $\alpha$

What am I doing wrong? Thanks in advance for the help.

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Honestly, I don't understand this formula:
$\sum Ma=(IG+m(rG)^2)$

hi etotheix!
etotheix said:
I take the moment about A …

no, you can't do that, the Iω formula for https://www.physicsforums.com/library.php?do=view_item&itemid=313" about either the centre of mass or the centre of rotation …

A is neither

(because A will start accelerating upwards as soon as the string is cut)
Also … I though that $(aG)y = rG*\alpha$, and $rG = 1m$ in this case, so $(aG)y$ must be equal to $\alpha$

no, a = rα assumes that r is the distance to the centre of rotation, and (again) that isn't A, so it isn't

you'll need to use the https://www.physicsforums.com/library.php?do=view_item&itemid=189" force, before and after (just call it F)
Quinzio said:
Honestly, I don't understand this formula:
$\sum Ma=(IG+m(rG)^2)$

it's the parallel axis theorem in disguise

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Thank you very much tiny-tim! It is very clear now.