# Dynamics: rogid bodies force and acceleration

## 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.

Last edited by a moderator:

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

tiny-tim
Homework Helper
hi etotheix! 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)
Honestly, I don't understand this formula:
$\sum Ma=(IG+m(rG)^2)$

it's the parallel axis theorem in disguise Last edited by a moderator:
Thank you very much tiny-tim! It is very clear now.