Dynamics: rogid bodies force and acceleration

[etotheix]

Homework Statement

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

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.

 

[Quinzio]

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

 

[tiny-tim]

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

 

[etotheix]

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