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:
[itex]\sum Ma=(IG+m(rG)^2)[/itex]

it's the parallel axis theorem in disguise

 

[etotheix]

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

 

[rwisz]

Your approach to finding the angular acceleration is correct, but the mistake you made is in the calculation of the moment of inertia (IG). In your calculation, you used the formula for the moment of inertia of a thin rod rotating about its end (IG = (1/12)ml^2), but in this problem, the rod is rotating about its center of mass (point G). Therefore, the correct formula to use for the moment of inertia is IG = (1/12)ml^2 + mr^2, where r is the distance from the center of mass to the end of the rod. In this case, r = 1m, so the correct formula becomes IG = (1/12)(4kg)(2m)^2 + (4kg)(1m)^2 = 5/3 kgm^2. Using this value for IG in your calculation will give you the correct answer of 14.7 rad/s^2.

As for the value of (aG)y, it is indeed equal to rG*\alpha, but in this case, rG is not equal to 1m. Since the rod is rotating about its center of mass, rG is equal to half the length of the rod, which is 1m. Therefore, (aG)y = (1m)(7.3575 rad/s^2) = 7.3575 m/s^2, which is the same as the value in the solutions.

In summary, the mistake you made was using the wrong formula for the moment of inertia. Make sure to always check which point the object is rotating about before using a formula for the moment of inertia.