Bar coming to a rest (vertical position)

[Krappy]

Bar collision with ground

Homework Statement

A uniform bar of mass m and length l has no angular velocity when the extremity A hits the ground without bouncing. If the angle α is 30º, what is the magnitude of v1, so that the bar rotates, around A, until it immobilizes on the vertical position.

http://img15.imageshack.us/img15/4899/pictrt.png

Homework Equations

\frac{dL}{dt} = \tau
\frac{dp}{dt} = F

The Attempt at a Solution

After the collision one can solve applying conservation of energy. From now I'm only considering the collision.
Now, my teacher says that the impulse by gravity is negligible and so, using conservation of angular momentum:

m v_1 \sin {30º} \dfrac{l}{2} + 0 = mv_{CG}\dfrac{l}{2} + I_{CG} \omega_2 \Rightarrow v_1 = \frac{4}{3}l\omega_2 = \frac{8}{3}v_{CG}

If we apply conservation of linear momentum in the direction perpendicular to the bar, the only force acting on this direction is gravity and if we neglect it (as above), we have:

\Delta p = F \Delta t = 0

And so, I would conclude that the velocity perpendicular (vCG above) remains constant and so

v_{CG} = v_1 \sin \alpha = \dfrac{v_1}{2}Am I missing something? If yes, what?EDIT: I was thinking and maybe, my error was thinking that the ground reaction had to in the same direction of the bar. Now I don't believe that this is necessarily true. Is this the error?

Thank You

 

[tiny-tim]

Hi Krappy!

I don't understand why you didn't stop here …

m v_1 \sin {30º} \dfrac{l}{2} + 0 = mv_{CG}\dfrac{l}{2} + I_{CG} \omega_2 \Rightarrow v_1 = \frac{4}{3}l\omega_2 = \frac{8}{3}v_{CG}

… you now have initial ω as a function of v₁, what else is there to do?

(and yes, you can't apply conservation of momentum separately, since you have no idea which direction the reaction force is, unless you deduce it from the ω equation which you already have)

now apply conservation of energy ​

 

[Krappy]

Hi tiny-tim! Thanks for answering.

I was confused, because if I applied angular momentum conservation I would get a different result of applying linear momentum conservation. What I was missing and later found (which you confirmed) is that I couldn't apply the latter because I don't know the reaction force (initially I thought that it was directed along the bar, but I now I know it isn't.)

Thank you