- #1
Krappy
- 16
- 0
Bar collision with ground
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
[tex]\frac{dL}{dt} = \tau[/tex]
[tex]\frac{dp}{dt} = F[/tex]
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:
[tex]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}[/tex]
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:
[tex]\Delta p = F \Delta t = 0[/tex]
And so, I would conclude that the velocity perpendicular (vCG above) remains constant and so
[tex]v_{CG} = v_1 \sin \alpha = \dfrac{v_1}{2}[/tex]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
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
[tex]\frac{dL}{dt} = \tau[/tex]
[tex]\frac{dp}{dt} = F[/tex]
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:
[tex]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}[/tex]
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:
[tex]\Delta p = F \Delta t = 0[/tex]
And so, I would conclude that the velocity perpendicular (vCG above) remains constant and so
[tex]v_{CG} = v_1 \sin \alpha = \dfrac{v_1}{2}[/tex]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
Last edited by a moderator: