Cube collision (with a pivot) and Angular momentum

[fblin]

Hi everybody some help needed here!
A cube, mass M move with v₀ collide with small object "fixed" to the surface in Point P that make it pivot over its side


So far I think I understand:
During the impact there is no conservation of KE
L Angular Momentum : there is an unknown force F (Impulse) on the pivot point
p Linear Momentum : there is a Impulse = F*t in the pivot point (or I should have to consider the mass of the earth)
Known data:
mass M, velocity v₀ or v, Side 2a (note that is 2a just for simplification), cm center of mass, r_cm vector distance from P to cm, rp distance from P to P (only for completeness)
Before (around Point P of pivot):
\vec{L} = \vec{r_{cm}}\times M\vec{v_{0}}
in this case |r_{cm}||v_{0}|sin(\theta)=av_0
L_{i} = aMv_{0}

After:

\vec{L_{f}} = \vec{r_{p}} \times \vec{F}*t + \vec{r_{cm}} \times M\vec{v_{cm}} +I\vec{\\w}

Where I is the Moment of Inertia of the cube and

v=\\wr_{cm} and r_{cm} = a\sqrt{2}
Note that r_{p} =0 so

L_{f} = 0*F*t + a\sqrt{2}M\\wa\sqrt{2}+I\\w So

L_{i} = L_{f}
aMv_{0} = 2a^2M\\w +I\\w

<br /> \\w = \frac{aMv_{0}}{2a^2M+I}<br />


After the collision It is posible to use Conservation of energy:
E = KE+ V = mgh and KE is 0 and the higher point (the min velocity).

\frac{1}{2}Mv_{cm}+\frac{1}{2}I\\w^2 = Mga(\sqrt{2}-1)



It that ok?

 

[ideasrule]

What does the question ask for?

 

[fblin]

I have doubts with KE after the collision.
Is posible that in
\frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\\w^2 = Mga(\sqrt{2}-1)
The term
\frac{1}{2}Mv_{cm}^2 is OK? or It's enough with \frac{1}{2}I\\w^2 = Mga(\sqrt{2}-1)?

I am not clear so some help would be great

 

[fblin]

I think I have the answer.
It depends on I (moment of inertia)
If I is taken from the point of pivot P then I is including all the efect of the movement of the CM (center of mass). On the other hand, if I is referred to the CM then It is necessary to use :
\frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\\w^2 = Mga(\sqrt{2}-1)
including the KE of the CM.

same is true for the angular momentum L:
\vec{L_{f}} = +I\vec{\\w}
If I is referred to P then \vec{r_{cm}} \times M\vec{v_{cm}} is not necessary (It is included in I)
Notice that v_{cm}=a\sqrt{2}\\w and adding it in the \vec{L_f} and in KE It should be, by Parallel axis theorem, equivalent. So
I_p = I_{cm} + Md^2\ in\ this \ case \ I_p =I_{cm}+M2a^2}