Cube collision (with a pivot) and Angular momentum

AI Thread Summary
The discussion focuses on the dynamics of a cube colliding with a fixed point, causing it to pivot. Key points include the non-conservation of kinetic energy during the impact and the application of angular momentum principles, where an impulse force acts at the pivot. The equations for initial and final angular momentum are established, leading to the calculation of angular velocity after the collision. The conversation also addresses the conservation of energy post-collision, questioning the inclusion of kinetic energy terms related to the center of mass. Clarification is sought on whether to consider the kinetic energy of the center of mass in the energy equation, depending on the reference point for the moment of inertia.
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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, rcm 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?
 
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What does the question ask for?
 
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
 
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}
 
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