# Cube collision (with a pivot) and Angular momentum

1. Jun 19, 2009

### 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, 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

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

$$\\w = \frac{aMv_{0}}{2a^2M+I}$$

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?

Last edited: Jun 19, 2009
2. Jun 19, 2009

### ideasrule

What does the question ask for?

3. Jun 20, 2009

### 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

4. Jun 21, 2009

### 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 refered 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 refered 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
$$I_p = I_{cm} + Md^2\ in\ this \ case \ I_p =I_{cm}+M2a^2}$$