# B How does an angled rod behave upon collision?

1. Mar 15, 2017

### Michal Fishkin

Hi! I am new to the forums and this is my first question.
I am investigating the properties of this type of chain.

In order to understand it better, I need to know the physics behind a falling angled rod.

I understand that the other end speeds up to conserve momentum, but I do not understand how to calculate it. The paper I read on it stated that
If one lets a tilted rod fall onto a table, its other end speeds up on hitting (by 50% for a uniform rod with small θ having plastic impact: VB+ = 3V/2).
I also understand that for very small angles and lengths I could assume that end B is traveling in a straight line.

What would be the derivation behind the 50% increase in speed?
Are there other important aspects I am missing?
Does the angle affect the velocity?

2. Mar 15, 2017

### Khashishi

Can you give a citation for the paper?

3. Mar 15, 2017

### Michal Fishkin

4. Mar 16, 2017

### Khashishi

Cool, I wasn't aware of such physics, so I learned a bunch working it out.

Before the collision, the momentum is $p_i=mv$
After the collision, the linear momentum is $p_f=mv_b/2$ and the angular momentum about the center of mass is $L_a=I \omega$, where
$I = mL^2/12$ and $\omega = v_b/L$ (taking small angle approximations). So
$L_a = mv_bL/12$
We need to consider the impulse $p_c$ needed at A to slow down the linear momentum and also produce the angular momentum.
Clearly, $p_c = p_f-p_i$
Also, the impulse creates a change in angular momentum of $-p_c L/2$ which we can set equal to the final angular momentum $L_a$
We've got two equations and two unknowns:
$-p_c L/2 = mv_bL/12$
$p_c = mv_b/2 - mv$
Eliminate $p_c$ by substitution
$-(v_b/2 m - vm) L/2 = mL/12 v_b$
Clean up:
$v_b/2 - v = -v_b/6$
$v_b = 3v/2$

5. Mar 16, 2017

### Khashishi

If the angle is not small, you can't use the approximation $\omega = v_b/L$. You can probably work out the effects.

6. Mar 16, 2017

### Michal Fishkin

Brilliant! Thank you so much for the explanation. :) Very helpful!
I have a followup question:

I came to the derivation for acceleration of the free end of a rod with one end on the table and the other one in the air, angled at x.
a=3/2×g×cos^2(θ)
This means that the maximum acceleration would result from extremely small angles, so if I were trying to design a super-fast chain the small angle assumption would work.
Under this assumption, in the zig-zig ladder, the first free end would fall 50% faster than gravity, and then the second free end would fall 50% than that, and so on, producing a compounding acceleration effect - is this an accurate way to describe the 'sucking' motion of the chain?

7. Mar 16, 2017

### Khashishi

That's way too oversimplified. When the chain pulls on a higher link, the higher link pulls back. Overall, the increase in falling speed is slight.

8. Mar 17, 2017

### 256bits

Hi.
The original paper assumes plastic impact.
Any chance an elastic impact would change the result?

9. Mar 17, 2017

### Khashishi

Of course. The plastic collision is what keeps point A stuck to the ground after the collision. If you calculate the energy, you will notice that some is lost in the collision.

For an elastic collision, your two equations are conservation of energy and the equation that connects impulse and angular momentum change.

10. Mar 17, 2017

### Michal Fishkin

Ah! Thanks so much, I'm not sure why I'm having trouble visualizing this problem in particular.
So in the design of a super fast chain would have as many bars as possible.
Thanks so much again!