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I've been analyzing "ball hits a rod in space" type collisions, where speeding ball transfers part of its kinetic energy during elastic collision to the motionless rod, making it gain linear and angular momentum. There are many videos explaining such a scenario, I think I understood basic concept.

I tried to write a motion equations of similar, yet bit more complicated situation, where motionless rod B is hit by rod A. Rod A has only linear momentum ($V_a$), is going to hit rod B at a distance r off the center of mass of rod B, at the angle of α. Masses and lengths (and therefore moments of inertia) of both rods are given. I'd expect, that - after the elastic collision - both rods are going to have nonzero linear momentums and angular momentums.

rod hits rod collision

By my understanding, to describe this collision, 4 equations of motion are needed:

  1. conservation of kinetic energy
  2. conservation of linear momentum
  3. conservation of angular momentum in relation to the center of mass of rod A
  4. conservation of angular momentum in relation to the center of mass of rod B

However I've some troubles with points 3 and 4.

  1. $\frac{1}{2}m_av_a^2 = \frac{1}{2}m_av_a'^2 + \frac{1}{2}I_aω_a'^2 + \frac{1}{2}m_bv_b'^2 + \frac{1}{2}I_bω_b'^2$
  2. $m_av_a = m_av_a' + m_bv_b'$
  3. ???
  4. ???

I'd appreciate some help :)

edit1. fixed 1 and 2 eqs.

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edit2. okay, I've spent some time on this and here are my thoughts:

I can define an impulse $J_b$, that affects rod B, and is equal to the local linear momentum gained. And the same can be done for rod A. Sum of those two impulses is equal to 0.

$$J_a = \Delta p_a = m_a * (v_a' - v_a)$$ $$J_b = \Delta p_b = m_b * (v_b' - v_b)$$ $$J_a + J_b = 0$$

Those impulses are the source of angular momentum. Considering distribution of $J_a$ vector, we can finally define equation 3 and 4.

  1. $$I_a\omega_a = J_a * cos(\alpha)l = m_a (v_a' - v_a)*cos(\alpha)l$$
  2. $$I_b\omega_b = J_b * r = m_b v_b'*r$$

What do you think? I'd appreciate if anyone could evaluate the correctness of my reasoning. :)

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  • $\begingroup$ You do not need #1, since not all contacts conserve energy. What you need is the contact law describing the relative separation speed after the contact in terms of the speed before the contact and the coefficient of restitution. $\endgroup$
    – John Alexiou
    Jun 2 at 16:13
  • $\begingroup$ You have $\omega_A'$ drawn in the wrong sense. Body A is going to kind of rotate about the contact point in a clock-wise fashion. $\endgroup$
    – John Alexiou
    Jun 2 at 17:27

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Note: Your equation 2 has assumed $v_a'$ towards the right, and I will do the same for my equations. Also, I'm assuminng $d$ as the perpendicular distance between the centres of mass of the two rods (perpendicular to their lines of motion). $d=r+\frac{l}{2}cos(\alpha)$

  1. $m_b v_a d = m_b d (v_a' - v_b') + I_b \omega_b' + I_a \omega_a'$

  2. $m_a v_a d = m_a d (v_b' - v_a') + I_b \omega_b' + I_a \omega_a'$

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  • $\begingroup$ I'm sorry, I don't know, where did you took those equations from, but these don't strike correct for me. The distance between two centers of mass is not $d = r + \frac{1}{2}cos(\alpha)$ and equations you proposed make the arrangement unsolvable... $\endgroup$
    – Hypasist
    Jul 29, 2020 at 22:18
  • $\begingroup$ @Hypasist, I have made some edits to the answer to clarify my derivation. I've just applied conservation of angular momentum with respect to the two centres of masses, as you havewritten in the question. $\endgroup$
    – dnaik
    Jul 30, 2020 at 2:06

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