Not so simple problem using momentum, energy and angular velocity...?

Question

I have an object in free space (no gravity) with angular momentum $ = \omega_i $, and some velocity vector $=\vec{V_i}$. To simplify we will say it has a mass-less rigid rod length $ = \ell $, connecting two small masses both of mass $ = M $. The masses are small in the sense of a radius equal to the rod radius both much smaller than $ \ell $. Of course for simplicity keep this in the 2D plane. I know precession can play a role, but accounting for it can make the problem easier, I've already done that.

Now, we want to change the velocity of the object by $ \Delta V $. We must do this by taking a bit of the matter of mass $ = m $ off the object and get it moving away at a velocity $ = \vec{V_m} $. While keeping $M>>m$ What will be the most energy efficient mechanism to do this? What will be the most momentum efficient mechanism to do this?

The final answer, like with the precession related solution, gives some $ \vec{V_m} $ value parallel or perpendicular to $\vec{V_i}$, and some mass "$ m $" proportional to "$ M $".

The goal will be to obtain maximum $ \Delta V $ with minimum energy or momentum. While keeping $M>2m$

Right now I am getting different solutions based on minimizing energy vs. momentum, is this logical? Why? This has helped me get to my solution thus far. Thanks JCooper and Maksim Zholudev.

EDIT: The answer should be at least partially derived using formulas. Some relation between $ \Delta V $ and the input energy or momentum must be shown.

Answer 1

The answer is simple. The optimum strategy to slow (accelerate) the system is to remove equal amounts of mass from each small mass at the same time and in the same (opposite) direction as the current motion of the pair of point masses connected by the rigid rod. By symmetry the momentum change is then all in translational momentum in the direction anti-parallel (parallel) to the existing motion. Consequently the entire energy introduced into the system will be coupled into a decrease (increase) in velocity. If unequal masses are removed, or equal masses are removed in different directions, from the two point masses then there will be an asymmetric change and some energy will be coupled into a change in angular momentum of the pair of masses. This will reduce the magnitude of the desired change in translational velocity since some of the invested energy will have gone into a change in angular momentum. That energy is then unavailable to be used in optimally changing the translational velocity of the pair of small masses which was originally sought.

Answer 2

This should get you going in the right direction. I am using different variables so you can understand the concepts behind impacts and collisions (or explosions).

Two rigid bodies are attached at common point A located a distance $\vec{r}_{A1}$ and $\vec{r}_{A2}$ from their respective centers of mass. With the CM velocities $\vec{v}_{1}$ and $\vec{v}_{2}$ such that the common point velocity is equal $$\begin{aligned}\vec{v}_{A} & =\vec{v}_{1}+\vec{\omega}_{1}\times\vec{r}_{A1}\\ \vec{v}_{A} & =\vec{v}_{2}+\vec{\omega}_{2}\times\vec{r}_{A2} \end{aligned}$$ They might have common rotational speed $\vec{\omega}=\vec{\omega}_{1}=\vec{\omega}_{2}$, or not. It is not a requirement. Two contacting gears for example have comon velocity at a point, but different rotational speeds.

An impulse of magnitude $J$ is applied towards the second mass with direction $\hat{n}$ separating the rigid bodies. The step response of the two bodies is\begin{aligned}\Delta\vec{v}_{1}&=m_{1}^{-1}\left(-J\,\hat{n}\right)&\Delta\vec{v}_{2}&=m_{2}^{-1}\left(+J\,\hat{n}\right)\\\Delta\vec{\omega}_{1}&=I_{1}^{-1}\vec{r}_{A1}\times\left(-J\,\hat{n}\right)&\Delta\vec{\omega}_{2}&=I_{2}^{-1}\vec{r}_{A2}\times\left(+J\,\hat{n}\right)\end{aligned} Combined the separating velocity of point A along the impulse direction $\hat{n}$ is\begin{aligned} \Delta v_{A}^{rel}&=\hat{n}\cdot\left(\Delta\vec{v}_{A2}-\Delta\vec{v}_{A1}\right)\\&=\hat{n}\cdot\left(\left(\Delta\vec{v}_{2}+\Delta\vec{\omega}_{2}\times\vec{r}_{A2}\right)-\left(\Delta\vec{v}_{1}+\Delta\vec{\omega}_{1}\times\vec{r}_{A1}\right)\right)\\&=\left(m_{1}^{-1}+m_{2}^{-1}+\hat{n}\cdot\left(I_{1}^{-1}\left(\vec{r}_{A1}\times\hat{n}\right)\times\vec{r}_{A1}+I_{2}^{-1}\left(\vec{r}_{A2}\times\hat{n}\right)\times\vec{r}_{A2}\right)\right)\, J \end{aligned} The above is used to find what impulse J is needed to achieve a given separation velocity at A$$\boxed{ J=\frac{\Delta v_A^{rel}}{m_{1}^{-1}+m_{2}^{-1}+\hat{n}\cdot\left(I_{1}^{-1}\left(\vec{r}_{A1}\times\hat{n}\right)\times\vec{r}_{A1}+I_{2}^{-1}\left(\vec{r}_{A2}\times\hat{n}\right)\times\vec{r}_{A2}\right)} }$$ The impulse $J$ is used in the velocity step response to find the final velocities for bodies 1 and 2 .