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

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.

• This is like a reverse impact problem. Instead of two masses impacting and sticking together, the explode apart. – ja72 Nov 12 '13 at 16:17
• Thanks for that note ja72 I forgot to make sure no one just makes m=M. hahaha Good One – Luke Burgess Nov 12 '13 at 16:20
• If you want to consider the rotational changes also, you need to specify the mass moments of inertia, and the center of mass locations for the two bodies. – ja72 Nov 12 '13 at 16:31
• You can simplify the problem if you do not consider rotations and use only a 1D case. Once you clarify the 1D you can extend the question to 2D or 3D. Rotations add the dependency of the location of $m$ and the combined response to an impulse along a line not passing through the center of mass. – ja72 Nov 12 '13 at 17:46
• Here is great resource for learning the details of an impact calculation (cs.cmu.edu/~baraff/sigcourse/notesd2.pdf). In your case you have to run it backwards in time with coefficient in resitution of $\epsilon=0$. – ja72 Nov 12 '13 at 17:51

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 .