0

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.

Community
  • 1
Luke Burgess
  • 531
  • 1
    This is like a reverse impact problem. Instead of two masses impacting and sticking together, the explode apart. – John Alexiou 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. – John Alexiou Nov 12 '13 at 16:31
  • Both bodies have equal mass "$M$" the center of mass is at $\ell \over 2$. – Luke Burgess Nov 12 '13 at 16:33
  • If one thinks they can make the problem more simple by changing the center of mass, do so. Any way you can just show a solution to the idea of this problem is fine by me. – Luke Burgess Nov 12 '13 at 16:36
  • Is $m$ part of $M$, or in addition to $M$. – John Alexiou Nov 12 '13 at 17:25
  • 1
    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. – John Alexiou Nov 12 '13 at 17:46
  • 1
    Here is great resource for learning the details of an impact calculation (http://www.cs.cmu.edu/~baraff/sigcourse/notesd2.pdf). In your case you have to run it backwards in time with coefficient in resitution of $\epsilon=0$. – John Alexiou Nov 12 '13 at 17:51
  • ja72, once more, whatever will make it more simple to solve. – Luke Burgess Nov 12 '13 at 19:07
  • Hi Luke Burgess. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic Nov 12 '13 at 22:01
  • I thought the other tags were more significant, but if it simply must be defined as homework... – Luke Burgess Nov 12 '13 at 23:12

2 Answers2

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.

user33665
  • 11
  • Thank you for your input, I think you are most likely correct about the energy. Don't forget inelastic collisions demonstrate energy and momentum are not always symmetrical. I do not think energy and momentum are symmetrical in this problem. If you disagree please demonstrate with formulas why that assumption is wrong. – Luke Burgess Nov 12 '13 at 19:18
1

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 .

John Alexiou
  • 38,341