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We have two girls, with mass ($M$). They become close to each other in speed of $V$. The distance between them is $3L$.

I was asked to calculate the Angular velocity ($\omega$) of the two girls. So I set the rotation axis in the middle (where the distance between the two girls and the axis is $1.5L$, and I calculate the Angular velocity using this equation: $\omega=v/r$, where $r=1.5L$, and I got that $\omega={2v\over 3L}$.

As I understood, the answer is correct, but this is not the correct way. What is my mistake? (sorry about my english)

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The photo is here: img2.timg.co.il/forums/1_159160419.jpg –  Adam Sh Feb 3 '12 at 22:54
Please do not write "w" for "$\omega$" :-). You can use LaTeX math in $...$ signs. So you can write $\omega$. –  queueoverflow Feb 3 '12 at 23:20
What? Do the girls run and grab the bar at the same time, transferring their momentum? Is the bar free, or pivoted by some point? Please explain in more detail what is your question. –  ja72 May 14 at 17:34

1 Answer 1

They do not come closer to each other according to the picture. They always keep the distance of the $3L$ since they hold onto that bar that is going to rotate counter clockwise.

I think your answer $$\omega = \frac{2}{3} \frac{v}{L}$$ is fine. This works since $v$ and $r$ are perpendicular, with a 90° angle in between.

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But I have one problem with my answer: If I would set the rotation axis in a distance of 2/3 fron the one, and 1/3 from the other - meaning from one girl the distance is L and from the second, the distance is 2L. I would get a diffrent $w$. My way is working beacuse we are talking on center of mass? –  Adam Sh Feb 4 '12 at 7:57
Yes, the symmetry says that they rotate around the center. If one girl is faster than the other, they will have different $\omega$, that is true. –  queueoverflow Feb 4 '12 at 14:38
@AdamSh Centre of mass does not come into play here. Using the fact that they have same speed, we write $\omega r=\omega(3L-r)$ from which we get that rotating axis is in middle. Also, angular speed of girl wrt anything on bar will be same. Just count the speed of frame you are sitting in and make calculations. You will get same answer. –  Awesome May 14 at 15:38

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