# Changing Momentum p -> L -> -p?

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1. Nov 5, 2015

### SikorskyUH60

Ok, so as far as I understand it, it is impossible to turn linear momentum (p) into rotational momentum (L), but I don't quite understand why. The main thought experiment I have in my head is this:

A ball in space is traveling with a momentum mbVb, and gravity and friction are assumed to be zero. It impacts an arm, which it sticks to, and the arm is connected to a frictionless, fixed axle. This cause the arm, with the ball attached, to rotate about the axle with angular momentum ω(Ib+Iarm) and zero linear momentum, due to the fixed axle (I'm guessing that is where it's wonky, if the axle is fixed to something such as a planet [while ignoring gravity], I assume the planet itself would take over the linear momentum?). After a 180o rotation, the arm releases the ball, which now has the linear momentum mb(-Vb) or (-p).

This makes perfect sense in my head, but from what I understand it is incorrect. Assuming the axle is fixed to the aforementioned planet, how would you determine the momentum at different points in time? Would the ball still indeed have an equal magnitude momentum before and after it met the arm/axle? Would the planet take on the linear momentum until the ball is released at which point the planet would return to whatever momentum it had prior to contact with the ball?
Alternatively, if the axle is not fixed at all, would the center of mass of the ball/arm/axle system gain the linear momentum, but only until the ball was released?

EDIT: Also, what if the arm only has a 180o field of movement about the axle, such that the arm stopped rotating at the moment the ball released?

Last edited: Nov 5, 2015
2. Nov 5, 2015

### A.T.

That's an approximation, for elastic collisions with M >> m and time of interaction -> 0. I don't think you can assume this to hold exactly in your much more complicated case.

No. If the ball's momentum vector changes, then the planet's momentum vector must change too.

Same as above.

3. Nov 5, 2015

### SikorskyUH60

Thank you for the quick response! Quick clarification, though:
Would the planet's final momentum vector (assuming initial to be zero, and time of interaction -> 0) still be equal to the final momentum of the ball, but in the opposite direction? In other words, would the two post-separation bodies be moving in opposite directions, but with equal momentum?

Finally, would the planets final momentum vector be equal to or less than if the ball had simply struck the planet (assume the collision is elastic)?

4. Nov 5, 2015

### A.T.

No. The planets change in momentum is equal but opposite as the ball's change in momentum.

If the ball goes:
1 -> -1
Then the planet goes:
0 -> 2

The planets final forward momentum when the ball returns back is greater than if they stuck.

5. Nov 5, 2015

### CWatters

Part of the problem is that you are changing your system boundary half way through. You start off talking about a system comprising a ball and saying that "it" can't convert it's linear momentum to rotational. Then you change the system to one comprising a ball and an arm fixed to an axle/planet etc.

The key to understanding is to start off with a system that comprises the ball AND the arm/axle/planet and stick to it throughout. If you analyse that system before and after the impact you would find that both linear and rotational momentum is conserved.

For example... The ball appears to collide with one end of the arm (eg not the centre of mass of the arm/axel/planet). So before the collision there was some rotation in the system that you didn't account for - because you only considered the ball not the whole system.