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This question follows from this post here.

Can linear momentum convert into angular monetum?

I would want to partly disagree that they cannot be transformed into one another.

Say for example we have a ball. We throw it such that it has a forward (top) spin (here "forward" is parallel to the ball's translational velocity, ie. its tangential velocity on top of the ball at any given point is in the same direction as its motion). Upon hitting the ground and bouncing, the ball will move faster in the direction it was moving due to the top spin. How did the ball gain extra translation motion? It should have come from somewhere.

To answer this question, I came up with 2 scenarios.

1. Some of the ball's angular momentum is transformed to linear momentum: When the ball hit the ground, the ball would experience a change in its spin and angular momentum. Since, momentum is not lost, it has to go somewhere - it transform to linear momentum (tangentially) - Whats perfect about this answer is that, the puzzle pieces of the ball's angular momentum being lost (spin slowed) and it suddenly gaining translational velocity, match up.

2. Obviously, with the first answer - doing the math, the domains don't add up... Here's the second answer: What if they still do convert, but indirectly? This answer I have not thought of completely, as I could not find a proper medium or force for which they could convert through. In my mind, I think about it something like this: $$Ang \to R \to Translational$$ or vice-versa, where $R$ is the medium. In the ball scenario, we can express the change in translational momentum as $F\triangle t$, where the force would the medium. We know it has gained translational momentum because $v_{Intial} < v_{Final}$ with respect to the collision with the ground. Hence, the $F\triangle t > 0$, when considered with same direction of motion.

Question:

The answers are theoretical and display my ideas. My question is, why cannot angular momentum be converted to linear momentum (and vice-versa)(other than their dimensions/domain)? Even if the answer is no, how can one explain the ball scenario above?

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    $\begingroup$ Both linear and rotational momentum are conserved separately, thus it's not possible to convert one into the other. $\endgroup$
    – Kenneth Goodenough
    Commented Dec 11, 2021 at 8:20

2 Answers 2

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You can't attack an interaction using conservation of momentum if you assume that one of the objects in the interaction has infinite mass and/or an infinitely distant center of mass. The angular momentum of the spinning ball goes into the angular momentum of the ball-Earth system, which can be closely approximated by treating the ball as a point mass with an initial angular velocity of 0 about an initially non-rotating sphere with a large but finite moment of inertia.

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  • $\begingroup$ True (+1)... but how does the ball gain translational velocity after hitting the ground (and this only occurs if one applies a top spin)? $\endgroup$
    – Dstarred
    Commented Dec 11, 2021 at 11:37
  • $\begingroup$ Any change in the ball's translational momentum corresponds to an equal and opposite change in the Earth's translational momentum, and any change in the total angular momentum of the ball (spin angular momentum vector summed with orbital angular momentum) corresponds to an equal and opposite change in the Earth's angular momentum. $\endgroup$
    – g s
    Commented Dec 11, 2021 at 17:25
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    $\begingroup$ Kenneth's comment above is true but might be confusing you more. It's very possible to convert angular momentum into two equal and opposite linear momenta such that the total system angular momentum and linear momentum are unchanged. Note that even straight-line trajectory linear translation with respect to a noncolinear point implies an angular velocity about that point, hence angular momentum. The product $\omega r$ is constant since the velocity is constant, hence the angular momentum doesn't change even though the angular velocity is always changing. $\endgroup$
    – g s
    Commented Dec 11, 2021 at 18:45
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    $\begingroup$ @James Same issue as the OP. Assign the wall a large but finite mass and apply Newton's 3rd. $\endgroup$
    – g s
    Commented Oct 1 at 6:51
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    $\begingroup$ @James That's correct. $\endgroup$
    – g s
    Commented Oct 1 at 15:18
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Not sure if I am missing a point here but doesn't a slingshot do exactly that?

You rotate a mass in a circular path and it has angular momentum = mrw But rw = tangential velocity v

When you release the sling the body exits in linear tangential path with velocity v and linear momentum mv..

EDIT : This answer is an incorrect understanding of the issues. Please ignore.

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  • $\begingroup$ At the point the mass leaves in the tangential direction the conservation of momentum moves the sling and user . which go in the opposite direction and that reaction should also add up to conserve angular momentum $\endgroup$
    – anna v
    Commented Dec 11, 2021 at 13:57
  • $\begingroup$ Please illustrate how that is true. The user will remain standing. and the sling 'rope' will temporarily continue rotating in the original direction. $\endgroup$
    – Wreckless
    Commented Dec 11, 2021 at 14:01
  • $\begingroup$ If the rope rotates its because the user provided by his momentum the impulse due to not stopping the rotational motion. Moemntum conservation goes -mv to the mv of the shot, but the person and the earth to which the user is attached are so massive that it easily takes up the momentum -mv. That -mv by definition has an angular momentum to the rotation center, and it is the negative of the one leaving. $\endgroup$
    – anna v
    Commented Dec 11, 2021 at 14:49
  • $\begingroup$ see youtube.com/watch?v=FaQRNj0WpYU $\endgroup$
    – anna v
    Commented Dec 11, 2021 at 14:50
  • $\begingroup$ The you tube video is nothing other than a demo of slingshots. However, further reading shows that I have misunderstood this concept and my answer above is incorrect. $\endgroup$
    – Wreckless
    Commented Dec 11, 2021 at 21:53

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