Angular momentum is defined as $L = r \times p$ where r is the center of rotation and p is the momentum. Since angular momentum is conserved, if r decreases then p must increase. And since p is m*v the velocity must increase. But that would change the momentum while keeping the angular momentum constant. Aren't both angular and linear momentum supposed to be conserved unless acted upon by an external force?

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    $\begingroup$ What makes $r$ decrease? $\endgroup$ – Hritik Narayan Jun 1 '15 at 16:31

Angular momentum is conserved. If you have ever been in an office rolly-chair, you might (admit it) have spun around in it. If you stick out your legs, you will actually slow down. If you tuck them in, you will speed up.

This is just a consequence of the equation $$L = r \times p$$

Since $L$ is conserved, as $r$ -> $0$, $p$ -> $\infty$, and as $r$ -> $0$, $p$ -> $\infty$

Therefore, these forms of momentum are still conserved. This is only why an ice skater spins faster when they tuck their feet and arms together, and they slow down when they stick a leg out.

  • $\begingroup$ So the momentum changes as the ice skater moves his arms in and out, yet it is conserved? Would you mind elaborating on that? $\endgroup$ – lawls Jun 1 '15 at 17:12
  • $\begingroup$ @lawls The angular momentum is conserved. I see. I messed up in the answer. edited $\endgroup$ – Jimmy360 Jun 1 '15 at 17:32
  • $\begingroup$ Hm, but isn't momentum supposed to be conserved too? I don't see what I'm missing. $\endgroup$ – lawls Jun 1 '15 at 17:36
  • $\begingroup$ @lawls You have changed the system. $\endgroup$ – Jimmy360 Jun 1 '15 at 17:37
  • $\begingroup$ @lawls Just to clear things up, the linear momentum, as Mukhopadhyay pointed out is NOT conserved for change in radius. Of course it is conserved (and just the total momentum) if the radius does not change (and there's no dissipation, like friction). The angular momentum, however is always conserved in a central field of force, no matter if the central force is external or internal. So when you change the radius, you change it with a central force field (you pull back towards the center), thus angular momentum conservation is true, but it is an external force that changes linear momentum. $\endgroup$ – user3237992 Jun 1 '15 at 17:40

Linear momentum is, as the name suggests, LINEAR hence in case of rotation you need to take care of the tangential motion and not the curved motion which is nicely explained by angular momentum. When the radius increases an external force is used to push the body away and that is the reason why p is not conserved. This is because the p depends on a single linear dimension that of tangent. But when you increase radius in the perpendicular distance the vector is laterally displaced hence the value of p is unchanged. But since the entire frame is non-inertial and you want to conserve angular momentum as the first preference the tangential motion has to reduce. Hence the vector is laterally displaced so that the direction is constant but the magnitude changes. So you are changing the frame of reference while thinking about the p and L.

  • $\begingroup$ No, they are not nessecarily. Considering a system of points of masses, only the total momentum is conserved (if there are no external forces), the momenta of each points may change. $\endgroup$ – user3237992 Jun 1 '15 at 17:08

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