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enter image description here

A point mass with mass $m$, distance $r$ from circle and constant tangential velocity and constant angular velocity is swung around a circle. ($p$ is linear momentum)

Angular momentum is radius x linear momentum. It is conserved.

If $r$ is increased, linear momentum decreases. However, shouldn't linear momentum be conserved as well? Where does linear momentum go?

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Your setup implicitly ignores the motion of the object in the center. For instance, in the solar system, the Sun applies a force that keeps the planets moving in roughly circular orbits. The planets apply equal and opposite forces back on the Sun.

Newton's Third Law says that every force from object A on object B is accompanied by a force from object B on object A. In fact, a force is a flow of linear momentum, and without these equal and opposite flows, linear momentum is not conserved (as pointed out by my2cts and WarreG).

For simplicity, let's just consider the Sun and Jupiter (which contain much of the mass of the inner solar system). If you swing a yo-yo in a circle on a string, you need to apply a force toward the center of the circle (a centripetal force) on the yo-yo to keep it moving in a circle. If you cut the string that transfers the force, the yo-yo will fly outward in straight line. So the Sun applies the force to keep Jupiter in its nearly circular orbit, rather than flying off in a straight line. Jupiter likewise applies an opposite force on the Sun. However, the Sun is much more massive than Jupiter, so it accelerates very little in comparison (acceleration = force/mass). The Sun orbits in a little circle around the center of mass of the solar system, while Jupiter also orbits this center of mass. This video should give you the idea.

So one way to answer your question is just to say that momentum isn't conserved if you have unmatched forces (you violate Newton's Third Law). Another answer is that momentum is conserved, you just need to consider the momentum of the central object. As the Sun pulls on Jupiter and changes Jupiter's momentum, Jupiter pulls back on the Sun and changes the Sun's momentum. In the center of mass reference frame, the momentum of the Sun always exactly cancels the momentum of Jupiter. Jupiter makes a big circle with a big velocity and the Sun makes a little circle with a small velocity. As the Sun becomes more and more massive compared to Jupiter, the Sun's circle gets smaller and smaller (and has a smaller and smaller velocity), until it's barely noticeable. That's the situation you are describing.

On the other hand, if the two objects have more similar mass, the motion of the mass closer to the center becomes more obvious. Here is a video of Pluto and Charon orbiting their center of mass, presumably to scale.

For objects of equal mass, the equal and opposite linear momenta should be clear. I've made a little image in Inkscape for two equal mass objects showing the momentum and forces for equal mass objects in circular orbits about their center of mass:

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  • $\begingroup$ Which program did you use to draw this image? $\endgroup$ – M.N.Raia Jan 13 at 23:38
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    $\begingroup$ Inkscape (it's free and open source!) $\endgroup$ – WaterMolecule Jan 14 at 14:57
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"Linear momentum is conserved when there is no force acting on the system". So far so good. As the mass is moving in a circle, there is a force acting on it and linear momentum is not conserved. If the force vanishes, the bass will move on a straight line and both linear and angular momentum are conserved.

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  • $\begingroup$ Let's see if I am understanding this correctly: Linear momentum is not conserved due to the external centripetal force acting on the ball. (Direction changes) However, the magnitude of linear momentum is still the same. If the radius is increased, the linear momentum is still not conserved (because of the direction) The magnitude of the new linear momentum will be lower, but still constant. (But why is it lower?) $\endgroup$ – helpme Jan 14 at 20:52
  • $\begingroup$ The linear momentum of the point mass is not conserved. The total linear momentum is. As explained by @WaterMolecule there is the rest of the system to consider. As long as the two are connected energy and momentum will be exchanged. That is where the momentum goes. $\endgroup$ – my2cts Jan 14 at 23:24
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Linear momentum is conserved when there is no force acting on the system. If you increase the radius, you will have to exert a force on the system. If this could be done without a force then you could accelerate the particle to high speed and then increase the radius to infinity. This would violate conservation of energy (and angular momentum).

In a circle, I like to imagine that the area of the triangle made by $r$ and $p$ must be conserved just like in Kepler's Second Law. So when you double the radius, the linear momentum has to half to conserve the area.

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    $\begingroup$ That’s wrong, in fact essentially backwards — if the particle just flies forward with no force on it, its radius will increase. And if you keep the radius constant as the particle moves in a circle, there is a force, namely the centripetal force. $\endgroup$ – knzhou Jan 13 at 21:24

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