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I have a point mass connected to a string with negligible mass. The point mass has mass $m$ and is moving at a velocity $V$. The string is of length $r$, and it is keeping the point mass tied to a center of rotation.

Suppose we were to suddenly cut the string. It is easy to see that the point mass will continue in a straight line tangential to the path at the moment it was cut at the same velocity $V$ that it originally had. And so, this implies a conservation of linear momentum.

But what about angular momentum, $I\omega=mr^2\omega$, how does it stay constant? Or does the point mass somehow lose angular momentum all together?

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The concept of angular momentum only makes sense when we specify a rotation axis. So we will pick the axis passing through the initial center of rotation.

When the string is cut, the point mass has a linear momentum $p = mv = mr\omega$. One can define the angular momentum of a particle about an axis as $\vec{L} = \vec{d} \times \vec{p}$ where $\vec{d}$ is the vector from the axis of rotation to the particle (I used $d$ to avoid confusion with the $r$ you specified). The magnitude of angular momentum is then $L = dmr\omega \sin\theta$, but $\sin\theta$ can be shown from geometry to be $r/d$.

Thus the solution is $L = dmr\omega \frac{r}{d} = mr^2 \omega$, which does not depend on the distance the point particle is from the axis and is the same as the initial angular momentum. Hence angular momentum is conserved.

Now as for where the linear momentum came from. The object always had linear momentum of magnitude $p = mv = mr\omega$, however if you remember, $\vec{F} = \frac{d\vec{p}}{dt}$, and since there was a tension force on the point mass while it was rotating the direction of the momentum was changing. When you cut the string the direction is chosen and remains constant, and linear momentum is conserved.

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