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Can I assume a uniform motion of a point mass with velocity $v$ for a point in time to be a rotation around a point $P$ that has a distance $r$ to the point mass, so that the angular velocity of it is $w=v/r$ (only for an infinitely short time span)?

The motion of the point mass would then be tangential to the circle.

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    $\begingroup$ It depends on what you do with that assumption. For some purposes, it will work. For others it will not. $\endgroup$ – Ben51 Feb 18 at 13:24
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Not in general. You can describe always a motion in terms of rotation around a fixed point but that won't always be a uniform rotation i.e. it might have accelerations, radial displacements etc.

What you can do is very locally approximate your motion to a rotation.

We can write the position $\vec{r}=(x(t), y(t))$ of your particle in polar coordinates (assuming motion is on a plane for simplicity) with respect to the origin as

$$\vec{r}=(\rho(t), \theta(t))$$ using the transformations $$\rho=\sqrt{x^2+y^2}$$ and $$\theta=atan2(y, x)$$ (see here for the definition of atan2)

You could also do the same for any point $P=(x_0, y_0)$ substituing $x\to x-x_0$ and $y\to y-y_0$.

Then of course $$\omega=d\theta/dt$$ but you already see that this involves a derivative of the atan2 function so it is not in general going to be a constant speed as you ask in your question.

Now however, if your displacement is really small you could locally approximate it to a rotation $\hat{R}$ so that $$\vec{r}(t+dt)=\hat{R}\vec{r}$$ where $\hat{R}$ is a rotation matrix.

If the displacement is very small, you could use the infinitesimal form of the rotation (assuming $P$ is the origin) $$\hat{R}=\begin{bmatrix}cos(d\theta) & -sin(d\theta) \\ sin(d\theta) & cos(d\theta)\end{bmatrix}\approx\begin{bmatrix}1 & -d\theta\\d\theta & 1\end{bmatrix}$$ where I used $cos(d\theta)\approx 1$ and $sin(d\theta)\approx d\theta$ for very small angles, so that, applying the matrix

$$\vec{r}(t+dt)=\hat{R}\vec{r}$$ becomes $$\vec{r}(t+dt)=\begin{bmatrix}x(t+dt) \\y(t+dt)\end{bmatrix}=\begin{bmatrix}1 & -d\theta\\d\theta & 1\end{bmatrix}\begin{bmatrix}x(t) \\y(t)\end{bmatrix}$$

which we can separate in two equations

$$x(t+dt)=x(t)-y(t)d\theta$$ $$y(t+dt)=x(t)d\theta+y(t)$$

which however assumes the distance from the origin is constant (so you have an error in the radial direction is $\rho$ is actually changing).

If we rewrite that as $$dx=x(t+dt)-x(t)=-y(t)d\theta$$ $$dy=y(t+dt)-y(t)=x(t)d\theta$$ you get that $$dr^2=dx^2+dy^2=d\theta^2(y^2+x^2)=d\theta^2\rho^2(t)$$ meaning $$d\theta=dr/\rho(t)$$ so that this whole thing is valid if $$d\theta\ll 1$$ i.e. $$dr\ll \rho(t)$$ meaning that you can do it only if your total displacement $dr$ is much smaller than the radius of the circle with which you are approximating your motion. Also, this is valid at one timestep $dt$ and the more time-steps you do, the less this approximation is going to be valid if your radius changes consistently over time.

Now, in terms of speed, becuase $dr=vdt$ we get $$d\theta=dr/\rho(t)=vdt/\rho(t)$$ so that $$\omega = d\theta/dt = v / \rho$$ as expected, but again: it is only valid for an infinitesimal displacement. As you move away from the circle or even if you rotate around the origin at non-constant angular acceleration, this approximation sooner or later fails. Especially, as mentioned, you are completely neglecting displacement in the radial position, so if your motion is completely radial this is completely wrong as $d\theta = 0$ meaning you would get $\vec{r}(t+dt)=\hat{R}\vec{r}(t)=\vec{r}(t)$ i.e. you are not moving at all.

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  • $\begingroup$ Ok wow that was very detailed. Thanks, but I think I didn't quite understand that. It's a little difficult for me language wise, cause I'm German. I was kinda thinking about a straight motion with a constant velocity. The Center of the rotation would then move parallel with the original motion. So that I could use Angular Momentum for a point mass, moving in a straight line, hitting another pointmass in a rotating motion $\endgroup$ – Paulemeister Feb 18 at 20:17
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    $\begingroup$ You can use angular momentum, which is conserved for a particle moving in a straight line. But you cannot change the center of rotation every time. Just choose one point (probably the center of the other rotating pointmass) and compute the angular momentum from there. $\endgroup$ – JalfredP Feb 19 at 9:30
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The linear motion of an object can be used to determine one contribution to the angular momentum of the object relative to a point which does not lie in line with the velocity vector. Also, with an extended object, it is sometimes useful to work with an instantaneous axis of rotation. For that, the position vector from the axis to the center of mass must be perpendicular to both the axis and the velocity vector of the center of mass, and the angular velocity vector for the center of mass relative to the axis must be equal to the angular velocity vector for motion about the center of mass.

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