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Consider a system of two masses that rotates with constant angular velocity. When a force contracts the system the velocitie of the two masses increase. I understand this in terms of conservation of angular momentum but I would like to understand how does the force that cause the contraction accelerates the two masses.

enter image description here

Using polar coordinates, the force is central, therefore radial. This means that $v_r$ of the two masses increase, while $v_{\theta}$ should remain constant. During the contraction the motion is a spiral, so the velocity is not perpendicular to the force, hence the magnitude of the velocity vector changes. But at the end, when the system is compressed the two masses follow a circular motion which is faster than the one at the beginning. This means that $v_{\theta}$ has somehow increased, but how?

The increase of the magnitude of the velocity does not imply the increase of the component perpendicular to the radial direction. This increase seems impossible to me since the force itself is radial.

How can $v_{\theta}$ increase during the motion?

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  • $\begingroup$ Look at the relation between velocity and angular velocity. $\endgroup$ Commented Apr 11, 2016 at 9:47
  • $\begingroup$ Angular momentum is a conserved quantity. Angular velocity? No. $\endgroup$ Commented Apr 11, 2016 at 10:54

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Let's look at the hodograph of a constant radius & constant velocity motion.

hodograph

Left: trajectory of one of the masses. Right: hodograph, i.e. locus of the velocity vectors.

Now, let's look closer at how the velocity changes during a small time interval $\mathrm dt$.

hodograph zoomed

A force is needed to rotate it (difference between the brown and red arrows). If you exert a larger force, you see that:

  • the velocity increases in norm (the magenta arrow is longer)
  • the velocity rotates faster (the angle red-magenta is bigger than red-brown)

The key to understanding the phenomenon is realizing the radial & orthoradial directions are not fixed: the radial direction at time $t$ will soon be the orthoradial direction at some time $t'$. Thus, when you say "the radial force changes $v_r$", in fact you should say "the radial force changes both $v_r$ and $v_θ$".

For a more formal explanation, note the acceleration along $\hat r$ and $\hat θ$ is not the derivative of the amplitude of velocity along $\hat r$ and $\hat θ$. Indeed, $\vec v=v_r \hat r+v_θ \hat θ$, so $\vec a=(\dot{v_r}-v_θ\dot θ)\hat r+(v_r\dot θ+\dot{v_θ})\hat θ$, that is $a_r=\dot{v_r}-v_θ\dot θ$ and $a_θ=v_r\dot θ+\dot{v_θ}$. Hence $a_θ=0$ does not imply $v_θ=\text{const.}$, rather $\dot{v_θ}=-v_r\dot θ$: because of rotation ($\dot θ≠0$), radial velocity ($v_r$) is "converted" into variation of orthoradial velocity ($\dot{v_θ}$).

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  • $\begingroup$ Thanks for the anwer! By "orthoradial" you mean the $\hat{\theta}$ direction? If so, and if I got your point, even if $a_{\theta}=0$ by definition of central force ($F || \hat{r}$), $v_{\theta}$ changes in time because the direction of $\hat{\theta}$ is not fixed. Would you be so kind as to give some further explanation about the change in $\vec{v_{\theta}}$? I can understand that the direction of this vector changes, but I still don't see how a central (and so radial) force can change the magnitude of $\vec{v_{\theta}}$, since it is always perpendicular to it, by definition. $\endgroup$ Commented Apr 13, 2016 at 12:48
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    $\begingroup$ Yes, orthoradial means $\hat θ$ [in French, at least :-)]. Figuratively I've done my best and can't say better than "the magenta arrow is longer than the red one". More formally, see the last paragraph I've just added. $\endgroup$ Commented Apr 13, 2016 at 19:29

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