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Assuming two hollow cyliders $C_{\text{in}}$ and $C_{\text{out}}$ with radius $r_{\text{in}}$ and $r_{\text{out}}$ such that $r_{\text{in}} < r_{\text{out}}$. Each of the cylinder is uniform and of length $L$. Now assuming $C_{\text{in}}$ is symmetrically placed inside $C_{\text{out}}$ without them touching. Both of them are rotating at $k$ rotations per second along x-axis in our reference frame in space. The x axis is their center of mass.

Does Conservation of Angular Momentum imply that no matter what physical/mechanical/chemical process these 2 cylinders try they would not be able to change their rotational speed from $k$ rpm to anything different.

We are asuming there is no 3rd object to change their angular momentum or apply torque or external force etc.

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  • $\begingroup$ How are the cylinders going to "try" anything if they're not allowed to touch? $\endgroup$
    – Peter Shor
    Commented Oct 2, 2021 at 12:10
  • $\begingroup$ Any sort of force sir under the laws of nature, EM, gravity. Anyhting we can think of. $\endgroup$
    – TheoryQuest1
    Commented Oct 2, 2021 at 12:53
  • $\begingroup$ What is conserved is the angular momentum, not the angular velocity. The angular velocity can always change in order to maintain the angular momentum constant. So if the cylinder get dilated because something else, maybe stress/strain due to rotation, its moment of inertia will change and the angular velocity will too. $\endgroup$
    – LSS
    Commented Oct 2, 2021 at 13:32
  • $\begingroup$ @LSS thank you. I understand that part. Perhaps I should have been clearer. The Q is using any sort of forces of nature we know of can these 2 cylinders (in our reference frame) bring themselves to a halt. Of course at the time of measuring their final angular velocity they are assumed to have the same physical dimentions. $\endgroup$
    – TheoryQuest1
    Commented Oct 2, 2021 at 13:40
  • $\begingroup$ @TheoryQuest1 A halt? No, of course not. You have $\vec{L} = I\vec{ω}$, so if $\vec{ω} = 0$ then $\vec{L} = 0$. It doesn't matter what your inertia tensor is. $\endgroup$
    – gmz
    Commented Oct 6, 2021 at 9:36

2 Answers 2

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If we take the two cylinders as one system, and if: $$\sum \vec\Gamma_{\text{external}}=0$$ where, $\Gamma$ is torque ($\vec\Gamma=\vec r\times \vec F$) acting on system about any axis, then we have:

$$\sum L=\text{constant}$$, where $L$ is angular momentum of individual bodies.

Now here, first equation doesn't simply implies that $\sum F_{\text{external}}=0$, but rather $$\sum_i(\vec r_i \times \vec F_i)=0$$, where all $\vec F_i$s are external.

Hope you got the got the main idea/concept!

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  • $\begingroup$ thanks. just to reiterate. There two cylinders are one system. The sum of external torques (since there are no exteral bodies at all) in this setup is 0. Thus sum of this system of two cylinders cannot be changed (by any internal mechanism or forces of any kind). $\endgroup$
    – TheoryQuest1
    Commented Oct 3, 2021 at 8:41
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The angular momentum will only change , if you have a force. So for example if you have air in between the two cylinders it has to move at different speed near the outside cylinder, after some time the slower movement of the inner will brake the outer one, so they will move as one cylinder with the same angular speed.

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  • $\begingroup$ we have already assumed there is no air (the space part). And no external object or force. Any force of any kind (EM, gravity, strong nuclear, weak nuclear etc, anything permitted by the laws of physics) can only originate from one or both cylinders. Is it possible for them to bring themselves to a halt in our frame? $\endgroup$
    – TheoryQuest1
    Commented Oct 2, 2021 at 13:50
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    $\begingroup$ the short answer is NO $\endgroup$
    – trula
    Commented Oct 2, 2021 at 13:55

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