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If you have a spinning disk coated in a layer of dry ice with some initial angular momentum, when the dry sublimates, do the leaving particles of gaseous carbon dioxide leave with any of the momentum related to the spin of the disk, or do they only leave with energy from their increase in temperature? The result being a decrease in mass of the spinning disk system while allowing it to retain the same angular momentum, leading to an increase in its speed.

Or do they take with them the portion of the system's angular momentum with which they started, allowing the system to continue spinning at constant speed, in a similar manner to what happens when you life a soup can off of a spinning platform?

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  • $\begingroup$ I agree. That's the verdict I reached. Has anyone taken it to experiment though $\endgroup$ – rjm27trekkie Jun 12 '17 at 23:42
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I saw this question in a textbook, and it got the answer wrong.

The speed of the turntable does NOT change as the dry ice sublimates.

Here's how to think about it: what if you had a spinning metal turntable and cut it in half horizontally producing two thinner turntables? Each is still spinning around the same axis. Would either half speed up or slow down? Of course not. What if you removed one of those halves, would the other half speed up or slow down? No. The dry ice problem is exactly the same.

What happens to the angular momentum of the dry ice? It remains. Each dry ice molecule flies off tangentially (moving in a straight line, continuing the velocity it had at the moment it separated from the disk) and the angular momentum of the entire cloud of molecules remains the same. The cloud as a whole is both expanding and rotating, even though each molecule is moving in a straight line. (And maybe those molecules collide with something, e.g. air, and that changes their motion, but this is irrelevant to the turntable.)

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When particles exit the surface, they would continue in the direction they were headed prior to release. That is, tangent to the rim, in the direction of rotation. So their momentum would be subtracted from the disk, slowing it down.

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  • $\begingroup$ It is like a reverse plastic collision where momentum is exchanged. $\endgroup$ – John Alexiou Dec 18 '16 at 8:18
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    $\begingroup$ Don't you mean that it would maintain the same speed as momentum must be conserved. If the CO2 particles take with them their momentum and not just their mass, then the disk should stay at the same speed. This would be analogous to lifting a food can off of a spinning platform spinning at some constant speed. When this is done, the platform continues to spin at the same speed. $\endgroup$ – rjm27trekkie Dec 19 '16 at 0:36
  • $\begingroup$ If they took just their speed somehow, then the disk would slow down. But that is impossible as the CO2 particles have mass. $\endgroup$ – rjm27trekkie Dec 19 '16 at 2:41
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If you imagine standing on the edge of the spinning disk and watching particles sublimating, you would see them flying out horizontally (ignore the vertical direction) equally in all directions on a long enough timescale. This is because the particles' outgoing velocities are based on random thermal motion when they left the surface of the solid.

Since the time-averaged group has no net velocity in the horizontal direction, we conclude that their escape produced no net horizontal force on the disk by the impulse-momentum theorem (otherwise, we would observe dry ice blocks flying of in a single direction when they were exposed to air). From this we see that the time-averaged group of particles exerts no torque on the disk, and the disk does not slow down.

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  • $\begingroup$ forky40, I agree. If I rotate at a constant angular velocity on a frictionless turntable with bricks held at arms length, and I drop those bricks straight down in my reference frame, I am positive that my rotation rate will not slow down. $\endgroup$ – David White Dec 19 '16 at 7:08
  • $\begingroup$ Sounds like you've recently dropped some bricks on the floor. I agree with this answer, the question is why would my textbook propose this example as a scenario where the sublimating particles did not leave with their tangential velocity, rather they rose straight up and the disk sped up. An error or is there a possibility that this is a special case? Hence why i made the thread. $\endgroup$ – rjm27trekkie Dec 19 '16 at 23:05

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