Rotating cylindrical spaceship

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SUMMARY

The energy of a rotating cylindrical spacecraft is not conserved when a point-like astronaut climbs a spoke connecting the walls to the center. While conservation of angular momentum applies, conservation of rotational energy does not. As the astronaut climbs, the center of mass of the spacecraft-astronaut system shifts, affecting the rotational dynamics. The equations governing this scenario include L_0 = I_0 ω_0 = I_1 ω_1 and K_1 = (I_0/I_1)K_0, indicating that the astronaut must expend energy to climb, resulting in an increase in rotational energy due to a decrease in moment of inertia.

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guillefix
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Is the energy of a rotating cylindrical spacecraft conserved when a point-like astronaut climbs up a spoke connecting the walls with the center of the cylinder?
If so, when I calculate the fractional change in apparent gravity at the walls when the astronaut reaches the middle I get different answers using conservation of angular momentum and conservation of energy

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I can't see that link but...

There are all sorts of things going on in the case you describe. For one thing the center of mass of the spacecraft -man combination moves as he climbs the spoke. So the point about which the spacecraft rotates will move.

If the man is climbing against the artificial gravity he is expending energy (provided by his food).

Plenty of scope to make an error somewhere.
 
I assume this is not really homework. If it is, please do not read my post :smile:

If the spacecraft and astronaut are isoloated, you can assume conservation of angular momentum, but not conservation of rotational energy. If you write up angular momement before and after "the climb" and equate them you have

[tex]L_0 = I_0 \omega_0 = I_1 \omega_1 = L_1[/tex]

where [itex]I_0[/itex] and [itex]I_1[/itex] is the moment of inertia before and after (here assuming the direction of the rotational axis is unchanged). Note, that the second equal comes from assumed conservation of angular momentum. This implies that

[tex]\omega_1 = \frac{I_0}{I_1}\omega_0[/tex]

If you now write up rotational energy before and after you get

[tex]K_0 = \frac{1}{2}I_0\omega_0^2[/tex]

and

[tex]K_1 = \frac{1}{2}I_1\omega_1^2[/tex]

Combining the last three equations you then get that

[tex]K_1 = \frac{I_0}{I_1}K_0[/tex]

that is, the rotational energy will increase because the moment of inertia decreases when he climbs towards the center. This also fits the intuition that he has to provide energy in order to "climb up" the ladder. If he were to drop down instead the moment of inertia increases and the rotational energy of the system would decrease, which again fits the intuition that he would be able to gain (internal) energy dropping down.
 

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