Rotating cylindrical spaceship

AI Thread Summary
The discussion centers on the conservation of energy and angular momentum in a rotating cylindrical spacecraft as an astronaut climbs a spoke. It is noted that the center of mass of the spacecraft-astronaut system shifts during the climb, affecting the rotation point. While angular momentum is conserved, rotational energy is not, as the astronaut expends energy climbing against artificial gravity. The equations derived show that as the astronaut moves toward the center, the moment of inertia decreases, leading to an increase in rotational energy. The analysis concludes that energy must be provided by the astronaut to climb, aligning with the principles of physics involved.
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

Details here: http://openstudy.com/study#/updates/51366339e4b093a1d94a782b
 
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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

L_0 = I_0 \omega_0 = I_1 \omega_1 = L_1

where I_0 and I_1 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

\omega_1 = \frac{I_0}{I_1}\omega_0

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

K_0 = \frac{1}{2}I_0\omega_0^2

and

K_1 = \frac{1}{2}I_1\omega_1^2

Combining the last three equations you then get that

K_1 = \frac{I_0}{I_1}K_0

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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