2007: Problem 27

Topic: Rigid Bodies
Concepts: Artificial gravity, Conservation of angular momentum


Solution:

Let ω\omega be the initial angular velocity of the space station. In the rotating frame, all objects experience a (fictitious) centrifugal force of magnitude mac=mω2rma_c=m\omega^2r which looks like gravity mgmg if we have

g=ω2rg=\omega^2r

After two astronauts climb into the center, the space station rotates at angular velocity ω\omega’ so the artificial gravity becomes g=ω2rg’=\omega’^2r. We can find ω\omega’ by conserving angular momentum since no external torque acts on the space station.

We have

Li=Iiω=2Nmr2ωL_i=I_i\omega=2Nmr^2\omega
Lf=Ifω=2(N1)mr2ωL_f=I_f\omega’=2(N-1)mr^2\omega’

Since Li=LfL_i=L_f,

ωω=NN1\frac{\omega’}{\omega}=\frac{N}{N-1}

Finally,

gg=ω2ω2=(NN1)2\frac{g’}{g}=\frac{\omega’^2}{\omega^2}=\left(\frac{N}{N-1}\right)^2

so the answer is E.

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