本题目来源于试卷: 2007美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A space station consists of two living 2g4bh7*fnlum: bx7rpkv;s2n f 6 v3ccmodules a hpyziqi,7:x /ttached to a central hub on opposite sides of the hub by long corridors of equal length. Each living module contains N astronauts of equal mass. The mass of the space station is negligible compared to the mass of the astronauts, and the size of the central hub and living modules is negligible compared to the length of the corridors. At the beginning of the day, the space station is rotating so that the astronauts feel as if they are in a gravitational field of s y, /qizp:xhi7trength $g$. Two astronauts, one from each module, climb into the central hub, and the remaining astronauts now feel a gravitational field of strength $g^{\prime}$. What is the ratio $g^{\prime}/g$ in terms of N?
A. $2N/(N-1)$
B. $N/(N-1)$
C. $\sqrt{(N-1)/N}$
D. $\sqrt{N/(N-1)}$
E. none of the above
参考答案: E
本题详细解析:
The apparent gravityzw( fwjo-zsrm+ :2 g*l $g$ is the centripetal acceleration $g = \omega^2 r$.
Therefore, the ratio $g^{\prime}/g = (\omega^{\prime}/\omega)^2$.
Since there are no external torques, angular momentum $L = I\omega$ is conserved.
So, $I\omega = I^{\prime}\omega^{\prime}$, which gives $\omega^{\prime}/\omega = I / I^{\prime}$.
Initially, there are $2N$ astronauts (N in each of two modules) at radius $r$.
The initial rotational inertia is $I = (2N)mr^2$.
Finally, two astronauts (one from each module) move to the hub ($r=0$), so $2N-2$ astronauts remain at radius $r$.
The final rotational inertia is $I^{\prime} = (2N-2)mr^2 = 2(N-1)mr^2$.
The ratio of angular velocities is $\omega^{\prime}/\omega = I / I^{\prime} = \frac{2Nmr^2}{2(N-1)mr^2} = \frac{N}{N-1}$.
Finally, the ratio of apparent gravity is $g^{\prime}/g = (\omega^{\prime}/\omega)^2 = \left(\frac{N}{N-1}\right)^2$.
This result is not among options (a)-(d), so the correct answer is (e).
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