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 uniform disk, a thin hoop, and a uniform sphere, wl;4zfpsto/)x (: k * fr fxt/bg1aojdj71+ edall with the same mass and s s cc.l2-gg0dco i0c dk28kik7ame outer radius, are each free to rotate about a fixed axis through its center. Assum 2si .cggol0 dcki 7ck-2kc 0d8e the hoop is connected to the rotation axis by light spokes. With the objects starting from rest, identical forces are simultaneously applied to the rims, as shown. Rank the objects according to their kinetic energies after a given time $t$, from least to greatest. 
A. disk, hoop, sphere
B. sphere, disk, hoop
C. hoop, sphere, disk
D. disk, sphere, hoop
E. hoop, disk, sphere
参考答案: E
本题详细解析:
All objects experience the same fo.7) oc -e)b h ys8:ykqkpwb6dorce $F$ at the same radius $R$, so the torque $\tau = FR$ is identical for all three.
Angular acceleration $\alpha = \tau / I$.
Angular velocity after time $t$ is $\omega = \alpha t = (\tau t) / I$.
Rotational kinetic energy $K = \frac{1}{2}I\omega^2 = \frac{1}{2}I \left( \frac{\tau t}{I} \right)^2 = \frac{(\tau t)^2}{2I}$.
Since $\tau$ and $t$ are identical, the kinetic energy $K$ is inversely proportional to the rotational inertia $I$ ($K \propto 1/I$).
The rotational inertias are: $I_{hoop} = mR^2$, $I_{disk} = \frac{1}{2}mR^2$, $I_{sphere} = \frac{2}{5}mR^2$.
Ordering the inertias from greatest to least: $I_{hoop} > I_{disk} > I_{sphere}$ (since $1 > 1/2 > 2/5$).
Therefore, the kinetic energies from least to greatest are in the opposite order: $K_{hoop} < K_{disk} < K_{sphere}$.
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