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 thin, uniform rod has x7oi8( .dgewol4eyb ;,v *)g5maiize+ qfbi1on7 ( dymass $m$ and length $L$. Let the acceleration due to gravity be $g$. Let the rotational inertia of the rod about its center be $md^{2}$. The rod is suspended from a point a distance $kd$ from the center, and undergoes small oscillations with an angular frequency $\beta\sqrt{\frac{g}{d}}$. Find an expression for $\beta$ in terms of $k$.
A. $1+k^{2}$
B. $\sqrt{1+k^{2}}$
C. $\sqrt{\frac{k}{1+k}}$
D. $\sqrt{\frac{k^{2}}{1+k}}$
E. none of the above
参考答案: E
本题详细解析:
This is a physical pendulum.
T*n2ik*8kbs*ngif :hzgsxvy0+ nko4 *he angular frequency $\omega$ for small oscillations is given by $\omega = \sqrt{\frac{mg\ell}{I_p}}$, where $\ell$ is the distance from the pivot to the center of mass and $I_p$ is the rotational inertia about the pivot.
Here, the pivot-to-CM distance is $\ell = kd$.
We find $I_p$ using the parallel axis theorem: $I_p = I_{cm} + m\ell^2$.
Given $I_{cm} = md^2$ and $\ell = kd$, we have:
$I_p = md^2 + m(kd)^2 = md^2 + mk^2d^2 = md^2(1+k^2)$.
Now substitute $\ell$ and $I_p$ into the angular frequency formula:
$\omega = \sqrt{\frac{mg(kd)}{md^2(1+k^2)}} = \sqrt{\frac{gk}{d(1+k^2)}}$.
We can rearrange this as $\omega = \sqrt{\frac{k}{1+k^2}} \sqrt{\frac{g}{d}}$.
The problem gives $\omega = \beta\sqrt{g/d}$.
By comparing the two expressions, we find $\beta = \sqrt{\frac{k}{1+k^2}}$.
This expression is not listed in options (a)-(d), so the correct answer is (e).
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