本题目来源于试卷: 2012美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
Three point masses $m$ are attached together by identical springs. When placed at rest on a horizontal surface the masses form a triangle with side length $l$. When the assembly is rotated about its center at angular velocity $\omega$, the masses form a triangle with side length $2l$. What is the spring constant $k$ of the springs?
A. $2m\omega^2$
B. $\frac{2}{\sqrt{3}}m\omega^{2}$
C. $\frac{2}{3}m\omega^{2}$
D. $\frac{1}{\sqrt{3}}m\omega^{2}$
E. $\frac{1}{3}m\omega^{2}$
参考答案: C
本题详细解析:
Assume the spring's natt io,sw;;e*es qvs3 3yt3k so9ural length is $l_0 = l$.
When rotating, the side length is $s = 2l$. The extension of each spring is $x = 2l - l_0 = l$.
The force from each spring is $F_s = kx = kl$.
Each mass $m$ is at a radius $r$ from the center of rotation. For an equilateral triangle of side $s$, the distance from center to vertex is $r = s / \sqrt{3}$. Here, $r = 2l / \sqrt{3}$.
The centripetal force required is $F_c = m\omega^2 r = m\omega^2 (2l / \sqrt{3})$.
This centripetal force is provided by the vector sum of the two spring forces acting on the mass. The angle between the two springs is $60^\circ$. The net force $F_{net}$ points radially inward and is $F_{net} = 2 F_s \cos(30^\circ) = 2 (kl) (\sqrt{3}/2) = kl\sqrt{3}$.
Set $F_{net} = F_c$:
$kl\sqrt{3} = m\omega^2 (2l / \sqrt{3})$
$k\sqrt{3} = 2m\omega^2 / \sqrt{3}$
$k (\sqrt{3} \cdot \sqrt{3}) = 2m\omega^2$
$3k = 2m\omega^2$
$k = \frac{2}{3}m\omega^{2}$
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