2012美国US F=MA物理竞赛 (id: 5ff1a1110)

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]
A rigid hoop can rotate about tn 8,2cfv aq7t y/xtnz0he center. Two ml hxr*dx8 nr7/assless strings are attached to the hoop, one at A, the other at B. These strings are tied together at the center of the hoop at O, and a weighxlh *d8rx n7r/t G is suspended from that point. The strings have a fixed length, regardless of the tension, and the weight G is only supported by the strings. Originally OA is horizontal.

Now, the outer hoop will start to slowly rotate $90^{\circ}$ clockwise until OA will become vertical, while keeping the angle between the strings constant and keeping the object static. Which of the following statements about the tensions $T_{1}$ and $T_{2}$ in the two strings is correct?

A. $T_{1}$ always decreases.
B. $T_{1}$ always increases.
C. $T_{2}$ always increases.
D. $T_{2}$ will become zero at the end of the rotation.
E. $T_{2}$ first increases and then decreases.


参考答案:  D


本题详细解析:
Let $T_1$ and $T_2$ be the magnitudes of the tensions. Let the weight $G$ hang in the $-y$ direction. Let the angle of rotation (clockwise) be $\alpha$, from $0^\circ$ to $90^\circ$. From the diagram, A starts at $180^\circ$ ($-x$ axis) and B starts at $45^\circ$ ($+x, +y$ quadrant). The angle between them is $135^\circ$. As the hoop rotates clockwise by $\alpha$: A's new position angle is $180^\circ + \alpha$. B's new position angle is $45^\circ + \alpha$. The forces on O are $\vec{T}_1$ (towards A), $\vec{T}_2$ (towards B), and $\vec{G}$ (down). $\vec{F}_{net} = \vec{T}_1 + \vec{T}_2 + \vec{G} = 0$. $F_x: T_1 \cos(180^\circ+\alpha) + T_2 \cos(45^\circ+\alpha) = 0 \implies -T_1 \cos\alpha + T_2 \cos(45^\circ+\alpha) = 0$. $F_y: T_1 \sin(180^\circ+\alpha) + T_2 \sin(45^\circ+\alpha) - G = 0 \implies -T_1 \sin\alpha + T_2 \sin(45^\circ+\alpha) = G$. From $F_x$: $T_1 = T_2 \frac{\cos(45^\circ+\alpha)}{\cos\alpha}$. Substitute into $F_y$: $- (T_2 \frac{\cos(45^\circ+\alpha)}{\cos\alpha}) \sin\alpha + T_2 \sin(45^\circ+\alpha) = G$ $T_2 \frac{\sin(45^\circ+\alpha)\cos\alpha - \cos(45^\circ+\alpha)\sin\alpha}{\cos\alpha} = G$ $T_2 \frac{\sin((45^\circ+\alpha) - \alpha)}{\cos\alpha} = G \implies T_2 \frac{\sin(45^\circ)}{\cos\alpha} = G$ $T_2 = \frac{G \cos\alpha}{\sin(45^\circ)} = G\sqrt{2} \cos\alpha$. At the start ($\alpha=0^\circ$): $T_2 = G\sqrt{2} \cos(0^\circ) = G\sqrt{2}$. At the end ($\alpha=90^\circ$): $T_2 = G\sqrt{2} \cos(90^\circ) = 0$.

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