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2007美国US F=MA物理竞赛 (id: b6a6ee194)

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admin 发表于 2025-12-20 23:13:49 | 显示全部楼层 |阅读模式
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 point object of magfjgy4/ ,,h9o30go fe*auc1a h uwgv3m3(glv zwis87hy*g;5j6 2oom j5olrzss $m$ is connected to a cylinder of radius $R$ via a massless rope. At time $t=0$ the object is moving with an initial velocity $v_{0}$ perpendicular to the rope, the rope has a length $L_{0},$ and the rope has a non-zero tension. All motion occurs on a horizontal frictionless surface. The cylinder remains stationary on the surface and does not rotate. The object moves in such a way that the rope slowly winds up around the cylinder. The rope will break when the tension exceeds $T_{max}$ Express your answers in terms of $T_{max}$ m, $L_{0},$ R, and $v_{0}$. What is the length (not yet wound) of the rope?


A. $L_{0}-\pi R$
B. $L_{0}-2\pi R$
C. $L_{0}-\sqrt{18}\pi R$
D. $\frac{{mv_0}^2}{T_{max}}$
E. none of the above


参考答案:  D


本题详细解析:
This question asks for the unwz 19x/ 0qwnrujound length of the rope, $r$, at the instant it breaks. As determined in the analysis for problem 34, the tension $T_{max}$ provides the centripetal force for the object (moving at a constant speed $v_0$) at a radius $r$: $T_{max} = \frac{mv_0^2}{r}$. Solving this equation for $r$ gives: $r = \frac{mv_0^2}{T_{max}}$.

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