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 ma w-ey1x nkgy4b1d e)9r ss 1 p:-d0 r6ntdprmxy,nu. g-ff8gmd* g $m$ 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 angular momentum of the object with respect to the axis of the cylinder at the instant that the rope breaks? 
A. $mv_{0}R$
B. $\frac{m^{2}{v_{0}}^{3}}{T_{max}}$
C. $mv_{0}L_{0}$
D. $\frac{T_{max}R^{2}}{v_{0}}$
E. none of the above
参考答案: B
本题详细解析:
Because the rope "slowly winds up" and the tension force is always directed alonl(s:zo 0fk0jo 9fqv 4ag the rope (perpendicular to the velocity), the tension does no work on the object.
Since the surface is frictionless, net work is zero, and kinetic energy is conserved.
Thus, the speed o a9 4ofkjvsf:lo0 q(0zf the object remains $v_0$ at all times.
At the instant the rope breaks, the tension $T_{max}$ is providing the centripetal force for the object's circular motion.
Let $r$ be the unwound length of the rope at this instant.
$T_{max} = \frac{mv^2}{r} = \frac{mv_0^2}{r}$.
Solving for $r$, we find the length of the rope at breaking: $r = \frac{mv_0^2}{T_{max}}$.
The angular momentum $L$ is given by $L = mvr$.
Substituting $v = v_0$ and $r = \frac{mv_0^2}{T_{max}}$:
$L = m(v_0)\left( \frac{mv_0^2}{T_{max}} \right) = \frac{m^2 v_0^3}{T_{max}}$.
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