Topic: Dynamics
Concepts: Angular momentum, Circular motion, Work-energy theorem

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Solution:

Note that the velocity is always perpendicular to the tension force by conservation of string, since having a parallel component would mean the total string length is changing (not just the unwound portion). Because perpendicular forces do no work, the object has a constant kinetic energy and hence constant speed $v=v_0$.

The angular momentum $L$ of the object when the rope breaks is given by

where $r$ is the length of the unwound rope at this instant. To relate $r$ to $T_{\text{max}}$, note that the object approximately moves in a circle of radius $r$ when the rope breaks so

Thus,

so the answer is B.

Topic: Energy
Concepts: Kinetic energy, Work-energy theorem

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Solution:

Since we found in the previous problem that the speed of the object is constant $v=v_0$, the kinetic energy when the rope breaks is

so the answer is A.

Topic: Dynamics
Concepts: Circular motion

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Solution:

We found (in problem 34) that the length of the unwound rope when its breaks is

so the answer is D.