At a winter fair a 62.5-kg stunt man is shot from a horizontal cannon that rests

Question

At a winter fair a 62.5-kg stunt man is shot from a horizontal cannon that rests at live edge of a frozen lake. The human projects is cast onto the smooth ice, slides some distance, and grabs the end of a long rope whose other end is attached to a pivot that is firmly anchored to the ice. The rope initially lies perpendicular to the stunt man's line of motion, and when he grabs it, he starts sliding in circuit motion about the pivot and continues to revolve until he comes to rest. A tensometer on the rope indicates a tension of 873 N at the beginning of the circular motion. With a coefficient of kinetic friction of 0.0501, how many revolutions does the stunt man make? Take g = 9.81 m/s2. Assume the rope supplies all the centripetal force.

Explanation / Answer

First: centrip force = mv^2 / r And: kinetic energy = (1/2) m v^2 Also: work done by friction = force of friction * distance = umg*N 2pi r Where "N" is the number of revolutions. The idea is that the KE must equal the work done by the friction force, for him to stop. So... (1/2) m v^2 = 2pi umg N r or mv^2 = 4pi umg Nr Plug this into the first equation: centrip force = 4pi umg Nr / r Or: centrip force = 4pi umg N solve for N 873 = 4pi * 0.0501 * 62.5 * 9.8 * N N = 2.26 is the number of revolutions he will make

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