At a winter fair a 75.3-kg stunt man is shot from a horizontal cannon that rests

Question

At a winter fair a 75.3-kg stunt man is shot from a horizontal cannon that rests at the edge of a frozen lake. The human projectile 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 circular motion about the pivot and continues to revolve until he comes to rest. A tensometer on the rope indicates a tension of 877 N at the beginning of the circular motion. With a coefficient of kinetic friction of 0.0419, how many revolutions does the stunt man make? Take g = 9.81 m/s2. Assume the rope supplies all the centripetal force.

Explanation / Answer

Here ,

coefficient of friction , u = 0.0419

acceleration , a = - u * g

a = -9.8 * 0.0419

a = -0.411 m/s^2

as T = mv^2/R

877 = 75.3 * w^2 * R

w^2 * R = 11.65

angle = (w^2/(a/R))

angle = w^2 * R/a

angle = 11.65/0.411

angle = 28.33 rad

number of revolutions = 28.33/2pi

number of revolutions = 4.51

the number of revolutions is 4.51

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