At a winter fair a 74.9-kg stunt man is shot from a horizontal cannon that rests

Question

At a winter fair a 74.9-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 897 N at the beginning of the circular motion. With a coefficient of kinetic friction of 0.0499, 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,

mass of the stuntman , m = 74.1 kg

tension in the rope ,t = 871 N

coefficient of kinetic friction , uk = 0.0495

let the initial speed be v

m*v^2/r = Tension

74.1 *v^2/r = 871

v = 3.43 *r^1/2

distnace travelled be d

deaccleration , a = - u*g

a = - 0.4851 m/s^2

v^2 - u^2 = 2 * a*s

3.43^2 *r = 2 * 0.4851*d

d = 12.13 *r

the number of revolution , n = d/(2*pi*r)

n = 1.93 revolutions

the number of revolutions are 1.93 revolutions

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