Two friends hold on to a rope, one at each end, on a smooth, frictionless ice su

Question

Two friends hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about an axis through the center of the rope and perpendicular to the ice. The mass of one friend is 82.0 kg and the other has a mass of 125.0 kg. The rope of negligible mass is 3.0 m long and they move at a speed of 4.80 m/s. (a) What is the magnitude of the angular momentum of the system comprised of the two friends? (b) They now pull on the rope and move closer to each other so that the rope between them is now half as long. Determine the speed with which they move now. (c) The two friends have to do work in order to move closer to each other. How much work did they do?

Explanation / Answer

a) For system of particles the angular momentum is

P =M*Vcm

where M is the totoal mass of the system and Vcm is the velocity of Center of mass

in this case the cm is fixed so P=0

b)

previously v=r*w

w =v/r = 4.8/1.5 = 3.2 rad/sec

now v= rw

here r is 1.5/2 = 0.75m

v =0.75*3.2 = 2.4 m/sec

c) work done = change in kinetic energy

intial KE = 0.5 *(82*4.8^2 +125*4.8^2) =2384.64 J

final KE = 0.5*(83*2.4^2 +125*1.4^2) = 361.54

change in KE = 2023.1 J

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