Two identical twins hold on to a rope, one at each end, on a smooth, frictionles

Question

Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the center of the rope (the center of mass of the two-body system) and perpendicular to the ice. The mass of each twin is 84.0 kg. The rope of negligible mass is 3.0 m lang and they move at a speed of 6.00 m/s. center of mass (a) what is the magnitude, in kg·m2/s, of the angular momentum of the system comprised of the two twins? kg m2/s They now pull on the rope and move closer to each ather so that the rope between them is now half as long. Determine the speed, in m/s, with which they move now. (b) m/s (c) The two twins have to do work in order to move closer to each other. How much work, in joules, did they do?

Explanation / Answer

a) r = 3.00 / 2 = 1.5 m

L = 2 m v r [as both have same angular momentum]

L = 2 * 84 * 6 * (3.00/2)

= 1512 kg.m2 /s

b) length of rope reduces to half r = 1.5 / 2 = 0.75 m

Lf = Li

2 * 84 * 0.75 * vf = 1512

vf = 12 m/s

c) work done W = Kf - Ki

W = 2(1/2 m vf2 -1/2 m vi2 ) = m(vf2 - vi2 )

= 84 [ 122 - 62 ]

work done = 9072 J

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