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 80 kg and the other has a mass of 80 kg. The rope of negligible mass is 2.5 m long and they move at a speed of 5.40 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

Here ,

a)

as angular momentum = m*v*r

net angular momentum of friends = 80 * (2.5/2) * 5.4 + 80 * (2.5/2) * 5.4

net angular momentum of friends = 1080 Kg.m/s

b)

as the angular momentum is conserved

net angular momentum of friends = 80 * (1.25/2) * v + 80 * (1.25/2) * v

80 * (1.25/2) * v + 80 * (1.25/2) * v = 1080

v = 10.8 m/s

their speed will be 10.8 m/s

c)

work done by them = change in kientic energy

work done by them = 2 * 0.5 *80 * 10.8^2 - 2 * 0.5 * 80 * 5.4^2

work done by them = 6998 J

work done by them is 6998 J

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