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 79.0 kg and the other has a mass of 125.0 kg. The rope of negligible mass is 2.5 m long and they move at a speed of 4.40 m/s.

(a) What is the magnitude of the angular momentum of the system comprised of the two friends? (kg · m2/s)

(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. (m/s)

(c) The two friends have to do work in order to move closer to each other. How much work did they do? (J)

Explanation / Answer

Distance of each friend from centre of the rod r = 1.25 m

Mass of the first friend m = 79 kg

Mass of the second friend M = 125 kg

Now mass of the total system M1 = m + M = 125 + 79 = 204 kg

Velocity of the system v = 4.40 m/s

a) Now angular momentum of the system is given by

L = M1 * v * r

204 * 4.4 * 1.25 = 1122 kgm2/s

b) Angular momentum is conserved even if distance decreases so

M1 * V * R = 1122

where V = ?

R = r/2 = 1.25/2 = 0.625 m

So 204 * V * 0.625 = 1122

V = 1122/ ( 204 * 0.625 )

= 8.8 m/s

c) Work done by the friends is given by the change in kinetic energy

W = 1/2 * M1 * ( V2 - v2 )

= 1/2 * 204 * ( 8.82 - 4.42 )

= 102 * ( 77.44 - 19.36 )

= 5924.16 J

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