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 77.0 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 4.80 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

length of rod , d = 2.5 m

a)

for the centre of mass is x m from 77 kg

x = 125 * 2.5/(125 + 75)

x = 1.5625 m

as the angular momentum = m * v * r

Total angular momentum = 125 * 1.5625 * 4.8 + 75 * (2.5 - 1.5625) * 4.8

Total angular momentum = 1275 Kg.m^2/s

the magnitude of the angular momentum of the system comprised of the two friends is 1275 Kg.m^2/s

b)

for half as long length

for the centre of mass is x m from 77 kg

x = 125 * 1.25/(125 + 75)

x = 0.781 m

let the final velocity is v

Total angular momentum = 125 * 0.781 * v + 75 * (1.25 - 0.781) * v

125 * 0.781 * v + 75 * (1.25 - 0.781) * v = 1275

solving for v

v = 9.601 m/s

the speed of the persons is 9.6 m/s

c)

Using work energy theorum

Work done by friends = 0.5 * (125 + 75) * 9.6^2 - 0.5 * (125 + 75) * 4.8^2

Work done by friends = 6912 J

the Work done by friends in pulling is 6912 J

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