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Quick Question about Rotational Energy

  1. Jun 6, 2007 #1
    Two astronauts (Fig. P11.51), each having a mass of 77.0 kg, are connected by a 11.0 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.40 m/s.

    (a) Treating the astronauts as particles, calculate the magnitude of the angular momentum.
    4573.8 kgm2/s
    (b) Calculate the rotational energy of the system.

    What is the equation for rotational energy of a system....I thought it was Krot = L^2/(2I) which for this problem would have been (4573.8^2)/(2 x 144 x 5.5^2) but this is not correct. I though 2 I would be equal to 2MR as the astronauts are like a hollow disk rotating in the air
  2. jcsd
  3. Jun 6, 2007 #2
    wow..never mind 2x77 is 154 and not 144 lol

    thanks for looking
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