Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 weeks ago

Physics Rotational Equilibrium question?

Two astronauts, each having a mass of 81.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 4.80 m/s. Treating the astronauts as particles, calculate each of the following.

(a) the magnitude of the angular momentum of the system

(b) the rotational energy of the system

By pulling on the rope, the astronauts shorten the distance between them to 5.00 m.

(c) What is the new angular momentum of the system?

(d) What are their new speeds?

(e) What is the new rotational energy of the system?

(f) How much work is done by the astronauts in shortening the rope?

2 Answers

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  • JOHN
    Lv 7
    3 weeks ago
    Favorite Answer

     (a) This is 81.0 x 4.80 x 5.0 + 81.0 x 4.80 x 5.0 = 3880kgm²/s

    (b) This is 2 x 0.5 x 81.0 x 4.80² = 1866J

    (c) The two pulls on the rope are internal forces of the system and leave the angular momentum unchanged.

    (d) Momentum being conserved, their speed must be doubled if the length of the rope is halved; so new speed is 9.60m/s

    (e) What is the new rotational energy of the system?

    Since their speed is doubled the system KE is quadrupled, making it 7464J

    (f) The increase in KE is sourced from the work done by the astronauts in pulling on the rope; this is 7464J - 1866J = 5599J.

  • 3 weeks ago

    a) angular momentum = wi = v/x * m x^2 = mvx= 162*4.80 * 5  kg m^2 /s

    b) energy = 1/2 i w^2 = 1/2 mx^ * (v/x)^2  = 1/2 m v^2 ( surprise ) = 1/2 * 162 * 4.8^2

    c) angular momentum is conserved.

    d) Look at the equation I derived in the first line  mvx  if this value must remain constant then halving x must double v

    e) Ek = 1/2 m v^2 as v has doubled the energy is now four times greater.

    f) must have added three times the original energy by work to get a total 4 times greater.

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