Two astronauts, each having a mass M , are connected by a rope of length d havin

Question

Two astronauts, each having a mass M, are connected by a rope of length d having negligible mass. They are isolated in space, orbiting their center of mass at speeds v. (Use any variable or symbol stated above as necessary.)

By pulling on the rope, one of the astronauts shortens the distance between them to d/10. (c) What is the new angular momentum of the system?
Lf =

(d) What are the astronauts' new speeds?
vf =

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

(f) How much chemical potential energy in the body of the astronaut was converted to mechanical energy in the system when he shortened the rope?
W =

Explanation / Answer

a) angular momentum = moment of inertia(I)*angular velocity(w). angular velocity of each= v/(d/2) = 2v/d moment of inertia = M*d^2 / 4 + M*d^2 / 4 = M*d^2 / 2 [calculated with respect to centre of mass] so net angular momentum = Li = Mvd b) rotational energy = (1/2)I*w^2 [for each person] K = (1/2)*(1/4)M*d^2*(4v^2)/d^2 + (1/2)*(1/4)M*d^2*(4v^2)/d^2 =Mv^2 c)as there are no external torques present, net angular momentum = constant. so Lf = Li = Mvd d) to find the new angular velocity, let it be wf then total moment of inertia = M*(d/20)^2 + M*(d/20)^2 = Md^2 / 200 so Md^2 / 200 * wf = Mvd ==> wf = 200v/d if their speed is vf, wf = vf/(d/20) = 20vf/d ==> 20vf/d = 200v/d ==> vf = 10v e) new rotational energy = (1/2)*Md^2/ 400 * (200v/d)^2 + (1/2)*Md^2/ 400 * (200v/d)^2 = 100Mv^2 f)W = change in rotational kinetic energy ==> W = 100Mv^2 - Mv^2 = 99Mv^2

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