Two astronauts, each having a mass of 78.0 kg, are connected by a 10.0 m rope of

Question

Two astronauts, each having a mass of 78.0 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 5.20 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?

Explanation / Answer

Initial speeds v1i = v2i = 5.20 m/s Separation between them r1i = r2i = d/2 = 10.0 m / 2 = 5.0 m
a) Li = m1v1ir1i + m2v2ir2i = 2(78)(5.2)(5) = 4056 kgm^2/s b) Ki=1/2m1v1i2 + 1/2m2v2i2 = 2(1/2)(78)(5.2)2 = 2109.12 J c) Conservation of angular momentum: Li = Lf So same answer as part a) 4056 kgm^2/s
d) Lf =2mrfvf vf = (4056)/ (2(78)(2.5)) = 10.4 m/s e) Kf = 2(1/2)mvf2 = (78)(10.42) = 8436.48 J f) W = Kf - Ki = 8436.48 - 2109.12 = 6327.36

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