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, orbiting their center of mass at speeds of 5.10 m/s. (a) Treating the astronauts as particles, calculate the magnitude of the angular momentum. (b) Calculate the rotational energy of the system. (c) By pulling on the rope, one of the astronauts shortens the distance between them to 5.00 m. What Is the new angular momentum of the system? (d) What are the astronauts' new speeds? (e) What is the new rotational energy of the system? (f) How much work does the astronaut do in shortening the rope?

Explanation / Answer

(a) L = 2mvr = (2*78.0kg) * 5.10m/s * 5m = 3978 kg·m²/s

(b) KE = ½mv² = ½ * (2*78.0kg) * (5.10m/s)² = 2028.8 J

(c) No EXTERNAL torques are applied, so new L = old L = 3978 kg·m²/s

(d) old m*v*r = new m*v*r
3978 kg·m²/s = 2*78.0kg * v * 2.50m
v = 10.2 m/s

(e) KE = ½ * 2*78.0m/s * (10.2m/s)² = 8115.1 J

(f)The difference between (b) and (e) is the work done by the astronaut pulling on the rope.

W = 8115.1-2028.8= 6086.3 J

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