Question

Two astronauts, each having a mass of 70.0 kg, are connected by a 9.0 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.50 m/s.

(a) Treating the astronauts as particles, calculate the magnitude of the angular momentum. (kg·m²/s)

(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? (kJ)

Answer 1

Since there is no external force on the system and the force on each astronaut passes through the centre of the circle, the torque will be zero and angular momentum will be conserved. We use this to find the solution as shown below

a) the middle since both have same, the center of mass of rope, i.e. at r=a= 4.5m from is in each Li= murt mur = zmur vermogu 1 = 2(70kg) (5.5m/s) (4.5m) 1. = 13465 kg.m²ls V= speed me mass 6) KE; = 2 mu² + 1 / 2 mu? - mu? = 62013) (8:5mils) 2 - 2117.515 foom conservation of angular momentum If = 2i = 3465) Agimeys c) d) 2 new radius r = 5m= 2.5m So, if = amviri => 34651cg.m/ = 2(70kg) v' (2.5m) » v' = 19.9 mis New energy REF = 2x2 muiz & mule = (2014) (@.amks) a = 6860.75 W = AKE (from world energy theorem) - KEF - Kei = 6860.75 - 2117.55 = 4743.25