Two astronauts, each having a mass of 79.0 kg,are connected by a 10.0 m rope of
ID: 1725741 • Letter: T
Question
Two astronauts, each having a mass of 79.0 kg,are connected by a 10.0 m rope of negligible mass. They areisolated in space, moving in circles around the point halfwaybetween them at a speed of 5.90 m/s. Treating the astronauts asparticles, 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 betweenthem 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 therope?
Thanks for any help Two astronauts, each having a mass of 79.0 kg,are connected by a 10.0 m rope of negligible mass. They areisolated in space, moving in circles around the point halfwaybetween them at a speed of 5.90 m/s. Treating the astronauts asparticles, 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 betweenthem 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 therope?
Thanks for any help
Explanation / Answer
Given that the mass of astronaut ism = 79.0 kg separation between the eachastronaut is d = 10 m speed of each astronautis v = 5.90 m/s ---------------------------------------------------------------- The separation between the astronaut and center of rotationis r = d/2 = 5.0 m (a) the angular momentum of the system is L1 = 2*m*v*r = 4661kg.m2/s2 (b) rotational kinetic energy is K1 =(1/2)I2 =(1/2)[ 2mr2 ][ v / r]2 = mv2 =--------- J If the new separation between the persons is d = 5.00 m thenr = 2.50 m Since there is no external torque on thesystem so the angular momentum is conserved L1 =L2 L2 = L1 =4661 kg.m^2/s^2 (d) But the angular momentum of the system is L2 =2*m*v*r = 4661 kg.m2/s2 v = 4661 / 2*m*r = ---------- m/s ( where r = 2.50 m ) (e) The new rotational energy is K2 =(1/2)I2 = (1/2)[ 2mr2 ][ v / r]2 = mv2 = --------- J ( f ) work done W = change in kinetic energy = K1 - K2 = ------ J