wo astronauts, each having a mass of 78.0 kg, are connected by a d = 11.0-m rope

Question

wo astronauts, each having a mass of 78.0 kg, are connected by a d = 11.0-m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 4.90 m/s.

(a) Treating the astronauts as particles, calculate the magnitude of the angular momentum of the two-astronaut system.
Kg · m2/s

(b) Calculate the rotational energy of the system.
kJ

(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?
kg · m2/s

(d) What are the astronauts' new speeds?
m/s

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

(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?
kJ

Explanation / Answer

a) angular momentum= IW where I is moment of inertia and W is angular velocity here I for one body is mr2 where r is distance from centre and m is mass of body that is astronaut

W is equal to v/r where v is velocity and r is distance from centre

so angular momentum for one astronaut= IW=mr2 *(v/r) =m*v*r

here both astronaut have same mass and distance from centre is also same so

for system angular momentum =2*m*v*r=2*78*4.9*(11/2) because d is distance between two but we need r that is distance from centre that is 11/2=5.5m

so for system angular momentum =2*m*v*r=2*78*4.9*(11/2) =4204.2 Kg · m2/s

b) rotational energy of system=1/2 *I*W2 + 1/2 *I*W2 where I is moment of inertia and W is angular velocity

putting I=m*r2 and W=v/r

we get rotational energy of system =1/2 *m*r2 *(v/r)2 +1/2 *m*r2 *(v/r)2 =1/2 *m*v2 +1/2 *m*v2 =m*v2

so after solving we get rotational energy of system=m*v2 =78*4.92 =1872.78J

to convert it into KJ divide by 1000 so  rotational energy of system in KJ=1.872KJ

c) angular momentum of system will remain constant as calculated in part a) due to conservation of angular momentum so now also angular momentum is same as in part a) that is 4204.2 Kg · m2/s

d) to calculate new speeds use formula for angular momentum of system=2*m*v*r here in this case distance between them is 5 so r will be 5/2 =2.5m

2*78*v*2.5=4204.2

from above v=10.78m/s

e) from part b rotational energy of system m*v2 here v=10.78m/s

than rotational energy of system=78*10.782 =9064.2552J=9.064KJ

f) difference in energy of system in part b) and part e)=9.064-1.872=7.192KJ of total chemical energy of two astronaut converted into mechanical energy of system so for each astronaunt 7.192/2 =3.596KJ of chemical energy is converted into mechanical energy of system

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