Two astronauts (Fig. P11.51), each having a mass of 70.0 kg, are connected by a

Question

Two astronauts (Fig. P11.51), each having a mass of 70.0 kg, are connected by a 11.0 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.00 m/s.

Two astronauts (Fig. P11.51), each having a mass of 70.0 kg, are connected by a 11.0 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.00 m/s. Treating the astronauts as particles, calculate the magnitude of the angular momentum. Calculate the rotational energy of the system. 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? What are the astronauts' new speeds? Your response differs from the correct answer by more than 10%. Double check your calculations, m/s What is the new rotational energy of the system? How much work does the astronaut do in shortening the rope?

Explanation / Answer

Here ANGULAR MOMENTUM will remain Conserved

Therefore

I1W1 = I2W2

Therefore

2*(m*d/2)^2*v/(d/2) = 2*(m*d'/2)^2*v'/(d'/2)

2*(m*d/2)*v = 2*(m*d'/2)*v'

dv = d'v'

11*5 = 5*v'

v' = 11 m/sec

Now Rotational Kinetic Energy = 0.5*I*w^2

= 0.5*(2*70*2.5^2)*(11/2.5)^2

= 8470 J

Therefore

Work Done = Change in Rotational Energy = 8470 - 1750

= 6720 J

= 6.720 kJ

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