Physics, 12.08.2019 22:10 gabrieljerron

Two astronauts, each having a mass of 75.0 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 5.00 m/s. treating the astronauts as particles, calculate (a) the magnitude of the angular momentum of the system and (b) the rotational energy of the system. by pulling on the rope, one of the astronauts shortens the distance between them to 5.00 m. (c) 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?

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Answer from: jamielytle

a. The magnitude of the angular momentum of the system is $3750 \;Kgm^2/s$

b. The rotational energy of the system is 1875 Joules.

c. The new angular momentum of the system is $3750 \;Kgm^2/s$

d. To calculate the astronauts’ new speeds is $10 \;m/s^2$

e. The new rotational energy of the system is 7500 joules.

f. The amount of work the astronaut did in shortening the rope is 5625 Joules.

Given the following data:

Mass = 75.0 kgDistance = 10.0-mVelocity = 5.00 m/s.

a. To calculate the magnitude of the angular momentum of the system:

First of all, we would determine the distance of each astronaut from the center of mass of the rope.

$D_c = \frac{1}{2} \times distance\\\\D_c = \frac{1}{2} \times 10\\\\D_c = 5\;meter$

Mathematically, angular momentum of the system is given by:

$W_m = 2mvD_c\\\\W_m = 2 \times 75 \times 5 \times 5\\\\W_m = 3750 \;Kgm^2/s$

b. To calculate the rotational energy of the system:

$R_E = 2 \times \frac{1}{2} MV^2\\\\R_E = 2 \times \frac{1}{2} \times 75 \times 5^2\\\\R_E = 75 \times 25\\\\R_E = 1875 \;Joules$

c. To calculate the new angular momentum of the system:

Based on the law of conservation of momentum, the initial angular momentum is equal to the final or new angular momentum.

New angular momentum = $3750 \;Kgm^2/s$

d. To calculate the astronauts’ new speeds:

Note: By pulling on the rope, one of the astronauts shortens the distance between them to 5.00 m.

$D_n = \frac{1}{2} \times distance\\\\D_n = \frac{1}{2} \times 5\\\\D_n = 2.5\;meter$

From the new angular momentum:

W_n = 2mV_2D_n

$3750 = 2 \times 75 \times V_2 \times 2.5\\\\3750 = 375V_2\\\\V_2 = \frac{3750}{375} \\\\V_2 = 10 \; m/s^2$

e. To calculate the new rotational energy of the system:

$R_N = 2 \times \frac{1}{2} MV_2^2\\\\R_N = 2 \times \frac{1}{2} \times 75 \times 10^2\\\\R_N = 75 \times 100\\\\R_N = 7500 \;Joules$

f. To find how much work the astronaut did in shortening the rope:

$Work\;done = R_N - R_E\\\\Work\;done = 7500 - 1875$

Work done = 5625 Joules

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