Two astronauts, each having a mass of 75.0 kg, are connected by a 9.5 m rope of
ID: 1403632 • Letter: T
Question
Two astronauts, each having a mass of 75.0 kg, are connected by a 9.5 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 4.80 m/s. (a) Treating the astronauts as particles, calculate the magnitude of the angular momentum. (b) Calculate the ro4atlonal 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?
Explanation / Answer
Given ,
Mass m = 75 Kg
Length of rope l = 9.5m
Speed v = 4.8m/s
a) Magnitude of Angular Momentum = mvl/2
Magnitude of Angular Momentum = (75* 4.8 * 5 )/2
Magnitude of Angular Momentum = 1710 Kgm^2/s
b) Rotational Energy of system = 2 * 1/2 mv^2
Rotational Energy of system = 2 * 1/2 75 * 4.8^2
Rotational Energy of system = 1728 J
c) According to Momentum Conservation Theorem =
Final Angular Momentum = Initial Angular Momentum
Therefore, Magnitude of Angular Momentum = 1710 Kgm^2/s
d) Let Astronaut's New Speed = v'
Using Momentum Conservation -
Final Angular Momentum = mv'l'/2
where l' = 5m
Final Angular Momentum = Initial Angular Momentum
mv'l'/2 = mvl/2
v' * 5 = 4.8 * 9
v' = (4.8 * 9) /5
v' = 8.64 m/s
Astronaut's New Speed v' = 8.64 m/s
e)
New Rotational Energy of system = 2 * 1/2 mv'^2
New Rotational Energy of system = 75 * 8.64^2
New Rotational Energy of system = 5598.72 J