a space station, consisting of a long thin uniform rod of mass 3.5 x 106 kg and
ID: 1520704 • Letter: A
Question
a space station, consisting of a long thin uniform rod of mass 3.5 x 106 kg and length 372 meters, with two identical uniform hollow spheres, each of mass 2.0 x 106 kg and radius 119 meters, attached at the ends of the rod, as shown below. Note that none of the diagrams shown is drawn to scale!
(b) Suppose once again that the space station begins at rest, not rotating. This time, instead of using rocket engines attached to the spherical end modules, we will have small probes periodically launched from two points on the rod-shaped part of the station as shown. The probes will launch in pairs in opposite directions, each individual probe of identical mass 1950 kg and launched at a speed of 20400 m/s with respect to the space station. The launch points are each located at the same distance 140 meters from the center of the rod, on opposite sides of the rod. Each time a pair of probes is launched, some angular momentum is imparted to the station, increasing its spin rate. Question: how many such pairs of probes must be launched, so that the station’s angular velocity will reach the required value of 0.15 rad/s? Answer: _____________.___ launched pairs
Explanation / Answer
For this problem we will calculate the angular momentum of the space station and gives each pair of probe
L probe
L = r x p
L = r m v
L = 140* 1950* 20400
L = 5.569 10^9 Kg m²/ s
Lpar = 2 L
Lpar = 1.113 1010 Kg m²/s
The angular momentum of the space station
Lt = I w
The moment of inertia of the bar, moment of inertia of the sphere and the theorem parallel axes
I = Ir + 2 (Ie + me (L/2 + Re)2
I = 1/12 mL L2 + 2(2/3 me Re2 + me (L/2 + Re)2 )
I = 1/12 (3.5 x 10^6) 3722 + 2 ( 2 106 1192 +2 106 (372/2 + 119)2
I = 2.452 1011 Kg m²
Lt = 2.452 1011* 0.15
Lt = 3.679 1010 Kg m²/s
Number of pairs launched
Lt = n Lpar
n = Lt / Lpar
n = 3.679 1010/1.113 1010
n = 3.3
result 3 pair of spheres
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