B) a space station, consisting of a long thin uniform rod of mass 3.3 x 10^6 kg
ID: 2077379 • Letter: B
Question
B) a space station, consisting of a long thin uniform rod of mass 3.3 x 10^6 kg and length 405 meters, with two identical uniform hollow spheres, each of mass 2.2 x 10^6 kg and radius 189 meters, attached at the ends of the rod, as shown below. Note that none of the diagrams shown is drawn to scale! launch point launch point axis (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 2055. kg and launched at a speed of L m/s with respect to the space H. 2h30o station. The launch points are each located at the same distance J 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 F. e rad/s? Answer: launched pairsExplanation / Answer
Moment of Inertia of the system is
I = Moment of Inertia of Rod + 2*(Mpment of Inertia of Spheres)
= (M*L^2 / 12) + 2*[ { (2/3)Mr^2 } + {M * (r + l/2)^2 } ]
= [ { (3.3*10^6) 405^2 } / 12 ] + 2*[ { (2/3) * (2.2*10^6)* 189^2 } + { (2.2*10^6) *(189+ 202.5)^2 } ]
= 4.346*10^11 kgm^2
Let the pair be launched n times
Therefore By conservation of momentum
n * 2mvr = I*
n * 2*2055*26300*101 = ( 4.346*10^11) * 0.14
n = 5.57 = 6 times
Answer is 6 times launched pairs
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