One way to provide artificial gravity (i.e., a feeling of weight) on long space
ID: 1608279 • Letter: O
Question
One way to provide artificial gravity (i.e., a feeling of weight) on long space voyages is to separate a spacecraft into two parts at the ends of a long cable, and set them rotating about their center of mass. A craft has been separated into two parts with a mass of 88800 kg each, at the ends of a cable with their centers of mass 35 m apart, rotating around the center point of the cable with a period of 183.5 seconds.
If the cable is reeled in so that the the centers of the two pieces are now only 26.6 m apart, what will be the new period?
1pts
What happens to the angular momentum L and kinetic energy K of the system consisting of both pieces of the space ship when the two pieces are pulled closer to the center?
L remains the same, K increases
L increases, K remains the same
L increases, K increases
L remains the same, K decreases
L and K remain the same
1pts
Explanation / Answer
Its given that,
m1 = m2 = 88800 Kgs
distance between the center of masses = d = 35 m
period = T1 = 183.5 sec
When distance bet com is = d' = 26.6 m ; Period = T = to be calculated.
Despite of the changes stated above in the question, the angular momentum of the syste should be conserved.
We know that, Angular momentum L is given by
L = I w (where I is moment of inertia and omega is angular vel)
I = m r2 and w = 2 pi f = 2 pi / T
L(intial) = m (d/2)^2 2 pi / T1 = m (35/2)^2 2 pi / 183.5
Similarly,
L(final) = m (d'/2)^2 2 pi / T2 = m (26.6/2)^2 2 pi / T2
L(intial) = L(final)
m (35/2)^2 2 pi / 183.5 = m (26.6/2)^2 2 pi / T2 ( m, 2 pi , 1/4 cancel out)
(35)^2/185.5 = (26.6)^2/ T2 gives us
T2 = 26.6 x 26.6 x 185.5/35^2 = 107.14 s
Hence, T2 = 107.14 s
# L remains conserved and since T2 decrreases, w will increase and hence KE will also increase.
L remains the same, K increases
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