A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.
1 answer:
Answer:
correct option is b. 31.3 m/s
Explanation:
given data
artificial gravity a1 = 1 g
artificial gravity a2 = 2 g
diameter = 100 m
radius r= 50 m
speed v1 = 22.1 m/s
solution
As acceleration is ∝ v²
so we can say
.....................1
put here value
solve it
v2 = × 22.1
v2 = 31.25 m/s
so correct option is b. 31.3 m/s
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1) Equilibrium of forces:
where
is the weight of the person
is the weight of the scaffold
Re-arranging, we can write the equation as
(1)
2) Equilibrium of torques:
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using
and replacing T1 with (1), we find
from which we find
And then, substituting T2 into (1), we find
1) Equilibrium of forces:
where
Re-arranging, we can write the equation as
2) Equilibrium of torques:
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using
from which we find
And then, substituting T2 into (1), we find