Weight change in rotating space station

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SUMMARY

The forum discussion centers on calculating the apparent weight of an astronaut in a rotating space station with a radius of 25 meters, aiming to achieve an apparent gravity of 0.7g. The period of rotation required for this condition is determined to be approximately 12 seconds. The astronaut's weight increases significantly when running at 5 m/s, with calculations suggesting an apparent weight of 982.83 N, derived from the formula for apparent weight in a rotating frame, incorporating both centrifugal and Coriolis effects.

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indie452
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Homework Statement



station has radius 25m and rotate with w.
we were asked to find the period of rotation required to produce apparent gravity 0.7g i worked this out to be 11.99s = 12s. using a(0.7g) = v^2 /r

we are then asked if astronaut weighs 75Kg and runs at 5m/s in the direction of rotation what is his apparent weight?

The Attempt at a Solution



I know the weight will increase but I am not sure by how much

i tried finding the coriolis acc. = 2wv' v' is velocity in rotating frame
this is relative to the centrifugal force (which in this case is 0.7g) so i divided by this amount and got:

2*2pi/12*5 = 5.24m/s/s
5.24/0.7g = 0.763

i was wondering if this means that the weight increases by 76% so it changes from:
before = 75*0.7g = 515.025N
after = 907.98N

i thought this might be too much but then again he is running at 5m/s
is this right?
 
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ok I've looked through some of y books and found that it says that for a guy standing in a spinning wheel the apparent weight i m*w^2*r and so i thought that i could just change w to be w[o]+w[5] where w[0] is him still and w[5] is him running at 5m/s

so apparent weight = 75*(0.524 + 0.2)^2*25 = 982.83N

2pi/12=0.524

is this right? cause it seems ore logical
 

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