What is the weight of the space tourist?

I get 8.31 m, too. x = r * [(√(2rh-h^2))/(r-h) - acos((r-h)/r)]

For derivation, see: http://answers.yahoo.com/question/index;_ylt=Ask8g...

Here is my drawing: http://i278.photobucket.com/albums/kk114/Remo_Avir...

As for time:

t= d/v

d = √(2rh-h^2) ... This is the distance traveled in the non-rotating frame of reference.

v = (r-h)ω

t= √((2rh-h^2)/(r-h)ω

ω = 2π rads/min

t = √(2rh-h^2)/(r-h) *1/(2π rads/min)

===> t = 10.676 sec

*Track has the correct equation:

t= √(2rh-h^2)/(r-h) *1/(2π rads/min)

t = √(500)/(20) *1/(2π rads/min) .... Track is correct up to here

t = √(500)/(40π) min

t = √(500)* 3/2π sec = 10.676 sec

He just lost something in the final step.

The system as a whole is in freefall about the earth, so you can neglect all that so long as the station is sufficiently small that you don't have significant tidal effects.

So the weight is simply centrifugal force:

F = m omega^2 r = m (2 pi / T)^2 r

You have the mass, radius, and period. Plugnchug. Then do it again for the smaller radius.

For the next part, you can neglect the coriolis force on the falling man, since it doesn't really affect his drop time. So just assume he falls straight down with acceleration:

d^2 r / dt^2 = omega^2 r

The solution to the differential equation is:

r(t) = A exp (omega t) + B exp (- omega t)

Use the initial position and velocity to find A and B

r(0) = A + B = radius of bunk

r'(0) = omegaA - omega B = 0

So clearly A = B = 1/2 radius of bunk

r(t) = radius of bunk * cosh (omega t)

So the time to fall is:

t = arc cosh (radius of wall / radius of bunk) / omega

BONUS FUN QUESTION for those who don't want to neglect coriolis forces:

Where does he land?

EDIT--roz, much better now. ;)

Track, yeah, you're right. It's a second order effect, but the h/R ratio is such that it's going to show up. The exponential solutions are only correct to first order (ie, h<<R). When I re-did the problem to figure out generally where and when he lands, I ended up doing it your way. Are your numbers off by a factor of four in the last step? If you fix that, the answer above me (pretty much a duplicate of mine) is only off by about 15%. If the bed weren't quite so high, you'd see very good agreement.

----The tourist's weight is straightforward in this case (simply stepping on the scale) because it's just the centrifugal force. Coriolis only contributes when he's moving. His weight when he IS moving is slightly interesting, which is why I wrote the problem that Remo linked (but I can't).

** His first night he falls out of his bunk bed

** that is 10 m above the "floor."

** How long does it take for him to hit the floor?

Bekki:

**** You can neglect the coriolis

**** force on the falling man, since it doesn't really

**** affect his drop time.

Wrong. Coriolis force DOES afffect drop time, very significantly. Correct equations in rotating frame of reference is

d²R/dt² = Ω² R + 2Ω x R dφ/dt

1/m d[Momentum]/dt = d[R dφ/dt]/dt = -2ΩR dR/dt

Initially, of course, dφ/dt = 0, but immedetly after that the man gains radial speed dR/dt, and Coriolis force -2Ω x dR/dt begins to deflect the fall from vertical, and the man gains velocity in tangential direction, and this velocity R dφ/dt in turn creates coriolis force in RADIAL direction, slowing down the fall. According to your equation the man falls exponetially. Can you belive that yourself?

The correct and easy solution is simply use non-rotating frame of reference:

the tourist detaches from the bed with velocity

Ω Ro = 2π/T * 20m

and must travel the distance to outer cylinder

√30²-20² = √500 m

and it takes time

√500 / (2π/T * 20) = √5/π T = √5/π minutes

R=30 m

T=60 s

What is the weight of the space tourist?

F=ma=mω²R, where ω=π/30 rad/s (if we neglect the height of the scales).

F=100*30*(π/30)²=32.90 N

The reading on the scales is F/g=32.90/9.81=3.35 kg

How long does it take for him to hit the floor?

t=[arccosh(1.5)]/ω=

Edit:

a=ω²r, where r=0 is at the axis of the cylinder

r"=ω²r, r"-ω²r=0

k²-ω²=0 ==> k=±ω ==>

r=Ae^(ωt)+Be^(-ωt)

r(0)=20 ==> A+B=20 ..... (1)

r'(0)=0 ==> ω(A-B)=0 .....(2)

From (1) and (2) ==>

A=B=10

r=10[e^(ωt)+e^(-ωt)]=20*cosh(ωt) ==>

t=[arccosh(r/20)]/ω

r=30 m ==> t=9.19 s

l believe that when in outer space...the tourist becomes "weightless".

Like Astronauts and that!!!

Almost like floating in a swimming pool...the water supports your body and you are free to move.