A proposed space station includes living quarters in a circular ring 50.0 m in diameter. At what angular speed should the ring rotate so the occupants feel that they have the same weight as they do on Earth?

Question

mos1635 · Answer

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N=W earth
Fcenter=m*g
m*V^2/R=m*g
V=sqr(gR)
ω*R=(gR)^1/2
ω=.....

anonymous · Answer

To have earth like gravity the frequency of the space ship is increased to such an extent that it equals to the value of 'g'

$$f=1/2\pi \sqrt{g/R}$$
and since :$$w=2\pi f$$

so do the math :)

anonymous · Answer

so i would have 
mv^2/r = mg
both masses would cancel.
so v = sqrt(gr)
correct?

anonymous · Answer

giving me 15.7 rad/s

mos1635 · Answer

dos not much mine
let me chek my calc....

anonymous · Answer

I'm doing an online assignment and it says i'm incorrect but i don't know why

anonymous · Answer

from my equations, the answer comes out to be 15.7 rad/sec too, should be correct.

mos1635 · Answer

ω=sqr (9.8/50)
ω=sqr 0.196
ω=0.442718 rad/sec
am i doing something wrong???

mos1635 · Answer

oooo pellet
R=25!!!!!!

anonymous · Answer

50 is diameter, not radius

mos1635 · Answer

0.62609

anonymous · Answer

you are correct! how did you derive that equation again?

mos1635 · Answer

N=W earth 
Fcenter=m*g 
m*V^2/R=m*g 
V=sqr(gR)
 ω*R=(gR)^1/2 
$$\omega =\sqrt{g/R}$$

anonymous · Answer

thanks a lot!!