Answer:
1742.24106 revolutions per day
Explanation:
v = Velocity
d = Diameter = 1.1 km
r = Radius = [tex]\dfrac{d}{2}=\dfrac{1.1}{2}=0.55\ km[/tex]
g = Acceleration due to gravity = 9.81 m/s²
g = 0.9 g
The centrifugal force will balance the gravitational force
[tex]F_c=mg\\\Rightarrow \dfrac{mv^2}{r}=m0.9g\\\Rightarrow v=\sqrt{\dfrac{0.9gmr}{m}}\\\Rightarrow v=\sqrt{0.9gr}\\\Rightarrow v=\sqrt{0.9\times 9.81\times 0.55\times 10^3}\\\Rightarrow v=69.68464\ m/s[/tex]
[tex]\dfrac{1}{T}=\dfrac{v}{2\pi r}\\\Rightarrow \dfrac{1}{T}=\dfrac{69.68464}{2\pi 0.55\times 10^3}\times 24\times 60\times 60\\\Rightarrow \dfrac{1}{T}=1742.24106\ rev/day[/tex]
The rotation speed is 1742.24106 revolutions per day
Answer:
1728 rev/sec
Explanation:
The expression for the gravitational force is given by
F_g=mg_1
g_1=0.9g (Given)
therefore,
[tex]F_g=0.9mg[/tex]
The centripetal force balances this gravitational force to keep the space station in equilibrium.
Hence we can write
[tex]\frac{mv^2}{r} =0.9mg[/tex]
Rearrange the above equation in terms of velocity
[tex]v=\sqrt{\frac{0.9mgr}{m} }[/tex]
⇒[tex]v=\sqrt{0.9gr}[/tex]
putting the values we get
[tex]v=\sqrt{\frac{0.9(9.81)(1100)}{2} }[/tex]
v=69.65 m/sec
the rotational speed can be calculated as or frequency of rotation
[tex]f= \frac{v}{2\pi r}[/tex]
putting values we get
[tex]f= \frac{69.65}{2\pi 5500}[/tex]
f= 0.02 rev/sec
meaning 0.02 rev per second
therefore no. or revolution per day
= 0.02×24×3600= 1728 rev/sec
