Answer:
[tex]T= 11.0003[/tex]s
Explanation:
From the question we are told that
The outer ring with a radius of 30 m
inner Gravity Approximately 9.80 m/s'
Outer Gravity Approximately 5.35 m/s.
Generally the equation for centripetal force is given mathematically as
Centripetal acceleration enables Rotation therefore?
[tex]\omega ^2 r =Angular\ acc[/tex]
Considering the outer ring,
[tex]\omega ^2 r = 9.8[/tex]
[tex]\omega ^2= \frac{9.8}{30}[/tex]
[tex]\omega = \sqrt{\frac{9.8}{30}}[/tex]
[tex]\omega= 0.571 rad/s[/tex]
Therefore solving for Period T
Generally the equation for solving Period T is mathematically given as
[tex]T= \frac{2\pi}{\omega}[/tex]
[tex]T= \frac{2\pi}{0.571 rad/s}[/tex]
[tex]T= 11.0003[/tex]s
