Answer:2.55 rad/s
Explanation:
Given
Diameter of ride=5 m
radius(r)=2.5 m
Static friction coefficient range=0.60-1
Here Frictional force will balance weight
And limiting frictional force is provided by Centripetal force
[tex]f=\mu N=\mu m\omega ^2\cdot r[/tex]
weight of object=mg
Equating two
f=mg
[tex]\mu m\omega ^2\cdot r=mg[/tex]
[tex]\omega ^2=\frac{g}{\mu r}[/tex]
[tex]\omega =\sqrt{\frac{g}{\mu r}}[/tex]
[tex]\omega =2.55 rad/sec[/tex]