To solve the problem, it is necessary to apply the related concepts to Newton's second law as well as the Normal and Centripetal Force experienced by passengers.
By Newton's second law we understand that
[tex]F = mg[/tex]
Where,
m= mass
g = Gravitational Acceleration
Also we have that Frictional Force is given by
[tex]F_r = \mu N[/tex]
In this particular case the Normal Force N is equivalent to the centripetal Force then,
[tex]N = \frac{mv^2}{r}[/tex]
Applying this to the information given, and understanding that the Weight Force is statically equivalent to the Friction Force we have to
[tex]F = F_r[/tex]
[tex]mg = \mu N[/tex]
[tex]mg = \mu \frac{mv^2}{r}[/tex]
Re-arrange to find v,
[tex]v= \sqrt{\frac{gr}{\mu}}[/tex]
[tex]v = \sqrt{\frac{(9.8)(2)}{0.25}}[/tex]
[tex]v = 8.9m/s[/tex]
From the last expression we can realize that it does not depend on the mass of the passengers.
