Ferris Wheel Part 2

On each passenger works the gravitational force $mg$ in downward direction.

In order to remain in their seat, the passenger must travel in a circle of radius $r$ at angular speed $\omega$ . This requires a centripetal net force of magnitude $m\omega^2 r$ directed toward the center of the wheel. In particular, $F_{net,x} = -m\omega^2 r\cos\theta;\ \ \ F_{net,y} = -m\omega^2 r \sin\theta.$

The normal force $N$ by the seat ensures that the latter equation is true: $N - mg = F_{net,y}\ \ \ \ \therefore\ \ \ \ N = m(g - \omega^2 r\sin\theta).$

The friction $f$ force must provide the horizontal component of the net force. However, friction is limited to $|f| \leq \mu |N|$ . Thus friction can only keep the passenger in his seat as long as $m\omega^2 r\cos\theta \leq \mu m(g - \omega^2 r \sin\theta).$ Rewrite this as $\omega^2 r(\cos\theta + \mu \sin\theta) \leq \mu g.$ In general, $a\cos\phi + b\sin\phi = \sqrt{a^2 + b^2}\cos(\phi - \zeta),$ where $\tan \zeta = b/a$ ; applying this here, we find $\omega^2 r\sqrt{1 + \mu^2}\cos(\theta - \zeta) \leq \mu g,$ where $\tan \zeta = \mu$ . The left-hand side is maximal for $\theta = \zeta$ , making the cosine equal to one. Thus the condition is true for all direction angles $\theta$ if $\omega^2 r\sqrt{1 + \mu^2} \leq \mu g.$ Solve the equality to find the extreme case. We get $\omega^2 = \frac{\mu g}{r \sqrt{1 + \mu^2}}$ and the answer follows by taking the square root.