Ferris Wheel part 3

A Ferris wheel with radius \(r\) is rotating at constant angular speed \(\omega\). The position of a passenger at the rim of the wheel may be described by a direction angle \(\theta\):

  • \(\theta = 0^\circ\) at the right side of the wheel

  • \(\theta = -90^\circ\) at the bottom of the wheel

  • \(\theta = +90^\circ\) at the top of the wheel

Each passengers sits on a level (horizontal) seat. The coefficient of static friction between passenger and seat is \(\mu\). However, the seats are so slippery that the friction is not sufficient to keep the passengers in their seats. Between \(\theta = -5^\circ\) and \(\theta = +25^\circ\) they must hold on to a bar to stay in their seats.

If \(r = \SI{20}{\meter}\), how fast is the wheel turning, in rotations per minute?

(For the acceleration due to gravity, use \(g = \SI{9.81}{\meter/\second^2}\).)

Bonus for those who think this still too easy: Calculate the maximum minimum force the passengers need to exert on the bar. ("Minimum" at any given moment, "maximum" over the entire ride.)


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