# 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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