Ferris Wheel part 3

Classical Mechanics Level 5

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

  • θ = 0 ∘ \theta = 0^\circ at the right side of the wheel

  • θ = − 9 0 ∘ \theta = -90^\circ at the bottom of the wheel

  • θ = + 9 0 ∘ \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 μ \mu . However, the seats are so slippery that the friction is not sufficient to keep the passengers in their seats. Between θ = − 5 ∘ \theta = -5^\circ and θ = + 2 5 ∘ \theta = +25^\circ they must hold on to a bar to stay in their seats.

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

(For the acceleration due to gravity, use g = 9.81 m / s 2 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.)


The answer is 2.83566.

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

Arjen Vreugdenhil
Jul 26, 2017

At the given angles θ − = − 5 ∘ \theta_{-} = -5^\circ and θ + = + 2 5 ∘ \theta_{+} = +25^\circ , the static friction is maximal: f = μ N f = \mu N .

The net force on a passenger is the centripetal force m ω 2 r m\omega^2 r . This consists of the downward force of gravity m g mg , the upward normal force N N , and a sideways friction force f f . For the y y - and x x -components we have m ω 2 r sin ⁡ θ = m g − N ; m ω 2 r cos ⁡ θ = f = μ N . m\omega^2 r\sin\theta = mg - N;\ \ \ \ m\omega^2 r\cos\theta = f = \mu N. Eliminate N N to find m ω 2 r ( cos ⁡ θ + μ sin ⁡ θ ) = μ m g . m\omega^2 r(\cos\theta + \mu\sin\theta) = \mu mg. This we write as ω 2 r 1 + μ 2 cos ⁡ ( θ − ζ ) = μ g , tan ⁡ ζ = μ . \omega^2 r\sqrt{1 + \mu^2}\cos(\theta - \zeta) = \mu g,\ \ \ \ \ \tan\zeta = \mu. Because cosine is an even function, the solutions θ \theta lie symmetrically around ζ \zeta . Thus ζ = θ + + θ − 2 , Δ = θ − ζ = θ + − θ − 2 . \zeta = \frac{\theta_{+} + \theta_{-}}{2},\ \ \ \ \Delta = \theta - \zeta = \frac{\theta_{+} - \theta_{-}}2. From this we can calculate μ = tan ⁡ − 1 ζ \mu = \tan^{-1}\zeta . Note also that μ / 1 + m u 2 = sin ⁡ ζ \mu/\sqrt{1+mu^2} = \sin\zeta . Thus we obtain the equation ω 2 r = μ g 1 + μ 2 cos ⁡ Δ = g sin ⁡ ζ cos ⁡ Δ , \omega^2 r = \frac{\mu g}{\sqrt{1+\mu^2}\cos \Delta} = g \frac{\sin \zeta}{\cos \Delta}, ω = g sin ⁡ 1 2 ( θ + + θ − ) r cos ⁡ 1 2 ( θ + − θ − ) . \omega = \sqrt{\frac{g\sin \tfrac12 (\theta_{+}+\theta_{-})}{r\cos \tfrac12 (\theta_{+}-\theta_{-})}}.

With the given values ω = 9.81 sin ⁡ 1 0 ∘ 20 cos ⁡ 1 5 ∘ = 0.296 95 rad / s , \omega = \sqrt{\frac{9.81\sin 10^\circ}{20\cos 15^\circ}} = \SI{0.29695}{\radian/\second}, and multiplication by 60 / π 60/\pi gives ω = 2.83566 rpm . \omega = \boxed{2.83566}\ \text{rpm}.

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Solution of Bonus question:

Let S S be the force exerted on the bar with direction angle ϕ \phi . The equations for the y y - and x x -components of force become m ω 2 r sin ⁡ θ = m g − N + S cos ⁡ ϕ ; m ω 2 r cos ⁡ θ = μ N + S sin ⁡ ϕ , m\omega^2 r \sin\theta = mg - N + S\cos\phi;\ \ \ \ \ m\omega^2 r\cos\theta = \mu N + S\sin\phi, assuming of course that static friction is of maximal help in keeping the passenger in place. Eliminate N N : m ω 2 r ( cos ⁡ θ + μ sin ⁡ θ ) − S ( cos ⁡ ϕ + μ sin ⁡ ϕ ) = μ m g , m\omega^2 r(\cos\theta + \mu\sin\theta) - S(\cos\phi + \mu\sin\phi) = \mu mg, ω 2 r cos ⁡ ( θ − ζ ) − S m cos ⁡ ( ϕ − ζ ) = g sin ⁡ ζ , \omega^2 r\cos(\theta - \zeta) - \frac S m \cos(\phi - \zeta) = g \sin\zeta, with tan ⁡ ζ = μ \tan \zeta = \mu .

To minimize S S we must maximize cos ⁡ ( ϕ − ζ ) \cos(\phi - \zeta) , i.e. the bar must be pushed in direction ϕ = ζ \phi = \zeta ; the equation becomes S m = ω 2 r cos ⁡ ( θ − ζ ) − g sin ⁡ ζ . \frac S m = \omega^2 r\cos(\theta - \zeta) - g\sin\zeta. This is maximal in the situatin θ = ζ \theta = \zeta , with value S m ) max = ω 2 r − g sin ⁡ ζ = g sin ⁡ ζ ( 1 cos ⁡ Δ − 1 ) . \left.\frac S m\right)_\text{max} = \omega^2 r - g\sin\zeta = g\sin\zeta\left(\frac 1{\cos\Delta}-1\right). With the given values, S m ) max = 0.060 N / kg . \left.\frac S m\right)_\text{max} = \SI{0.060}{\newton/\kilo\gram}.

Arjen Vreugdenhil - 3 years, 10 months ago

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