An amusement park ride shown in Fig. 6.3 consists of a 20 ft diameter cylindrical room that turns about its axis. People stand again st the rough cylindrical wall. After the room has reached a certain angular speed, the floor drops from under the riders. What must be the angular speed of the room

Question

An amusement park ride shown in Fig. 6.3 consists of a 20 ft diameter cylindrical room that turns about its axis. People stand again st the rough cylindrical wall. After the room has reached a certain angular speed, the floor drops from under the riders. What must be the angular speed of the room to assure that a person will not slide on the wall? The design coefficient of static friction is μ = 0.4 .

Accepted Answer

To assess the safe angular speed design, we seek a no-slip Coulomb condition sufficient to assure that a rider does not slide on the wall of the rotating room. The free body diagram of a rider represented as a center of mass object P is shown in Fig 6.3a. The rider's weight is [latex]\mathbf{W}=-W \mathbf{k}, ext { and } \mathbf{N}=-N \mathbf{e}_r[/latex] and [latex]\mathbf{f}=f_\phi \mathbf{e}_\phi + f_z \mathbf{k}[/latex] are the normal and the tangential frictional forces exerted by the wall. Thus, the total force F on a rider in a cylindrical frame that turns with the room is

[latex]\mathbf{F}(P, t)=\mathbf{N} + \mathbf{f}+\mathbf{W}=-N \mathbf{e}_r + f_\phi \mathbf{e}_\phi + \left(f_z-W ight) \mathbf{k}[/latex] (6.19a)

For the safety of a rider, we require that the rider remain at rest relative to the wall. Then by (6.4) in which [latex]\dot{\phi}=\omega,[/latex] or by (4.48) in which [latex]\boldsymbol{\omega}_f=\omega \mathbf{k}[/latex], it follows
that [latex]m \mathbf{a}_P=-m r \omega^2 \mathbf{e}_r .[/latex] Equating this to the force in (6.19a), we obtain the scalar equations of motion

[latex]\mathbf{F} \equiv F_r \mathbf{e}_r + F_\phi \mathbf{e}_\phi + F_z \mathbf{e}_z=m\left[\left(\ddot{r} - r \dot{\phi}^2 ight) \mathbf{e}_r + \frac{1}{r} \frac{d}{d t}\left(r^2 \dot{\phi} ight) \mathbf{e}_\phi + \ddot{z}\mathbf{e}_z ight] .[/latex] (6.4)

[latex]N=m r \omega^2, \quad W=f_z, \quad f_\phi=0 .[/latex] (6.19b)

In the steady rotation of the room, no circumferential component [latex]f_\phi[/latex] of the frictional force is exerted on the rider by the wall; and the second of these relations shows that the rider will not slide down the wall if the Coulomb condition [latex]W=f_z \leq f_c=\mu N[/latex] holds. Therefore, with the first equation in (6.19b), the design criterion for safety of the riders is given by [latex]\mu m r \omega^2 \geq W[/latex]. That is,

[latex]\omega \geq \sqrt{\frac{g}{r \mu}},[/latex] (6.19c)

equality holding when slip is imminent; the smallest value [latex]\omega^*=\sqrt{g / r \mu}[/latex] being the critical angular speed of the room . The result is independent of the weight of the rider; so all persons, fat or thin, will stay on the wall, provided that their coefficient of friction with the wall is not less than the design value chosen for μ.
For the given conditions r = 10 ft and μ = 0.4 , the critical angular speed is [latex]\omega^*=2.84 \mathrm{rad} / \mathrm{sec}[/latex], which is about 27 rpm. Thus, to secure the safety of the riders, the room must spin at a rate greater than 27 rpm.

Question 6.6: An amusement park ride shown in Fig. 6.3 consists of a 20 ft...

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