### 2 Amusement rides

The design of amusement rides is intended to make the forces experienced by passengers as exciting as possible. High speeds are not enough. The production of accelerations of up to four times that due to gravity and occasional feelings of near-weightlessness are deliberate design goals. The accelerations may not only be in the forward or backward directions from the passengers’ perspective but also sideways. Sideways accelerations are more limited than backwards or forwards ones, since they are less welcome to passengers and also pose particular problems for the associated structures. Upward accelerations of greater than $g$ are avoided because of the safety risk and the need for reliable constraints to prevent passengers ‘floating’ out of the carriages. The rates at which the forces, or accelerations, change are important in producing the overall sensation also. The rate of change of acceleration is called jerk . Even the rate of change of jerk, called jounce , may be of interest. On roller-coasters the height, the tightness and the twistiness of turns in the track are the main parameters that influence the thrill of riding on them. Another factor that relates to thrill or discomfort is the degree of mismatch between what is observed and what is felt. For example it is odd to feel as though one’s weight is acting in some other direction than the perceived vertical. In this regard the magnitude of the force that is experienced may not be as important as its direction . Although we consider only two dimensions in this Workbook, the design of roller-coasters requires calculations and considerations in 3D. We shall consider some particular examples of forces experienced by passengers. We start by exploring the forces that contribute to what we feel when riding in a vehicle. Then we shall look at the forces experienced by passengers on amusement rides ranging from rotors to roller-coasters.

Figure 21 :

#### 2.1 Fearsome forces

The experiences we have on amusement rides include those of being subjected to linear and sideways accelerations that are sensed by the balancing system close to our ears and can make us feel giddy. Potentially more ‘enjoyable’ are sensations of unfamiliar compressive forces that act on our bodies through the vehicle in which we are travelling.

Figure 22 :

Imagine that you are sitting on a stool, which is sufficiently tall so that your feet are not touching the ground. What do you feel? Your weight is acting vertically downward. However you feel an upward force, which is the normal reaction of the stool to your weight. This reaction is pushing you upward (see Figure 21). Of course the total force on you is zero. Consequently there is a difference in this case between the total force on you, which includes gravity, and the force that you experience which excludes gravity.

On the other hand if you are sitting facing forwards in a vehicle that is accelerating forwards on a flat track then you will experience the same acceleration as the vehicle, through the seat which pushes you forwards (see Figure 22). The forward force from the seat combines with the normal reaction to give a resultant that is not vertical. This simple example suggests that the force experienced by a passenger during an amusement ride can be calculated by adding up all the component forces except for the passenger’s weight.

##### Example 11
1. Calculate the horizontal and vertical components of the force $\underset{̲}{F}$ experienced by a passenger of mass 100 kg seated in a rollercoaster carriage that starts from rest and moves in a straight line on a flat horizontal track with a constant acceleration such that it is moving at $40\text{m}{\text{s}}^{-1}$ after 5 s.
2. Deduce the magnitude and direction of the force experienced by the passenger.
##### Solution
1. Consider first the acceleration of the passenger’s seat. The coordinate origin is chosen at the start of motion. The $x$ -axis is chosen along the direction of travel, with unit vector $\underset{̲}{i}$ , and the $y$ -axis is vertical with $y$ positive in the upward direction, with unit vector $\underset{̲}{j}$ . The acceleration $\underset{̲}{a}$ may be calculated from

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\underset{̲}{a}=\frac{d\underset{̲}{v}}{dt}=a\underset{̲}{i}$

or $\underset{̲}{v}=at\underset{̲}{i}+\underset{̲}{c}.$ When $t=0,\underset{̲}{v}=\underset{̲}{0}$ , so $\underset{̲}{c}=\underset{̲}{0}$ . When $t=5,\underset{̲}{v}=40\underset{̲}{i}$ . Hence $40=5a$ or $a=8$ . The acceleration in the direction of travel is $8\text{m}{\text{s}}^{-2}$ , so the component of the force $F$ in the direction of travel is given by $ma=100×8\phantom{\rule{1em}{0ex}}\mathrm{\text{N}}=800$ N. The seat exerts a force on the passenger that balances the force due to gravity i.e. the passenger’s weight. So the vertical component of the force on the passenger is $mg=100×9.81\phantom{\rule{1em}{0ex}}\mathrm{\text{N}}=981$ N. Hence the total force experienced by the passenger may be expressed as

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\underset{̲}{F}=800\underset{̲}{i}+981\underset{̲}{j}.$

This is the total force exerted by the vehicle on the passenger. The force exerted by the passenger on the vehicle is the direct opposite of this i.e. $-800\underset{̲}{i}-981\underset{̲}{j}$ .

2. The magnitude of the total force is $\sqrt{80{0}^{2}+98{1}^{2}}\phantom{\rule{1em}{0ex}}\mathrm{\text{N}}\approx 1300$ N and the total force experienced by the passenger is at an angle with respect to the horizontal equal to ${tan}^{-1}\left(981∕800\right)\approx 5{1}^{\circ }$ .

Here we considered the sudden application of a constant acceleration of $8\text{m}{\text{s}}^{-2}$ which will cause quite a jerk for the passengers at the start. On some rides the acceleration may be applied more smoothly.

##### Example 12

Calculate the total force experienced by the passenger as a function of time if the horizontal component of acceleration of the vehicle is given by

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left\{\begin{array}{cc}8sin\left(\frac{\pi t}{10}\right)\hfill & 0\le t\le 5\hfill \\ \hfill \\ 8\hfill & t>5\hfill \end{array}\right\$

##### Solution

With this horizontal component of acceleration, the component of force experienced by the passenger in the direction of motion is (see Figure 23)

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left\{\begin{array}{ccc}800sin\left(\frac{\pi t}{10}\right)\hfill & \mathrm{\text{for}}\hfill & 0\le t\le 5\hfill \\ \hfill \\ 800\hfill & \mathrm{\text{for}}\hfill & t>5\hfill \end{array}\right\$

Figure 23 :

The vertical component remains constant at 981 N, as in Example 11. The total force may be written

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\underset{̲}{F}=\left\{\begin{array}{cc}800sin\left(\frac{\pi t}{10}\right)\underset{̲}{i}+981\underset{̲}{j}\hfill & 0\le t\le 5\hfill \\ \hfill \\ 800\underset{̲}{i}+981\underset{̲}{j}\hfill & t>5\hfill \end{array}\right\$

The idea of an amusement ride called the ‘Rotor’ is to whirl passengers around in a cylindrical container at increasing speed. When the rotation is sufficiently fast the floor is lowered but the passengers remain where they are supported by friction against the wall. Given a rotor radius of 2.2 m and a coefficient of friction of 0.4 calculate the minimum rate of revolution when the floor may be lowered.

The reaction of the wall on the passenger will have the same magnitude as the force causing motion in a circle i.e. $\frac{m{v}^{2}}{r}=R$ . The vertical friction force between the passenger and the wall is $\mu R$ . The passenger of mass will remain against the wall when the floor is lowered if $\mu R\ge mg$ . Hence it is required that $\frac{\mu m{v}^{2}}{r}\ge mg$ or $v\ge \sqrt{\frac{rg}{\mu }}$ . Since $v=\omega r$ , where $\omega$ is the angular velocity, the minimum required angular velocity is $\omega =\sqrt{\frac{g}{\mu r}}$ and the corresponding minimum rate of revolution $n=\frac{\omega }{2\pi }\sqrt{\frac{g}{\mu r}}$ . Hence with $r=2.2$ and $\mu =0.4$ , the rate of revolution must be at least 0.53 revs per sec or at least 32 revs per min.

In an amusement ride called the ‘Yankee Flyer’, the passengers sit in a ‘boat’, which stays horizontal while executing a series of rotations on an arm about a fixed centre. Given that the period of rotation is 2.75 s, calculate the radius of rotation that will give rise to a feeling of near weightlessness at the top of each rotation.

To achieve a feeling of ‘near-weightlessness’ near the top of the rotation, the force on the passenger towards the centre of rotation must be nearly equal and opposite to the reaction force of the seat on the passengers i.e. $\frac{m{v}^{2}}{r}=mg$ . This means that $r=\frac{{v}^{2}}{g}$ . Since $v=\omega r$ , this requires that $r=\frac{g}{{\omega }^{2}}$ . The period $T=\frac{2\pi }{\omega }=2.75$ s, so $\omega =2.285\text{rad}{\text{s}}^{-1}$ . Hence $r=1.879$ m.

##### Example 13

An amusement ride carriage moves along a track at constant velocity in the horizontal ( $x$ -) direction. It encounters a bump of horizontal length $L$ and maximum height $h$ with a profile in the vertical plane given by

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}y\left(x\right)=\frac{h}{2}\left(1-cos\frac{2\pi x}{L}\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}0\le x\le L.$

Calculate the variation of the vertical component of force exerted on a passenger by the seat of the carriage with horizontal distance ( $x$ ) as it moves over the bump.

##### Solution

Figure 24 :

Figure 24 shows a graph of $y\left(x\right)$ against $x$ for $h=2$ and $L=100$ . Since the component of velocity of the vehicle in the $x$ -direction is constant, then after time $t$ , the horizontal distance moved, $x$ , is given by $x=ut$ as long as $x$ is measured from the location at $t=0$ . Consequently the $y$ -coordinate may be written in terms of $t$ rather than $x$ , giving

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}y\left(t\right)=\frac{h}{2}\left(1-cos\frac{2\pi ut}{L}\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}0\le t\le \frac{L}{u}.$

The vertical component of velocity is given by differentiating this expression for $y\left(t\right)$ with respect to $t$ .

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}ẏ\left(t\right)=\frac{h}{2}\frac{2\pi u}{L}sin\left(\frac{2\pi ut}{L}\right)=\frac{h\pi u}{L}sin\left(\frac{2\pi ut}{L}\right).$

The vertical acceleration is given by differentiating this again.

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}ÿ\left(t\right)=\frac{h\pi u}{L}\frac{2\pi u}{L}cos\left(\frac{2\pi ut}{L}\right)=\frac{2h{\pi }^{2}{u}^{2}}{{L}^{2}}cos\left(\frac{2\pi ut}{L}\right).$

Two forces contribute to the vertical force exerted by the seat on the passenger: the constant reaction to the passenger’s weight and the variable vertical reaction associated with motion over the bump. The magnitude of the total vertical force $R$ N exerted on the passenger by the seat is given by

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}R=mg+mÿ\left(t\right)=mg+m\frac{2h{\pi }^{2}{u}^{2}}{{L}^{2}}cos\left(\frac{2\pi ut}{L}\right)=mg\left(1+\frac{2h{\pi }^{2}{u}^{2}}{g{L}^{2}}cos\left(\frac{2\pi ut}{L}\right)\right).$

Some horizontal force may be needed to keep the vehicle moving with a constant horizontal component of velocity and ensure that the net horizontal component of acceleration is zero. Since the horizontal component of velocity is constant, there is no horizontal component of acceleration and no net horizontal component of force. Consequently $\underset{̲}{R}=R\underset{̲}{j}$ represents the total force exerted on the passenger. Figure 25 shows a graph of $R∕mg$ for $h=2$ m, $L=100$ m and $u=20\text{m}{\text{s}}^{-1}$ .

Figure 25 :

#### 2.2 Banked tracks

Look back near the end of Section 34.2 subsection 1 which considered the forces on a cyclist travelling around a circular bend of radius $R$ . We were concerned with the way in which cyclists and motorcyclists bank their vehicles to create a ‘new vertical’ along the direction of the resultant force. This counteracts the torque that would otherwise encourage the rider to fall over when cornering. Clearly, passengers in four wheeled vehicles, railway trains and amusement park rides are not able to bank or tilt their vehicles to any significant extent. However what happens if the road or track is banked instead? If the road or track is tilted or banked at angle $\theta ={tan}^{-1}\left(\frac{{v}^{2}}{gr}\right)$ to the horizontal, then, at speed $v$ around the circular bend, it is possible to obtain the same result as that achieved by tilting the cycle or motorcycle (see Figure 26).

Figure 26 :

##### Example 14

Calculate the angle at which a track should be tilted so that passengers in a railway carriage moving at a constant speed of $20\text{m}{\text{s}}^{-1}$ around a bend of radius 100 m feel the resultant ‘reaction’ force as though it were acting vertically through their centre line.

##### Solution

The angle of the resultant force on passengers if the track were horizontal is given by

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\theta ={tan}^{-1}\left(\frac{{v}^{2}}{gr}\right),$

where $v=20,r=100$ and $g=9.81$ .

Hence $\theta =22.1{8}^{\circ }$ and the track should be tilted at $22.1{8}^{\circ }$ to the horizontal for the resultant force to act at right-angles to the track.

##### Example 15

The first three seconds of an amusement ride are described by the position coordinates

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\begin{array}{c}x\left(t\right)=10-10cos\left(0.5t\right)\hfill \\ y\left(t\right)=20sin\left(0.5t\right)\hfill \end{array}\phantom{\rule{1em}{0ex}}\left(0

in the horizontal plane where $x$ and $y$ are in m.

1. Calculate the velocity and acceleration vectors.
2. Hence deduce the initial magnitude and direction of the acceleration and the magnitude and direction of the acceleration after three seconds.
##### Solution

Figure 27 :

In this Example, the path of the ride is not circular (see Figure 27). In fact it is part of an ellipse. If we choose unit vectors $\underset{̲}{i}$ along the $x$ -direction and $\underset{̲}{j}$ along the $y$ -direction, and origin at $t=0$ , the position vector may be written

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\underset{̲}{r}\left(t\right)=\left(10-10cos\left(0.5t\right)\right)\underset{̲}{i}+20sin\left(0.5t\right)\underset{̲}{j}.$

The velocity vector is obtained by differentiating this with respect to time.

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\underset{̲}{v}\left(t\right)=5sin\left(0.5t\right)\underset{̲}{i}+10cos\left(0.5t\right)\underset{̲}{j}.$

The acceleration vector is obtained by differentiating again.

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\underset{̲}{a}\left(t\right)=2.5cos\left(0.5t\right)\underset{̲}{i}-5sin\left(0.5t\right)\underset{̲}{j}$

At $t=0,\underset{̲}{a}\left(t\right)=2.5\underset{̲}{i}$ . So the initial acceleration is $2.5\text{m}{\text{s}}^{-2}$ in the $x$ -direction.

At $t=3,\underset{̲}{a}\left(t\right)=2.5cos\left(1.5\right)\underset{̲}{i}-5sin\left(1.5\right)\underset{̲}{j}=0.177\underset{̲}{i}-0.487\underset{̲}{j}$ .

So after three seconds the acceleration is $4.99\text{m}{\text{s}}^{-2}$ at an angle of $8{8}^{\circ }$ in the negative $y$ -direction. This means a sideways acceleration of about 0.5 $g$ towards the inside of the track and almost at right-angles to it.