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The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 \(\mathrm{m}\) . Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 \(\mathrm{s}\) ). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 \(\mathrm{N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

Short Answer

Expert verified
(a) 5.24 m/s. (b) 798 N at top, 966 N at bottom. (c) 14.22 s. (d) 1709 N at bottom.

Step by step solution

01

Calculate the Circumference of the Ferris Wheel

The diameter of the Ferris wheel is given as 100 meters. The circumference can be calculated using the formula: \( C = \pi \times \text{diameter} \). Substitute the given diameter: \( C = \pi \times 100 \). Therefore, \( C = 100\pi \) meters.
02

Find the Speed of the Passengers

The Ferris wheel completes one revolution every 60 seconds. Using the circumference found in Step 1, we find the speed using the formula: \( v = \frac{C}{T} \), where \( T \) is the period. So, \( v = \frac{100 \pi}{60} \approx 5.24 \) m/s.
03

Calculate Apparent Weight at the Top of the Ferris Wheel

The apparent weight at the highest point is the normal force acting upwards. It can be calculated using \( F_{\text{apparent, top}} = mg - mv^2/r \). Given \( m = \frac{882}{9.81} \approx 90 \text{ kg} \), \( v^2 \approx 27.46 \), and \( r = 50 \), we find \( F_{\text{apparent, top}} = 90 \times 9.81 - \frac{90 \times 27.46}{50} \approx 797.64 \) N.
04

Calculate Apparent Weight at the Bottom of the Ferris Wheel

At the lowest point, the apparent weight is the normal force acting upwards and it is calculated as \( F_{\text{apparent, bottom}} = mg + mv^2/r \). Substitute the given and calculated values: \( F_{\text{apparent, bottom}} = 90 \times 9.81 + \frac{90 \times 27.46}{50} \approx 966.36 \) N.
05

Find the Revolution Period for Zero Apparent Weight at the Top

At the top, zero apparent weight occurs when the centripetal acceleration equals gravitational acceleration: \( g = v^2/r \). Solving the equation: \( 9.81 = \frac{4\pi^2r}{T^2} \) gives \( T = \sqrt{\frac{4\pi^2 \times 50}{9.81}} \approx 14.22 \) s.
06

Determine Apparent Weight at the Bottom with Zero Weight at the Top

With the new period, speed is recalculated: \( v = \frac{2\pi \times 50}{14.22} \approx 22.1 \) m/s. The apparent weight at the bottom becomes \( F_{\text{apparent, bottom}} = mg + mv^2/r = 90 \times 9.81 + \frac{90 \times 488.41}{50} \approx 1708.74 \) N.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
In circular motion, an object traveling in a circular path constantly changes direction, requiring a continuous force directed towards the center of the circle. This force is called the centripetal force. On a Ferris wheel, this force ensures that the passengers do not fly off the ride as it spins. The formula for calculating centripetal force is \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is the speed, and \( r \) is the radius of the circle.
- **Centripetal vs. Centrifugal:** While centripetal force pulls objects towards the center, centrifugal is often used to explain the apparent "pull" outward. However, centrifugal force isn't an actual force, it's just the sensation of inertia.
- **Gravity's Role:** Gravity provides part of the centripetal force when you're at the top of the Ferris wheel. At the bottom, it acts with centripetal force, increasing the sensation of weight. This interplay is crucial in calculations involving circular motion like on a Ferris wheel.
Apparent Weight
Apparent weight refers to the force a person feels as their weight, which can change throughout the motion of a Ferris wheel due to the acceleration acting on them. When you're at the top of a Ferris wheel, your apparent weight decreases because the centripetal force works against gravity. On the ground, or at the bottom of the wheel, you feel heavier due to gravity and centripetal force pointing in the same direction.
- **At the Top:** Apparent weight is calculated as \( mg - \frac{mv^2}{r} \), where \( mg \) is the gravitational force and the second term is the centripetal force effect. This results in a lighter feeling.
- **At the Bottom:** Apparent weight is \( mg + \frac{mv^2}{r} \), adding to the gravitational force. You feel heavier because the normal force minus centripetal force equals your calculated weight. These variations in apparent weight can range from feeling almost weightless at the top to feeling noticeably heavier at the bottom.
Ferris Wheel Physics
The Ferris wheel is a perfect example of rotational dynamics, as it demonstrates principles involving circular motion, force, and period. Its simplicity helps in visualizing how forces keep an object moving in a circular path. As the wheel rotates, the physics involved includes the calculation of speed and period which dictates how quickly a revolution is completed. Each arm or spoke of the Ferris wheel contributes to ensuring balance and structural integrity.
- **Structural Design:** The strength and number of arms impact how forces are distributed, enabling the wheel to carry passengers safely.
- **Speed and Design:** The wheel's circular design means forces act consistently around its circumference, allowing for the precise calculation of passenger motion and apparent weight. Understanding how these forces interact helps ensure the wheel's safe and predictable operation. Bracings and supports deal with both the gravitational and rotational forces to prevent structural deformation.
Revolution Period
The period of revolution is the time taken for one complete rotation of the Ferris wheel. This affects how you perceive motion on the ride. The period is directly influenced by the speed of rotation and the size of the wheel. As the Cosmoclock 21 Ferris wheel completes one rotation every 60 seconds under normal condition, its previous period can be calculated using the circumference and speed.
- **Formula:** The period \( T \) is derived from \( v = \frac{C}{T} \), where \( C \) is the circumference of the wheel, and \( v \) is the speed of the passenger.
- **Zero Weight at the Top:** To achieve zero apparent weight at the top, a specific period is required where gravitational acceleration matches centripetal acceleration (\( g = \frac{v^2}{r} \)). Adjusting the period can alter the forces experienced by the passengers.
Understanding these elements helps in creating thrill settings for theme parks while ensuring safety, by calculating the right mix of speed and rotation to give varied experiences.

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Most popular questions from this chapter

A stockroom worker pushes a box with mass 11.2 \(\mathrm{kg}\) on a horizontal surface with a constant speed of 3.50 \(\mathrm{m} / \mathrm{s}\) . The coefficient of kinetic friction between the box and the surface is 0.20 . (a) What horizontal force must the worker apply to maintain the motion? (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?

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