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Ch 05: Applying Newton's Laws
Chapter 5, Problem 5

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 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 s). (d) What then would be the passenger's apparent weight at the lowest point?

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1
Identify the radius of the Ferris wheel by dividing the diameter by 2. Since the diameter is 100 m, the radius (r) will be 50 m.
Calculate the angular velocity (\(\omega\)) of the Ferris wheel. Angular velocity can be calculated using the formula \(\omega = \frac{2\pi}{T}\), where T is the period of revolution. Given that the Ferris wheel makes one revolution every 60.0 s, use T = 60.0 s.
Determine the centripetal acceleration (\(a_c\)) at the lowest point using the formula \(a_c = \omega^2 \times r\). Substitute the values of \(\omega\) and r obtained from the previous steps.
Calculate the apparent weight of the passenger at the lowest point. The apparent weight can be found using the equation \(W_{apparent} = mg + ma_c\), where m is the mass of the passenger, g is the acceleration due to gravity (approximately 9.8 m/s^2), and \(a_c\) is the centripetal acceleration calculated in step 3.
Interpret the result. If the centripetal acceleration is less than the acceleration due to gravity, the apparent weight will be greater than the actual weight. If it is more, the apparent weight will be less than the actual weight.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Acceleration

Centripetal acceleration is the acceleration directed towards the center of a circular path that an object follows. It is necessary for maintaining circular motion and is calculated using the formula a_c = v^2 / r, where v is the tangential speed and r is the radius of the circular path. In the context of the Ferris wheel, this acceleration affects the apparent weight of passengers as they move through different points in the rotation.
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Apparent Weight

Apparent weight refers to the sensation of weight experienced by an object or person, which can differ from actual weight due to acceleration. When an object is in a non-inertial frame, such as a Ferris wheel, the forces acting on it can change its apparent weight. At the lowest point of the Ferris wheel, the apparent weight is the sum of the gravitational force and the centripetal force required to keep the passenger moving in a circular path.
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Forces in Circular Motion

In circular motion, two primary forces are at play: gravitational force and the normal force. At the lowest point of the Ferris wheel, the normal force exerted by the seat on the passenger must counteract the gravitational force and provide the necessary centripetal force for circular motion. Understanding how these forces interact is crucial for calculating the apparent weight of the passenger at that point.
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Related Practice
Textbook Question
A small car with mass 0.800 kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00 m (Fig. E5.45). If the normal force exerted by the track on the car when it is at the top of the track (point B) is 6.00 N, what is the normal force on the car when it is at the bottom of the track (point A)?
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Textbook Question
The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 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 s). (b) A passenger weighs 882 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?
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Textbook Question
The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 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 s).(c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?
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Textbook Question
flat (unbanked) curve on a highway has a radius of 170.0 m. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum coefficient of static friction that will prevent sliding?
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Textbook Question
You throw a baseball straight upward. The drag force is proportional to υ2. In terms of g, what is the y-component of the ball's acceleration when the ball's speed is half its terminal speed and (a) it is moving up?
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Textbook Question
You throw a baseball straight upward. The drag force is proportional to υ2. In terms of g, what is the y-component of the ball's acceleration when the ball's speed is half its terminal speed and (b) It is moving back down?
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