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Ch 05: Applying Newton's Laws
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 5, Problem 54b

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100100 m. Its name comes from its 6060 arms, each of which can function as a second hand (so that it makes one revolution every 60.060.0 s). A passenger weighs 882882 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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Determine the radius of the Ferris wheel. Since the diameter is given as 100 m, the radius \( r \) is half of the diameter: \( r = \frac{100}{2} \).
Calculate the angular velocity \( \omega \) of the Ferris wheel. The Ferris wheel completes one revolution in 60.0 seconds. Use the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of rotation.
Determine the centripetal acceleration \( a_c \) experienced by the passenger. Use the formula \( a_c = \omega^2 r \), where \( \omega \) is the angular velocity and \( r \) is the radius.
At the highest point, the apparent weight is the normal force exerted by the seat on the passenger. The net force at the highest point is \( F_{net} = mg - N \), where \( N \) is the normal force (apparent weight), \( m \) is the mass of the passenger, and \( g \) is the acceleration due to gravity. Solve for \( N \) using \( F_{net} = m a_c \).
At the lowest point, the apparent weight is again the normal force. The net force at the lowest point is \( F_{net} = N - mg \). Solve for \( N \) using \( F_{net} = m a_c \). Note that the mass \( m \) can be found using \( m = \frac{W}{g} \), where \( W \) is the weight of the passenger (882 N).

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Key Concepts

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

Apparent Weight

Apparent weight refers to the normal force experienced by an object in a non-inertial frame of reference, such as when it is in an accelerating system. In the context of a Ferris wheel, a passenger's apparent weight changes due to the centripetal acceleration acting on them as the wheel rotates. At the highest point, the gravitational force and the centripetal force combine, while at the lowest point, they oppose each other.
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Centripetal Acceleration

Centripetal acceleration is the acceleration directed towards the center of a circular path that keeps an object moving in that path. It 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. For the Ferris wheel, this acceleration affects the forces acting on the passenger, influencing their apparent weight at different points of the ride.
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Forces in Circular Motion

In circular motion, two main forces are typically at play: gravitational force and the normal force. The gravitational force acts downward, while the normal force can vary depending on the position in the circular path. At the highest point of the Ferris wheel, the normal force is reduced due to the gravitational force assisting in providing the necessary centripetal force, whereas at the lowest point, the normal force increases to counteract gravity and provide the required centripetal force.
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Related Practice
Textbook Question

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100100 m. Its name comes from its 6060 arms, each of which can function as a second hand (so that it makes one revolution every 60.060.0 s). Find the speed of the passengers when the Ferris wheel is rotating at this rate.

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Textbook Question

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100100 m. Its name comes from its 6060 arms, each of which can function as a second hand (so that it makes one revolution every 60.060.0 s). 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

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100100 m. Its name comes from its 6060 arms, each of which can function as a second hand (so that it makes one revolution every 60.060.0 s). What then would be the passenger's apparent weight at the lowest point?

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Textbook Question

In another version of the 'Giant Swing' (see Exercise 5.505.50), the seat is connected to two cables, one of which is horizontal (Fig. E5.515.51). The seat swings in a horizontal circle at a rate of 28.028.0 rpm (rev/min). If the seat weighs 255255 N and an 825825-N person is sitting in it, find the tension in each cable.

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Textbook Question

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates 'artificial gravity' at the outside rim of the station. If the diameter of the space station is 800800 m, how many revolutions per minute are needed for the 'artificial gravity' acceleration to be 9.809.80 m/s2?

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