Ch 05: Applying Newton's Laws

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo

Young & Freedman Calc 14th Edition

Ch 05: Applying Newton's Laws

Problema 54c

Capitolo 5, Problema 54c

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). What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?

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Understand the problem: The passenger's apparent weight at the highest point being zero means that the normal force acting on the passenger is zero. This occurs when the centripetal force required to keep the passenger in circular motion is entirely provided by the gravitational force. We need to find the time for one revolution under this condition.

Write the equation for centripetal force at the highest point: The centripetal force is provided by gravity, so we can write $ F_c = F_g $, where $ F_c = \frac{m v^2}{r} $ and $ F_g = m g $. Here, $ m $ is the mass of the passenger, $ v $ is the tangential speed, $ r $ is the radius of the Ferris wheel, and $ g $ is the acceleration due to gravity.

Simplify the equation: Cancel $ m $ from both sides of the equation to get $ \frac{v^2}{r} = g $. Rearrange to solve for $ v $, the tangential speed: $ v = \sqrt{g r} $.

Relate the tangential speed to the period of revolution: The tangential speed $ v $ is related to the period $ T $ by the equation $ v = \frac{2 \pi r}{T} $. Substitute $ v = \sqrt{g r} $ into this equation to get $ \sqrt{g r} = \frac{2 \pi r}{T} $.

Solve for the period $ T $: Rearrange the equation to isolate $ T $: $ T = \frac{2 \pi r}{\sqrt{g r}} $. Simplify the expression to get $ T = 2 \pi \sqrt{\frac{r}{g}} $. Here, $ r $ is half the diameter of the Ferris wheel ($ r = 50 \ \text{m} $), and $ g $ is the acceleration due to gravity ($ g \approx 9.8 \ \text{m/s}^2 $).

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Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path, directed towards the center of the circle. In the context of the Ferris wheel, this force is provided by the gravitational force acting on the passengers and the normal force from the seat. At the highest point, if the apparent weight is zero, the normal force is zero, meaning the gravitational force alone provides the necessary centripetal force.

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Intro to Centripetal Forces

Apparent Weight

Apparent weight is the sensation of weight experienced by an object or person, which can differ from actual weight due to acceleration or other forces acting on it. In this scenario, when the passenger's apparent weight is zero at the highest point of the Ferris wheel, it indicates that they are in free fall, experiencing no normal force from the seat, which is a result of the centripetal acceleration required for circular motion.

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Circular Motion and Period

Circular motion refers to the movement of an object along the circumference of a circle, characterized by a constant distance from a central point. The period is the time taken to complete one full revolution. For the Cosmo Clock 21, the period can be calculated using the relationship between the radius, gravitational acceleration, and the conditions at the highest point, where the forces balance to allow for zero apparent weight.

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Intro to Circular Motion

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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). Find the speed of the passengers when the Ferris wheel is rotating at this rate.

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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). What then would be the passenger's apparent weight at the lowest point?

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