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1. Intro to Physics Units1h 29m
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5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
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8. Centripetal Forces & Gravitation

Centripetal Forces

Physics

8. Centripetal Forces & Gravitation

Centripetal Forces

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3:26 minutes

Problem 54d

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

Verified step by step guidance

Determine the centripetal acceleration of the passenger at the lowest point. The centripetal acceleration is given by the formula:

a c_{r} = v^{2} / r

, where

v

is the tangential speed and

r

is the radius of the Ferris wheel.

Calculate the tangential speed

v

of the passenger. The tangential speed is related to the period of rotation

T

and the radius

r

by the formula:

v = \frac{2 π r}{T}

. Use the diameter of the Ferris wheel to find the radius

r = \frac{100}{2} = 50 m

, and the period

T = 60 s

Determine the net force acting on the passenger at the lowest point. At this point, the forces acting on the passenger are the gravitational force

F g_{= m g}

(downward) and the normal force

F n

(upward). The net force is the centripetal force required to keep the passenger moving in a circular path:

F c = m a c_{r}

Relate the normal force to the net force. At the lowest point, the normal force is greater than the gravitational force because it must provide the centripetal force in addition to balancing the passenger's weight. The relationship is:

F n = F g + F c

, or equivalently:

F n = m g + m a c_{r}

Substitute the expressions for

a c_{r}

and

v

into the equation for

F n

. This gives:

F n = m g + m \frac{{\frac{2 π r}{T}}^{2}}{r}

. Simplify this expression to find the apparent weight of the passenger at the lowest point.

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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 forces acting on the passengers as they move along the circular trajectory.

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. It is influenced by the net forces acting on the object, including gravitational force and any additional forces due to motion, such as centripetal force. 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 circle.

Forces in Circular Motion

In circular motion, several forces interact to keep an object moving along a curved path. The key forces include gravitational force, which pulls objects downward, and the normal force, which acts perpendicular to the surface. At the lowest point of the Ferris wheel, the normal force is greater than the gravitational force, resulting in an increased apparent weight for the passengers due to the need for centripetal force to maintain circular motion.

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Related Practice

Textbook Question

(II) At what rate must a cylindrical spaceship rotate (Fig. 6–32) if occupants are to experience simulated gravity of 0.70 g? Assume the spaceship’s diameter is 28 m, and give your answer as the time needed for one revolution.

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Multiple Choice

A small 4kg block is tied to the end of 3m string and slides around in a circle on a frictionless table. Suppose the string will break if the tension exceeds 50N. Find the maximum speed the block can have without breaking the string.

In another version of the 'Giant Swing' (see Exercise $5.50$ ), the seat is connected to two cables, one of which is horizontal (Fig. E $5.51$ ). The seat swings in a horizontal circle at a rate of $28.0$ rpm (rev/min). If the seat weighs $255$ N and an $825$ -N person is sitting in it, find the tension in each cable.

The 'Giant Swing' at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable $5.00$ m long, and the upper end of the cable is fastened to the arm at a point $3.00$ m from the central shaft (Fig. E $5.50$ ). Find the time of one revolution of the swing if the cable supporting a seat makes an angle of $30.0 degrees$ with the vertical.

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

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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. E $5.45$ ). 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 $A$ )?