/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 The cosmo Clock 21 Ferris wheel ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 54

The cosmo Clock 21 Ferris wheel in Yokohama, 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
The speed of the passengers when the Ferris wheel is rotating at one revolution every 60 seconds is approximately \(5.235 \mathrm{m/s}\). The passenger's apparent weight at the highest and lowest point are approximately \(784 \mathrm{N}\) and \(980 \mathrm{N}\) respectively. The time for one revolution when the passenger's apparent weight at the highest point becomes zero is approximately \(89.41 \mathrm{s}\). The passenger's apparent weight at the lowest point at this situation is approximately \(1235 \mathrm{N}\).

Step by step solution

01

Calculate the Speed of Passengers

First, we find the speed of the passengers by using the formula for the speed of an object in circular motion: \(v = \frac{2\pi r}{T}\), where \(r\) is the radius of the ferris wheel and \(T\) is the period of one rotation. Here, the radius of the ferris wheel is \(50 \mathrm{~m}\), as radius is half the diameter and the time period \(T = 60.0 \mathrm{~s}\). Substituting these values, we get the speed of the passengers.
02

Find the Apparent Weight at the Highest and Lowest Points

Next, we find the apparent weight of the passenger at the highest and the lowest points. At the highest point, the direction of gravitational force and the centripetal force are the same. The apparent weight equals the true weight plus the weight equivalent of the centripetal force (\(w_{app} = mg + ma_c\)), where \(m\) is the mass, \(g\) is the acceleration due to gravity and \(a_c\) is the centripetal acceleration. At the lowest point, the direction of gravitational force and the centripetal force are opposite. The apparent weight equals the true weight minus the weight equivalent of the centripetal force (\(w_{app} = mg - ma_c\)).
03

Find the Time for One Revolution if the Apparent Weight is Zero

Now, we find the time for one revolution where the apparent weight at highest point is zero by using the formula of centripetal acceleration \(a_c = \frac{v^2}{r}\). Setting the apparent weight at the highest point (i.e., the true weight plus the weight equivalent of the centripetal force) equal to zero, we can find a new speed, and thus a new period.
04

Calculate the Apparent Weight at the Lowest Point

Finally, with the new speed found in the previous step, the apparent weight at the lowest point can be calculated using the same formula used in step 2, except inserting the new speed obtained in the previous step.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is a crucial concept when studying circular motion, such as the motion of a passenger on a Ferris wheel.
Imagine a passenger moving in a circle along the wheel. To keep the passenger moving in a circular path, a force must act inward toward the center of the circle. This force is directed inward and is due to the centripetal acceleration, which keeps the passenger from flying off the wheel.
Some important points about centripetal acceleration:
  • It is not a separate force; it鈥檚 the resultant acceleration that prompts an object to move along a circular path.
  • The formula for centripetal acceleration is given by: \[ a_c = \frac{v^2}{r} \] where \(v\) is the speed of the object and \(r\) is the radius of the circle.
  • In our problem, this helps determine the apparent weight of the passenger at different points of the Ferris wheel.
Understanding this concept is key to solving problems involving circular motion, ensuring that you can apply the right formula and reasoning to find you're seeking.
Apparent Weight
The apparent weight of an object is perceived weight, which can differ from its actual gravitational weight.
When a passenger is on the Ferris wheel, they will experience changes in their apparent weight.
Let's see why this happens:
  • At the **highest point** of the Ferris wheel, the apparent weight is the result of both gravitational force and centripetal force acting in the same direction. The formula can be expressed as: \[ w_{app} = mg + ma_c \] - **'\(m\)'** is the mass of the passenger. - **'\(g\)'** is the acceleration due to gravity. - **'\(a_c\)'** is the centripetal acceleration.
  • Conversely, at the **lowest point**, these forces act in opposite directions. The passenger feels lighter as part of the gravitational force is counteracted by the centripetal force: \[ w_{app} = mg - ma_c \]
Additionally, if the Ferris wheel's speed were increased to the point that the passenger feels weightless at the top, the apparent weight would be zero. Understanding these variations in apparent weight helps in recognizing how forces in circular motion alter our perception of weight.
Rotation Period
The rotation period of a Ferris wheel refers to the time it takes for the wheel to make a complete revolution.
This duration directly influences the speed of the passengers and, consequently, their apparent weight.
Here's how it is determined:
  • To find the **period** when the passenger is at the point of zero apparent weight at the top, we use the condition that centripetal force equals gravitational force. This gives: \[ mg = ma_c \] Simplifying this delivers a critical speed and subsequent rotation period.
  • The formula links the period '**\(T\)**' with speed '**\(v\)**' and radius '**\(r\)**' of the Ferris wheel: \[ v = \frac{2\pi r}{T} \] Knowing this relationship helps adjust the wheel's period to achieve specific conditions like zero apparent weight at the highest point.
Understanding how to calculate and adjust rotation period is essential for safely designing rides and predicting the resulting forces experienced by passengers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(2.00 \mathrm{~kg}\) box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to \(v(t)=\left(2.00 \mathrm{~m} / \mathrm{s}^{2}\right) t+\left(0.600 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\). What is the tension in the rope when the velocity of the box is \(9.00 \mathrm{~m} / \mathrm{s} ?\)

Friction in an Elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{~m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of \(36.0 \mathrm{~kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

Friction and Climbing Shoes. Shoes made for the sports of bouldering and rock climbing are designed to provide a great deal of friction between the foot and the surface of the ground. Such shoes on smooth rock might have a coefficient of static friction of 1.2 and a coefficient of kinetic friction of 0.90 . For a person wearing these shoes, what's the maximum angle (with respect to the horizontal) of a smooth rock that can be walked on without slipping? (a) \(42^{\circ} ;\) (b) \(50^{\circ} ;\) (c) \(64^{\circ} ;\) (d) larger than \(90^{\circ}\).

A horizontal wire holds a solid uniform ball of mass \(m\) in place on a tilted ramp that rises \(35.0^{\circ}\) above the horizontal. The surface of this ramp is perfectly smooth, and the wire is directed away from the center of the ball (Fig. P5.64). (a) Draw a free-body diagram of the ball. (b) How hard does the surface of the ramp push on the ball? (c) What is the tension in the wire?

\(\mathrm{}\) A \(25.0 \mathrm{~kg}\) box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .(\) a) As \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid \(5.0 \mathrm{~m}\) along the loading ramp?
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Engineering Physics

Read Explanation

Dynamics

Read Explanation

Thermodynamics

Read Explanation

Linear Momentum

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.