/*! 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 71 A student of weight \(667 \mathr... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 71

A student of weight \(667 \mathrm{~N}\) rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force \(\vec{N}\) on the student from the seat is \(556 \mathrm{~N}\). (a) Does the student feel "light" or "heavy" there? (b) What is the magnitude of \(\vec{N}\) at the lowest point? (c) What is the magnitude \(N\) if the wheel's speed is doubled?

Short Answer

Expert verified
a) The student feels lightest at the highest point. b) The normal force at the lowest point is 778 N. c) If speed is doubled, the normal force is 223 N at the highest point and 1111 N at the lowest point.

Step by step solution

01

Understanding Forces at the Highest Point

At the highest point of the Ferris wheel, the forces acting on the student include gravitational force (weight) and the normal force. The gravitational force, which points downward, is given as 667 N, while the normal force points upward and is given as 556 N. Since gravity pulls the student down, if the normal force is less than the weight, the student will feel 鈥榣ight.鈥�
02

Determine Feeling at the Highest Point

Given the normal force (556 N) is less than the weight (667 N), the student will feel 鈥榣ight鈥� at the highest point, because the normal force is less than the gravitational force.
03

Calculate Net Force at the Highest Point

The net force at the highest point can be found using the centripetal force equation: \[ F_{\text{net}} = mg - N_{highest} \ = 667 \text{ N} - 556 \text{ N} = 111 \text{ N} \]
04

Determine Normal Force at the Lowest Point

At the lowest point, the direction of forces changes. The equation becomes: \[ N_{lowest} - mg = F_{\text{net}} \ N_{lowest} - 667 \text{ N} = 111 \text{ N} \ N_{lowest} = 667 \text{ N} + 111 \text{ N} = 778 \text{ N} \]
05

Calculate Normal Force When Speed is Doubled

When the wheel's speed is doubled, the centripetal force quadruples (since centripetal force is proportional to the square of the speed). Therefore, the new centripetal force is: \[ F_{\text{net,new}} = 4 \times 111 \text{ N} = 444 \text{ N} \]At the highest point: \[ N_{highest,new} = mg - F_{\text{net,new}} = 667 \text{ N} - 444 \text{ N} = 223 \text{ N} \]At the lowest point: \[ N_{lowest,new} = mg + F_{\text{net,new}} = 667 \text{ N} + 444 \text{ N} = 1111 \text{ N} \]

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Key Concepts

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

Normal Force
The normal force is the support force exerted upon an object in contact with another stable object. In the context of a Ferris wheel, it's the force exerted by the seat onto the student. When calculating the normal force, we need to consider how it changes at different points in the ride: at the highest point, it counters part of the gravitational pull, making the student feel lighter. At the lowest point, it adds to the gravitational pull, making the student feel heavier.
Gravitational Force
Gravitational force is the force with which the Earth pulls objects toward its center. This force acts downward and is calculated as the product of mass and gravitational acceleration, which is approximately \(9.8 \text{ m/s}^2\). For the student weighing \(667 \text{N}\), this force remains constant throughout the Ferris wheel ride. Understanding the balance between gravitational force and other forces such as the normal force is key in determining how light or heavy the student feels at different points.
Centripetal Acceleration
Centripetal acceleration is the acceleration that keeps an object moving in a circular path and points toward the center of the circle. This results from the net force acting towards the center. For the Ferris wheel, the centripetal acceleration is crucial for maintaining circular motion. When the wheel speeds up, centripetal acceleration increases. This directly affects the net force calculations and the magnitudes of the normal force at different points of the ride.
Net Force
Net force is the overall force acting on an object. For an object in circular motion, it's the force required to keep it moving along the curved path. In our Ferris wheel example, net force is the result of balancing the gravitational force and normal force. At the highest point, net force is given by: \(F_{\text{net}} = mg - N \). At the lowest point, it is calculated as: \(N - mg = F_{\text{net}}\). Understanding these relationships helps us calculate how the student feels at different points and how this experience changes with speed variations.
Ferris Wheel Physics
Ferris wheel physics combines principles of circular motion with forces. Key factors include the gravitational and normal forces. As the Ferris wheel rotates, the positions (highest and lowest) determine how these forces interact. For instance, at the top, the normal force makes the student feel light. At the bottom, a higher normal force results in a feeling of heaviness. Changes in the wheel's speed affect the centripetal force required, altering the net and normal forces, and thus the ride experience. These principles create the unique sensations experienced during the circular journey.

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Most popular questions from this chapter

A bolt is threaded onto one end of a thin horizontal rod, and the rod is then rotated horizontally about its other end. An engineer monitors the motion by flashing a strobe lamp onto the rod and bolt, adjusting the strobe rate until the bolt appears to be in the same eight places during each full rotation of the rod (Fig. 6-75). The strobe rate is 2000 flashes per second; the bolt has mass \(30 \mathrm{~g}\) and is at radius \(3.5 \mathrm{~cm}\). What is the magnitude of the force on the bolt from the rod?

While two forces act on it, a particle is to move at the constant velocity \(\vec{v}=(3 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(4 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) One of the forces is \(\vec{F}_{A}=(2 \mathrm{~N}) \hat{\mathrm{i}}+(-6 \mathrm{~N}) \hat{\mathrm{j}}\). What is the other force?

We know that as an object passes through the air, the air exerts a resistive force on it. Suppose we have a spherical object of radius \(R\) and mass \(m\). What might the force plausibly depend on? \- It might depend on the properties of the object. The only ones that seem relevant are \(m\) and \(R\). \- It might depend on the object's coordinate and its derivatives: \(\vec{r}, \vec{v}, \vec{a}, \ldots\) \- It might depend on the properties of the air, such as the density, \(\rho\). (a) Explain why it is plausible that the force the air exerts on a sphere depends on \(R\) but implausible that it depends on \(m\). (b) Explain why it is plausible that the force the air exerts depends on the object's speed through it, \(|\vec{v}|\), but not on its position, \(\vec{r}\), or acceleration, \(\vec{a}\). (c) Dimensional analysis is the use of units (e.g., meters, seconds, or newtons) associated with quantities to reason about the relationship between the quantities. Using dimensional analysis, construct a plausible form for the force that air exerts on a spherical body moving through it.

the terminal speed of a sky diver is \(160 \mathrm{~km} / \mathrm{h}\) in the spread-eagle position and \(310 \mathrm{~km} / \mathrm{h}\) in the nosedive position. Assuming that the diver's drag coefficient \(C\) does not change from one position to the other, find the ratio of the effective cross-sectional area \(A\) in the slower position to that in the faster position.

Suppose the coefficient of static friction between the road and the tires on a Formula One car is \(0.6\) during a Grand Prix auto race. What speed will put the car on the verge of sliding as it rounds a level curve of \(30.5 \mathrm{~m}\) radius?
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