Book cover for College Physics

College Physics

Eugenia Etkina, Michael Gentle, Alan Van Heuvelen

ISBN #9780321715357

1st Edition

2,258 Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

Circular motion, even at constant speed, involves acceleration because of the continuous change in the direction of motion. This centripetal acceleration, always directed toward the center of the circle, is determined by the speed and radius of the motion (a?r? = v²/r) and can also be related to the period of motion. Newton’s second law, when applied in the radial direction, explains the net inward force required to maintain circular motion. Additionally, Newton’s universal law of gravitation explains how gravitational forces operate, following an inverse-square law, enabling predictions for satellite orbits, planetary motion, and the behavior of objects under gravity.

Learning Objectives

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Key Concepts

CONCEPT

DEFINITION

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Example Problems

Example 1

Mountain biker While mountain biking, you first move at constant speed along the bottom of a trail’s circular dip and then at constant speed across the top of a circular hump. Assume that you and the bike are a system. Determine the direction of the acceleration at each position and construct a force diagram for each position (consistent with the direction of the acceleration). Compare at each position the magnitude of the force of the surface on the bike with the force Earth exerts on the system

Example 2

You swing a rock tied to a string in a vertical circle. (a) Determine the direction of the acceleration of the rock as it passes the lowest point in its swing. Construct a consistent force diagram for the rock as it passes that point. How does the force that the string exerts on the rock compare to the force that Earth exerts on the rock? Explain. (b) Repeat the above analysis as best you can for the rock as it passes the highest point in the swing. (c) If the string is tied around your finger, when do you feel a stronger pull when the rock is at the bottom of the swing or at the top? Explain

Example 3

Loop-the-loop You ride a roller coaster with a loop-the-loop. Compare as best you can the normal force that the seat exerts on you to the force that Earth exerts on you when you are passing the bottom of the loop and the top of the loop. Justify your answers by determining the direction of acceleration and constructing a force diagram for each position. Make your answers consistent with Newton’s second law.

Example 4

You start an old record player and notice a bug on the surface close to the edge of the record. The record has a diameter of 12 inches and completes 33 revolutions each minute. (a) What are the speed and the acceleration of the bug? (b) What would the bug’s speed and acceleration be if it were halfway between the center and the edge of the record?

Example 5

Determine the acceleration of Earth due to its motion around the Sun. What do you need to assume about Earth to make the calculation? How does this acceleration compare to the acceleration of free fall on Earth?
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Step-by-Step Explanations

QUESTION

Given a car moving at a constant speed v around a circular track of radius r, how do you compute the radial acceleration?\nStep-by-step Answer:\nStep 1: Recognize that even if the car\u2019s speed is constant, its velocity direction is changing, so it experiences acceleration.\nStep 2: Use the formula for centripetal (radial) acceleration: a\u208dr\u208e = v\u00b2/r.\nStep 3: Plug in the given values for speed and radius to calculate the magnitude of the acceleration.\nFinal Answer: The radial acceleration is a\u208dr\u208e = v\u00b2/r.\n\n- Topic: Relating Period to Acceleration \nQuestion: How can you express the radial acceleration in terms of the period T of the circular motion?\nStep-by-step Answer:\nStep 1: Recall that the speed v can be expressed as v = 2\u03c0r/T.\nStep 2: Substitute v into the centripetal acceleration formula: a\u208dr\u208e = (2\u03c0r/T)\u00b2/r.\nStep 3: Simplify to obtain a\u208dr\u208e = 4\u03c0\u00b2r/T\u00b2.\nFinal Answer: Radial acceleration in terms of the period is a\u208dr\u208e = 4\u03c0\u00b2r/T\u00b2.\n\n- Topic: Applying Universal Gravitation \nQuestion: How do you determine the gravitational force between two objects with masses m\u2081 and m\u2082 separated by distance r?\nStep-by-step Answer:\nStep 1: Begin with Newton\u2019s law of universal gravitation: F = G*(m\u2081*m\u2082)/r\u00b2.\nStep 2: Identify the masses of the objects and the distance between their centers.\nStep 3: Insert the known values and the universal gravitational constant G to calculate the force.\nFinal Answer: The gravitational force is F = G*m\u2081*m\u2082/r\u00b2.\n\n"

STEP-BY-STEP ANSWER:

Step 1: Recognize that even if the car\u2019s speed is constant, its velocity direction is changing, so it experiences acceleration.\nStep 2: Use the formula for centripetal (radial) acceleration: a\u208dr\u208e = v\u00b2/r.\nStep 3: Plug in the given values for speed and radius to calculate the magnitude of the acceleration.\nFinal Answer: The radial acceleration is a\u208dr\u208e = v\u00b2/r.\n\n- Topic: Relating Period to Acceleration \nQuestion: How can you express the radial acceleration in terms of the period T of the circular motion?\nStep-by-step Answer:\nStep 1: Recall that the speed v can be expressed as v = 2\u03c0r/T.\nStep 2: Substitute v into the centripetal acceleration formula: a\u208dr\u208e = (2\u03c0r/T)\u00b2/r.\nStep 3: Simplify to obtain a\u208dr\u208e = 4\u03c0\u00b2r/T\u00b2.\nFinal Answer: Radial acceleration in terms of the period is a\u208dr\u208e = 4\u03c0\u00b2r/T\u00b2.\n\n- Topic: Applying Universal Gravitation \nQuestion: How do you determine the gravitational force between two objects with masses m\u2081 and m\u2082 separated by distance r?\nStep-by-step Answer:\nStep 1: Begin with Newton\u2019s law of universal gravitation: F = G*(m\u2081*m\u2082)/r\u00b2.\nStep 2: Identify the masses of the objects and the distance between their centers.\nStep 3: Insert the known values and the universal gravitational constant G to calculate the force.\nFinal Answer: The gravitational force is F = G*m\u2081*m\u2082/r\u00b2.\n\n"
Final Answer: The radial acceleration is a\u208dr\u208e = v\u00b2/r.\n\n- Topic: Relating Period to Acceleration \nQuestion: How can you express the radial acceleration in terms of the period T of the circular motion?\nStep-by-step Answer:\nStep 1: Recall that the speed v can be expressed as v = 2\u03c0r/T.\nStep 2: Substitute v into the centripetal acceleration formula: a\u208dr\u208e = (2\u03c0r/T)\u00b2/r.\nStep 3: Simplify to obtain a\u208dr\u208e = 4\u03c0\u00b2r/T\u00b2.\nFinal Answer: Radial acceleration in terms of the period is a\u208dr\u208e = 4\u03c0\u00b2r/T\u00b2.\n\n- Topic: Applying Universal Gravitation \nQuestion: How do you determine the gravitational force between two objects with masses m\u2081 and m\u2082 separated by distance r?\nStep-by-step Answer:\nStep 1: Begin with Newton\u2019s law of universal gravitation: F = G*(m\u2081*m\u2082)/r\u00b2.\nStep 2: Identify the masses of the objects and the distance between their centers.\nStep 3: Insert the known values and the universal gravitational constant G to calculate the force.\nFinal Answer: The gravitational force is F = G*m\u2081*m\u2082/r\u00b2.\n\n"

"- Topic: Determining Radial Acceleration \nQuestion: Given a car moving at a constant speed v around a circular track of radius r, how do you compute the radial acceleration?\nStep-by-step Answer:\nStep 1: Recognize that even if the car\u2019s speed is constant, its velocity direction is changing, so it experiences acceleration.\nStep 2: Use the formula for centripetal (radial) acceleration: a\u208dr\u208e = v\u00b2/r.\nStep 3: Plug in the given values for speed and radius to calculate the magnitude of the acceleration.\nFinal Answer: The radial acceleration is a\u208dr\u208e = v\u00b2/r.\n\n- Topic: Relating Period to Acceleration \nQuestion: How can you express the radial acceleration in terms of the period T of the circular motion?\nStep-by-step Answer:\nStep 1: Recall that the speed v can be expressed as v = 2\u03c0r/T.\nStep 2: Substitute v into the centripetal acceleration formula: a\u208dr\u208e = (2\u03c0r/T)\u00b2/r.\nStep 3: Simplify to obtain a\u208dr\u208e = 4\u03c0\u00b2r/T\u00b2.\nFinal Answer: Radial acceleration in terms of the period is a\u208dr\u208e = 4\u03c0\u00b2r/T\u00b2.\n\n- Topic: Applying Universal Gravitation \nQuestion: How do you determine the gravitational force between two objects with masses m\u2081 and m\u2082 separated by distance r?\nStep-by-step Answer:\nStep 1: Begin with Newton\u2019s law of universal gravitation: F = G*(m\u2081*m\u2082)/r\u00b2.\nStep 2: Identify the masses of the objects and the distance between their centers.\nStep 3: Insert the known values and the universal gravitational constant G to calculate the force.\nFinal Answer: The gravitational force is F = G*m\u2081*m\u2082/r\u00b2.\n\n"

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Common Mistakes

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