University Physics Volume 1 Release Notes 2017...University Physics Volume 1 Release Notes 2017...

University Physics Volume 1 Release Notes 2017

Publish Date:

March 16, 2018

Revision Number:

UP1-2016-002(05/18)-MJ

Page Count Difference:

In the newest edition of University Physics Volume 1, there is 16 less pages when compared to the last version. This is attributed to padding changes in the document as well as errata revisions.

Errata: Below is a table containing submitted errata, and the resolutions that OpenStax has provided for this latest text.

Location Detail Resolution Notes Error Type

Preface Update contributor list.

Our reviewers accepted this change. Other

Preface

On https://cnx.org/contents/[email protected]:Gofkr9Oy@14/Preface, the link to www.oercommons.org/hubs/OpenStax goes to https://cnx.org/www.oercommons.org/hubs/OpenStax which is a 404. The link probably just needs to be updated in the CNXML.

Our reviewers accepted this change.

Broken link

Ch 1: Units and Measurement, Sec 2: Units and Standards

Chapter 1 Section 2 (pg 17, pdf file pg 25)- the link: " https://openstaxcollege.org/l/21redefkilo " redirects the user to the wrong page (http://physics.nist.gov/cuu/Units/units.html). The page it redirects the user to is the page for the previous link on page 16 (pdf page 24). If the link could be updated to the correct page, that would be great. Thanks!

Revise the URL for https://openstax.org/l/21redefkilo to: https://www.nist.gov/pml/productsservices/redefining-kilogram

Broken link

Ch 1, Section 6, Example 1.7

Under "Significance," the word "inn" is used where it should just be "in."

This typo has been resolved in webview format. Typo

Ch 1: Units and The next-to-last bullet point says, Our reviewers Other

Measurement, Sec 6: Significant Figures

"When multiplying or dividing measured values, the final answer can contain only as many significant figures as the least-precise value." The problem lies in the meaning of "least precise." As introduced in the text, I would say a measurement of 456.7 m is less precise than a measurement of 0.0089 m. Yet if I multiplied these numbers, the result would have two significant figures, like the more precise value. The rule stated in the text of Section 1.6 seemed accurate. I would modify the summary point as follows, to agree with the main text: "When multiplying or dividing measured values, the final answer can contain only as many significant figures as the value with the least number of significant figures."

accepted this change.

factual inaccuracy in content

Ch 1, Section 7

The last sentence in this section says, "In many ways, it is in answering questions such as these science that progresses." It appears that the words "science" and "that" should be transposed.

This typo has been fixed on Webview format. Typo

Ch 1, Section 7

In the second bullet point under "Strategy," there is a sentence that says, "Drawing a sketch be very useful at this point as well." It seems to be missing either the word "can" or "may."

This typo has been fixed on webview format. Typo

Ch 1: Units and Measurement, Additional Problems, Exercise 87

I found the answer key in the online version to have the following errors. Please let me know if you have any questions. Ch 1: #87 should have an uncertainty of 0.1cm^3, not 0.2cm^3.

Our reviewers accepted this change. Typo

Ch 1 problem 77 solution

Chapter 1 problem 77 solution says "7.557^2". It should say "7.557 cm^2".

This error has been corrected in web view, and the next edition of the pdf will reflect the correction. Typo

Ch 2 Problem 28 (g) Chapter 2 Problem 28 (g)does not show the vector addition task

This error has been corrected in web Typo

(based on solution manual that the vector summation task is C-2D+3F).

view and will be reflected in the next edition of the pdf.

Ch 2, Section 1

When the concept of a unit vector is first introduced, the very first u-hat should be bold and not italicized, to agree with the formatting used consistently thereafter. Right now, the first u-hat appears bold but italicized in the PDF and not bold or italicized in the web view.

Formatting corrected as suggested. Typo

Ch 2, Section 2

At the end of the next-to-last line in Example 2.7 (titled "Takeoff of a Drone"), there is a quotation mark where I think there should be a colon.

Corrected; this quotation mark was revised to a colon. Typo

Ch 2, Section 2

Near the end of the first sentence in the subsection called "Vectors in Three Dimensions," I believe it should say "coordinate z gives a vertical position above or below the plane" instead of "coordinate z gives a vertical positions above or below the plane."

Corrected; the s was removed from "positions" in "coordinate z gives a vertical positions above or below the plane." Typo

Ch 2, Section 2

In Example 2.4 (titled "Magnitude and Direction of the Displacement Vector"), the second sentence should begin, "What are the magnitude and direction..."

Corrected "is" to "are" in Example 2.4. Typo

Ch 2, Section 2

In the caption for Figure 2.19 (fourth figure in this section), a space needs to be inserted between "(I)" and "have" in the second sentence. Inserted space. Typo

Ch 2, Section 2

Shortly after Figure 2.18 (third figure in this section), Equation 2.16 is referenced. Immediately following the equation reference, there is an unmatched right parenthesis that should simply be deleted.

Deleted unmatched parenthesis. Typo

Ch 2: Vectors, Sec 1: Scalars and Vectors, Exercise 28

The resultant for part g is missing. It should read "C - 2D + 3F".

Our reviewers accepted this change. None

Ch 2: Vectors, Sec 2: Coordinate Systems and Components of a

The θ's in Equation 2.16 (right before Figure 2.18, third figure in this section) were recently replaced


Other factual inaccur

Vector with θ_A's. If you read only the text immediately preceding this equation, that change seems correct. However, if you continue reading after the equation and figure, you find that the following discussion and figure *specifically depend* upon the use of θ in the equation and not θ_A. The original choice was intentional. I recommend switching back to θ in Equation 2.16, and ideally introducing the equation a little differently so that it does not seem like a mistake when you first read it. (Actually, θ_A would be fine in the part of the equation before the arrow; it's the relationship after the arrow that's problematic.) One further note: If the subscript A's remain, they should be italicized to match the formatting of θ_A in the rest of the text.

acy in content

Ch 2: Vectors, Sec 2: Coordinate Systems and Components of a Vector, Example 2.7

"We identify b(...) and e(...)" b and e are completely wrong and unrelated to the rest of the problem... The equations are right, which made it extra confusing to the students...


Ch 2: Vectors, Sec 2: Coordinate Systems and Components of a Vector, Example 2.7

Right above "2.7 Check Your Understanding", it says D=sqrt((0.90 km)^2+(1.90 km)^2+(0.15 km)^2)=4.44km. 4.44 km is incorrect. The answer should be 2.11 km, not 4.44 km.


Ch 2: Vectors, Sec 2: Coordinate Systems and Components of a Vector, Exercise 37

The figure that Exercise 37 refers to is missing.


Ch 2: Vectors, Sec 3: Algebra of Vectors, Example 2.10

The names of the dogs should be changed to avoid potential cultural insensitivity.


General/pedagogical suggestion or question

Ch 2: Vectors, Sec 3: In Example 2.13 (titled Our reviewers Typo

Algebra of Vectors, Example 2.13

"Displacement of a Jogger"), there are three places where it mentions vector D_TB but clearly means vector D_AT. I have highlighted these in the attached screenshot.


Ch 2: Vectors, Sec 3: Algebra of Vectors, Example 2.13

In Example 2.13 (titled "Displacement of a Jogger"), the sentence before the last equation in the solution is finding the width and height of a step. It erroneously says "the step height is w" instead of h.


Ch 2: Vectors, Sec 3: Algebra of Vectors, Examples 2.9 and Examples 2.12

Examples 2.9 and 2.12 both reference Equation 2.24 as the equation for getting the components of a resultant vector. This seems unhelpful, since it really just defines the equivalence of vectors by components. There are at least two options for improving this. It would technically be correct to reference Equation 2.25 instead, but I suspect that students find that one a little overwhelming for most applications. So my suggestion is to take the set of three equations between 2.24 and 2.25 (the one starting with R_x=A_x+B_x) and make that into a numbered equation to reference in the examples.



Ch 2: Vectors, Sec 3: Algebra of Vectors, Exercise 61

In part (a), the Boeing 747 is said to be located 30 degrees north of west (ie. west of the control tower, or negative with respect to the x axis), but the i component of the position vector provided in the solution is positive (12.278 i). 12.278 appears to be correct, but the sign should be negative. The DC-3 position vector has the correct sign for i (it is also west of the control tower, so the sign is negative) but the amount is incorrect. The DC-3 is at 3000 m and 5 degrees above the horizon, which would place it 34.290 km from the control tower, not .262 km from the control tower. This makes sense, since if the 747 (2500 m and 10 degrees above the horizon) is


12.279 km from the control tower, the DC-3 at roughly the same altitude but lower on the horizon must be further away than the 747 (since the angle of the DC-3 is half the angle of the 747 and both are at about the same altitude, the DC-3 should be at least twice as far away).

Ch 2: Vectors, Sec 4: Products of Vectors

Immediately before Equation 2.37 where the distributive property for cross products is introduced, the text mentions that this is similar to the dot product and references Equation 2.31. However, the distributive property for dot products is actually stated in Equation 2.32.


Ch 2: Vectors, Sec 4: Products of Vectors, Example 2.18

The solution portion of Example 2.18 (titled "The Torque of a Force") refers to Equation 2.11 where it clearly means to reference Equation 2.35.


Ch 2: Vectors, Sec 4: Products of Vectors, Example 2.18

The second sentence of Example 2.18 (titled "The Torque of a Force") says, "The distance R from the nut to the point where force vector F is attached and is represented by the radial vector R." I am not 100% certain of the intended phrasing, but I think you could just eliminate "and."


Ch 2: Vectors, Sec 4: Products of Vectors, Exercise 63

The problem starts with "Assuming the +x-axis is horizontal to the right for the vectors in the following figure, find the following scalar products..." The solutions provided for (a) the component of vector A along vector C, and (b) the component of vector C along vector A, are 8.66 and 10.39, respectively. However, vectors A and C are orthogonal, so neither has a component along the other, and the answer should be zero for both (a) and (b). The answers provided are correct only if vector C is assumed to have an


angle of 60 degrees instead of -60 degrees as shown in the diagram.


(1) Answers provided in the solution for these parts have 4 or 5 significant figures, while the angles of the original vectors are given with only 2 significant figures and the magnitudes are at most 3 significant figures. The significant figures of the answer should match the input with the least number of sig figures; (2) It appears that the answers provided were obtained after rounding intermediate calculations (like converting a vector given as an angle and magnitude to i, j, and k components), instead of rounding only the final answer; (3) Also, several parts appear to be calculated incorrectly: answer (b) should be zero, since vectors A and F are antiparallel (just as the answer for h is zero because the vectors are parallel), answer (d) should be -2.4x10^2 k, answer (e) should be 4.0 k, answer (f) should be -3.0 k, and answer (g) should be 15 k.



The answer for part c should be 20,000 or 20*10^3



The textbook answer provided for (b) is 173,194. However, it should be zero, since A and F are antiparallel and their cross product is zero.


Ch 2: Vectors, Additional Problems, Exercise 73

As shown on the attached drawing, the angle for the polar coordinate of the point (-x,y) should be PI-theta. The answer for (a) should be (r, PI-theta)



The answers should be dPM = 6.2 nmi = 11.4 km and dNP = 7.2 nmi = 13.3 km

Revise the solution to "d_PM = 6.2 nmi = 11.4 km, d_NP = 7.2 nmi = 13.3 km". None


The distance should be 270 m and the direction should be 4.2 degrees north of west


Ch 2: Vectors, Challenge Problems, Exercise 89

Please check the solution for Chapter 2, Problem 89 in both the student and instructor solution guides; Isn't the correct answer 4.6 N? In addition, the subscript for G should reflect the fact that the quantity requested is the component of G parallel to H, not perpendicular (suggest || or H for the subscript).


Ch 3 section 3.5

In the caption to Figure 3.26, "1.67 m/s2" should be replaced with "1.67 m/s^2".


Ch 3, Section 2

There are two mistakes in the Strategy section of Example 3.3 (titled "Instantaneous Velocity Versus Average Velocity"). 1) At the beginning, "give" should be "gives." 2) Toward the end, it should be Equation 3.3 to calculate the average velocity, not Equation 3.6.


Ch 3, Section 3

In the third sentence of the caption for Figure 3.16 (seventh figure in the section), "tangents lines" should be corrected to "tangent lines."

This typo has been fixed and the change will be reflected immediately on the web-view version of the text. Typo

Ch 3, Section 3

In line (c) of the solution for Example 3.6 (titled "Calculating Instantaneous Acceleration"), there is no space between the number and the unit for each of the four accelerations.


Ch 3, Section 3

There are two errors in the paragraph between Figures 3.14 and 3.15. First, the second sentence says the graph comes from Example 3.3, but it is actually from Example 3.4. Next, the word "is" must be removed from the paragraph's last sentence.


Ch 3, Section 3

In the caption of Figure 3.14 (fifth figure in the section), there is a spot where t_0 is referenced, but the zero is not subscript as it should be.


Ch 3, Section 4 The units are messed up in the next-to-last line of the Solution

Our reviewers accepted this

Other factual

section in Example 3.9 (Calculating Final Velocity). It should be (m^2)/(s^2), as in the line that follows it.

change. inaccuracy in content

Ch 3, Section 4

In the caption of Figure 3.18, a space needs to be inserted between the number and unit: 60km/h.


Ch 3: Motion Along a Straight Line, Sec 2: Instantaneous Velocity and Speed, Example 3.2

In the solution of Example 3.2 (titled "Finding Velocity from a Position-Versus-Time Graph"), it would make more sense for the displacement during the second time interval to be calculated "0.5 m - 0.5 m" rather than "0.0 m - 0.0 m" since the position is constant at 0.5 m.

In the solution to Example 3.2 Finding Velocity from a Position-Versus-Time Graph, revise the time interval for 0.5 s to 1.0 s to "0.5 m - 0.5 m/1.0 s - 0.5 s". None

Ch 3: Motion Along a Straight Line, Sec 2: Instantaneous Velocity and Speed, Example 3.4

In the Solution section of Example 3.4 (titled "Instantaneous Velocity Versus Speed"), the second line should not have a letter in front of it, and the third line should have "b." in front of it. That is, the first two lines are solving part (a) of the problem, and the third line is solving part (b). There is no part (c).

In the solution to Example 3.4, remove the letter b from in front of the second line. Revise the letter c to the letter b in front of the third line. Typo

Ch 3: Motion Along a Straight Line, Sec 3: Average and Instantaneous Acceleration

A few words are missing in the paragraph immediately following the box that defines average acceleration. In the first sentence, it should say "velocity in meters per second" rather than "velocity in meters." In the next-to-last sentence, the word "and" needs to be inserted so that it reads "reverses direction, and continues her run."

In the paragraph following the definition of average acceleration, revise the 1st sentence as follows: Because acceleration is velocity in meters per second divided by time in seconds, ... Revise the 3rd sentence as follows: For example, if a runner traveling at 10 km/h due east slows to a stop, reverses direction, and continues her Typo

run at 10 km/h due west, ...

Ch 3: Motion Along a Straight Line, Sec 3: Average and Instantaneous Acceleration, Subsec: Instantaneous Acceleration

In the second sentence at the beginning of the discussion on Instantaneous Acceleration, it says "we calculate the average velocity between two points in time" and let the time interval approach zero to get the instantaneous acceleration. However, it should say we calculate the average acceleration rather than the average velocity.

Revise the 2nd sentence in the 1st paragraph as follows: That is, we calculate the average acceleration between two points in time ...

Other factual inaccuracy in content

Ch 3: Motion Along a Straight Line, Sec 4: Motion with Constant Acceleration, Example 3.12

Solution section First line: The last part of the equation (specifically, "= 1/2 a t^2") needs to be removed. Second line: The t at the end should be squared. Third line: Put a space between the number and the unit of the acceleration. Significance section Change "may not be easy as" to "may not be as easy as."

In the solution to Example 3.12 Acceleration of a Spaceship, revise the solution as follows: 1) Remove the extra 1/2 a t^2 in the first line of the solution 2) Square the last t on the second line of the solution 3) Add a space between the number and unit of acceleration, for consistency 4) Change “may not be easy as” to “may not be as easy as” Typo

Ch 3: Motion Along a Straight Line, Sec 4: Motion with Constant Acceleration, Exercise 44

The problem says both "initial velocity of 30 m/s" and "at t=0, x=0 and v=0," so it contradicts itself on the initial velocity. One or the other should be deleted.

Revise the question as follows: A particle moves in a straight line with an initial velocity of 0 m/s and a constant acceleration of 30 m/s^2. If t = 0 at x = 0, what is the particle’s position at t = 5 s?


Ch 3: Motion Along a Straight Line, Sec 4: Motion with Constant Acceleration, Exercise

The problem states an acceleration of 2.04m/s^2, but the solution in the answer key at the end of the book assumes an acceleration of

In exercise 53, revise "2.04 m/s^2" to "2.40 m/s^2". Typo

53 2.40m/s^2. Ch 3: Motion Along a Straight Line, Sec 4: Motion with Constant Acceleration, Exercise 57

Answer to part a has units of s^2, it is an acceleration, so it should be m/s^2

Revise the units in the solution to part a of exercise 57 to "m/s^2". Typo

Ch 3: Motion Along a Straight Line, Sec 4: Motion with Constant Acceleration, Subsec: Displacement and Position from Velocity

In the sentence right after Equation 3.11, there is a spot where v is used when it should be v-bar (the average velocity). You can't say "v is just the simple average of the initial and final velocities" because v IS the final velocity.

In the 1st sentence after Equation 3.11, revise "v" to " v̅ " as in the first part of the sentence. Typo

Ch 3: Motion Along a Straight Line, Sec 5: Free Fall, Example 3.14

There are four places in Example 3.14 (Free Fall of a Ball) where a space needs to be inserted between a number and its unit. I have highlighted them in the attached screenshot.

Ensure there is a space inserted between all numbers and their units (4 instances). Typo


In the Significance section of Example 3.15, there is a sentence that says: "We are used to thinking of the effect of gravity is to create free fall downward toward Earth." This does not make sense (grammatically) as written. One possible correction might be, "We are used to thinking that the effect of gravity..."

In the Significance section, revise the 4th sentence to: We are used to thinking that the effect of gravity... Other


In the solution of Example 3.15, at the end of part (a), the unit of the initial velocity is stated as m/sec. The convention used elsewhere throughout the text for seconds is "s", so I'd recommend switching to m/s.

Revise "m/sec" to "m/s".



There are four places in Example 3.15 (Vertical Motion of a Baseball) where a space needs to be inserted between a number and its unit. I have highlighted them in the attached screenshot.

Ensure there is a space inserted between all numbers and their units (4 instances). Typo


Example 3.16, Solution, part (a), second line (the centered equation solving for y). There should either be a negative sign in both the numerator and denominator, or in neither. Eliminating the negative sign would be cleaner. The final

In part a of the Solution, remove the negative signs from the denominators of the fractions. None

answer is already correct, but it does not agree with the current equation.

Ch 3: Motion Along a Straight Line, Sec 5: Free Fall, Exercise 67

I found the answer key in the online version to have the following errors. Please let me know if you have any questions. Ch 3: #67 part (b) is missing a negative sign in front of the velocity.

In the solution to exercise 67 part b, add a negative sign before the second value "-23.8 m/s". Typo

Ch 3: Motion Along a Straight Line, Sec 5: Free Fall, Exercise 76

I suggest specifying whether the amount of time the hiker has to move is after *hearing* the rock break loose or *seeing* it. Seeing it makes more sense and is appropriately more challenging, in my opinion.

In part a of the question, revise "see" to "hear".


Ch 3: Motion Along a Straight Line, Sec 5: Free Fall, Figure 3.26

In the caption for Figure 3.26, the unit of the moon's gravitational acceleration should be m/s^2 (the 2 just needs to be changed to superscript).

In the caption for Figure 3.26, ensure the 2 in M/s^2 is superscript. Typo

Ch 3: Motion Along a Straight Line, Sec 6: Finding Velocity and Displacement from Acceleration, Check Your Understanding 3.8

A space needs to be inserted between the function and the unit for the acceleration given in Check Your Understanding 3.8.

Add a space between "10t" and "m/s^2". Typo

Ch 3: Motion Along a Straight Line, Sec 6: Finding Velocity and Displacement from Acceleration, Example 3.17

In the question, the time-dependent acceleration has incorrect units. To correct this, the power of the seconds needs to be 3, not 2. Unit errors continue from there. For example, in part b of the solution, the term (1/8)t^2 has units of seconds squared, so cannot be subtracted from 5.0 m/s. In part d, the unit for the result appears at the end as if by magic, after omitting the units from the rest of the line. (etc...) This is sloppy, and not helpful for teaching purposes.

Revise Example 3.17 as follows: In the 1st sentence, revise "a(t) = -1/4 m/s^2" to "a(t) = -1/4 m/s^3". Revise the solution to Example 3.17 as follows: b. v(t) = 0 = 5.0 m/s - 1/8 t^2 m/s^3 => t = 6.3 s c. x(t) ... = 5.0t m/s - 1/24 t^3 m/s^3 d. ... x(6.3) = 5.0(6.3 s) - 1/24(6.3 s) = 21.1 m. None

Ch 3: Motion Along a Straight Line, Sec 6:

At the end of the Strategy section in Example 3.17, it says, "we only

In the Strategy section of Example

Other factual

Finding Velocity and Displacement from Acceleration, Example 3.17

have to evaluate the position function at t = 0." This is incorrect, as seen later in part (d) of the solution. It could instead say, "we only have to evaluate the position function at the time when the velocity is zero."

3.17 Motion of a Motorboat, revise part (d) to "Since the initial position is taken to be zero, we only have to evaluate the position function at the time when the velocity is zero."

inaccuracy in content

Ch 3: Motion Along a Straight Line, Sec 6: Finding Velocity and Displacement from Acceleration, Example 3.17

In the question, the time-dependent acceleration has incorrect units. To correct this, the power of the seconds needs to be 3, not 2. Unit errors continue from there. For example, in part b of the solution, the term (1/8)t^2 has units of seconds squared, so cannot be subtracted from 5.0 m/s. In part d, the unit for the result appears at the end as if by magic, after omitting the units from the rest of the line. (etc...)

Revise Example 3.17 as follows: In the 1st sentence, revise "a(t) = -1/4 m/s^2" to "a(t) = -1/4 m/s^3". Revise the solution to Example 3.17 as follows: b. v(t) = 0 = 5.0 m/s - 1/8 t^2 m/s^3 => t = 6.3 s c. x(t) ... = 5.0t m/s - 1/24 t^3 m/s^3 d. ... x(6.3) = 5.0(6.3 s) - 1/24(6.3 s) = 21.1 m.

Incorrect calculation or solution

Ch 3: Motion Along a Straight Line, Sec 6: Finding Velocity and Displacement from Acceleration, Subsec: Kinematic Equations from Integral Calculus

The first sentence under the heading "Kinematic Equations from Integral Calculus" is: "Let’s begin with a particle with an acceleration a(t) is a known function of time." This does not work grammatically. One possible correction could be, "Let’s begin with a particle with an acceleration a(t) which is a known function of time."

Revise the 1st sentence as follows: Let’s begin with a particle with an acceleration a(t) which is a known function of time. Other

Ch 3 review problem 96

The problem says "the Jacob", which should be corrected to "Jacob".


Ch 4: Motion in Two and Three Dimensions, Challenge Problems, Exercise 101

Solution is incorrect due to misapplication of chain rule. See attached solution.


Ch 4: Motion in Two and Three Dimensions, Sec 2: Acceleration

In part 3 the k-hat component is given as -6.3, I believe the correct answer should be +3.75 (-1.25 - (-


Vector, Exercise 29 )5) Ch 4: Motion in Two and Three Dimensions, Sec 2: Acceleration Vector, Exercise 29

In the answer key for Chapter 4, #29, the units are given as cm, they should be m.


Ch 4: Motion in Two and Three Dimensions, Sec 3: Projectile Motion, Exercise 33 Change "and initial" to "an initial".

Revise "and initial" to "an initial". Typo

Ch 4: Motion in Two and Three Dimensions, Sec 3: Projectile Motion, Exercise 35

In Exercise 35, the figure referred to is missing.


Ch 4 Sec 3: Projectile Motion

There is an extra space between the word Principia and the period. Also two lines below that sentence there is an extra spacing between the lines.



On the last line of page 183. In "(meaning to seek")" a quotation mark is missing.



Page 190, the for the line from the top says "of /her". It should be "her".



On page 205, problem 45. a. The solution is h = Voy^2/29 = 24^2/2/9.8 The solutions in the back of the book says the answer is 23.4m. The actual number should be 29.4m b. The solution is t = voy/g = 24/9.8 The solutions say the answer is 3s, but it should be 2.4s e. Also check the solution for x at t = 4s. It should be 72m (18*4); not 22.4m

Revise the solution to exercise 45 as follows: a. h = 29.4 m b. t = 2.4 s ... x = 43.2 m e. ... t = 4.0 s y = 17.6 m x = 72 m



The acceleration of gravity on Mars is given in m/s. It should be in m/s^2.



The problem needs a statement that the ball hits (or goes barely over or barely under) the crossbar at the top of the goal.

Revise the question as follows: A soccer goal is 2.44

Other factual inaccuracy in

m high. A player kicks the ball at a distance 10 m from the goal at an angle of 25°. The ball hits the crossbar at the top of the goal. What is the initial speed of the soccer ball?

content


Answer key states the range equation, and then states theta =15.0 degrees. Plugging in the numbers from the problem gives 15.9 degrees.



"Aaron Rogers" should be spelled "Aaron Rodgers".

Revise "Rogers" to "Rodgers". Typo


Parts b and c seem pointless. There's nothing fundamental about the components in some coordinate system, and components aren't vectors and don't really have directions. Asking whether the vector v ever is parallel to, antiparallel to, and perpendicular to the vector a, and if so, where, might make more sense. (Answers: No, No, and Yes, at the top.)



Ch 4: Motion in Two and Three Dimensions, Sec 5: Relative Motion in One and Two Dimensions, Figure 4.28

In Figure 4.28 the vector V_ET is pointing in the wrong direction. It needs to be reversed.

In Figure 4.28, reverse the direction of the vector V_ET to point upwards.


Ch 5: Newton's Laws of Motion, Sec 1: Forces, Figure 5.6

Panel (a) shows 4 forces, panel (b) only shows 2, that is wrong if the point is to show students how to come up with a FBD



Ch 5: Newton's Laws of Motion, Sec 6: Common Forces, Subsec: Real Forces and Inertial Frames

I think the Coriolis force discussion gets the wrong sign. I think the following statements in the book are wrong because they say the satellite feels a force to the west (should be east, the satellite curves to the east): "... if a satellite is heading due north



above Earth’s Northern Hemisphere, then to an observer on Earth, it will appear to experience a force to the west that has no physical origin. Instead, Earth is rotating toward the east and moves east under the satellite. In Earth’s frame, this looks like a westward force on the satellite ... In the example of the satellite, the reaction force would have to be an eastward force on Earth." The issue: the Coriolis force in the northern hemisphere is to the east (not west) for a satellite moving north. Thus ocean currents in the northern hemisphere are clockwise. Figure 10 shows a counterclockwise hurricane, but that is because it is a low pressure storm (high pressure storms rotate clockwise in the northern hemisphere).

Ch 5: Newton's Laws of Motion, Additional Problems, Exercise 82

Chapter 5, HW problem 82. The problem is impossible as stated. The "answer" shows FA with positive x and y components, but it's supposed to point in the third quadrant? Basically, this problem is not consistent with the diagram.

In exercise 82, revise "0.20 m/s^2" to "20 m/s^2". Revise solution accordingly.



HW solutions, Chap 5 number 83. If v = kx^2, the acceleration isn't 2kx. The units don't work. You forgot the chain rule.



Chapter 5, problem 90. Answer given is 1.13 m/s^2. I believe it should be 6.36 m/s^2. 65N cos -30 + 5kg(9.8 m/s^2)cos 240 = (5 kg)(a) a = 6.36 m/s^2

Revise the solution of exercise 90 to "6.36 m/s^2".


Ch 5 Review Question 15 (p. 255)

Chapter 5, Question 15: in "Can the force on it TO be acting to the left?" TO should be removed.


Ch 5: Newton's Laws of Motion, Challenge

Chapter 5, HW problem 94. Answer given is 139 N at angle of

Revise the solution to exercise 94 to

Incorrect

Problems, Exercise 94 112.5 degrees. I agree with direction, but believe the magnitude should be 57.4 N. delta v = 11.481 m/s at 112.5 degrees. accel = delta v divided by delta t of 2 s, so 5.74 m/s^2 at 112.5 degrees. force is ma, so 10 kg mass means a force of 57.4 N at 112.5 degrees.

"57.4 N at 22.5 degrees west of north".

calculation or solution

Ch 6: Applications of Newton's Laws, Additional Problems, Exercise 103

For the velocity, there needs to be a 1/m included in the y-direction term. For the position, the y-direction should go as 1/6m, not 1/60m



First, we should be told what direction the system is moving so we can assign a direction to the force of friction. I get an answer of 3.52 m/s^2 for a, rather than 5.4m/s^2.



Question asks for coefficient of static friction (0.75), but the answer gives the coefficient of kinetic friction (0.60).


Ch. 6: Applications of Newton's Laws, Sec 2: Friction, Example 6.10

Sloppy notation mixing vector P and scalar fk in the same equation. A scalar cannot be subtracted from a vector. P-vector should be replaced by the magnitude P without vector sign.




Back of the book gives 774N for part b, it's 0.774N.



Answer of 14 meters would give an acceleration beyond what is possible given the coefficient of static friction. See attached.


Ch. 6: Applications of Newton's Laws, Sec 1: Solving Problems with Newton’s Laws, Example 6.1

The first equation of the solution states: sum of forces_x=T_1x-T2x . The minus sign should be a plus sign since T1x itself is a negative quantity. The subsequent equations all have the same mistake. It is not correct that T1_x=T2_x; only the


absolute values of the components are equal. When the components are calculated explicitly, T1x should be - T1 cos 30 degrees.

Ch 6: Applications of Newton's Laws, Additional Problems, Exercises 75 and 119 Same question was listed twice.


Ch 6: Applications of Newton's Laws, Challenge Problems, Exercise 133

Parachutist has 10 seconds of free fall before opening parachute. So v should equal 9.8*10= 98m/s. But answer in back of book is 53.9m/s. Distance fallen should be 1/2 a t^2, or 490m, but answer in back is 328m. If F=-bv and b=0.75, then the terminal velocity should be 1070m/s, which would make this a very bad parachute. Something is clearly wrong with the question and/or its answer.


Ch 6: Applications of Newton's Laws, Sec 1: Solving Problems with Newton’s Laws, Exercise 45

Answer to part a should be 4.43 m/s^2. Note that this is consistent with the Answer Key for part b.


Ch 6: Applications of Newton's Laws, Sec 2: Friction, Exercise 53

Wx and f vectors are switched in the diagram.


Ch 6: Applications of Newton's Laws, Sec 2: Friction, Exercise 55

I found the answer key in the online version to have the following errors. Please let me know if you have any questions. Thank you. Ch 6: #55 part (a) should be 0.737m/s^2, not 1.69m/s^2,


Ch 6: Applications of Newton's Laws, Sec 3: Centripetal Force, Exercise 77

"The maximum speed at which the car traverse the curve" should be "the maximum speed at which the car CAN traverse the curve"


Ch. 6: Applications of Newton's Laws, Sec 2: Friction, Example 6.11

The question states that the skier is going down at constant velocity. The solution states that the friction is less than the component of weight along the slope so that the skier accelerates...


Ch 6: Applications of Newton's Laws, Sec 4: Drag Force and Terminal Speed,

Answer to part b is given as 90N. The setup has the value of 60N when accelerating downwards at 3.8m/s^2, but in part b we are


Exercise 39 moving upwards, but slowing down at 3.8m/s^2, which is a downwards acceleration of 3.8m/s^2, so we should get back the original result.

Ch 6: Applications of Newton's Laws, Sec 4: Drag Force and Terminal Speed, Exercise 93 Tension should be 199N, not 189N.


Ch 6: Applications of Newton's Laws, Sec 4: g Force and Terminal Speed, Exercise 81

Getting 11.8m/s rather than 25m/s for the terminal velocity of the squirrel.


Ch 6.4 Drag Force and Terminal Speed

The word "by" is missing in a sentence. Change: "Two situations for which the frictional force can be represented this equation..." into: "Two situations for which the frictional force can be represented BY this equation..."

Thank you for the feedback. We've corrected this error. Typo

Ch 7: Work and Kinetic Energy, Introduction, Figure 7.1

The introductory sentence conveys a significant misconception. "A sprinter exerts her maximum power to do as much work on herself as possible in the short time that her foot is in contact with the ground." An object cannot do work on itself. While a sprinter is not a simple object, and parts of her can do work on other parts, the sentence as a whole comprises a misconception. A relatively small but significant change could correct this. Please review the remainder of the chapter to ensure this issue does not pervade.



Ch 7: Work and Kinetic Energy, Sec 1: Work

When describing how friction forces relate to work the book states: The work done against a force may also be viewed as the work required to overcome this force, ... Some of what is said is wrong and confusing. "Work required to overcome a force" is a confusing phrase. When one lifts a box, gravity is doing negative work on



the box, but I wouldn't say the work done by gravity is overcoming the force I am applying to the box. Also, in general, a static friction force can't do negative work on an object and kinetic friction force can't do positive work on an object. (In the example given, when you are slowing down while walking and place your foot on the ground, your foot was moving forward as it touched the ground. Thus kinetic friction slows your foot down and brings it to a stop.)

Ch 7: Work and Kinetic Energy, Sec 1: Work

In Volume 1, Chapter 7 (pg. 336), the Openstax textbook states “In physics, work represents a type of energy.” Work is not a type of energy; work is the transfer of energy by an applied force over some displacement. Objects and systems have energy, but they do not have work. Objects and systems do work by applying a force over some distance on another object or system.

Revise the 1st two sentences of section Work as follows: "In physics, work is done on an object when energy is transferred to the object. In other words, work is done when a force acts on something that undergoes a displacement from one position to another."


Ch 7: Work and Kinetic Energy, Sec 1: Work, Figure 7.3

It is conceptually and factually wrong to suggest that the muscle fibers are doing work when contracted to hold up an object. It's much more complicated than that, and we shouldn't mislead students into thinking that there are little muscle fibers moving around and doing work. It's a bunch of (bio) chemical reactions that keep the fibres contracted. There is no "Force" and no "displacement", so I think it's misleading to imply that work is being performed inside a contracted muscle.



Ch 7: Work and Kinetic Energy, Sec 3: Work-Energy Theorem, Example 7.9

In addition to randomly using potential energy, which has not been introduced yet, the work done by gravity has the incorrect sign. The first equation in the solution


should be (y_1-y_2), otherwise, you have a positive quantity equal a negative quantity. The change in potential energy is the negative of the work done. The work done is the change in the kinetic energy.... Section 7.3, example 7.9, change the first equation in the solution from (y_2 - y_1) to (y_1 - y_2)

Ch 7: Work and Kinetic Energy, Sec 3: Work-Energy Theorem, Example 7.9

The work done by gravity is calculated as mgh, but gravitational potential energy, and conservative forces, have not been introduced yet. With the tools given so far, this is pulled out of thin air, and students have complained that they don't know where the equations come from. You should make the ramp linear, and then actually calculate the work done by gravity on the ramp and along the half-circle... Section 7.3, example 7.9, change the example problem so that it doesn't include gravitational potential energy and conservative forces which have not been taught yet



Ch 7.3 Work-Energy Theorem: https://cnx.org/contents/[email protected]:KPQmVMyj@7/73-Work-Energy-Theorem

Second paragraph, starting with "Let’s start by looking at..." has a double comma (,,) after equation F_net = m (dv/dt). Remove one comma.


Ch 8 , Figure 8.3

In the University Physics 1 text book on page 363, Figure 8.3 is captioned with what appears to be a suicide joke. http://www.sprc.org/sites/default/files/migrate/library/SuicideAmongCollegeStudentsInUS.pdf Maybe rethink the caption?

The figure and caption have been removed. Other

Ch 8, Section 8.1, Figure 8.3, p. 363

This is not an error exactly, but I thought the caption for Figure 8.3 was insensitive. I realize it is supposed to be a physics joke based on potential (energy), but this joke also relies on messages

The figure and caption have been removed.

General/pedagogical suggestion or questio

people use to tell people not to commit suicide ("Don't jump--you have so much potential.") The joke came off as making light of a very serious issue and seemed inappropriate in a university textbook. At worst, I think the caption could even be triggering for those who may struggle with suicidal ideation or have made suicide attempts. If possible, I would like to suggest changing the caption to something less insensitive and potentially harmful.

n

Ch 8: Potential Energy and Conservation of Energy, Additional Problems, Exercise 69

Work done by friction should be negative.


Ch 8: Potential Energy and Conservation of Energy, Additional Problems, Exercise 85

No mass given for the hockey puck, can't give numerical answers without that info.


Ch 8: Potential Energy and Conservation of Energy, Sec 2: Conservative and Non-Conservative Forces

In the shaded box about Conservative Force at the beginning of this section, the last equation has an E where it should have an F.


Ch 8: Potential Energy and Conservation of Energy, Sec 3: Conservation of Energy, Exercise 39

The work done by friction is given as positive, it should be negative.



Change "mupwards" to "m upwards" (where m stands for meters).



Work seems to be calculated incorrectly. In any case, it should be negative rather than positive.


Ch 8: Potential Energy and Conservation of Energy, Sec 4: Potential Energy Diagrams and Stability,

In the back of the book answers, the derivative for Ae^(-ax^2) is wrong, it dropped a 2 from the chain rule. Also, the sketch for the solution of part a is given in the


Exercise 51 problems section rather than in the answers section.

Ch 8: Potential Energy and Conservation of Energy, Sec 5: Sources of Energy, Exercise 54

"DeLorean" is misspelled the second of the three times it appears.



The following link needs a new target: (https://openstaxcollege.org/l/21topfailvideo)


Broken link


The answers given make more sense if the spring constant is 100N/m rather than the given 100N/cm.



In Chapter 8 problem 68 at the end of the chapter, the angle theta needs to be specified as 30 degrees to obtain the solution given in the instructor's guide. Otherwise, the final answer should be left in terms of theta.


Ch 8 Review

Found in webview only, not in the PDF: The Chapter Review section for Chapter 8 has an empty section header, "Challenge Problems". In other words, the header is there, with no exercises underneath it. This may be a recipe-based issue since University Physics vol. 1 is a baked book.

The collation instruction for Challenge Problems, of which there are none in the chapter, has been removed from the introduction module. Other

Ch 9: Linear Momentum and Collisions, Additional Problems, Exercise 111

Distance to the moon is given as 3.84E5 m. It's 3.84E5 km.


Ch 9: Linear Momentum and Collisions, Sec 1: Linear Momentum, Exercise 21

Students are asked to calculate "magnitude of its average linear momentum" for planet moving in a circular orbit. The solution given is the mass multiplied by a velocity. If we are truly looking at an average momentum shouldn't we get zero? The velocity vector at any part of the orbit should be cancelled by the velocity at a point 180


degrees away. The language on this needs to be tightened up.

Ch 9: Linear Momentum and Collisions, Sec 3: Conservation of Linear Momentum, Exercise 39

This is a problem that has raised issues before. In this problem a truck is moving at a certain speed, drops a load of gravel, and we are asked about its final velocity. The stated answer to the problem assumes that the gravel somehow transfers its momentum to the truck. I think this is a poor assumption. Any single piece of gravel should obey the First Law when dropped from a moving truck, and maintain its horizontal velocity. It will keep its momentum. If you believe that the gravel when it acts in a group somehow transfers momentum to the truck, then it would have to push the truck forwards. But then wouldn't the gravel on the edge near the truck accelerate towards the truck? This doesn't seem physical. I think the best answer is that the speed of the truck remains unchanged as it drops the gravel. It will lose mass, sure, but as it does it loses the corresponding momentum associated with the gravel, and hence no change in speed.


Ch 9: Linear Momentum and Collisions, Sec 4: Types of Collisions, Exercise 51

If we are to believe that the total mass (kid +sled) starts at 35kg and then an additional 35kg is added on, then the speed should be cut in half (1.75m/s). Since it isn't in the answer key, then the sled must have a mass independent of the first 35kg child, in which case the mass of the sled should be given in the statement of the problem.



The problem asks "What is the speed of the sled and the boulder after the collision?" "What is the speed" (singular) suggests there's only one speed, so the sled and boulder stick together. However, the solution manual gives two speeds, which seem consistent



with a perfectly elastic collision. If the sled and boulder collide perfectly elastically, the problem needs to say so. Also, the problem doesn't give the boy's speed as he flies over the boulder. I suspect it's supposed to still be 10 m/s, but in that case "he is *propelled* over the boulder" seems misleading. He flies over the boulder at his full original speed. The problem needs to be reworded to give all the needed information and the answer needs to match the corrected problem.


The inaccuracies are ornithological. The falcon in the illustration is a Prairie Falcon and the dove is a Common Wood Pigeon. First, the masses are given as 1.8 kg and 0.65 kg, whereas the ranges of mass are 0.500-0.975 kg for the falcon and 0.300-0.615 kg for the pigeon. Second, the Prairie Falcon lives in North America, while the Common Wood Pigeon lives in Europe and western Asia. Finally, falcons are technically not hawks. Deleting the illustration would solve both problems with it, but it looks good and the artist's work would be wasted. You could also change the masses. Dividing both by 2 would leave the solution the same. Maybe readers who recognized the birds could think of the falcon as belonging to an Old World falconer. In any case, "hawk" should be changed to "falcon" (and "dove" could be changed to "pigeon"). This has lower priority than a physics error, but physicists might imagine whether they'd want to correct minor physics errors in a biology textbook.



Ch 9: Linear Momentum and Collisions, Sec 4:

P in the y direction should be -79.6 kg m/s, not -80.6 kg m/s.


Types of Collisions, Exercise 57


The masses of the cars need to be given, or there needs to be some statement about them such as m_A = m_B.

Revise the 1st sentence of the question as follows: Two cars of the same mass approach an extremely icy four-way perpendicular intersection.


Ch 9: Linear Momentum and Collisions, Sec 6: Center of Mass, Example 9.16

We are given the mass of the earth as 5.97*10^24 but in the equation it is writen as 5.98*10^24. According to NASA the 5.97*10^24 is correct while the 5.98 is not.




We are given r of the moon= 3.82*10^5 m while in the equation we plug it into uses it as 3.82*10^8 m. Either change the original to say km or change the original to 10^8 to match the one used in the equation.




I would like to point out (as a solid-state physicist) that while the computation of the center of mass of the given unit cell is done correctly, it has no physical meaning since the unit cell may be chosen centered around any point whatsoever in the system. There is no meaningful concept of a center of mass of a unit cell (this is related to the lack of meaning of an electrical dipole moment per unit cell, well known in the theory of ferroelectrics), and indeed the symmetry argument deployed to bolster the idea of the center of mass being in the center of the unit cell equally well implies the center of mass is at the Na ion, or the Cl ion, or the midpoint between an Na and Cl ion, for example. As a result, I think this example is misleading and unfortunately does not constitute a useful application of the concept of center of mass. I recommend removing it.

In Example 9.17, replace the Significance section with the following: "Although this is a great exercise to determine the center of mass given a Chloride ion at the origin, in fact the origin could be chosen at any location. Therefore, there is no meaningful application of the center of mass of a unit cell beyond as an exercise."


Ch 9, problem 73

"Find the center of mass of cone" should be "Find the center of mass of *a* cone". resolved Typo

Ch 10: Fixed-Axis Rotation, Additional Problems, Exercise 115

Potential energy calculated for rod+ball seems incorrect. See attached.


Ch 10: Fixed-Axis Rotation, Sec 1: Rotational Variables

There are errors which appear to be the result of defining $\theta$ and $\mathbf s$ as vectors and an incorrect statement which appears to be the result of confusing the magnitude of a vector being constant with the vector being constant. I have attached a pdf explaining in detail. Redefine $\theta$ and $\mathbf s$ , and correct the statement about the magnitude of a vector



Ch 10: Fixed-Axis Rotation, Sec 2: Rotation with Constant Angular Acceleration, Exercise 39

Book answer is roughly half of the answer you get from kinematics. See attached.

Revise the solution to part b to "12,600 rad". None

Ch 10: Fixed-Axis Rotation, Sec 3: Relating Angular and Translational Quantities, Exercise 51

Details in answer key note a 40kg child, the problem as stated has a 30 kg child. Mass drops out of the question, so the answer will remain unchanged (child doesn't fall off), but the numbers should be adjusted.


Ch 10: Fixed-Axis Rotation, Sec 3: Relating Angular and Translational Quantities, Exercise 52

Chapter 10, Probl. 52: "0.3m" should be "0.3 m".


Ch 10: Fixed-Axis Rotation, Sec 6: Torque, Exercise 77

Answer in back of book has a typo for part c. 2*1.8*sin(70)=3.38, not 3.28. This means that the total torque is also off by 0.1kgm^2 in the answer key.


Ch 10: Fixed-Axis Rotation, Sec 6: Torque, Figure 10.31

The caption for the first figure in Section 10.6 calls the force in part (b) of the figure F', but in the image the force vector is labeled as F. I recommend changing the label in the image to F'.


Ch 10: Fixed-Axis The mass of the merry-go-round is In Example 10.16, Typo

Rotation, Sec 7: Newton’s Second Law for Rotation, Example 10.16

first given as 200 kg in the problem description, but all work and calculations use a mass of 50 kg.

revise the mass given for the merry-go-round from "200.0 kg" to "50.0 kg".

Ch 10: Fixed-Axis Rotation, Sec 7: Newton’s Second Law for Rotation, Exercise 88

Part b) asks "Through what angle does the wheel move through...?" One "through" should be deleted.


Ch 11: Angular Momentum, Additional Problems, Exercise 77

Answers to part b and c appear to be calculated based upon L rather than delta-L.


Ch 11: Angular Momentum, Additional Problems, Exercise 89

Answer given in the answer key assumes that the moments of inertia for each cylinder are identical, whereas we are only given that the masses are equal. Wording should be clarified.


Ch 11: Angular Momentum, Sec 1: Rolling Motion, Exercise 33

The first line in the answer key appeals to conservation of energy when the hollow cylinder is going up the ramp. But given the numbers in the problem, energy is apparently not conserved. There seems to be some difficulty with the word "height", it is possibly used as distance about ground and/or distance along the ramp in a confusing way.


Ch 11: Angular Momentum, Sec 2: Angular Momentum

In Volume 1, chapter 11 (pg. 564), the textbook states “….just as the total linear momentum in the universe is conserved, so is the total rotational motion conserved. We call the total rotational angular momentum, …” These statements are confusing. Rotational motion is not the same as angular momentum. Angular momentum does not only describe objects that are rotating. All objects that move have angular momentum. What is useful about conserved quantities is that they are conserved for a closed system, not that they are conserved in the scale of the universe.

Revise the 1st two sentences of the 2nd paragraph of section Angular Momentum as follows: "The answer is in a new conserved quantity, since all of these scenarios are in closed systems. This new quantity, angular momentum, is analogous to linear momentum."


Ch 11: Angular The introduction states: Our reviewers General

Momentum, Sec 2: Angular Momentum

Why does Earth keep on spinning?... The answer to these questions is that just as the total linear motion (momentum) in the universe is conserved, so is the total rotational motion conserved. We call the total rotational motion angular momentum, the rotational counterpart to linear momentum. For some reason, undefined superfluous concepts have been introduced: total linear motion, total angular motion. These are totally ambiguous concepts - for example the total linear motion could just as easily be referring to the total displacement (whatever that would mean). Oddly the version of this in the College Physics book is okay. It says: Why does Earth keep on spinning? What started it spinning to begin with? And how does an ice skater manage to spin faster and faster simply by pulling her arms in? Why does she not have to exert a torque to spin faster? Questions like these have answers based in angular momentum, the rotational analog to linear momentum.


/pedagogical suggestion or question

Ch 11: Angular Momentum, Sec 2: Angular Momentum

The result presented in this section is specific to rigid bodies that are ROTATIONALLY SYMMETRIC *and* rotating about the symmetry axis. In general the angular momentum vector does NOT point along the angular velocity axis. Since there is no mention of rotational symmetry in the section, it is just horribly wrong. Suggestion 1. Change the heading to "Angular momentum for cylindrically symmetric objects" and give the correct logic in the proof: the sentence ending "on the

Add clarification that the rigid body discussed is cylindrically symmetric.


opposite side of the rigid body" should become "on the opposite side of the body, because it is cylindrically symmetric." Suggestion 2, to improve the pedagogy: Prior to this section, it should be stated that the angular momentum of a rotating body can point in a variety of directions that may or may not be parallel to the angular velocity. Then state that the text will focus on cylindrically symmetric objects, and point out that these are a large category, including wheels, planets, galaxies etc.

Ch 11: Angular Momentum, Sec 2: Angular Momentum, Exercise 41

The answer given for part b is problematic. We are told that L depends only upon the perpendicular distance to the ground, no matter where the plane is. However, if the student models an airplane flight at constant altitude as travel along an arc rather than a constant distance over an infinite plane (geometric plane, not airplane), then the angular momentum will change. When I fly from San Francisco to New York, it is not in a straight line.


Ch 11: Angular Momentum, Sec 4: Precession of a Gyroscope

In this book, as in many other introductory physics texts, it is stated that "But when riding the bicycle at a good pace, it is harder to tip it over because we must change the angular momentum vector of the spinning wheels." That this claim is false has been known since 1970, when David Jones reported on a bicycle he built that had a second wheel on its front fork, clear of the ground, that spun backwards and so canceled out the angular momentum of the front wheel. It was not harder to ride, at any speed, than an ordinary bicycle. [Physics Today vol 23, pp 3440] The problem of bicycle stability is much more complicated, see for example


http://www.nature.com/news/the-bicycle-problem-that-nearly-broke-mathematics-1.20281

Ch 12, Section 12.3, Problem 49

The unit N/m^2 is in italics. It should be in roman.

Thank you for the feedback. We've corrected this error. Typo

Ch 13: Gravitation, Additional Problems, Exercise 63

The calculated acceleration seems to be off. See attached file.



Answer in the back of the book seems to be off. See attached solution.



Problem #73 relies upon information in problem #72, which is poorly stated. We are told that the diameter ranges from 578 to 458km, and then asked to assume a sphere of RADIUS 520km, which would in fact double the size of the object in question. In #73 we are asked about an orbit 10km from the surface, but the answer assumes a radius of 260km rather than 270km.


Ch 14: Fluid Mechanics, Sec 4: Archimedes’ Principle and Buoyancy

The following link needs a new target: (https://openstaxcollege.org/l/21archNASA)


Broken link

Ch 14: Fluid Mechanics, Sec 4: Archimedes’ Principle and Buoyancy, Exercise 71

I found the answer key in the online version to have the following errors. Please let me know if you have any questions. Ch 14: #71 part (b) should be 0.72, not 0.68.


Ch 14: Fluid Mechanics, Sec 7: Viscosity and Turbulence, Exercise 38

The figure referenced in Exercise 38 is missing.


Ch 15: Oscillations, Sec 6: Forced Oscillations, Equation 15.29

This is ok in the online view but the PDF (print) still contains the error in denominator of equation 15.29. In the PDF (print) version the denominator is still inaccurate with the mass not being squared within the radical sign. As printed, the units cannot work our correctly. The error exists in the Odigia online resources as well.

Thank you for the feedback! We update our PDF versions twice every year and this change will be reflected in our newest revision. Typo

Ch 16: Waves, Sec 2: The first equation of the paragraph This typo has been Typo

Mathematics of Waves titled "The Linear Wave Equation" - (x.t) should be (x,t) Thanks, Kane

resolved in the webview format.

Ch 16: Waves, Sec 4: Energy and Power of a Wave

I quote from the text: "In the case of the two-dimensional circular wave, the wave moves out, increasing the circumference of the wave as the radius of the circle increases. If you toss a pebble in a pond, the surface ripple moves out as a circular wave. As the ripple moves away from the source, the amplitude decreases. The energy of the wave spreads around a larger circumference and the amplitude decreases proportional to 1/r , and not 1/r^2, as in the case of a spherical wave." This is incorrect. the amplitude of a spherical wave decreases as 1/r not 1/r^2 (see Giancoli "Physics for Scientists and Engineers" 4th Edition section 15-3.



Ch 16: Waves, Sec 5: Interference of Waves, Exercise 91

I found the answer key in the online version to have the following errors. Please let me know if you have any questions. Ch 16: #91 should be 0.63, not 0.76.


Ch 17: Sound, Sec 2: Speed of Sound, Subsec: Speed of Sound in Various Media

Under "Speed of Sound in Various Media," the equation for the speed of sound in a fluid uses beta β for the bulk modulus. I believe this should be a capital B, as in Chapters 12 and 16.


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