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Errata for

An Introduction to Mechanics by Kleppner and Kolenkow compiled by Andrew Goetz

I make no claims to completeness or correctness. If you have any comments or corrections please email me. • Page 13, line 11: Change the first ˆj to ˆi. • Page 78, line -14: Change “Eq. (2)” to “Eq. (3)” (twice) and change “Eq. (3)” to “Eq. (4).” • Page 83, line 7: Change the period after “down” to a comma. • Page 102, line 9: In the denominator of the integrand, change R2 to R2 . • Page 105, Problem 2.17: In part (b), we should assume that tan θ > µ, not tan θ < µ. In part (c), we should assume that tan θ < 1/µ, otherwise there is no maximum acceleration. Going along with this, in the solution for θ = π/4, we need to assume that µ < 1. • Page 106, Problem 2.20: In the answer, delete the coefficient 2 from the term M2 M3 in the denominator. • Page 107, Problem 2.30: In the diagram, change ra to rA . • Page 126, line 5: Change R0 to R. • Page 126, line 10: Change re and rm to re and rm . • Page 126, line -12: Change re , rm , and Rem to re , rm , and Rem . • Page 126, last displayed equation: Change R to Rem (twice.) • Page 129, line 1: Change “length of the spring” to “displacement of the spring.” • Page 137, line 5: Delete one of the two periods after “rocket.” • Page 137, line -7: The expression on this line is missing the term ∆m∆v (which goes to zero in the limit as ∆t → 0, so maybe K&K purposely left it out.) The full expression should be M ∆v + (∆m)u + ∆m∆v. • Page 149, Problem 3.17: The answer clue is completely screwed up. First of all, K&K have previously defined “weight” as the true gravitational force (see pages 84-85), so it doesn’t make sense for the units of W to be kilograms. The units of W should be newtons. This suggests changing “W = 10 kg” to “W = 10 N.” Even with this change, I believe the answer clue is still wrong. With the values given the garbage can could be supported at ground level but no higher, i.e. the solution should be hmax = 0. Also, with the given value of v0 , there are no values of W and dm/dt that give the solution hmax ≈ 17 m. So something needs to be changed. If I have done the problem correctly, I suggest the following clue: If v0 = 20 m/s, W = 5 N, dm/dt = 0.5 kg/s, then hmax = 15.3 m.

1

• Chapter 4: K&K are inconsistent in the notation they use in the limits of the path integral defining Rb work. Their original limits are ra and rb , but many times they will just write a and b; a F · dr instead Rr of rab F · dr. To change all instances of the former notation to the latter, change the limits in the following places: – Page 160, lines 8–12 – Page 160, line -3 – Page 163, line 14 – Page 166, lines 6–8 – Page 167, line -8 – Page 168, line 6 – Page 169, line -1 – Page 173, line 4 – Page 182, lines -7 and -6 Also on page 182, line -8, change “as the particle moves from a to b” to “as the particle moves from ra to rb .” • Page 157, line 6: Change re to Re . • Page 175, Example 4.14: In the second diagram, the label for the angle on the left is missing an α. It reads + θ instead of α + θ. • Page 192, line 11: Change “sketch a” to “sketch b.” • Page 193, line 3: Change v to v1 . • Page 200, Problem 4.28: In the reaction equation, I think the −2.8 MeV should be +2.8 MeV, since it is energy released by the reaction. This would accord with the reaction equation in Problem 4.9 in which +5 eV are released. • Page 200, Problem 4.29: Change v to v0 in the first line of part (b) and in the diagram. • Page 205, line 12: Change the first dy to dx. • Page 219, line -4: There is a fraction line missing in the first term of the expression on the right. 1

• Page 228, Problem 5.2: The answer is incorrect. It should be vf = (v02 + 2A/m + 3B/5m) 2 . • Page 228, Problem 5.3b: Change the equation for the force from F = Aθˆr to F = Aθˆr. • Page 252, graphic: This isn’t really an error, but the axes shown in this graphic are left-handed, which is very unusual for an American textbook. • Page 252, line 10: Change ˆ k to ˆj. • Page 261, line 6: The second R is missing a dot over it. • Page 264, lines 7–8: I don’t understand the middle two lines of this calculation at all. In particular I think the appearance of the vectors ρj so soon is incorrect. I believe the correct calculation would run as follows: P K = 21 mj vj2 P = 12 mj |˙r0j + V|2 P P P = 21 mj |˙r0j |2 + mj r˙ 0j · V + 21 mj |V|2 = 12 I0 ω 2 + 12 M V 2 .

2

• Page 266, line 10: Change τs to τz . • Page 268, line -8: For consistency, the torque being integrated should be τ0 . • Page 269, line -14: The letter n in “length” seems to be in bold for no reason whatsoever. • Page 276, line -3: The term for the kinetic energy, 21 l2 φ˙ 2 , should have an m in it:

1 2 ˙2 2 ml φ .

• Page 382, line 13: Add a factor of µ into the second term of the second equation, so that it reads ˙ 0 = µr2 θ¨ + 2µrr˙ θ. • Page 383, line -11: The reference should be to Eqs. (9.17a and b), not Eqs. (9.16a and b). • Page 392, equation 9.22: The first term should be (1 − 2 )x2 , not (1 − 2 )x. • Page 394, line 3: The displayed equation should be cos θa =

1 .

• Page 396, line -11: The second equation on this line is missing a negative sign. The line should be A=

C (−E)

or

E=−

C . A

• Page 403, line -4: Equation (2) is missing a factor of in its second term. The equation should be (1 − 2 )x2 − 2r0 x + y 2 = r02 . • Page 425, line 6: The reference should be to Eq. (10.24), not Eq. (10.23). • Page 426, graphic: For consistency of notation the label on the vertical axis should be hE(ω)i and not E(ω). • Page 427, line 8: “γ is given in units of δ −1 ” doesn’t make any sense to me. I believe the units should be s−1 . • Page 436, line 3: Change to

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An Introduction to Mechanics by Kleppner and Kolenkow compiled by Andrew Goetz

I make no claims to completeness or correctness. If you have any comments or corrections please email me. • Page 13, line 11: Change the first ˆj to ˆi. • Page 78, line -14: Change “Eq. (2)” to “Eq. (3)” (twice) and change “Eq. (3)” to “Eq. (4).” • Page 83, line 7: Change the period after “down” to a comma. • Page 102, line 9: In the denominator of the integrand, change R2 to R2 . • Page 105, Problem 2.17: In part (b), we should assume that tan θ > µ, not tan θ < µ. In part (c), we should assume that tan θ < 1/µ, otherwise there is no maximum acceleration. Going along with this, in the solution for θ = π/4, we need to assume that µ < 1. • Page 106, Problem 2.20: In the answer, delete the coefficient 2 from the term M2 M3 in the denominator. • Page 107, Problem 2.30: In the diagram, change ra to rA . • Page 126, line 5: Change R0 to R. • Page 126, line 10: Change re and rm to re and rm . • Page 126, line -12: Change re , rm , and Rem to re , rm , and Rem . • Page 126, last displayed equation: Change R to Rem (twice.) • Page 129, line 1: Change “length of the spring” to “displacement of the spring.” • Page 137, line 5: Delete one of the two periods after “rocket.” • Page 137, line -7: The expression on this line is missing the term ∆m∆v (which goes to zero in the limit as ∆t → 0, so maybe K&K purposely left it out.) The full expression should be M ∆v + (∆m)u + ∆m∆v. • Page 149, Problem 3.17: The answer clue is completely screwed up. First of all, K&K have previously defined “weight” as the true gravitational force (see pages 84-85), so it doesn’t make sense for the units of W to be kilograms. The units of W should be newtons. This suggests changing “W = 10 kg” to “W = 10 N.” Even with this change, I believe the answer clue is still wrong. With the values given the garbage can could be supported at ground level but no higher, i.e. the solution should be hmax = 0. Also, with the given value of v0 , there are no values of W and dm/dt that give the solution hmax ≈ 17 m. So something needs to be changed. If I have done the problem correctly, I suggest the following clue: If v0 = 20 m/s, W = 5 N, dm/dt = 0.5 kg/s, then hmax = 15.3 m.

1

• Chapter 4: K&K are inconsistent in the notation they use in the limits of the path integral defining Rb work. Their original limits are ra and rb , but many times they will just write a and b; a F · dr instead Rr of rab F · dr. To change all instances of the former notation to the latter, change the limits in the following places: – Page 160, lines 8–12 – Page 160, line -3 – Page 163, line 14 – Page 166, lines 6–8 – Page 167, line -8 – Page 168, line 6 – Page 169, line -1 – Page 173, line 4 – Page 182, lines -7 and -6 Also on page 182, line -8, change “as the particle moves from a to b” to “as the particle moves from ra to rb .” • Page 157, line 6: Change re to Re . • Page 175, Example 4.14: In the second diagram, the label for the angle on the left is missing an α. It reads + θ instead of α + θ. • Page 192, line 11: Change “sketch a” to “sketch b.” • Page 193, line 3: Change v to v1 . • Page 200, Problem 4.28: In the reaction equation, I think the −2.8 MeV should be +2.8 MeV, since it is energy released by the reaction. This would accord with the reaction equation in Problem 4.9 in which +5 eV are released. • Page 200, Problem 4.29: Change v to v0 in the first line of part (b) and in the diagram. • Page 205, line 12: Change the first dy to dx. • Page 219, line -4: There is a fraction line missing in the first term of the expression on the right. 1

• Page 228, Problem 5.2: The answer is incorrect. It should be vf = (v02 + 2A/m + 3B/5m) 2 . • Page 228, Problem 5.3b: Change the equation for the force from F = Aθˆr to F = Aθˆr. • Page 252, graphic: This isn’t really an error, but the axes shown in this graphic are left-handed, which is very unusual for an American textbook. • Page 252, line 10: Change ˆ k to ˆj. • Page 261, line 6: The second R is missing a dot over it. • Page 264, lines 7–8: I don’t understand the middle two lines of this calculation at all. In particular I think the appearance of the vectors ρj so soon is incorrect. I believe the correct calculation would run as follows: P K = 21 mj vj2 P = 12 mj |˙r0j + V|2 P P P = 21 mj |˙r0j |2 + mj r˙ 0j · V + 21 mj |V|2 = 12 I0 ω 2 + 12 M V 2 .

2

• Page 266, line 10: Change τs to τz . • Page 268, line -8: For consistency, the torque being integrated should be τ0 . • Page 269, line -14: The letter n in “length” seems to be in bold for no reason whatsoever. • Page 276, line -3: The term for the kinetic energy, 21 l2 φ˙ 2 , should have an m in it:

1 2 ˙2 2 ml φ .

• Page 382, line 13: Add a factor of µ into the second term of the second equation, so that it reads ˙ 0 = µr2 θ¨ + 2µrr˙ θ. • Page 383, line -11: The reference should be to Eqs. (9.17a and b), not Eqs. (9.16a and b). • Page 392, equation 9.22: The first term should be (1 − 2 )x2 , not (1 − 2 )x. • Page 394, line 3: The displayed equation should be cos θa =

1 .

• Page 396, line -11: The second equation on this line is missing a negative sign. The line should be A=

C (−E)

or

E=−

C . A

• Page 403, line -4: Equation (2) is missing a factor of in its second term. The equation should be (1 − 2 )x2 − 2r0 x + y 2 = r02 . • Page 425, line 6: The reference should be to Eq. (10.24), not Eq. (10.23). • Page 426, graphic: For consistency of notation the label on the vertical axis should be hE(ω)i and not E(ω). • Page 427, line 8: “γ is given in units of δ −1 ” doesn’t make any sense to me. I believe the units should be s−1 . • Page 436, line 3: Change to

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