,Instructor Manual for Classical Mechanics – John R. Taylor |
All Chapters Complete | Verified Questions & Answers | A+
Grade
2
Contents
1 The algebra and calculus of vectors 3
2 Velocity, acceleration and scalar angular velocity 27
3 Newton’s laws of motion and the law of gravitation 62
4 Problems in particle dynamics 76
5 Linear oscillations and normal modes 139
6 Energy conservation 179
7 Orbits in a central field 221
8 Non-linear oscillations and phase space 276
9 The energy principle 306
10 The linear momentum principle 335
11 The angular momentum principle 381
12 Lagrange’s equations and conservation principles 429
13 The calculus of variations and Hamilton’s principle 473
14 Hamilton’s equations and phase space 505
15 The general theory of small oscillations 533
16 Vector angular velocity 577
17 Rotating reference frames 590
18 Tensor algebra and the inertia tensor 615
19 Problems in rigid body dynamics 646
Chapter One
, The algebra and calculus
of vectors
Problem 1 . 1
In terms of the standard basis set {i , j, k}, a = 2 i — j — 2k, b = 3 i — 4 k and
c = i — 5 j + 3 k.
(i)Find 3 a + 2 b —4 c and | a —b |2.
(ii)Find |a |, |b | and a · b. Deduce the angle between a and b.
(iii)Find the component of c in the direction of a and in the direction of b.
(iv) Find a ×b, b ×c and (a ×b) ×(b ×c).
(v) Find a · (b×c) and (a×b) · c and verify that they are equal. Is the set a,
{ b, c }
right- or left-handed?
(vi) By evaluating each side, verify the identity a×(b×c) = (a · c) b — (a · b) c.
Solution
(i)
3 a + 2 b — 4 c = 3(2 i — j — 2k) + 2(3 i — 4 k) — 4(i — 5 j + 3 k)
= 8 i + 17 j — 26 k.
| a — b | 2 = (a — b) · (a — b)
= (—i — j + 2 k) · (—i — j + 2 k)
= (—1)2 + (—1)2 + 22 = 6.
(ii)
|a|2 = a · a
= (2 i — j — 2k) · (2 i — j — 2k)
= 2 2+ (—1) 2 + (—2)2 = 9.
Hence |a| = 3.
|b|2 = b · b
= (3 i — 4 k) · (3 i — 4 k)
= 3 2+ (—4)2 = 25.
Hence |b| = 5.
a · b = (2 i — j — 2k) · (3 i — 4 k)
= 2 × 3 + (—1) × 0 + (—2) × (—4)
⃝c Cambridge University Press, 2006
, = 14.
Chapter 1 The algebra and calculus of vectors 4
The angle a between a and b is then given by
a·b
cos a =
|a||b|
14 14
= = .
3×5 15
Thus a = tan—1 15
14 .
(iii) The component of c in the direction of a is
a
a=c·
c ·b
|a|
2 i — j — 2k
= (i — 5 j + 3 k) ·
|2 i — j — 2k|
1 × 2 + (—5) × (—1) + 3 × (—2)
=
3
1
= .
3
The component of c in the direction of b is
b
c ·b
b=c·
|b|
3i — 4k
= (i — 5 j + 3 k) ·
|3 i — 4 k|
1 × 3 + (—5) × 0 + 3 × (—4)
=
5
9
=— .
5
(iv)
a×b = (2 i — j — 2k)×(3 i — 4 k)
i j k
= 2 —1 —2
ˇ 3 0 —4 ˇ
= 4 — 0 i — (—8) — (—6) j + 0 — (—3) k
= 4 i + 2 j + 3 k.
⃝c Cambridge University Press, 2006
All Chapters Complete | Verified Questions & Answers | A+
Grade
2
Contents
1 The algebra and calculus of vectors 3
2 Velocity, acceleration and scalar angular velocity 27
3 Newton’s laws of motion and the law of gravitation 62
4 Problems in particle dynamics 76
5 Linear oscillations and normal modes 139
6 Energy conservation 179
7 Orbits in a central field 221
8 Non-linear oscillations and phase space 276
9 The energy principle 306
10 The linear momentum principle 335
11 The angular momentum principle 381
12 Lagrange’s equations and conservation principles 429
13 The calculus of variations and Hamilton’s principle 473
14 Hamilton’s equations and phase space 505
15 The general theory of small oscillations 533
16 Vector angular velocity 577
17 Rotating reference frames 590
18 Tensor algebra and the inertia tensor 615
19 Problems in rigid body dynamics 646
Chapter One
, The algebra and calculus
of vectors
Problem 1 . 1
In terms of the standard basis set {i , j, k}, a = 2 i — j — 2k, b = 3 i — 4 k and
c = i — 5 j + 3 k.
(i)Find 3 a + 2 b —4 c and | a —b |2.
(ii)Find |a |, |b | and a · b. Deduce the angle between a and b.
(iii)Find the component of c in the direction of a and in the direction of b.
(iv) Find a ×b, b ×c and (a ×b) ×(b ×c).
(v) Find a · (b×c) and (a×b) · c and verify that they are equal. Is the set a,
{ b, c }
right- or left-handed?
(vi) By evaluating each side, verify the identity a×(b×c) = (a · c) b — (a · b) c.
Solution
(i)
3 a + 2 b — 4 c = 3(2 i — j — 2k) + 2(3 i — 4 k) — 4(i — 5 j + 3 k)
= 8 i + 17 j — 26 k.
| a — b | 2 = (a — b) · (a — b)
= (—i — j + 2 k) · (—i — j + 2 k)
= (—1)2 + (—1)2 + 22 = 6.
(ii)
|a|2 = a · a
= (2 i — j — 2k) · (2 i — j — 2k)
= 2 2+ (—1) 2 + (—2)2 = 9.
Hence |a| = 3.
|b|2 = b · b
= (3 i — 4 k) · (3 i — 4 k)
= 3 2+ (—4)2 = 25.
Hence |b| = 5.
a · b = (2 i — j — 2k) · (3 i — 4 k)
= 2 × 3 + (—1) × 0 + (—2) × (—4)
⃝c Cambridge University Press, 2006
, = 14.
Chapter 1 The algebra and calculus of vectors 4
The angle a between a and b is then given by
a·b
cos a =
|a||b|
14 14
= = .
3×5 15
Thus a = tan—1 15
14 .
(iii) The component of c in the direction of a is
a
a=c·
c ·b
|a|
2 i — j — 2k
= (i — 5 j + 3 k) ·
|2 i — j — 2k|
1 × 2 + (—5) × (—1) + 3 × (—2)
=
3
1
= .
3
The component of c in the direction of b is
b
c ·b
b=c·
|b|
3i — 4k
= (i — 5 j + 3 k) ·
|3 i — 4 k|
1 × 3 + (—5) × 0 + 3 × (—4)
=
5
9
=— .
5
(iv)
a×b = (2 i — j — 2k)×(3 i — 4 k)
i j k
= 2 —1 —2
ˇ 3 0 —4 ˇ
= 4 — 0 i — (—8) — (—6) j + 0 — (—3) k
= 4 i + 2 j + 3 k.
⃝c Cambridge University Press, 2006
