Classical mechanics(1)

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  • CLASSI CAL MECHANI CS

    SOLUTIONS MANUAL

    R. Douglas Gregory

    November 2006

    Please report any errors in these solutionsby emailing

    cm:solutions@btinternet:com

  • 2Contents1 The algebra and calculus of vectors 3

    2 Velocity, acceleration and scalar angular velocity 27

    3 Newtons 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 Lagranges equations and conservation principles 429

    13 The calculus of variations and Hamiltons principle 473

    14 Hamiltons 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 calculusof vectors

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 4

    Problem 1 . 1In terms of the standard basis set fi ; j ;kg, a D 2i j 2k, b D 3i 4k andc D i 5j C 3k.

    (i) Find 3aC 2b 4c and ja b j2.(ii) Find ja j, jb j 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 ab, bc and .ab/.bc/.(v) Find a .bc/ and .ab/ c and verify that they are equal. Is the set fa; b; cg

    right- or left-handed?(vi) By evaluating each side, verify the identity a.bc/ D .a c/b .a b/c.

    Solution

    (i)3aC 2b 4c D 3.2i j 2k/C 2.3i 4k/ 4.i 5j C 3k/

    D 8i C 17j 26k:

    ja b j2 D .a b/ .a b/D .i j C 2k/ .i j C 2k/D .1/2 C .1/2 C 22 D 6:

    (ii)jaj2 D a a

    D .2i j 2k/ .2i j 2k/D 22 C .1/2 C .2/2 D 9:

    Hence jaj D 3.

    jbj2 D b bD .3i 4k/ .3i 4k/D 32 C .4/2 D 25:

    Hence jbj D 5.

    a b D .2i j 2k/ .3i 4k/D 2 3C .1/ 0C .2/ .4/D 14:

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 5

    The angle between a and b is then given by

    cos D a bjaj jbjD 14

    3 5 D14

    15:

    Thus D tan1 1415

    .

    (iii) The component of c in the direction of a is

    c ba D c ajaj

    D .i 5j C 3k/

    2i j 2kj2i j 2kj

    D1 2C .5/ .1/C 3 .2/

    3

    D 13:

    The component of c in the direction of b is

    c bb D c bjbj

    D .i 5j C 3k/

    3i 4kj3i 4kj

    D1 3C .5/ 0C 3 .4/

    5

    D 95:

    (iv)ab D .2i j 2k/.3i 4k/

    D i j k2 1 2

    3 0 4

    D 4 0i .8/ .6/j C 0 .3/kD 4i C 2j C 3k:

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 6

    bc D .3i 4k/.i 5j C 3k/

    D i j k3 0 4

    1 5 3

    D 0 20i 9 .4/j C .15/ 0kD 20i 13j 15k:

    Hence

    .ab/.bc/ D .4i C 2j C 3k/.20i 13j 15k/

    D i j k4 2 320 13 15

    D .30/ .39/i .60/ .60/j C .52/ .40/kD 9i 12k:

    (v)a .bc/ D .2i j 2k/ .20i 13j 15k/

    D 2 .20/C .1/ .13/C .2/ .15/D 3:

    .ab/ c D .4i C 2j C 3k/ .i 5j C 3k/D 4 1/C 2 .5/C 3 3D 3:

    These values are equal and this verifies the identity

    a .bc/ D .ab/ c:

    Since a .bc/ is positive, the set fa;b; cg must be right-handed.(vi) The left side of the identity is

    a.bc/ D .2i j 2k/.20i 13j 15k/

    D i j k2 1 220 13 15

    D 15 26i .30/ 40j C .26/ 20kD 11i C 70j 46k:

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 7

    Since

    .a c/b D

    2 1C .1/ .5/C .2/ 3bD bD 3i 4k;

    .a b/ c D

    2 3C .1/ 0C .2/ .4/cD 14c D 14.i 5j C 3k/D 14i 70j C 42k;

    the right side of the identity is

    .a c/b .a b/c D .3i 4k/ .14i 70j C 42k/D 11i C 70j 46k:

    Thus the right and left sides are equal and this verifies the identity.

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 8

    Problem 1 . 2Find the angle between any two diagonals of a cube.

    FIGURE 1.1 Two diagonals of a cube.

    E

    A

    C

    D

    O

    B

    a

    SolutionFigure 1.1 shows a cube of side a; OE and AD are two of its diagonals. Let O

    be the origin of position vectors and suppose the points A, B and C have positionvectors ai , aj , ak respectively. Then the line segment

    !OE represents the vector

    ai C aj C ak

    and the line segment!AD represents the vector

    .aj C ak/ ai D ai C aj C ak:

    Let be the angle between OE and AD. Then

    cos D .ai C aj C ak/ .ai C aj C ak/jai C aj C akj j ai C aj C akj

    D a2 C a2 C a2p3ap

    3a D 1

    3:

    Hence the angle between the diagonals is cos1 13

    , which is approximately 70:5.

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 9

    Problem 1 . 3ABCDEF is a regular hexagon with centre O which is also the origin of positionvectors. Find the position vectors of the vertices C , D, E, F in terms of the positionvectors a, b of A and B.

    FIGURE 1.2 ABCDEF is a regularhexagon.

    A

    BC

    D

    E F

    O a

    b

    Solution

    (i) The position vector c is represented by the line segment!OC which has the

    same magnitude and direction as the line segment!AB. Hence

    c D b a:

    (ii) The position vector d is represented by the line segment!OD which has the

    same magnitude as, but opposite direction to, the line segment!OA. Hence

    d D a:

    (iii) The position vector e is represented by the line segment!OE which has the

    same magnitude as, but opposite direction to, the line segment!OB. Hence

    e D b:

    (iv) The position vector f is represented by the line segment!OF which has the

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 10

    same magnitude as, but opposite direction to, the line segment!AB. Hence

    e D .b a/ D a b:

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 11

    Problem 1 . 4Let ABCD be a general (skew) quadrilateral and let P , Q, R, S be the mid-pointsof the sides AB, BC , CD, DA respectively. Show that PQRS is a parallelogram.

    SolutionLet the points A, B, C , D have position vectors a, b, c, d relative to some originO . Then the position vectors of the points P , Q, R, S are given by

    p D 12.aC b/; q D 1

    2.bC c/; r D 1

    2.c C d/; s D 1

    2.d C a/:

    Now the line segment!PQ represents the vector

    q p D 12.bC c/ 1

    2.aC b/ D 1

    2.c a/;

    and the line segment!SR represents the vector

    r s D 12.c C d/ 1

    2.d C a/ D 1

    2.c a/:

    The lines PQ and SR are therefore parallel. Similarly, the lines QR and PS areparallel. The quadrilateral PQRS is therefore a parallelogram.

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 12

    Problem 1 . 5In a general tetrahedron, lines are drawn connecting the mid-point of each side withthe mid-point of the side opposite. Show that these three lines meet in a point thatbisects each of them.

    SolutionLet the vertices of the tetrahedron be A, B, C , D and suppose that these points haveposition vectors a, b, c, d relative to some origin O . Then X , the mid-point of AB,has position vector

    x D 12.aC b/;

    and Y , the mid-point of CD, has position vector

    y D 12.c C d/:

    Hence the mid-point of XY has position vector

    12.x C y/ D 1

    2

    12.aC b/C 1

    2.c C d/

    D 1

    4

    aC b C c C d:

    The mid-points of the other two lines that join the mid-points of opposite sides ofthe tetrahedron are found to have the same position vector. These three points aretherefore coincident. Hence the three lines that join the mid-points of opposite sidesof the tetrahedron meet in a point that bisects each of them.

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 13

    Problem 1 . 6Let ABCD be a general tetrahedron and let P , Q, R, S be the median centres of thefaces opposite to the vertices A, B, C , D respectively. Show that the lines AP , BQ,CR, DS all meet in a point (called the centroid of the tetrahedron), which divideseach line in the ratio 3:1.

    SolutionLet the vertices of the tetrahedron be A, B, C , D and suppose that these points

    have position vectors a, b, c, d respectively, relative to some origin O . Then P , themedian centre of the face BCD has position vector

    p D 13.bC c C d/:

    The point that divides the line AP in the ratio 3:1 therefore has position vector

    aC 3p4

    D 14

    aC bC c C d:

    The corresponding points on the other three lines that join the vertices of the tetra-hedron to the median centres of the opposite faces are all found to have the sameposition vector. These four points are therefore coincident. Hence the four linesthat join the vertices of the tetrahedron to the median centres of the opposite facesmeet in a point that divides each line in the ratio 3:1. It is the same point as wasconstructed in Problem 1.5.

    c Cambridge University Press, 2006

  • Chapter 1 The algebra and calculus of vectors 14

    Problem 1 . 7A number of particles with masses m1;m2;m3; : : : are situated at the points withposition vectors r1; r2; r3; : : : relative to an origin O . The centre of mass G of theparticles is defined to be the point of space with position vector

    R D m1r1 Cm2r2 Cm3r3 C m1 Cm2 Cm3 C

    Show that if a