Analytical mechanics solutions manual fowles08

1 CHAPTER MECHANICS OF RIGID BODIES: PLANAR MOTION m and centered at  b b b b  − ,  , ( 0, ) , and  ,  … 2 2  2   b  m   b   m  xcm =  −     + +      = m         8.1 (a) For each portion of the wire having a mass y b ycm = x  b   m   b   m  b     + +     =  m         b (b) ds = xdy = ( b − y ) dy ycm = ycm = 1 b 2 ρ y b − y ( ) dy m ∫0 − ρ ∫ y =b y =0 (b2 − y ) d (b2 − y ) πb ρ 4b 3π 4b From symmetry, xcm = 3π ycm = (c) The center of mass is on the y-axis ds = xdy = ( by ) dy b ycm = ∫ ∫ b ycm = (d) ρ y ( by ) dy 2 ρ ( by ) dy = ∫ b ∫ b 0 y dy y dy 3b The center of mass is on the z-axis dv = π r dz = π ( x + y ) dz = π bzdz b zcm ∫ = ∫ ρ zπ bzdz b ρπ bzdz b ∫ z dz = ∫ zdz b zcm = b (e) The center of mass is on the z-axis α is the half-angle of the apex of the cone rD is the radius of the base at z = and r is radius of a circle at some arbitrary z in a plane parallel to the base r r tan α = D = , a constant b b−z dv = π r dz = π ( b − z ) tan α dz 1 m = ρ π rD2b = πρ b3 tan α 3 zcm ∫ = b zcm = ρ zπ ( b − z ) tan α dz πρ b3 tan α xcm b3 ∫ ( b z − 2bz b 2 + z ) dz b b 8.2 = ∫ ρ xdx = ∫ cx dx = ∫ ρ dx ∫ cxdx b 2b xcm = 8.3 The center of mass is on the z-axis Consider the sphere with the cavity to be made of a (i) solid sphere of radius a and mass M s , with its center of mass at z = , and (ii) a solid sphere the size of the cavity, with a mass − M c and center of mass at z = − The actual sphere with the cavity has a mass m = M s − M c and center of mass zcm 0= Ms   a   M c  −  + m zcm      a M s = π a3 ρ , M c = π   ρ 3 2 3  a   a    a    =    −  +  a −    zcm  a          zcm = a 14 2 m  b  b  8.4 (a) I z = ∑ mi R =   + +         i i Iz = mb (b) ds = rdθ dr , R = r sin θ I z = ∫ R ρ ds Iz = ρ ∫ r =b r =0 Iz = ρb 4 r rdr ∫ θ= π θ =− π ∫ π sin − π sin θ dθ θ dθ θ sin 2θ    ∫ sin θ dθ = −  ρb4  π  Iz =  −   2 m = ρπ b mb Iz = (π − ) 4π  x2  ds = hdx =  b −  dx b   Where the parabola intersects the line y = b , (c) x = ( by ) = ±b b b   x2  x4  I y = ∫ x ρ  b −  dx = ρ ∫  bx −  dx −b −b b  b    I y = ρb4 15 b  x2  m = ∫ ρ  b −  dx = ρ b −b b   I y = mb (d) dv = 2π RhdR h =b−z R = (x + y 2 ) = ( bz ) 1  b 2 dR =   dz 2 z  b I z = ∫ R ρ dv = ∫ bz ρ 2π ( bz ) 2 1 b ( b − z )   dz 2 z  b I z = πρ b ∫ ( bz − z ) dz = πρ b5 b m = ∫ ρ dv = ∫ ρ 2π ( bz ) 1 b ( b − z )   dz 2 z  b m = πρ b ∫ ( b − z ) dz = πρ b3 I z = mb (e) α is the half-angle of the apex of the cone rD is the radius of the base at z = and r is radius of a circle at some arbitrary z in a plane parallel to the base tan α = dv = 2π RhdR = 2π ( b − z ) RD zdR RD R = , a constant b b−z b b z R − ( ) D , dR = − RD dz , and the limits of integration for R = → R Since R = D b b correspond to z = b → 0 (b − z ) b I z = ∫ R ρ dv = ∫ I z = +2πρ RD4 b4 b RD2 ∫ ( b z − 3b z b 2 ρ 2π ( b − z ) RD z  − RD  dz b + 3bz − z ) dz = m = ρ π RD2b 3 I z = mRD2 10    b  πρ RD4b 10 8.5 Consider the sphere with the cavity to be made of a (i) solid sphere of radius a and and mass M s , with its center of mass at z = , and (ii) a solid sphere the size of the a cavity, with mass − M c and center of mass at z = − The actual sphere with the cavity has a mass m = M s − M c and center of mass zcm 4 a 74 m = M s − M c = π a3 ρ − π   ρ = πa ρ 3 2 83 M s = m and M c = m 7 From eqn 8.3.2, I s = Ic + I 2 a M s a2 = M c   + I 5 2 28    a 31 I =  m  a −  m  = ma 57    70 8.6 The moment of inertia about one of the straight edges is I z = ∫ R ρ dv where R = x + y From Appendix F … dv = r sin θ dr dθ dφ R = x + y = r sin θ Let a = radius of sphere z R θ Iz = ∫ r φ r =a r =0 y Iz = ρ π θ= π π φ= 2 θ =0 φ =0 ∫ ∫ r =a θ= ∫ ∫θ r =0 π =0 r sin θρ r sin θ dr dθ dφ r sin θ drdθ π ρπ a ∫ sin θ dθ 10   cos3 θ = − cos θ  sin θ d θ ∫   Iz = x ρπ a 30 14 m= π a ρ = π a3 ρ 83 I z = ma Iz = 8.7 For a rectangular parallelepiped: dv = hdxdy y R = ( x2 + y h is the length of the box in the z-direction 2 ) I z = ∫ R ρ dv = ∫ x=a ∫ (x y =b x =− a y =− b + y ) ρ hdxdy  2b3  2 I z = ρ h ∫  2bx +  dx = ρ hab ( a + b ) −a   m = ρ ( 2a )( 2b ) h = ρ abh a x 2b Iz = z m a + b2 ) ( 2a For an elliptic cylinder: y Again dv = hdxdy , and R = ( x + y ) On the surface, x2 y + =1 a b2 2b  y2 2 x = ± a 1 −   b  x z I z = ∫ R dv = ∫ 2a y =b y =− b ∫  y2  x = a 1−   b     y2  x =− a  1−   b    (x + y ) ρ hdxdy   2 2     y y 2  I z = ρ h∫ a 1 −  + 2a 1 −  y dy −b    b   b    b  aa b 2 2 2 = ρ h  ∫ ( b − y ) dy + ∫ ( b − y ) y dy  − − b b b  3b  From a table of integrals: 3 y 3 y 2 2 2 2 − = − + − + b sin −1 b y dy b y b y b y ) ( ) ( ) ∫( 8 b y b y b 2 2 −1 y ∫ ( b − y ) y dy = − ( b − y ) + ( b − y ) + sin b a  a2   b  I z = ρ h   b 4π  + π  = ρ hπ ab ( a + b ) b  3b    b m = ρ h (π ab ) Iz = m a + b2 ) ( For an ellipsoid: R2 = x2 + y dv = hdxdy , and on the surface, x2 y z + + =1 a b2 c2  x2 y2  z = ±c 1 − −  b   a y 2b x  x2 y2  h = z = 2c  − −  b   a x2 y In the xy plane + = a b  y2 2 x = ± a 1 −   b  2a z 1  y2  x = a 1−   b     x2 y2  2 I z = R ρ dv = ∫ ∫ x y ρ c + ( ) 1 − a − b2  dxdy y =− b  y2    x =− a  1−    y =b  b  1   2     ρ c y =b   a y   2 a y  2 Iz = a −  − x  x dx + y ∫  a −  − x  dx     ∫ ∫ a y =− b   b  b        From a table of integrals: 2 x k2x k4 x 2 2 − = − − + − + sin −1 k x x dx k x k x ) ( ) ( ) ∫( 8 k 1 x k 2 −1 x ∫ ( k − x ) dx = ( k − x ) + sin k 2 2ρc b  a4  y2  y2   a  − + − 1 Iz = y π      π  dy  b2   a ∫− b   b    2 4 b a  2y y  y  I z = ρπ ac ∫  1 − +  + y −  dy −b b b  b  4 I z = ρπ abc ( a + b ) 15 m For an ellipsoid, m = ρ π abc , so I z = ( a + b ) 8.8 (See Figure 8.4.1) Note that l ′ + l is the distance from 0′ to , defined as d From eqn 8.4.13, kcm = ll ′ kcm + l = ll ′ + l = l ( l ′ + l ) kcm + l = ld From eqn 8.4.9b, k = kcm + l2 k = ld From eqn 8.4.6, TD = 2π TD = 2π k2 gl d g Period of a simple pendulum: T = 2π 8.9 Period of real pendulum: TD = 2π a M I Mgl (eqn 8.4.5) Where I = moment of inertia l = distance to CM of physical pendulum a = distance to CM of bob b = radius of bob location of CM of physical pendulum: (a − b) + M − m a m ( ) m a b m = a−  + = a− l= (a + b) m + ( M − m) M 2 2 2M m m b  l = a 1 − −  2M 2M a  Moment of inertia: 2   I bob = ( M − m ) a + ( M − m ) b = ( M − m )  a + b  5   m  = Ma  −  M I rod  b  +    a  m 1 = m ( a − b ) = Ma  3  M  b  1 −    a    m   b2  m ∴ I = I bob + I rod = Ma  −  1 + +   M   a  M m b letting α = and β = M a  b  1 −    a    2 2  2  Ma (1 − α )  + β  + α (1 − β )        TD = 2π    α αβ    Mga 1 − −    2  (a) (b) 8.10 TD = T 2 (1 − α ) 1 + β  + α (1 − β )   ≈ 1−  α αβ  1 − −   2  M = 1kg a = 1.27 m β = 0.0394 m = 10 g α = 0.01 TD ≈ − α = 0.9992 12 T α to 1st order in α 12 b = 5cm (actually 0.9994 using complete expression) The period of the “seconds” pendulum is I T2 = 2π = 2s Mgl The period of the modified pendulum is I′  n  T ′ = 2π = 2  M ′gl ′  n − 20  where I ′ , M ′ , l ′ refer to parameters of pendulum with m attached and n ( = 24 × 60 × 60 ) is the number of seconds in a day Ml + mlm M where lm is the distance of the attached mass m from the pivot point I ′ = I + mlm2 l′ = −2 π I ′ π I + π mlm2  20  So 1 −  = = n  M ′gl ′ ( Ml + mlm ) g  = Mgl + π mlm2 ( Ml + mlm ) g ( Ml + mlm ) g  20  Thus 1 −  = n  ( Mgl + π mlm2 )   α lm  1 +  40 m l  α= Or − ≈  2 n  π α lm  M 1 +  gl   Solving for α gives the approximate result 10 40 l n lm m α= ≅ M  π lm  − 1   g  Letting lm = 1.3m ; l = 1.0m ; we obtain α ≅ 1.15 ⋅10−3 8.11 (a) I ⊥cm = ma (all mass in rim) I ⊥rim = ma + ma = 2ma ∴T = 2π (b) I z = I x + I y = I &cm = ma ∴ I &cm = I ⊥ rim 2a = 2π mga g ( = I ⊥cm ) ma 2 ma + ma = ma hence I &rim = 2 I 3a ∴T = 2π &rim = 2π mga 2g 8.12 G T a G mg x 8.13 mxcm = mg − T I cmω = aT  xcm = aω I cm = ma  I ω 12 x  mxcm = mg − cm = mg −  ma cm  = mg − mxcm a a5 a   xcm = g When two men hold the plank, each supports mg When one man lets go: mg − R = mxcm and R From table 8.3.1, I cm = ω = ml 12 Rl 12 R ⋅ = ml ml 10 l = I cmω 11 l 3R  xcm = ω = m  3R  mg − R = m   = 3R  m mg R= R  mg   =  xend = lω = l  ml m    xend = g 8.14 For a solid sphere: M s = π a ρ and I s = M s a ( kcm2 )s = a For subscript c representing a solid sphere the size of the cavity, from eqn 8.3.2: I s = I + Ic 2   2 4  a   a  31 I =  π a ρ  a −  π   ρ    = ⋅ ⋅ π a5 ρ 53       32  4 a m = π a3 ρ − π   ρ = ⋅ π a3 ρ 3 2 I 31 31 kcm = = ⋅ ⋅ a2 = a2 70 m 32 From eqn 8.6.11, for a sphere rolling down a rough inclined plane: g sin θ  xcm =  kcm  1+   a   x =  xs (k ) 1+ cm s a k2 + cm2 a 1+ = = 31 101 1+ 70 70  x 98 = xs 101 11 12 Energy is conserved: a 8.15 m2 m1 8.16 1  x  m1 x + m2 x + I   + m2 gx − m1 gx = E 2 2 I  + m2 xx  + xx  + m2 gx − m1 gx = m1 xx a ( m1 − m2 ) g  x= I m1 + m2 + a x While the cylinders are in contact: mv b f r = cm = mg cos θ − R r G R mvcm r = a + b , so = mg cos θ − R vcm a+b From conservation of energy: 2 a mg ( a + b ) = mvcm + Iω + mg ( a + b ) cos θ 2 G mg From table 8.3.1, I = ma 2 vcm ω= a     vcm mvcm +  ma    = mg ( a + b )(1 − cos θ ) 22  a  mvcm = mg (1 − cos θ ) a+b When the rolling cylinder leaves, R = : mg cos θ = mg (1 − cos θ ) cos θ = 3 θ = cos −1 8.17 mx = N my = N1 − mg 12 13 2 mvcm + Iω + mgy = mgyD 2 l l x = cos θ , x = − θ sin θ , 2 l l y = sin θ , y = θ cos θ , 2 l 2 θ cos θ + θsin θ l  y = −θ sin θ + θ cos θ 2 2 2  l  l  lθ = x + y =  − θ sin θ  +  θ cosθ  = vcm   2   x= ( ) ( ml , and ω = θ 12 l 2θ ml  l l m + θ + mg sin θ = mg sin θ D 12 2 l 2 θ = g ( sin θ D − sin θ ) I=  3g 2 θ =  ( sin θ D − sin θ )   l  − 1  3g 3g   3g  θ =  ( sin θ D − sin θ )    ( − cos θ )θ = − cosθ 2 l 2l   l  ml    3g   3g  N = mx = −  cos θ   ( sin θ D − sin θ ) + sin θ  −  cos θ    l   2l   Separation occurs when N = : 2  sin θ D − sin θ − sin θ = , θ = sin −1  sin θ D  3  8.18 Rx = mx Ry − mg = my l l l x = sin θ , x = θ cos θ ,  x = −θ sin θ + θ cosθ 2 l l l  y = cos θ , y = − θ sin θ , y = − θ cos θ + θsin θ 2 2 2 l l mvcm + Iθ + mg cos θ = mg 2 2 ml l I= vcm = θ , 12 ( ) ( 13 ) ) 14 m l 2θ ml  l + θ = mg (1 − cos θ ) 12 2  lθ = g (1 − cos θ )  3g 2 θ =  (1 − cosθ )   l  − 1  3g 3g   3g  θ =  (1 − cosθ )    sin θθ = sin θ 2 l 2l   l  ml  3g 3g  Rx = ( − sin θ )   (1 − cosθ ) + cosθ  sin θ      l   2l  3mg Rx = sin θ ( 3cosθ − ) ml   3g   3g  Ry = mg − cos θ   (1 − cos θ ) + sin θ  sin θ      l   2l  Ry = mg − 3mg  sin θ   cos θ − cos θ +    mg ( 3cosθ − 1) The reaction force constrains the tail of the rocket from sliding backward for Rx > : 3cos θ − > θ < cos −1 The rocket is constrained from sliding forward for Rx < : θ > cos −1 Ry = 8.19 mx = −mg sin θ − µ mg cos θ  x = − g ( sin θ + µ cos θ ) x G f G mg ω= µ g cos θ t a θ xt : Since acceleration is constant, x = xDt +  gt x = vDt − ( sin θ + µ cosθ ) 2 Meanwhile ( µ mg cos θ ) a = Iω = ma 2ω 5 µ g cos θ ω = a 14 15 The ball begins pure rolling when v = aω … µ g cos θ v = vD +  xt = vD − g ( sin θ + µ cos θ ) t = a t a 2vD t= g ( 2sin θ + µ cos θ ) At that time: 2vD2 g 4vD ( sin θ + µ cos θ ) − x= g ( 2sin θ + µ cosθ ) g ( 2sin θ + µ cos θ )2 x= 8.20 2vD2 ( sin θ + µ cos θ ) g ( 2sin θ + µ cos θ )2 mx = µ mg  x = µg x = µ gt , and x = µ gt 2 2 ma ω = − µ mga 5 µg ω = − a µg t ω = ωD − a Slipping ceases to occur when v = aω … µ gt = aω D − µ gt 2 aω D t= g Iω =  aω D  x = µg    µg  x= 2 a 2ω D2 49 µ g     Let the moments of inertia of A and B be I a  = M a a  and I b  = M bb      The angular velocity of A is α while that of B is β − α + φ (remember that in two dimensions, angular velocity is the rate of change of an angle between an line or direction fixed to the body and one fixed in space) For rolling contact, lengths traveled along the perimeters of the disks A and B must be equal to the arc length traveled along the track C aφ = b β = ( a + 2b )(α − φ ) 8.21 15 16 a ( a + 2b ) α 2b ( a + b ) After some algebra … the angular velocity of B is found to be … aα ω B = β − α + φ = 2b For A , we take moments about O and for B we take moments about its center Call TA and TB the components of the reaction forces tangent to A and B (the “upward-going” TA acts on disk B The “downward-going” TA acts on disk A) so that φ = Thus ( a + 2b )α (a + b) and β = K − TA a = I Aα (Torque on Disk A) α −TAb − TB b = − I B β − α + φ = − I B a 2b TA − TB = M B ( a + b ) α − φ = M B aα Eliminate TA and TB α K = ( 4b I A + M B a 2b + a I B ) 4b Integrating this equation gives : α Kt = ( 4b I A + M B a 2b + a I B ) 4b Putting ω A = α at t = tD gives ( ( ) (Torque on Disk B) ) (Force on Disk B) 4b KtD ωA = ( 4b2 I A + M B a 2b2 + a I B ) Putting in values for I A and I B gives KtD ωA =   a  2M A + M B    Since the angular velocity of B is ω B = β − α + φ = ωB = a ωA = 2b KtD   ab  2M A + M B    16 aα , we have 2b 17 From section 8.7 (see Figure 8.7.1), the instantaneous center of rotation is the l point O If x is the distance from the center of mass to O and is the distance from the center of mass to the center of percussion O′ , then from eqn 8.7.10 … Ml   l l I x   = cm =  = 12  M  12 2 M l x= 8.22 8.23 In order that no reaction occurs between the table surface and the ball, the ball must approach and recede from its collision with the cushion by rolling without slipping Using a prime to denote velocity and rotational velocity after the collision: Pˆ ′ = vcm − vcm m Pˆ ′ = ω cm − ( h − a ) ω cm I ′ = aω cm ′ The conditions for no reaction are vcm = aω cm and vcm aω cm −   Pˆ Pˆ = a ω cm − ( h − a )  m I   1 = a (h − a) m I I = ma 2ma h= +a = a 5ma d a= h= d 10 8.24 During the collision, angular momentum about the point is conserved: m′v l ′ = Iθ D m′vDl ′ I After the collision, energy is conserved: 2 l l Iθ − mg − m′gl ′ = − mg cos θ D − m′gl ′ cos θ D 2 2 2 m′ vD l ′  ml  = g (1 − cos θ D )  + m′l ′  I   θ = 17 18 I= ml + m′l ′2 2   ml   ml 2 ′ ′ ′ ′ vD = g cos m l m l θ − + + ( )   D   m′l ′     8.25 The effect of rod BC acting on rod AB is impulse + Pˆ ′ The effect of AB on BC is − Pˆ ′ mv1 = Pˆ + Pˆ ′ mv2 = − Pˆ ′ l l Iω1 = Pˆ − Pˆ ′ Iω = − Pˆ ′ 2 ml I= 12 ˆ ˆ P − P′ ω1 = ω = − Pˆ ′ ml ml l l vB = v1 − ω1 vB = v2 + ω 2 Pˆ + Pˆ ′ l   ˆ ˆ vB = −   P − P′ m  ml  Pˆ ′ l  ˆ  vB = − +  − P′  m  ml  Pˆ + Pˆ ′ − Pˆ − Pˆ ′ = − Pˆ ′ − 3Pˆ ′ ( ) ( ) ( ( ) ) Pˆ ′ = Pˆ Pˆ Pˆ ′ = 5Pˆ 4m Pˆ ω1 = 2ml Pˆ 4m 3Pˆ ω2 = − 2ml v1 = v2 = − Pˆ m vB = − 18