We can tell whether or not the points represent stable or unstable points of equilibrium by examining the phase space plots in the neighborhood of the equilibrium points. R. Douglas Gregory 6. I am about to finish a two semester sequence of Mechanics using this text and I find it does a very good job of explaining theory and ideas along with guiding you through the mathematics that make it all work. I c = m    +  = m  +   4 12   16 12      2 b a 2  b2 a 2  a From the parallel axis theorem, I = I c + m   = m  +  2  16 3  4MglI From equation 9.7.12, S 2 ≥ I s2 2 2 a b a  4mg   m  +   2   16 3  S2 ≥ m 2b 4 64 1 1 16  ga  b 2 a 2   2 16  ( 980 )( 20 )  12 202   2 −1 S ≥ 2   +  = 2   +   rad ⋅ s b  2  16 3   3  (1)  2  16 S ≥ 18, 294 rad ⋅ s −1 = 2910 rps 9.18 I s = I rim + I spokes + I hub ma 2 2 3 a ma 2 I spokes = mspokes = , assuming the spokes to be thin rods 3 12 I hub = 0 , assuming its radius is small I rim = mrim a 2 = 131 From the perpendicular axis theorem, equation 8.3.14, I = From equation 9.10.14, S 2 > Is 2 Imga I s ( I s + ma 2 ) 1  2 1   1 6 mga g 2  = S>   2  2  7 ma    19 a  2    + ma    12   For rolling without slipping, v = aS 1   30   2 1 6 × 32 ×      6 ga  2  2   ft ⋅ s −1 = 3.55 ft ⋅ s −1 v>  =  19   19 ×12    ma 2 If the spokes and hub are neglected, I s = 2 1  2 1   2 1 mga g    = S>   2  2  ma    3a  2    + ma    2   1   30   2 1 32 ×    ga  2   2   ft ⋅ s −1 = 3.65 ft ⋅ s −1  v>  =  3   3 × 12    G 9.19 From equations 9.3.5 with N = 0 … I1ω1 + ( I 3 − I 2 ) ω 2ω 3 = 0 I 2ω 2 + ( I1 − I 3 ) ω1ω 3 = 0 I 3ω 3 + ( I 2 − I1 ) ω1ω 2 = 0 Differentiating the first equation with respect to t: I1ω1 + ( I 3 − I 2 )(ω 2ω 3 + ω 2ω3 ) = 0 From the second and third equations: (I − I ) (I − I ) ω 2 = 3 1 ω1ω 3 , and ω 3 = 1 2 ω1ω 2 I2 I3 132 (I − I )  (I − I ) I1ω1 + ( I 3 − I 2 )  1 2 ω1ω 22 + 3 1 ω1ω 32  = 0 I2  I3   ( I − I )( I − I )   ( I − I )( I − I )  ω1 + K1ω1 = 0 , K1 = −ω 22  3 2 2 1  + ω 32  3 2 3 1  I1 I 3 I1 I 2     (a) For ω3 large and ω 2 = 0 , K1 > 0 so ω1 + K1ω1 = 0 is the harmonic oscillator equation. Thus, the potential energy of the central atom can be approximated as 6 V = ∑ cri −α i =1 2d 47 1 r1 = ( d − x ) + y 2 + z 2  2   2 − α α  2x x2 + y2 + z 2  2 + r = ( d − 2dx + x + y + z ) = d 1 −  d d2   1 n From Appendix D, (1 + x ) = 1 + nx + n ( n − 1) x 2 + … 2 2  α  2 x x + y 2 + z 2  1  α   α  r1−α = d −α 1 −  − +  +  −   − − 1 d2  2  d  2  2  2  −α 1 2 2 2 2 −2 −α 2 2 2 2 2 2 2  2 x  2 x3   2x   x + y + z   x + y + z   + term s in +  −  − 2         d2 d2 d3   d   d      x3   α  α   4 x2  αx α − 2 ( x 2 + y 2 + z 2 ) +  + 1  2 + terms in 3   r1−α = d −α 1 + d   d 2d 4  2  d   αx α α α   − 2 ( x 2 + y 2 + z 2 ) + 2  + 1 x 2  r1−α ≈ d −α 1 + d 2d d 2    1 1 r2 = ( − d − x ) + y 2 + z 2  2 =  d 2 + 2dx + x 2 + y 2 + z 2  2   2 − −α 2 r r2−α α  2 x x2 + y 2 + z 2  2 = d 1 + +  d d2    αx α α α   ≈ d −α 1 − − 2 ( x 2 + y 2 + z 2 ) + 2  + 1 x 2  d 2d d 2    −α α α   r1−α + r2−α ≈ d −α  2 − 2 ( x 2 + y 2 + z 2 ) + 2 (α + 2 ) x 2  d  d  Similarly: α α   r3−α + r4−α ≈ d −α  2 − 2 ( x 2 + y 2 + z 2 ) + 2 (α + 2 ) y 2  d  d  α α   r5−α + r6−α ≈ d −α  2 − 2 ( x 2 + y 2 + z 2 ) + 2 (α + 2 ) z 2  d  d   3α   α 2 2α  V ≈ cd −α 6 − 2 ( x 2 + y 2 + z 2 ) +  2 + 2  ( x 2 + y 2 + z 2 )  d d  d   −α −α − 2 2 2 2 2 ≈ 6cd + cd (α − α )( x + y + z ) V ≈ A + B ( x2 + y 2 + z 2 ) 4.20 48 z ˆ kB ( F = q E+v×B ˆjE ˆ0 iv ( ) ) ˆ × kB ˆ = iyB ˆ + ˆjy + kz ˆ − ˆjxB v × B = ix ˆ F = iqyB + ˆjq ( E − xB ) y mx = Fx = qyB qB x−x = y m x qB  my = Fy = qE − qxB = qE − qB  x + m   y  2 2 qE qBx  qB  eE eBx  eB  − − + −  y y=  y=− m m  m m m m eE eB +ωx , ω= y +ω2 y = − m m 1  eE  + ω x  + A cos (ω t + θ ) y = 2 − ω  m  y = − Aω sin (ω t + θ ) y = 0 , so θ = 0 1  eE  − +ωx  2  ω  m  1  eE  y = a (1 − cos ω t ) , +ωx  a = 2 − ω  m  qB x=x + y = x − ω y = x − ω a (1 − cos ω t ) m x = ( x − ω a ) + ω a cos ω t y = 0 , so A = − x = ( x − ω a ) t + a sin ω t x = a sin ω t + bt , mz = Fz = 0 z=zt+z =0 b = x −ωa 4.21 1 2 b mv + mqh = mg 2 2 2 v = g ( b − 2h ) y b h x mv 2 = −mg cos θ + R Fr = − b 49 h b h mv 2 mg mg R = mg − =  h − ( b − 2h )  = ( 3h − b ) b b b b the particle leaves the side of the sphere when R = 0 b b above the central plane h = , i.e., 3 3 cos θ = 4.22 1 2 mv + mgh = 0 2 at the bottom of the loop, h = −b 1 2 mv = mgb , 2 v = 2gb so h b mv 2 Fr = −mg + R = b 2 mv R = mg + = mg + 2mg = 3mg b v 4.23 From the equation for the energy as a function of s in Example 4.6.2, 1 1  mg  2 E = ms 2 +  s , 2 2  4A  s is undergoing harmonic motion with: "k " g 1 g ω= = = m 4A 2 A Since s = 4 A sin φ , φ increases by 2π radians during the time interval:  A = 2π  2  ω  g For cycloidal motion, x and z are functions of 2φ so they undergo a complete cycle every time φ changes by π . The phase space trajectories are given by solutions to the above equation 1  2 x4 y = ±  x 2 − + 2C  . This is a wonderful textbook. From equation 1.12.14 … 155 ( ) m ( rθ + 2rθ − rφ sin θ cosθ ) = F cosθ ) = F m ( rφsin θ + 2rφ sin θ + 2rθφ m r − rφ 2 sin 2 θ − rθ 2 = Fr 2 θ φ The equations agree. Motion is stable for initial rotation about the 3 axis if the 3 axis is the principal axis having the largest or smallest moment of inertia. For the particle to remain on a single horizontal circle, there 2 must be two roots with r = rD . Many texts leave out so many steps in the examples and derivations that students, especially those seeing the applications of differential equations for the first time, find it hard to follow and become discouraged. n π n 4 1 1  f ( t ) = sin ω t + sin 3ω t + sin 5ω t + … π 3 5  = 29 3.22 In steady state, x ( t ) = ∑ An e ( i nω t −φn ) n An = Fn m 1 (ω 2 − n 2ω 2 )2 + 4γ 2 n 2ω 2  2   4F , n = 1,3,5, . The result was the second Italian edition (Bollati-Boringhieri, 2002), which was the original source for the translation. From equation 9.1.29: G 2 ω ˆ ˆ ˆ L = ma 2 ⋅ i + j+k 3 3 From equation 9.1.32, 1 ω 2ma 2ω 2 2 2 ma 2ω 2 T= (1 + 1 + 1 ) = 3 2 3 3 3 (b) From equation 9.1.10, with the moments of ( ) 2ma 2 and the products of inertia equal to 0: 3 2ma 2 2 I= cos 2 α + cos 2 β + cos 2 γ ) = ma 2 ( 3 3 (c) For the x-axis being any axis through the center of the lamina and in the plane of the lamina, and the y-axis also in the plane of the lamina … I xx = I yy due to the similar geometry of the mass inertia equal to distributions with respect to the x- and y-axes. Qθ and Qφ are torques. Motion of Rigid Bodies in Three Dimensions. 11. The trajectories are symmetrically disposed about the x and y axes. vr = r , vφ = rφ sin α 1 2 1 mv = m r 2 + r 2φ 2 sin α 2 2 V = mgr cos α m 2 2 2 2 L = T −V = r + r φ sin α − mgr cos α 2 d  ∂L  ∂L ∂L = mrφ 2 sin 2 α − mg cos α = mr , = mr ,   dt  ∂r  ∂r ∂r 2 2 mr = mrφ sin α − mg cos α ∂L ∂L 2 2 φ sin α , =0 = mr ∂φ ∂φ ( T= ( ) ) 156 d mr 2φ sin α = 0 dt mr 2φ sin α = A = constant ( Say ) A2 − mg cos α mr 3 d dr 1 d 2 r r = r = r = 2 dr dt dr m A 2 dr − mg cos α dr d ( r 2 ) = 2 m r3 mr 2 A2 =− − mgr cos α + C 2 2mr 2 The constant of integration C is the total energy of the particle: kinetic energy due to the component of motion in the radial direction, kinetic energy due to the component of motion in the angular direction, and the potential energy. 8. This point defines the position of m2. There was an error retrieving your Wish Lists. 3. The same holds true for the trigonometric terms in f ( t ) . Similarly, all the i other products of inertia are zero. = − ω b Since the center of the turntable is fixed. Hanspeter Schaub and John L. Junkins January 1, 2002 Related products . Fcor (ω × v′ )horiz ( ) 5.13 horiz From Example 5.4.1 … 1 1  8h 3  2 ′ xh = ω   cos λ 3  g  and yh′ = 0 . Analytical mechanics, 7ed., Solutions manual. 2πρ R 2 R + r ds rR ∫R − r 2πρ R 2 = −G  R + r − ( R − r )  rR  4π R 2 ρ M = −G φ = −G R R For r < R , φ is independent of r. It is constant inside the spherical shell. 9.25 a b1 + b2 = 3   2 1 b1 = sin 30D = b2 2 a a b1 = b2 = 2 3 3 a 2 = H 2 + b22 a 2 2a 2 H = a −b = a − = 3 3 2 2 2 2 2 137 2 a 3 Thus, the coordinates of the 4 atoms are: H= ( 0, 0, H ) Oxygen: a   a    −b1 , , 0  ;  −b1 , − , 0  2   2   ( b2 , 0, 0 ) Hydrogen: Carbon: (a) The axes 1, 2, 3 are principal axes if the products of inertia are zero. Reviewed in the United States on April 12, 2006. Customers who bought this item also bought. Using T = 2π g T (c) g= λ A3 ω2 and λ = B=− 32ω 2 6 B A2 = A 192 for A = 3.20 π 4 , f ( t ) = ∑ cn einω t B = 0.0032 A n = 0, ±1, ±2, . Your purchase details will be hidden according to our website privacy and be deleted automatically. NO Test Bank included on this purchase. . (1) It is described by a second-order, linear differential equation with constant coefficients. THE LOGIC BOOK Fourth Edition  m1 (1 + m1 m2 )  (1 + m1 m2 )  Thus, solving for γ … v1′2 = 94 1 γ= m1 m2 ( 2T m1 ) 2 1 2 ( 2 m1 ) (1 + m1   T − Q m2 )   (1 + m1 m2 )  1 2 1 2 Finally… γ= 7.21 m1 m2 1 1  Q (1 + m1 m2 )  2 1 −  T   The time of flight, τ = constant—so τ = v1 τ  γ + γ 2 + α 2 − 1   1+α As an example, let v1 τ = 1 and we have r but from problem 7.19 above v1′ r = v1′ τ = r1 = γ α =1 1 α =2 r2 = γ + γ 2 + 3   3 1 α =4 r3 = γ + γ 2 + 15   5 1 α = 12 r4 = γ + γ 2 + 143   13  Below is a polar plot of these four curves. 8. analytical mechanics 7th edition solutions analytical mechanics 7th edition textbook solutions Selected References. 7.22 pp scattering p–D p – He p–C From eqn. Motion is unstable for the initial rotation mostly about the principal axis having the median moment of inertia. Then (a) (b) x = x − x 3 . 5. The potential is independent of the center of mass coordinates. 8.7.10 … Ml 2  1  l 2 l I x   = cm =  = 12  M  12 2 M l x= 6 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. Full download : https://goo.gl/AsycGa Analytical Mechanics 7th Edition Fowles Solutions Manual Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. letting which reduces to: r + ρ1 ≈ 10−3 and R ≈ 0.01mm (small raindrop). General Motion of a Particle in Three Dimensions. Therefore, when terms that cancel in the summations are discarded: 28 1 T  f ( t ) = c + ∑  ∫ 2T f ( t ) cos ( nω t ) dt  cos nω t − n T 2  T 1  + ∑  ∫ 2T f ( t ) sin ( nω t ) dt  sin nω t , − n T 2  1 T2 n = ±1, ±2, . R  Since l is constant, vmax occurs at r and vmin occurs at r1 , i.e. G G G G ˆ + ˆjy + kˆω × ix ˆ + ˆjy v = v′ + ω × r ′ = ix G v = iˆ ( x − ω y ) + ˆj ( y + ω x ) ( ) ( ) 1 G G m 2 mv ⋅ v = ( x − 2 xω y + ω 2 y 2 + y 2 + 2 yω x + ω 2 x 2 ) 2 2 L = T −V d  ∂L  ∂L = m ( x − ω y ) , x − ω y ) ,   = m ( ∂x dt  ∂x  ∂L ∂V = m ( yω + ω 2 x ) − ∂x ∂x ∂V m ( x − ω y ) = m ( yω + ω 2 x ) − ∂x 2 Fx = m ( x − 2ω y − ω x ) T= ∂L = m ( y + ω x ) , ∂y d  ∂L  y + ω x ) ,   = m ( dt  ∂y  144 ∂L ∂V = m ( − xω + ω 2 y ) − ∂y ∂y m ( y + ω x ) = m ( − xω + ω 2 y ) − Fy = m ( y + 2ω x − ω 2 y ) ∂V ∂y G G For comparison, from equation 5.3.2 ( A0 = 0 and ω = 0 ) G G G G G G G F = ma′ + 2mω × v′ + mω × (ω × r ′ ) G ˆ + ˆjy + 2m ω kˆ × ix ˆ + ˆjy + m ω kˆ × ω kˆ × ix ˆ + ˆjy  F = m ix   2 Fx = m ( x − 2ω y − ω x ) ( ) ( ) ( ) Fy = m ( y + 2ω x − ω 2 y ) 10.9 Choosing the axis of rotation as the z axis … G G G G ˆ + ω kˆ × ix ˆ ˆ + ˆjy + kz ˆ + ˆjy + kz v = v′ + ω × r ′ = ix G ˆ v = iˆ ( x − ω y ) + ˆj ( y + ω x ) + kz ( ) ( ) 1 G G m 2 mv ⋅ v = ( x − 2 xω y + ω 2 y 2 + y 2 + 2 yω x + ω 2 x 2 + z 2 ) 2 2 L = T −V d  ∂L  ∂L = m ( x − ω y ) x − ω y ) ,   = m ( ∂x dt  ∂x  ∂L ∂V = m ( yω + ω 2 x ) − ∂x ∂x ∂V m ( x − ω y ) = m ( yω + ω 2 x ) − ∂x 2 Fx = m ( x − 2ω y − ω x ) T= ∂L d  ∂L  = m ( y + ω x ) , y + ω x )   = m ( ∂y dt  ∂y  ∂L ∂V = m ( − xω + ω 2 y ) − ∂y ∂y ∂V m ( y + ω x ) = m ( − xω + ω 2 y ) − ∂y Fy = m ( y + 2ω x − ω 2 y ) d  ∂L  ∂L ∂L ∂V = mz , = mz , =−   dt  ∂z  ∂z ∂z ∂z ∂V mz = − = Fz ∂z G G From equation 5.3.2 ( A0 = 0 and ω = 0 ) … 145 G G G G G G G F = ma′ + 2mω × v′ + mω × (ω × r ′ ) G ˆ + 2mω kˆ × ix ˆ + mω kˆ × ω kˆ × ix ˆ  ˆ + ˆjy + kz ˆ + ˆjy + kz ˆ + ˆjy + kz F = m ix   2 Fx = m ( x − 2ω y − ω x ) ( ) ( ) ( ) Fy = m ( y − 2ω x − ω 2 y ) Fz = m z 10.10 1 m r 2 + r 2θ 2 2 1 2 V = k ( r − lD ) − mgr cos θ 2 m 2 2 2 k 2 L = T −V = r + r θ − ( r − lD ) + mgr cos θ 2 2 d  ∂L  ∂L = mr ,   = mr dt  ∂r  ∂r ∂L = mrθ 2 − k ( r − lD ) + mg cos θ ∂r ( T= r ) ( θ mr = mr 2 − k ( r − lD ) + mg cos θ ) ∂L ∂L = mr 2θ = −mgr sin θ ∂θ ∂θ d mr 2θ = − mgr sin θ dt ( ) a a ( 2θ + sin 2θ ) y = (1 − cos 2θ ) 4 4 at θ = 0 x& y = 0 1 T = m ( x 2 + y 2 ) V = mgy 2 a aθ x = θ + θ cos 2θ = (1 + cos 2θ ) 2 2 a y = θ sin 2θ 2 2 2 ma θ  mga 2 L = T −V = (1 + cos 2θ ) + sin 2 2θ  − (1 − cos 2θ )  8 4 ma 2θ 2 mga = (1 − cos 2θ ) [1 + 2 cos 2θ + 1] − 8 4 ma 2θ 2 mga cos 2 θ  − sin 2 θ  = 2 2  where we used the trigonometric identies … 2 cos 2 θ = 1 + cos 2θ and 2sin 2 θ = 1 − cos 2θ 10.11 (See Example 4.6.2) x= ( ) 146 s = a sin θ so s = aθ cos θ ms2 mg 2 1 2 1 2 mg L= − s = ms − ks where k = a 2 2a 2 2 The equation of motion is thus k g s + s = 0 or s+ s=0 -a simple harmonic oscillator m a Let Coordinates: 10.12 x = a cos ω t + b sin θ y = a sin ω t − b cos θ x = − aω sin ω t + bθ cos θ y = aω cos ω t + bθ sin θ L = T −V = 1 m ( x 2 + y 2 ) − mgy 2 m 2 2  a ω + b 2θ 2 + 2bθaω sin (θ − ω t )  − mg ( a sin ω t − b cos θ ) 2 d ∂L = mb 2θ + mbaω θ − ω cos (θ − ω t ) dt ∂θ ∂L = mbθaω cos (θ − ω t ) − mgb sin θ ∂θ ∂L d ∂L − = 0 is The equation of motion ∂θ dt ∂θ a g θ − ω cos (θ − ω t ) + sin θ = 0 b b Note – the equation reduces to equation of simple pendulum if ω → 0 . 9. . 6. G G G G 5.2 (a) Fcent = −mω × (ω × r ′ ) G G G For ω ⊥ r ′ , Fcent = mω 2 r ′ eˆr ′ ω = 500 s −1 = 1000π s −1 G 2 Fcent = 10−6 × (1000π ) × 5 eˆr = 5π 2 dynes outward Fcent mω 2 r ′ (1000π ) 5 = = = 5.04 × 104 Fg mg 980 2 (b) 5.3 G G G mg + T − mAD = 0 (See Figure 5.1.2) g − mg ˆj + T cos θ ˆj + T sin θ iˆ − m   iˆ = 0  10  mg T cos θ = mg , and T sin θ = 10 1 tan θ = , θ = 5.71D 10 mg = 1.005mg T= cos θ G 5.4 The non-inertial observer thinks that g ′ points downward in the direction of the hanging plumb bob… Thus 51 g G G G g ′ = g − AD = g ˆj − iˆ 10 For small oscillations of a simple pendulum: 1 T = 2π g′ 2 g g ′ = g +   = 1.005 g  10  2 T = 2π 5.5 (a) 1 1 = 1.995π 1.005g g f = − µ mg is the frictional force acting on the G A0 G f box, so G G G f − mA0 = ma′ (b) (a) G ( a′ is the acceleration of the box relative to the truck. The radial momenta are equal and opposite. tmin = 2π n1 ωx = 2π n2 ωy = 2π n3 ωz The minimum time occurs at n1 = 1 , n2 = 2 , n3 = 3 . This bar-code number lets you verify that you're getting exactly the right version or edition of a book. l2 =0, 2m defines the turning points. Analytical Mechanics 7th Edition Solution Manual - incar.tw . Dynamics of Systems of Particles. G From Table 8.3.1, z ω m m 2 2 I xx = ( 2b ) + ( 2c )  = ( b 2 + c 2 )  3 12  m m I yy = ( a 2 + c 2 ) , I zz = ( a 2 + b 2 ) 3 3 2 2 y b + c 0 0  2c I m  0  I =  0 a2 + c2 3 2a  0 0 a 2 + b 2   x 2b a  ω ω G b  ω= aeˆ1 + beˆ2 + ceˆ3 ) = 2 1 ( 2 2   (a + b + c ) c  ( a 2 + b2 + c 2 ) 2   G IG From equation 9.1.28, L = I ω b 2 + c 2 0 0  a  G mω   2 2 0 0   b  L= a +c 1  3 ( a 2 + b 2 + c 2 ) 2  0 0 a 2 + b 2   c  a ( b2 + c 2 )   G mω 2 2  L= b ( a + c ) 1   3 ( a 2 + b2 + c2 ) 2  2 2  c ( a + b )  1 G G From equation 9.1.32, T = ω T ⋅ L 2 135 a ( b2 + c 2 )   1 mω 2 b ( a 2 + c 2 )  T= a b c [ ] 2 3 ( a 2 + b2 + c2 )   c ( a 2 + b 2 )    T= mω 2  a 2 ( b 2 + c 2 ) + b 2 ( a 2 + c 2 ) + c 2 ( a 2 + b 2 ) 2 2 2   6(a + b + c ) T= mω 2 a 2b 2 + a 2 c 2 + b 2 c 2 ) 2 2 2 ( 3( a + b + c ) With the origin at one corner, from the parallel axis theorem: m ( b2 + c 2 ) 4m 2 2 I xx = + m (b2 + c2 ) = (b + c ) 3 3 4m 2 2 4m 2 I yy = a +c ), I zz = ( ( a + b2 ) 3 3 I xy = − ∫ xydm = − ∫ xy ρ dV I xy = − ρ ∫ x=2a x =0 ∫ y =2b y =0 ∫ z =2c z =0 xydxdydz = −8 ρ a 2b 2 c m = ρ ( 2a )( 2b )( 2c ) = 8 ρ abc , so I xy = − mab I xy = − mac , I yz = − mbc 4 2 2  3 (b + c )  I I = m  −ab    − ac  −ab 4 2 2 (a + c ) 3 −bc 9.23 (See Figure 9.7.1) Lz = ( Lx′ ) z + ( Ly′ ) z + ( Lz ′ ) z (    −bc    4 2 2  + a b ( ) 3 − ac ) Lz = Ly′ sin θ + Lz′ cos θ = Iφ sin θ sin θ + ( I s S ) cos θ 9.24 (See Figure 9.7.1) If the top precesses without nutation, it must do so at θ = θ D where V (θ D ) is a minimum of V (θ ) … dV (θ ) =0 dθ θ =θ D ( L − Lz′ cosθ ) V (θ ) = z 2 2 I sin θ 2 + mgl cos θ (See equation 9.8.7) 136 dV dθ − cosθ D ( Lz − Lz′ cos θ D ) + Lz′ sin 2 θ D ( Lz − Lz ′ cosθ D ) = − mgl sin θ D = 0 I sin 3 θ D 2 θ =θ D let γ = Lz − Lz′ cos θ D Then γ 2 cos θ D − γ Lz ′ sin 2 θ D + mglI sin 4 θ D = 0 and solving for γ 1     Lz ′ sin θ D  4mglI cos θ D 2  γ= 1 ± 1 −   2 cos θ D   L2z′     now Lz ′ is large since ψ is large and the precession rate is small, so we can expand the term in square root above and use the (-) solution since γ must be positive … 2  2mglI cos θ D  1 − 1 +  L2z′   2 mglI sin θ D γ = Lz − Lz′ cosθ D ≈ Lz′ From equations 9.7.2, 9.7.5 and 9.7.7 … Lz = Iφ 2 sin 2 θ 0 + I s φ cos θ 0 + ψ cosθ 0 L cos θ = I (φ cosθ + ψ ) cosθ γ≈ Lz ′ sin 2 θ D 2 cos θ D ( z′ 0 s ) 0 and … so … 0 γ = ( I sin 2 θ D + I s cos 2 θ D ) φ + I s ψ cosθ D − I s φ cos 2 θ D − I s ψ cosθ D = Iφ sin 2 θ D ≈ mglI sin 2 θ D mglI sin 2 θ D = Lz′ I s ψ + φ cos θ D ( ) and since ψ >> 0 , we can ignore the φ term in the denominator and we have … mgl φ ≈ I sψ mgl = 0 and φ = the top will precess without nutation Hence, if ψ large, θ θ =θ1 θ =θ1 I sψ at θ1 = θ D the place where V (θ ) = min . 1+ ε From equations 6.5.21a&b, r1 = r 1− ε r 1 55 × 10 6 mi 1 AU a = ( r + r1 ) = = × 2 1− ε 1 − 0.967 93 × 10 6 mi a = 17.92 AU 6.16 3 From equation 6.6.5, τ = ca 2 τ = 1yr ⋅ AU τ = 75.9 yr − 3 2 3 3 ×17.92 2 AU 2 From equation 6.5.21a and 6.5.19 … ml 2 α −1 ε = − 1= r0 kr0 ε= mr v 2 − 1 and k k = GMm 1  GM 2 v = ( ε + 1)   r  From Example 6.5.3 we can translate the factor GM into the more convenient GM = ae ve2 … with ae the radius of a circular orbit and ve the orbital speed … 1 2 1 a v   93 ×106 mi 2 v = 1.967 (ε + 1)  =  ( )  ve 6  55 × 10 mi   r  v0 = 1.824 ve Since l is constant … r1v1 = r v r 1− ε 1 − .967 × 1.824 ve = 0.0306 ve v1 = v = v = 1+ ε 1.967 r1 2 e e ve ≈ 2π ae τ = 2π × 93 ×106 mi = 66, 705 mph 1 yr × 365 day ⋅ yr −1 × 24 hr ⋅ day −1 v = 1.22 × 105 mph and v1 = 2.04 × 103 mph 71 6.17 From Example 6.10.1 … 1   2 r v 2 2 ε = 1 +  q 2 −  ( qd sin φ )  where q = and d = d ve ae    are dimensionless ratios of the comet’s speed and distance from the Sun in terms of the Earth’s orbital speed and radius, respectively (q and d are the same as the factors V and R in Example 6.10.1). Gravitation and Central Forces. Lagrangian Mechanics. Therefore, k1 x′ = k2 ( x − x′ ) x′ = k2 x k1 + k2  k x  And keq x = k1  2   k1 + k2  1 ω= 3.8  k1k2 2 =  m  ( k1 + k2 ) m  keq For the system ( M + m ) , − kX = ( M + m ) X The position and acceleration of m are the same as for ( M + m ) : k xm M +m   k k xm = A cos  t + δ  = d cos t M m M m + +   The total force on m, Fm = mxm = mg − Fr xm = − 23 Fr = mg + mk mkd k xm = mg + t cos M +m M +m M +m For the block to just begin to leave the bottom of the box at the top of the vertical oscillations, Fr = 0 at xm = −d : mkd 0 = mg − M +m g ( M + m) d= k 3.9 x = e −γ t A cos (ω d t − φ ) dx = −e −γ t Aω d sin (ω d t − φ ) − γ e −γ t A cos (ω d t − φ ) dt dx = 0 = ω d sin (ω d t − φ ) + γ cos (ω d t − φ ) maxima at dt tan (ω d t − φ ) = − γ ωd thus the condition of relative maximum occurs every time that t increases by ti +1 = ti + 2π 2π ωd : ωd For the i th maximum: xi = e −γ ti A cos (ω d ti − φ ) xi +1 = e −γ ti +1 A cos (ω d ti +1 − φ ) = e −γ 2π ωd xi 2π −γ xi = e ωd = eγ Td xi +1 3.10 (a) (b) (c) c = 3 s −1 2m 2 ω d = ω 2 − γ 2 = 16 s −2 γ= k = 25 s −2 m 2 ω r = ω d 2 − γ 2 = 7 s −2 ω2 = ∴ω r = 7 s −1 F 48 = Amax = m = 0.2 m Cωd 60.4 tan φ = 2γω r 2γω r ω r 7 = = = 2 2 2 (ω − ω r ) 2γ γ 3 ∴φ ≈ 41.4 24 3.11 17 2 β mx = 0 2 3 17 γ = β and ω 2 = β 2 2 2 2 2 2 2 ω r = ω − 2γ = 4 β mx + 3β mx + (a) Amax = (b) = ω d2 = ω 2 − γ 2 = 25 2 β 4 ∴ω d = 5 β 2 2A 15β 2 e −γ Td = 3.12 F 2mγω d ∴ω r = 2 β γ= 1 2 1 ln 2 = f d ln 2 Td 1 ω d = (ω 2 − γ 2 ) 2 (a) 1 2 2 So, ω = (ω + γ ) 2 d 1 1 1 1  2  γ 2  2   ln 2  2  2 f =  fd +    = f d 1 +     2π     2π    f = 100.6 Hz 1 (b) ω r = (ω d2 − γ 2 ) 2  2  γ 2  2   ln 2 2  2 fr =  fd −    = f d 1 −     2π     2π    f r = 99.4 Hz 3.13 Since the amplitude diminishes by e −γ Td in each complete period, n 1 e −γ Td = = e −1 e γ Td n = 1 ω 1 γ= = d Td n 2π n ( Now So ) ( ωd = ω 2 − γ 2 ω = (ω d2 + γ ) 1 2 2 ) 1 2 1 1 2  = ω d 1 + 2 2   4π n  25 2π 1 Td ω d ω  1 2 = = = 1 + 2 2  2π ω d  4π n  T ω For large n, Td 1 ≈ 1+ 2 2 T 8π n 1 3.14 ( (a) ω r = ω − 2γ 2 1 2 2 ) 2 2  ω   = ω 2 − 2    = 0.707ω  2    1  1 2 2 ω 1 −  ω −γ ωd 4 = (b) Q = =  = 0.866 2γ 2γ ω  2   2  ω  2   ( 2ω ) 2γω 2 2 (c) tan φ = 2 =  2  2 =− 2 3 ω −ω ω − 4ω 1 2 2 ( )  2   φ = tan −1  −  = 146.3 3 1 2  2 2 ω  2 2 (d) D (ω ) = (ω − 4ω ) + 4   ( 4ω 2 )  = 3.606 ω 2  2    F F A (ω ) = m = 0.277 D (ω ) mω 2 3.15 A (ω ) ≈ Amaxγ 1  (ω − ω ) 2 + γ 2  2   1 1 γ for A (ω ) = Amax , = 1 2 2  (ω − ω ) 2 + γ 2  2   2 (ω − ω ) + γ 2 = 4γ 2 ω − ω = ±γ 3 ω =ω ±γ 3 26 3.16 (b) Q = ω2 −γ 2 ωd = 2γ 2γ ω2 = 1 , LC γ= R 2L 2  1   R   − 2   LC   4 L   L  1 =  2 −  Q=  R   R C  4 2   2L  L ω  R (c) Q =   = C = R R  2γ  3.17 Fext = F sin ω t = Im  F eiω t  and x ( t ) is the imaginary part of the solution to: mx + cx + kx = F eiω t i.e.