Analytical mechanics solutions manual fowles07

Chapter Dynamics of Systems of Particles 7.1 z ∑ mi ri m i From eqn 7.1.1, rcm = ( ) 1 ( r1 + r2 + r3 ) = iˆ + ˆj + ˆj + kˆ + kˆ 3 ˆ rcm = i + ˆj + 2kˆ d 1 vcm = rcm = ( v1 + v2 + v3 ) = 2iˆ + ˆj + iˆ + ˆj + kˆ dt 3 vcm = 3iˆ + ˆj + kˆ From eqn 7.1.3, p = ∑ mi vi = v1 + v2 + v3 rcm = ( ) ( ( ) ) i p = iˆ + ˆj + kˆ 7.2 (a) From eqn 7.2.15, T = ∑ mi vi2 i T =  22 + 12 + (12 + 12 + 12 )  = (b) From Prob 7.1, vcm = iˆ + ˆj + kˆ 1 mvcm = × × ( + 22 + 12 ) = 2 (c) From eqn.7.2.8, L = ∑ ri × mvi ( ( ) ( i ) ) ( ) L =  iˆ + ˆj × 2iˆ  +  ˆj + kˆ × ˆj  +  kˆ × iˆ + ˆj + kˆ        ˆ ˆ L = −2k + −iˆ + ˆj − iˆ = −2iˆ + ˆj − 2k ( ) ( ) ( 7.3 ) v = vb − vg Since momentum is conserved and the bullet and gun were initially at rest: = mvb + Mvg vg = −γ vb , v = (1 + γ ) vb vb = γ= m M v 1+ γ y x vg = − 7.4 γv 1+ γ v  mv = m   + Mvblk 2 m vblk = γ v γ= M 2 2 1  v   γ v   Ti − T f = mv −  m   + M          Momentum is conserved: 2 1 γ2  = mv 1 − − M    m Ti − T f γ = − 4 Ti 7.5 v0 / At the top of the trajectory: ˆ cos 60 = iˆ v v = iv Momentum is conserved: ˆ v = ˆj  m   v  + m v2 im    ˆ − ˆj v v2 = iv  v  −  Direction: θ = tan −1   = 26.6 below the horizontal  v      v 2  Speed: v2 =  v +    = 1.118 v     7.6 When a ball reaches the floor, As a result of the bounce, mv = mgh v′ =ε v The height of the first bounce: mgh′ = mv′2 v ′2 ε v = = ε 2h 2g 2g Similarly, the height of the second bounce, h′′ = ε h′ = ε h h′ = v0 60 o θ ∞   Total distance = h + 2ε h + 2ε h + … = h  −1 + ∑ 2ε n  n=0   ∞ a , r , so the positive square root is used ( v′p = ) v′p = 0.9288 v v′px = v′py = v′p = 0.657 v 1 v v − v′p2 ) = (1 − 9288 ) ( 2 vα′ = 0.1853v v′p v′p 9288 = tan φ = = v′p 2v − v′p − 9288 v − −1 φ = tan 1.9134 = 62.41 vα′ x = vα′ cos φ = 0.086 v vα′ y = −vα′ sin φ = −0.164 v vα′ = 7.15 Conservation of energy: 1 11  m p v = m p v′p2 + 4m p vα′ +  m p v  2 42  2 16vα′ = 3v − 4v′p From the conservation of momentum eqn of Prob 7.14: 16vα′ = v − 2v v′p + v′p2 Subtracting: = −2v − 2v v′p + 5v′p2 2v ± 2v + 40v v ± 42 = 10 10 Using the positive square root, since v′p > : ( v′p = ) v′p = 0.7895 v v′px = v′py = v′p = 0.558 v 1  2 v vα′ =  v − v′p2  = (.75 − 7895 )   16 vα′ = 0.1780v From the conservation of momentum eqns of Prob 7.14: v′p 7895 tan φ = = 2v − v′p − 7895 φ = tan −1 1.2638 = 51.65 vα′ x = vα′ cos φ = 0.110 v vα′ y = −vα′ sin φ = −0.140 v 7.16 sin θ φ1 and θ are the scattering angles in γ + cosθ the Lab and C.M frames respectively m From eqn 7.6.16, for Q = 0, γ = m2 sin θ tan 45 =1 = + cos θ + cos θ = sin θ and squaring … 1 + cos θ + cos θ = − cos θ 16 15 cos θ + cosθ − = 16 1 15 − ± + = −.125 ± 696 cosθ = From eqn 7.6.14, Since < θ < 7.17 π , tan φ = θ = cos −1 571 ≈ 55.2 From eqn 7.6.14, tan φ = m From eqn 7.6.18, γ = m2 sin θ γ + cosθ  Q  m1   1 − 1 +   T  m2   − − 1    γ = 1 − 1 +   = 0.3015    sin θ tan 45 = 3015 + cosθ 3015 + cosθ = sin θ (since sin θ > cos θ , θ > 45 ) 2 3015 = sin θ − 2sin θ cosθ + cos θ Using the identity 2sin θ cos θ = sin 2θ sin 2θ = − 30152 = 0.9091 Since θ > 45 , 2θ > 90 : 2θ = sin −1 9091 = 114.62 θ = 57.3 P1’ 7.18 Conservation of momentum: P1 = P1′ cos φ + P2′ cos (ψ − φ φ ψ P1 ) = P1′ sin φ − P2′ sin (ψ − φ ) From Appendix B for sin (α + β ) and cos (α + β ) : P2’ P1 = P1′ cos φ + P2′ ( cosψ cos φ + sinψ sin φ ) = P1′ sin φ − P2′ ( sin ψ cos φ − cosψ sin φ ) P12 = P1′2 cos φ + P2′2 ( cos ψ cos φ + cosψ cos φ sin φ sinψ + sin ψ sin φ ) +2 P1′ P2′ ( cos φ cosψ + cos φ sinψ sin φ ) = P1′2 sin φ + P2′2 ( sin ψ cos φ − 2sin ψ cos φ cosψ sin φ + cos ψ sin φ ) −2 P1′ P2′ ( sin φ sinψ cos φ − cosψ sin φ ) Adding: P12 = P1′2 + P2′2 + P1′ P2′ cosψ Conservation of energy: P12 P1′2 P2′2 = + +Q 2m 2m 2m 1 P1′ P2′ cosψ Q= P12 − P1′2 − P2′2 ) = ( 2m 2m P′ P ′ cosψ Q= m ( ) 1 m1v12 T1′ = m1v1′ 2 2 T ′ v′ let r = = 12 … ratio of scattered particle to incident particle energy T1 v1 Looking at Figure 7.6.2 … v1′ ⋅ v1′ = ( v1′ − vcm ) ⋅ ( v1′ − vcm ) 7.19 T1 = v1′ = v1′ + vcm − 2v1′ vcm cos φ1 + 2v1′vcmγ hence v1′ = v1′ − vcm 2v′v γ v1′ v − + 2cm v1 v v1 v1′ = v1 but scattered particle are the same ∴r = where γ = cos φ1 cm v1′ m2 α = = v1 m2 + m1 + α …the center of mass speeds of the incident and …from equation 7.6.12 where α = m2 m1 vcm m1 = = Equation 7.6.11 v1 m2 + m1 + α Thus α2 2γ v1′ α2 2γ r2 r= − + = − + 2 2 1 + α v α + ( ) ( ) α α + + 1 1 + α + α ( ) ( ) ( ) ( ) Simplifying 2γ 12  − α  r− r + =0 1+α  1+α  Let x = r and solving the resulting quadratic for x γ 2   + γ − (1 − α )  x= 1+α 1+α  Squaring   2 2 2γ + α − + 2γ (γ + α − 1)  r=x =  (1 + α )     2 2 2 + − + + − γ α γ γ α ( )     And, after a little algebra, we get the desired solution ∆T1 2γ  = − γ + γ + α − 1   T1 + α (1 + α ) Now 7.20 ∆T1 = 1− r = 1− T1 (1 + α ) From Equation 7.6.15 … γ = m1v1 m1 v1 = v1′ ( m1 + m2 ) m2 (1 + m1 m2 ) v1′ v1 … v1′ Now we solve for  2T  v1 =   and now solving for v1′ starting with Equation 7.6.9 …  m1  m2 1 v1 we get … µ v1′2 = µ v12 − Q and using v1′ = 2 m1 + m2 1 T = m1v12 −Q = −Q (1 + m1 m2 ) (1 + m1 m2 )   T − Q  m1 (1 + m1 m2 )  (1 + m1 m2 )  Thus, solving for γ … v1′2 = 10 γ= m1 m2 ( 2T m1 ) 2 ( m1 ) (1 + m1   T m2 )  − Q  (1 + m1 m2 )  2 Finally… γ= 7.21 m1 m2 1  Q (1 + m1 m2 )  1 −  T   The time of flight, τ = constant—so τ = r but from problem 7.19 above v1′ v1 τ  γ + γ + α − 1   1+α As an example, let v1 τ = and we have r1 = γ α =1 r2 = γ + γ +  α =2  3 r3 = γ + γ + 15  α =4  5 r4 = γ + γ + 143  α = 12  13  Below is a polar plot of these four curves r = v1′ τ = 7.22 pp scattering p–D p – He p–C From eqn 7.7.6, Fu − Fg = mv + vm since v = constant, v = m = λ z = λv , λ = mass per unit length Fg = ( λ z ) g  v2  Fu = λ zg + ( λ v ) v = g λ  z +  g  Fu is equal to the weight of a length z + 11 v2 of chain g 7.23 m = π r3ρ m = 4π r ρ r ∝ π r z where v = z r = kz k a constant of proportionality r = r + kz From eqn 7.7.6, mg = mv + vm 4 π r ρ g = π r ρ z + 4π r ρ ( kz ) z 3 3kz g = z+ r 3z z=g− r z+ k 3z For r = , z = g − z A series solution is used for this differential equation: ∞ z = ∑ an z n n=0 z= dz dz dz dz d ( z ) = ⋅ =z = dt dz dt dz dz d (z ) = ∑ an nz n −1 dz n z2 = ∑ an z n −1 z n ∴ z = ∑ an nz n −1 = g − 3∑ an z n −1 n n For n = : a1 = g − 3a1 2 a1 = g For n ≠ : nan = −3an Since n is an integer, an = for n ≠ z2 = g z 32  g z = g −  g z = z7  12 7.24 From eqn 7.7.6, mg = mv + vm , where m and v refer to the portion of the chain hanging over the edge of the table m = λ z and v = z where λ is the mass per unit length of chain m = λ z and v = z dz dz dz dz d ( z ) = ⋅ =z = dt dz dt dz dz  d (z )  λ z g = λ z  + z (λ z )  dz  z= d (z2 ) z2 =g− z dz Because of the initial condition z = b ≠ , a normal power series solution to this differential equation (…as in Prob 7.22) does not work Instead, we use the Method of Frobenius … z= ∞ z = ∑ an z n + s n=0 d (z ) = ∑ an ( n + s ) z n + s −1 dz n z2 = ∑ an z n + s −1 z n z = ∑ an ( n + s ) z n + s −1 = g − ∑ an z n + s −1 n n Equality can be attained for an ≠ at n = and n = n ≠ 0,3 …otherwise an = a s = −a s = −2 z = ∑ an ( n − ) z n −3 = g − ∑ an z n −3 n n For n = , a3 = g − a3 2 a3 = g For all n ≠ , 3: an ( n − ) = − an an = , n ≠ 0,3 z = a z −2 + gz For n = , 13 At t = , z = , and z = b a gb 0= + b a = − gb3 b3 z = − g + gz z  b3  g At z = a , z = g  a −  = ( a − b3 ) a  3a   2g 2 z =  ( a − b3 )   3a  7.25 Initially, the upward buoyancy force balances the weight of the balloon and sand FB − ( M + m ) g = (1) Let m = m ( t ) − the mass of sand at time t where ≤ t ≤ t  t (2) m = m 1 −   t  The velocity of sand relative to the balloon is zero upon release so V = in equation 7.7.5 … there is no upward “rocket-thrust.” As sand is released, the net upward force is the difference between the initial buoyancy force, FB, and the weight of the balloon and remaining sand Let y be the subsequent displacement of the balloon, so equation 7.7.5 reduces to F = ma dv FB − ( M + m ) g = ( M + m ) dt and using (1) and (2) above we get ( M + m ) gt dv m gt = = −g + dt ( M + m ) t − m t (M + m )t − m t whose solution is: ( M + m ) gt ln 1 − m t  dy v= = − gt −   dt m  (M + m )t  g m   y = C − ∫  gt + ln (1 − kt )  dt , k= k t (M + m   gt tdt = C − gt − ln (1 − kt ) − g ∫ k − kt Integrating by parts 14 ) gt g   gt − ln (1 − kt ) − ∫  −1 +  dt k k  − kt  gt gt g = C − gt − ln (1 − kt ) + + ln (1 − kt ) k k k g gt = C − gt + (1 − kt ) ln (1 − kt ) + k k but y = at t = so C = gt g y = − gt + (1 − kt ) ln (1 − kt ) k k and at t = t =C− (a) (b) (c)  M gt  2M + m ) m + 2M ( M + m ) ln  ( 2m  M +m (M + m ) − m  gt  v= ( M + m ) ln  m  M  H= letting ε =    m