Analytical mechanics solutions manual fowles04

CHAPTER GENERAL MOTION OF A PARTICLE IN THREE DIMENSIONS Note to instructors … there is a typo in equation 4.3.14 The range of the projectile is … v sin 2α v sin 2α … NOT R=x= g g 4.1 ∂V ˆ ∂V ˆ ∂V −j −k ∂x ∂y ∂z ˆ ˆ + ˆjxz + kxy F = −c iyz (a) F = −∇V = −iˆ ( ) (b) F = −∇V = −iˆ2α x − ˆj β y − kˆ 2γ z (c) F = −∇V = ce−(α x + β y +γ z ) iˆα + ˆj β + kˆγ ( (d) F = −∇V = −eˆr ) ∂V ∂V ∂V − eˆθ − eˆφ ∂r r ∂θ r sin θ ∂φ F = −eˆr cnr n −1 4.2 (a) iˆ ∂ ∇× F = ∂x x ˆj ∂ ∂y y kˆ ∂ =0 ∂z z conservative iˆ ∂ ∇× F = ∂x y ˆj ∂ ∂y −x kˆ ∂ = kˆ ( −1 − 1) ≠ ∂z z2 conservative iˆ ∂ ∇× F = ∂x y ˆj ∂ ∂y x kˆ ∂ = kˆ (1 − 1) = ∂z z3 conservative (b) (c) (d) eˆr ∂ ∇× F = r sin θ ∂r − kr − n 4.3 eˆθ r ∂ ∂θ eˆφ r sin θ ∂ = conservative ∂φ (a) iˆ ∂ ∇× F = ∂x xy ˆj ∂ ∂y cx kˆ ∂ = k ( 2cx − x ) ∂z z3 ˆj ∂ ∂y cxz y2 kˆ ∂ ∂z x y 2cx − x = c= (b) iˆ ∂ ∇× F = ∂x z y also 4.4 (a)  x cx  = iˆ  − −  + y   y x cx − − =0 y y cz z + =0 y2 y2 ˆj  −  + kˆ  cz + z  y2   y y y c = −1 implies that E = constant = V ( x, y, z ) + mv 2 at the origin E = + mv 1 at (1,1,1) E = α + β + γ + mv = mv 2 2 c = −1 as it must v2 = v2 − (α + β + γ ) m  2 v = v − (α + β + γ )  m   (b) v2 − (α + β + γ ) = m 2 2 v =  (α + β + γ )  m  (c) mx = Fx = − ∂V ∂x mx = −α ∂V my = − = −2 β y ∂y ∂V mz = − = −3γ z ∂z 4.5 (a) ˆ + ˆjy F = ix ˆ + ˆjdy dr = idx on the path x = y : (1,1) ∫( 0,0 ) 1 1 0 0 F ⋅ dr = ∫ Fx dx + ∫ Fy dy = ∫ xdx + ∫ ydy = ˆ dr = idx dr = ˆjdy on the path along the x-axis: and on the line x = : (1,1) ∫( 0,0 ) 1 0 F ⋅ dr = ∫ Fx dx + ∫ Fy dy = F is conservative (b) ˆ − ˆjx F = iy on the path x = y : (1,1) ∫( 0,0 ) 1 0 1 0 F ⋅ dr = ∫ Fx dx + ∫ Fy dy = ∫ ydx − ∫ xdy (1,1) and, with x = y ∫( on the x-axis: ∫( and, with y = on the x-axis ∫( on the line x = : ∫( 0,0 ) (1,0 ) 0,0 ) (1,0 ) 0,0 ) (1,1) 1,0 ) 1 0 F ⋅ dr = ∫ xdx − ∫ ydy = 1 0 F ⋅ dr = ∫ Fx dx = ∫ ydx F ⋅ dr = 1 0 F ⋅ dr = ∫ Fy dy = ∫ xdy (1,1) ∫( and, with x = on this path 1,0 ) (1,1) ∫( 0,0 ) F ⋅ dr = ∫ dy = F ⋅ dr = + = F is not conservative 4.6 From Example 2.3.2, V ( z ) = −mg re2 ( re + z )  z V ( z ) = − mgre 1 +   re  From Appendix D, (1 + x ) −1 −1 = − x + x +…   z z2 V ( z ) = − mgre 1 − + + …   re re  mgz V ( z ) = −mgre + mgz − +… re With − mgre an additive constant,  z V ( z ) ≈ mgz 1 −   re  ∂ F = −∇V = − kˆ V ( z ) ∂z ˆ 1 − z + z  −   = − kmg    re    re ˆ 1 − z  F = − kmg   re   mx = Fx = , mz my = Fy =  2z  dz = − mg 1 −  dz re   h 2z  zdz = − g ∫ 1 −  dz z re    h2  − v 2z = − g  h −  re   ∫ v re v 2z h − re h + =0 2g  2z  mz = − mg 1 −  re   h= re 2re v 2z − re − g 2 h= re re 2v − 1− z 2 gre From Appendix D, ( h, z (1 + x ) = + h= re re v 2z v4 − + + z +… 2 g gre h≈ v 2z  v 2z  1 +  g  gre  re ) x x2 − +… −1 v2  v2  From Example 2.3.2, h = 1 −  g  gre  v2  v2  −1 And with (1 − x ) ≈ + x , h≈ +   g  gre  4.7 For a point on the rim measured from the center of the wheel: ˆ cos θ − ˆjb sin θ r = ib vt θ =ωt = , so r = −iˆv sin θ − ˆjv cos θ b Relative to the ground, v = iˆv (1 − sin θ ) − ˆjv cos θ For a particle of mud leaving the rim: y = −b sin θ and v y = −v cosθ So v y = v y − gt = −v cosθ − gt y = −b sin θ − v t cosθ − gt 2 At maximum height, v y = : and t=− v cos θ g  v cos θ h = −b sin θ − v  − g  2 v cos θ h = −b sin θ + 2g Maximum h occurs for   v cos θ   cos θ − g  −  g    dh 2v cos θ sin θ = = −b cos θ − 2g dθ sin θ = − gb v2 cos θ = − sin θ = v − g 2b v4 gb v − g 2b gb v + = 2+ v2 gv 2v 2g Measured from the ground, gb v ′ =b+ + hmax 2v 2g hmax =  gb  The mud leaves the wheel at θ = sin −1  −   v  4.8 x = R cos φ so t = and x = v x t = ( v cos α ) t R cos φ v cos α 1 and y = v y t − gt = ( v sin α ) t − gt 2 y = R sin φ R cos φ  R cos φ  R sin φ = ( v sin α ) − g  v cos α  v cos α  gR cos φ sin φ = tan α cos φ − 2v cos α α φ 2v cos α 2v cos α tan cos sin α φ − φ = ( ) ( sin α cos φ − cos α sin φ ) g cos φ g cos φ From Appendix B, sin (θ + φ ) = sin θ cos φ + cos θ sin φ R= 2v cos α R= sin (α − φ ) g cos φ dR 2v =0=  − sin α sin (α − φ ) + cos α cos (α − φ )  dα g cos φ  Implies that cos α cos (α − φ ) − sin α sin (α − φ ) = R is a maximum for From appendix B, cos (θ + φ ) = cos θ cos φ − sin θ sin φ so cos ( 2α − φ ) = 2α − φ = Rmax = π α= 2v π φ  π φ  cos  +  sin  −  g cos φ  2  2 π + φ π  π φ  π φ  π φ  Now sin  −  = cos  −  −   = cos  +   2  2    2v π φ  Rmax = cos  +  g cos φ  2 Again using Appendix B, cos 2θ = cos θ − sin θ = cos θ − v2   π 2v  π  1   Rmax = cos cos  + φ  + 1  +φ  +  =   2 g cos φ    2   g cos φ   π  Using cos  + θ  = − sin θ , 2  v Rmax = (1 − sin φ ) g (1 − sin φ ) Rmax = v g (1 + sin φ ) 4.9 (a) Here we note that the projectile is launched “downhill” towards the target, which is located a distance h below the cannon along a line at an angle φ below the horizon α is the angle of projection that yields maximum range, Rmax We can use the α results from problem 4.8 for this problem We simply have to replace the angle φ in the above φ result with the angle -φ, to account for the downhill h slope Thus, we get for the downhill range … Rmax 2v0 cos α sin (α + ϕ ) R= cos ϕ g The maximum range and the angle is α are obtained from the problem above again by v (1 + sin ϕ ) π replacing φ with the angle -φ … Rmax = and … 2α = − ϕ 2 g cos ϕ 2 v (1 + sin ϕ ) v0 h We can now calculate α … Rmax = = = g cos ϕ g (1 − sin ϕ ) sin ϕ  gh  1 +   v0  π  But, from the above … sin ϕ = sin  − 2α  = cos 2α = − 2sin α 2  gh  gh  Thus … − sin α =  +  v0  v0  Solving for sin ϕ … sin ϕ = gh v0 2sin α = gh = 1− 2 csc α v0  gh  1 +  =  v0  + gh v0  gh  Finally … csc α = 1 +   v0  (b) Solving for Rmax … Rmax = h h h = = sin ϕ − 2sin α − csc2 α Substituting for csc α and solving … v2 gh  Rmax = 1 +  g  v0  4.10 We can again use the results of problem 4.8 The maximum slope range from problem 4.8 is given by … v0 h = Rmax = g (1 + sin ϕ ) sin ϕ Solving for sin ϕ … sin ϕ = gh v0  gh  1 −   v0  Thus … cos ϕ sin ϕ We can calculate cos ϕ from the above relation for sin ϕ … xmax = Rmax cos ϕ = h  gh   gh  cos ϕ = (1 − sin ϕ ) = 1 − 2   −  v0   v0   Inserting the results for sin ϕ and cos ϕ into the above … xmax cos ϕ v0  gh  2 =h = −   sin ϕ g  v0  4.11 We can simplify this problems somewhat by noting that the trajectory is symmetric about a vertical line that passes through the highest point of the trajectory Thus we have the following picture … z zmax v0 h0 h1 α x0 δ x x1 R We have “reversed the trajectory so that h0 (= 9.8 ft), and x0 , the height and range within which Mickey can catch the ball represent the starting point of the trajectory h1 (=3.28 ft) is the height of the ball when Mickey strikes it at home plate δ is the distance behind home plate where the ball would be hypothetically launched at some angle α to achieve the total range R x1 (=328 ft) is the distance the ball actually would travel from home plate if not caught by Mickey (Note, because of the symmetry, v0 is the speed of the ball when it strikes the ground … also at the same angle α at which was launched We will calculate the value of x0 assuming a time-reversed trajectory!) v sin 2α 2v0 sin α cos α = (1) The range of the ball … R = g g R g R tan α −   2 2v0 cos α   g (3) The height at x1 … h1 = x1 tan α − ( x1 ) 2 2v0 cos α g tan α = and inserting this into (2) gives … From (1) … 2 R 2v0 cos α R R R zmax = tan α − tan α = tan α 4 4z Thus, R = max and inserting this expression and the first previously derived into (3) tan α (2) The maximum height … zmax = (4) h1 = x1 ( x tan α ) tan α − zmax Let u = x1 tan α and we obtain the following quadratic … u − zmax u + zmax h1 = and solving for u … h   u = zmax 1 ± (1 − h1 zmax )  and letting ε = , we get … zmax   u ≈ zmax ε = h1 or u ≈ zmax ( − ε ) = zmax ( − 0475 ) = 3.9 zmax This result is the correct one … 3.9 zmax = 0.821 ∴ α = 39.40 x1 Now solve for x0 using a relation identical to (4) … Thus, tan α = h0 = x0 ( x tan α ) tan α − zmax Again we obtain a quadratic expression for u = x0 tan α which we solve as before This time, though, the first result for u is the correct one to use … u = zmax ε ≈ h0 and we obtain … h x0 = = 11.9 ft tan α 4.12 The x and z positions of the ball vs time are 1 x = v t cos θ z = v t cos θ sin θ − gt 2 2 Since vx = v cos θ v2 The horizontal range is cos θ sin 2θ R= g dR The maximum range occurs @ =0 dθ 1 dR v   =  cos θ cos 2θ − cos θ sin θ sin 2θ  = 2 dθ g   1 Thus, cos θ cos 2θ = cos θ sin θ sin 2θ 2 2 Using the identities: cos θ = + cos 2θ and sin 2θ = 2sin θ cosθ We get: (1 + cosθ ) ( cos2 θ − 1) = sin θ sin θ cosθ = (1 − cos θ ) cosθ or (1 + cosθ ) ( 3cos θ Thus cos θ = −1 , − cosθ − 1) = cosθ = ( 1 ± 13 ) Only the positive root applies for the θ -range: ( ) cosθ = + 13 = 0.7676 Thus (b) for v = 25 m s −1 Rmax = 55.4 m 10 ≤θ ≤ θ = 39 51′ @ θ = 39 51′ π (a) The maximum height occurs at or v cos θ sin θ = gT v2 H= cos θ sin θ 2g or at dz =0 dt v cos θ sin θ T= g maximum at fixed θ dH =0 dα dH v  1  2 =  cos θ sin θ cosθ − cos θ sin θ sin θ  = dα g  2  Using the above trigonometric identities, we get 1 (1 + cosθ ) sin θ cosθ = sin θ sin θ = sin θ (1 − cos θ ) 2 or sin θ (1 + cos θ )( 3cos θ − 1) = The maximum possible height occurs @ The first two roots give minimum heights; the last gives the maximum Thus, H max = 18.9m @θ = cos −1 = 70 32′ There are 3-roots: 4.13 sin θ = , cos θ = −1 , cos θ = The trajectory of the shell is given by Eq 4.3.11 with r replacing x z g z = r − r2 where r = v cos θ z = v sin θ r 2r Thus, z = r tan θ − g r2 sec θ 2v Since sec2 θ = + tan θ We have: g r2 g r2 r z tan θ − tan θ + + =0 2v 2v (r,z) are target coordinates The above equation yields two possible roots: 1  2 2 ± − − tan θ = v v gzv g r ( )  gr   The roots are only real if v − gzv − g r ≥ The critical surface is therefore: v − gzv − g r = 4.14 If the velocity vector, of magnitude s , makes an angle θ with the z-axis, and its 11 projection on the xy-plane make an angle φ with the x-axis: x = s sin θ cos φ , and Fx = Fr sin θ cos φ = mx y = s sin θ sin φ , and Fy = Fr sin θ sin φ = my z = s cos θ , and Fz = − mg + Fr cos θ = mz Since Fr = −c2 s = −c2 ( x + y + z ) , the differential equations of motion are not 2 separable mx = −c2 s sin θ cos φ = −c2 sx dx dx ds dx m = m ⋅ = ms = −c2 sx dt ds dt ds dx c2 c = − ds = −γ ds , where γ = m x m x ln x − ln x = ln = −γ s x θ x z g γ x g  γx  From eqn 4.3.16,  +  max + ln 1 − max  = x  γ γ γ  x  u u From Appendix D: ln (1 − u ) = −u − − − … for u < 3  γx  γx γ xmax γ xmax − + terms in γ ln  − max  = − max − x  x 2x 3x  z xmax gxmax gxmax gxmax gγ xmax + − − − + terms in γ = γx γx 2x 3x x + xmax 3x 3x z ≈0 xmax − 2γ gγ xmax 3x  x 3x z  ≈− ± +  gγ  4γ  16γ x x  16γ z  ± xmax ≈ − 1 +  4γ 4γ  3g  Since xmax > , the + sign is used From Appendix D:  16γ z  8γ z  16γ z  −  1 +  = 1+  + terms in γ 3g  3g  3g   3x 3x x z x γ z + + − + terms in γ xmax = − 4γ 4γ g 3g 12 s y x = x e −γ s y = y e −γ s Similarly 4.15 z φ x z 8x z − γ +… g 3g z = v sin α and x z = v sin 2α : xmax = For xmax = v sin 2α 4v sin 2α sin α − γ +… g 3g 4.16 y x x = A cos (ω t + α ) , x = − Aω sin (ω t + α ) from x = , from x = A , y = B cos (ω t + β ) , α =0 x = A cos ω t y = −ω B sin (ω t + β ) 2 kB = ky + my 2 with B = 16 A2 + Then ω2 ( 9ω y = A , y = 3ω A and ω = A2 ) = 25 A2 B = 5A A = A cos β and 3ω A = −5ω A sin β 4  3 β = cos −1   = sin −1  −  = −36.9 5  5 y = A cos (ω t − 36.9 ) Since maximum x and y displacements are ± A and ±5A , respectively, the motion takes place entirely within a rectangle of dimension A and 10A ∆ = β − α = −36.9 − = −36.9 AB cos ∆ From eqn 4.4.15, tan 2ψ = A − B2 4 ( A)( A) cos ( −36.9 ) 10   = =− tan 2ψ = −24 A2 − ( A )  1 ψ = tan −1  −  = −9.2  3 4.17 mx = Fx = −  x = A cos   ∂V = − kx = −π mx ∂x  k t + α  = A cos (π t + α ) m  13 k : m ∂V = −4π mx ∂y y = B cos ( 2π t + β ) my = − ∂V = −9π mz ∂z z = C cos ( 3π t + γ ) mz = − Since x = y = z = at t = , α = β =γ = − π π  x = A cos  π t −  = A sin π t 2  x = Aπ cos π t Since v = x + y + z and x = y = z , v x = = Aπ v A= π v x= sin π t π y = B sin 2π t , y = Bπ cos 2π t v y = = 2π B v B= 2π v y= sin 2π t 2π z = C sin 3π t , z = 3Cπ cos 3π t v z = = 3Cπ v C= 3π v z= sin 3π t 3π Since ω x = π , ω y = 2π , and ω z = 3π the ball does retrace its path tmin = 2π n1 ωx = 2π n2 ωy = 2π n3 ωz The minimum time occurs at n1 = , n2 = , n3 = 14 tmin = 4.18 2π π =2 Equation 4.4.15 is tan 2ψ = AB cos ∆ A2 − B Transforming the coordinate axes xyz to the new axes x′y′z ′ by a rotation about the z-axis through an angleψ given, from Section 1.8: x′ = x cosψ + y sinψ , y′ = − x sinψ + y cosψ or, x = x′ cosψ − y′ sinψ , and y = x′ sinψ + y′ cosψ From eqn 4.4.10: x2 cos ∆ y x y − + = sin ∆ A AB B Substituting: x′2 cos ψ − x′y′ cosψ sinψ + y′2 sin ψ ) ( A cos ∆  x′ cosψ sinψ + x′y′ ( cos ψ − sin ψ ) − y′2 cosψ sinψ  −  AB  + ( x′2 sin ψ + x′y′ cosψ sin ψ + y′2 cos ψ ) = sin ∆ B For x′ to be a major or minor axis of the ellipse, the coefficient of x′y′ must vanish cosψ sinψ cos ∆ cosψ sinψ cos ψ − sin ψ ) + − − =0 ( A AB B2 From Appendix B, cosψ sinψ = sin 2ψ and cos ψ − sin ψ = cos 2ψ sin 2ψ cos ∆ cos 2ψ sin 2ψ − − + =0 A2 AB B2  cos ∆  tan 2ψ  −  = A  AB B AB cos ∆ tan 2ψ = A − B2 4.19 Shown below is a face-centered cubic lattice Each atom in the lattice is centered within a cube on whose faces lies another adjacent atom Thus each atom is surrounded by nearest neighbors at a distance d We neglect the influence of atoms that lie at further distances Thus, the potential energy of the central atom can be approximated as V = ∑ cri −α i =1 2d 15 r1 = ( d − x ) + y + z    − α α  2x x2 + y2 + z  + r = ( d − 2dx + x + y + z ) = d 1 −  d d2   n From Appendix D, (1 + x ) = + nx + n ( n − 1) x + … 2  α  x x + y + z   α   α  r1−α = d −α 1 −  − +  +  −   − − 1 d2   d     −α 2 2 −2 −α 2 2 2  x  x3   2x   x + y + z   x + y + z   + term s in +  −  −         d2 d2 d3   d   d      x3   α  α   x2  αx α − ( x + y + z ) +  + 1  + terms in   r1−α = d −α 1 + d   d 2d   d   αx α α α   r1−α ≈ d −α 1 + − ( x + y + z ) +  + 1 x  d 2d d 2    1 r2 = ( − d − x ) + y + z  =  d + 2dx + x + y + z    − −α r r2−α α  x x2 + y + z  = d 1 + +  d d2    αx α α α   ≈ d −α 1 − − ( x + y + z ) +  + 1 x  d 2d d 2    −α α α   r1−α + r2−α ≈ d −α  − ( x + y + z ) + (α + ) x  d  d  Similarly: α α   r3−α + r4−α ≈ d −α  − ( x + y + z ) + (α + ) y  d  d  α α   r5−α + r6−α ≈ d −α  − ( x + y + z ) + (α + ) z  d  d   3α   α 2α  V ≈ cd −α 6 − ( x + y + z ) +  +  ( x + y + z )  d d  d   −α −α − 2 2 ≈ 6cd + cd (α − α )( x + y + z ) V ≈ A + B ( x2 + y + z ) 4.20 16 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 qE qBx  qB  eE eBx  eB  y= − − + −  y  y=− m m  m m m m eE eB y +ω2 y = − +ωx , ω= m m  eE  y = − + ω x  + A cos (ω t + θ ) ω  m  y = − Aω sin (ω t + θ ) y = , so θ =  eE  − +ωx   ω  m   eE  a = − y = a (1 − cos ω t ) , +ωx  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 = , so A = − x = ( x − ω a ) t + a sin ω t x = a sin ω t + bt , mz = Fz = z=zt+z =0 b = x −ωa 4.21 b mv + mqh = mg 2 v = g ( b − 2h ) y b h mv Fr = − = −mg cos θ + R b x 17 h b h mv mg mg R = mg − =  h − ( b − 2h )  = ( 3h − b ) b b b b the particle leaves the side of the sphere when R = b b h = , i.e., above the central plane 3 cos θ = 4.22 mv + mgh = at the bottom of the loop, h = −b mv = mgb , v = 2gb so h b mv Fr = −mg + R = b 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  mg  E = ms +  s , 2  4A  s is undergoing harmonic motion with: "k " g g ω= = = m 4A A Since s = A sin φ , φ increases by 2π radians during the time interval:  A = 2π   ω  g For cycloidal motion, x and z are functions of 2φ so they undergo a complete cycle every time φ changes by π Therefore, the period for the cycloidal motion is one-half the period for s A T = T ′ = 2π g T′ = 2π 18