Analytical mechanics solutions manual fowles02

CHAPTER NEWTONIAN MECHANICS: RECTILINEAR MOTION OF A PARTICLE 2.1 ( F + ct ) m t F c x = ∫ ( F + ct ) dt = t + t m m 2m t F c 2 F c x=∫  t+ t  dt = t + t m m 2m  6m  (a) x = F sin ct m t F F F t x=∫ sin ct dt = − cos ct = (1 − cos ct ) m cm cm t F F x=∫ (1 − cos ct ) dt = 1 − sin ct  cm cm  c  (b) x = F ct e m F ct t F ct x= e = ( e − 1) cm cm F  ct  F ct x=  e − − t  = ( e − − ct ) cm  c c  cm (c) x = 2.2 dx dx dx dx = ⋅ =x dt dx dt dx dx x = ( F + cx ) dx m xdx = ( F + cx ) dx m 1 cx  x = F x+  m 2  (a) x = x 2 x =  ( F + cx )  m  (b) x = x dx = F e − cx dx m 1 F e − cx dx m F − cx F x =− e − 1) = ( (1 − e−cx ) cm cm xdx =  2F 2 x= − e − cx )  (  cm  dx (c) x = x = ( F cos cx ) dx m F xdx = cos cx dx m F sin cx x = cm  2F 2 sin cx  x=  cm  2.3 (a) V ( x ) = − ∫ x ( F + cx ) dx = − F x − x cx +C F − cx e +C c x F (c) V ( x ) = − ∫ F cos cx dx = − sin cx + C x c x (b) V ( x ) = − ∫ F e − cx dx = x 2.4 dV ( x ) = −kx dx x V ( x ) = ∫ kx dx = kx 2 (b) T = T ( x ) + V ( x ) (a) F ( x ) = − T ( x) = T −V ( x) = k ( A − x2 ) 2 kA (d) turning points @ T ( x1 ) → ∴ x1 = ± A (c) E = T = 2.5 x kx  1 kx V ( x ) = ∫  kx − dx = kx − A  A2  kx (b) T ( x ) = T − V ( x ) = T − kx + A2 (c) E = T (a) F ( x ) − kx + kx A2 so (d) V ( x ) has maximum at F ( xm ) → kx kxm − m2 = xm = ± A A 1 kA4 V ( xm ) = kA2 − = kA A2 If E < V ( xm ) turning points exist Turning points @ T ( x1 ) → let u = x1 1 ku E − ku + =0 A2 solving for u , we obtain   4E     u = A ± 1 −    kA     or  4E    x1 = ± A 1 − 1 −    kA    2.6 x = v ( x) = α x=− x α x2 x=− α2 x3 mα F ( x ) = mx = − x 2.7 2.8 F ≥ Mg sin θ F = mx = mx dx dx x = bx −3 dx = −3bx −4 dx F = m ( bx −3 )( −3bx −4 ) F = −3mb x −7 2.9  m m  (a) V = mgx = (.145kg )  9.8  (1250 ft )  3048  = 541J s  ft    (b) T = T= 1  mg  m g mv = mvt = m  = 2  c2  22 D (.145kg ) m   9.8  s   kg ( )(.22 ) ( )(.0366 ) m = 87 J   t  ∫ Fdx = ∫ −cv dx = −c ∫ v dt = −c ∫  −vt  τ   dt t  t   t   = cvt τ  −   + ∫   d    τ   τ   τ   t  t   = cvt τ  −   + ln cosh    τ   τ   t Now  τ   ≅ for t τ    t  t Meanwhile x = ∫ vdt = ∫  −vt    dt = vtτ ln cosh   τ  τ   t x ln cosh   =  τ  vtτ  m x = (1250 ft )  3048  = 381 m ft    m 2  kg 145 9.8 ( )     mg  m s2      = 34.72 vt =  =  kg s  (.22 )(.0732 )   c2    m   2    m 2 (.145kg )   = 3.543s τ = =   (.22 )(.0732 )2 kg  9.8 m    c2 g     m s     3.81 Fdx = 22 0732 34.72 3.543 − + ( )( ) ( ) ( )   = 454 J ∫ ( 34.72 )( 3.54 )   V − T = 541J − 87 J = 454 J 2.10 F 1F t t, x= m 2m F F For t1 ≤ t ≤ 2t1 : v = t1 , x = t1 , t = t1 m 2m For ≤ t ≤ t1 : v = F F 2F t1 + t1 ( t − t1 ) + ( t − t1 ) m 2m m F F F 5F At t = 2t1 : x = t1 + t1 + t1 = t1 m m 2m 2m dv dv dx dv c a= = ⋅ = v ⋅ = − v2 dt dx dt dx m − c v dv = − dx m v − xmax c ∫v v dv = ∫0 − mdx c −2v = − xmax m x= 2.11 xmax 2.12 2mv = c Going up: Fx = − mg sin 30 − µ mg cos 30 x = − g ( sin 30 + 0.1cos 30 ) = −5.749 m s2 v = v + at at the highest point v = so tup = − v = 0.174v s a 2 xup = v tup + atup2 = 0.174v − 087v = 0.087v m 2 Going down: x ′ = 0.087v , v ′ = , a′ = −9.8 ( 0.5 − 0.0866 ) ttotal 2.13 2 xdown = = 0.087v − 4.0513tdown tdown = 0.207v s = tup + tdown = 0.381v s At the top v = so e −2 kxmax = Coming down x = xmax g k g +v k and at the bottom x = g  v g g k 1) =   v = −  ( g k k g 2 +v  +v  k k  v= vt v (v t 2.14 +v 2 ) , vt = g mg = k c2 Going up: Fx = −mg − c2 v dv c2 a = v = − g − kv , k = dx m v x vdv ∫v − g − kv = ∫0 dx − v ln ( − g − kv ) = x v 2k g + kv = e −2 kx g + kv g g 2 v =  + v  e −2kx − k k  Going down: Fx = −mg + c2 v dv v = − g + kv dx v x vdv = ∫0 − g + kv ∫0 dx v ln ( − g + kv ) = x − x 2k k − v = e kx e −2 kx g g g  v = −  e −2 kx  e kx k k  2.15 Using dv = mg − c1v − c2 v dt t dt v dv = ∫0 m ∫0 mg − c1v − c2v m dx 2cx + b − b − 4ac ln = ∫ a + bx + cx b2 − 4ac 2cx + b + b2 − 4ac , −2c2 v − c1 − c1 + 4mgc2 t = ln 2 m c1 + 4mgc2 −2c2 v − c1 + c1 + 4mgc2 ( t c1 + 4mgc2 m ) 2c v + c + ( = ln ( 2c v + c − )( c − + 4mgc )( c + c1 + 4mgc2 c1 v 2 ) + 4mgc ) c1 + 4mgc2 c1 2 as t → ∞ , 2c2 vt + c1 − c1 + 4mgc2 = c1  c1  mg   +   + 2c2  2c2  c2    Alternatively, when v = vt , dv m = = mg − c1vt − c2 vt dt 2 vt = − c1  c1  mg   +   + 2c2  2c2  c2    vt = − 2.16 dv k = − x −2 dx m v x kdx ∫0 vdv = ∫b − mx 2 k 1 1 v =  −  m x b a=v 1 dx  2k  1    2k  b − x   v= =  −  =   dt  m  x b    mb  x    x 2  x t  mb   mb   x  ∫0 dt = ∫b  2k  b − x  dx =  2k  ∫1  b x  d  b   −   b x Since x ≤ b , say = sin θ b 1  mb3  sin θ ( 2sin θ cosθ dθ )  2mb3  2 t = =  ∫− π  ∫− π sin θ dθ k cos k θ      mb3  t =  π  8k  2.17 dv dv = mv = f ( x )i g ( v ) dt dx mvdv = f ( x ) dx g (v) m By integration, get v = v ( x ) = If F ( x,t ) = f ( x )i g ( t ) : dx dt d 2x d  dx  = m   = f ( x )i g ( t ) dt dt  dt  This cannot, in general, be solved by integration If F ( v,t ) = f ( v )i g ( t ) : m dv = f ( v )i g ( t ) dt mdv = g ( t ) dt f (v) m Integration gives v = v ( t ) dx = v (t ) dt dx = v ( t ) dt A second integration gives x = x ( t ) 2.18 c1 = (1.55 ×10−4 )(10−2 ) = 1.55 × 10−6 c2 = ( 0.22 ) (10−2 ) = 2.2 × 10−5 kg s kg s  1.55 ×10−6  (10−7 ) ( 9.8 )  1.55 × 10  vt = − +   + × 2.2 ×10−5  × 2.2 ×10−5  2.2 × 10−5    m vt = 0.179 s −6 (10 ) ( 9.8) = 0.211 m −7 Using equation 2.29, vt = 2.2 ×10−5 s 2.19 F ( x ) = − Aeα x = mx Let u = eα v or du = α eα v dv F ( v ) = − Aeα v = mv dv = du du = αv αe αu dv A = − dt αv e m du αA ∴ =− dt u m Integrating 1 A − = αt u u m and substituting eα v = u  A  ln 1 + eα v α t  α  m  (b) t = T @ v =  A  α v = ln 1 + eα v αT   m  A m 1 − e −α v  eα v = + eα v αT T= αA m dv A vdv A = − dx (c) v = v = − eα v αv dx m e m (a) v = v − again, let u = eα v du = α udv or dv = du αu v= α ln u 1  du α ln u  α u A = − dx Integrating and solving u m m x = 1 − (1 + α v ) e −α v  α A 2.20 d ( mv ) = mv + vm = mg dt but m = ρ π r m = ρ1π r v 4 so (1) π ρ r 3v + π ρ1r v = π = π ρ r g 3 F= Now ρ1 ≈ 10−3 ρ so, second term is negligible-small hence v ≈ g and v ≈ gt speed ∝ t but m = ρ 4π r r = ρ1 π r v or r ≅ ρ1 v 4ρ Hence r ≈ ρ1 gt and rate of 4ρ growth ∝ t The exact differential equation from (1) above is: 4ρ 4ρ r r + πρ1 πρ r = π ρ rg 3 ρ1 ρ1 3r ρ = g r 4ρ Using Mathcad, solve the above non-linear d.e letting which reduces to: r + ρ1 ≈ 10−3 and R ≈ 0.01mm (small raindrop) Graphs ρ show that v ∝ r ∝ t and r ∝ t 10