Analytical mechanics solutions manual fowles01

CHAPTER FUNDAMENTAL CONCEPTS: VECTORS 1.1 K K (a) A + B = (iˆ + ˆj ) + ( ˆj + kˆ) = iˆ + ˆj + kˆ K K A + B = (1 + + 1) = K K (b) A − B = 3(iˆ + ˆj ) − 2( ˆj + kˆ) = 3iˆ + ˆj − 2kˆ K K (c) A ⋅ B = (1)(0) + (1)(1) + (0)(1) = iˆ ˆj kˆ K K (d) A × B = 1 = iˆ(1 − 0) + ˆj (0 − 1) + kˆ(1 − 0) = iˆ − ˆj + kˆ 1 K K A × B = (1 + + 1) = K K K 1.2 (a) A ⋅ ( B + C ) = 2iˆ + ˆj ⋅ iˆ + ˆj + kˆ = (2)(1) + (1)(4) + (0)(1) = )( ( K K ) K ( A + B ) ⋅ C = ( 3iˆ + ˆj + kˆ ) ⋅ ˆj = (3)(0) + (1)(4) + (1)(0) = K K K (b) A ⋅ ( B × C ) = 1 = −8 K K K K K K ( A × B ) ⋅ C = A ⋅ ( B × C ) = −8 K K K K K K K K K (c) A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C = iˆ + kˆ − ˆj = 4iˆ − ˆj + 4kˆ K K K K K K ( ) ( ) K K K K K K ( A × B ) × C = −C × ( A × B ) = − ( C ⋅ B ) A − ( C ⋅ A) B  = − 0 ( 2iˆ + ˆj ) − ( iˆ + kˆ )  = 4iˆ + 4kˆ   K K 5a A ⋅ B (a )(a ) + (2a )(2a) + (0)(3a) 1.3 cos θ = = = AB a 14 5a 14a θ = cos −1 ≈ 53° 14 1.4 K (a) A = iˆ + ˆj + kˆ K K A = A⋅ A K (b) B = iˆ + ˆj K K B = B⋅B : body diagonal = iˆ ⋅ iˆ + ˆj ⋅ ˆj + kˆ ⋅ kˆ = : face diagonal = iˆ ˆj kˆ K K K (c) C = A × B = 1 1 K K A⋅ B 1−1 (d) cos θ = = =0 AB 1.5 ∴θ = 90D K K K B = B = A × C = AC sin θ K K A ⋅ C = AC cos θ = u K K A C = Cx + A u K = A+ A 1.6 B A u ∴ Cx = C cos θ = A ∴ C y = C sin θ = K K K K B× A u K B× A B  K K Cy = A +   A AB  A  B× A K K B× A A2 K dA ˆ d d d = i (α t ) + ˆj ( β t ) + kˆ (γ t ) = iˆα + ˆj β t + kˆ3γ t dt dt dt dt K d2A ˆ = j β + kˆ6γ t dt 2 K K = A ⋅ B = ( q )( q ) + ( 3)( − q ) + (1)( ) = q − 3q + 1.7 ( q − ) ( q − 1) = , q = or K K2 K K K K K K A + B = ( A + B ) ⋅ ( A + B ) = A2 + B + A ⋅ B 1.8 K K  A + B  = A2 + B + AB   K K Since A ⋅ B = AB cosθ ≤ AB , K K K K A+ B ≤ A + B K K K K K K A ⋅ B = AB cosθ = A B cosθ ≤ A B B cos θ ≤ B K K K K K K K K K Show A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C 1.9 iˆ K A × Bx Cx or kˆ K K Bz = ( Ax Cx + Ay C y + Az Cz ) B − ( Ax Bx + Ay By + Az Bz ) C Cz ˆj By Cy = ( Ax BxCx + Ay Bx C y + Az Bx Cz − Ax Bx Cx − Ay By Cx − Az Bz Cx ) iˆ + ( Ax By Cx + Ay By C y + Az By Cz − Ax Bx C y − Ay By C y − Az Bz C y ) ˆj + ( Ax Bz Cx + Ay Bz C y + Az Bz C z − Ax BxCz − Ay By Cz − Az Bz Cz ) kˆ ( A B C + A B C − A B C − A B C ) iˆ = + ( A B C + A B C − A B C − A B C ) ˆj + ( A B C + A B C − A B C − A B C ) kˆ y x y z x z y y x z z x x y x z y z x x y z z y x z x y z y x x z y y z iˆ Ax By Cz − Bz C y iˆ ( Ay Bx C y − Ay By Cx − Az Bz Cx + Az Bx Cz ) kˆ Az = + ˆj ( Az By Cz − Az Bz C y − Ax Bx C y + Ax By Cx ) Bx C y − By Cx + kˆ A B C − A B C − A B C + A B C ( x z x x x z y y z y z y) ˆj Ay Bz Cx − Bx Cz 1.10 y = A sin θ 1  Α =  xy  + y ( B − x ) = xy + yB − xy = AB sin θ 2  K K Α = A× B 1.11 iˆ K K K K A ⋅ ( B × C ) = A ⋅ Bx Cx 1.12 x G u kˆ Ax Bz = Bx Cz Cx ˆj By Cy Ay By Cy Bx Az Bz = − Ax Cz Cx By Ay Cy Bz K K K Az = − B ⋅ ( A × C ) Cz z G A G C x G B G K K Let A = (Ax,Ay,Az), B = (0,By,0) and C = (0,Cy,Cz) K K G G G Cz is the perpendicular distance between the plane A , B and its opposite u = B x C is G G G G directed along the x-axis since the vectors B , C are in the y,z plane u x = B x C = ByCz G G is the area of the parallelogram formed by the vectors B , C Multiply that area times the K K height of plane A , B = Ax to get the volume of the parallopiped G G G V = Ax u x = Ax By Cz = A • B x C ( ) 1.13 For rotation about the z axis: iˆ ⋅ iˆ′ = cos φ = ˆj ⋅ ˆj ′, kˆ ⋅ kˆ′ = iˆ ⋅ ˆj ′ = − sin φ ˆj ⋅ iˆ′ = sin φ For rotation about the y′ axis: iˆ ⋅ iˆ′ = cos θ = kˆ ⋅ kˆ′, iˆ ⋅ kˆ′ = sin θ kˆ ⋅ iˆ′ = − sin θ  cos θ I  T =  sin θ  ˆj ⋅ ˆj ′ = − sin θ   cos φ    − sin φ cosθ   sin φ cos φ 0   cos θ cos φ    =  − sin φ   sin θ cos φ cosθ sin φ cos φ sin θ sin φ − sin θ    cosθ  1.14 iˆ ⋅ iˆ′ = cos 30D = ˆj ⋅ iˆ′ = sin 30D = kˆ ⋅ iˆ′ = iˆ ⋅ ˆj ′ = − sin 30D = − ˆj ⋅ ˆj ′ = cos 30D = kˆ ⋅ ˆj ′ = ˆj ⋅ kˆ′ = iˆ ⋅ kˆ′ =   3  0  3+   2     Ax′       3  A  = −   y′   2    =  − 1   −1  Az ′       −1         K A = 3.232iˆ′ + 1.598 ˆj ′ − kˆ′ kˆ ⋅ kˆ′ = 1.15 Rotate thru φ about z-axis φ = 45D Rφ Rotate thru θ about x’-axis Rθ Rotate thru ψ about z’-axis θ = 45 ψ = 45D     Rφ =  −     2  1   Rθ =    0 −  D  0   0  1          2   0  2     Rψ =  − 0 2   1      1  +  2− 2 2  1  − + R (ψ ,θ , φ ) = Rψ Rθ Rφ =  − − 2 2 2  1  −  2  1 K   where x′ =   and 0   2 we have: ψ + β + α = K K Condition is: x′ = Rx K K Since x ⋅ x = After a lot of algebra: α = 1.16 Rψ    1 α     or   = R (ψ ,θ , φ )  β   0 γ         2 α  K   x =β  γ    2 − , β= + , γ = 4 K v = vτˆ = ctτˆ v2 c 2t K a = vτˆ + nˆ = cτˆ + nˆ b ρ b K , v = τˆ bc and c K K v ⋅a c bc cos θ = = = va bc 2c K a = cτˆ + cnˆ at t = θ = 45D 1.17 K ˆ ω sin (ω t ) + ˆj 2bω cos (ω t ) v ( t ) = −ib 1 K v = ( b 2ω sin ω t + 4b 2ω cos ω t ) = bω (1 + 3cos ω t ) K ˆ ω cos ω t − ˆj 2bω sin ω t a ( t ) = −ib K a = bω (1 + 3sin ω t ) 1.18 K v = 2bω ; t = 0, at at t= π , 2ω K v = bω K ˆ ω cos ω t − ˆjbω sin ω t + kˆ 2ct v ( t ) = ib K ˆ ω sin ω t − ˆjbω cos ω t + kˆ 2c a ( t ) = −ib 1 K a = ( b 2ω sin ω t + b 2ω cos ω t + 4c ) = ( b 2ω + 4c ) 1.19 K  ˆr + rθeˆθ = bke kt eˆr + bce kt eˆθ v = re K a =  r − rθ eˆr + rθ + 2rθ eˆθ = b ( k − c ) e kt eˆr + 2bcke kt eˆθ K K b k ( k − c ) e kt + 2b c ke kt v ⋅a = cos φ = 1 va kt 2 kt  2 2 2 be ( k + c ) be ( k − c ) + 4c k   ( cos φ = 1.20 ) ( ) k ( k + c2 ) (k +c 2 ) (k +c k = ) (k +c 2 ) , a constant K  − Rφ ) eˆ + ( Rφ + Rφ ) eˆ +  a = (R zeˆz R φ K a = −bω eˆR + 2ceˆz K a = ( b 2ω + 4c ) 1.21 1.22 K r ( t ) = iˆ (1 − e − kt ) + ˆje kt K ˆ − kt + ˆjkekt r ( t ) = ike K ˆ e − kt + ˆjk ekt r ( t ) = −ik K v = eˆr r + eˆφ rφ sin θ + eˆθ rθ π  π K  v = eˆφ bω sin  1 + cos ( 4ω t )   − eˆθ b ω sin ( 4ω t )  2  π K π  v = eˆφ bω cos  cos ( 4ω t )  − eˆθ bω sin ( 4ω t ) 8   2 K π  π v = bω cos  cos 4ω t  + sin 4ω t  8    Path is sinusoidal oscillation about the equator 1.23 K K v ⋅ v = v2 K K dv K K dv ⋅v + v ⋅ = 2vv dt dt K K 2v ⋅ a = 2vv K K v ⋅ a = vv 1.24 1.25 K d K K K dr K K K d K K  r ⋅ ( v × a )  = ⋅ (v × a ) + r ⋅ (v × a ) dt dt dt K K K K K K  dv K   K da   = v ⋅ ( v × a ) + r ⋅  × a  +  v ×   dt      dt K  K K  = + r ⋅ 0 + ( v × a ) d K K K K K K  r ⋅ ( v × a )  = r ⋅ ( v × a ) dt K K v = vτˆ and a = aτ τˆ + an nˆ K K v ⋅a K K v ⋅ a = vaτ , so aτ = v a = aτ2 + an2 , so an = ( a − aτ2 ) 1.26 For 1.14, aτ = −b 2ω cos ω t ⋅ sin ω t + b 2ω sin ω t ⋅ cos ω t + 4c 2t (b ω aτ = cos ω t + b 2ω sin ω t + 4c 2t ) 4c 2t (b ω 2 2 + 4c t )  2 2 16c 4t 2 a n =  b ω + 4c − 2  b ω + 4c t   b k ( k − c ) e kt + 2b c ke kt kt 2 = bke ( k + c ) For 1.15, aτ = kt 2 be ( k + c ) 1 2 an = b e kt ( k + c ) − b k e kt ( k + c )  = bce kt ( k + c )   1.27 v K K v = vτˆ , a = vτˆ + nˆ ρ v v3 K K v × a = v ⋅ an = v = ρ ρ 1.28 K ˆ sin θ + ˆjb cos θ rD P = ib K ˆ θ cos θ − ˆjbθ sin θ vrel = ib K ˆ θ cosθ − θ sin θ − ˆjb θ sin θ + θ cosθ arel = ib ( ) at the point θ = π K So, vrel = bθ = v ( ) K K , vrel = −v v aD = b b v2 v2 K Now, arel = vrelτˆ + rel nˆ = aDτˆ + nˆ ρ b θ = v b θ =  v4  K arel =  aD +  b   K K K K K K vP = v + vrel and aP = aD + arel  a  a  v2 v2 K aP = iˆ  aD + b  D cos θ − sin θ   − ˆjb  D sin θ + cosθ  b b b  b    2 v4 2v K aP = aD  + cos θ + 2 − sin θ  aD b aDb   K aP is a maximum at θ = , i.e., at the top of the wheel 2v −2sin θ − cosθ = aDb  v2    aDb  θ = tan −1  − 0  x -x   x x   x       = x x − x x = x  Therefore, x = 1.29 RR      0 1 0 1  0      The transformation represents a rotation of 45o about the z-axis (see Example 1.8.2) 10 1.30 (a) a θ b φ a = iˆ cos θ + ˆj sin θ b = iˆ cos ϕ + ˆj sin ϕ ( )( a ⋅ b = cos (θ − ϕ ) = iˆ cos θ + ˆj sin θ ⋅ iˆ cos ϕ + ˆj sin ϕ ) cos (θ − ϕ ) = cos θ cos ϕ + sin θ sin ϕ (b) ( ) ( b × a = kˆ sin (θ − ϕ ) = iˆ cos θ + ˆj sin θ × iˆ cos ϕ + ˆj sin ϕ ) sin (θ − ϕ ) = sin θ cos ϕ − cos θ sin ϕ 11