Solution manual for analytical mechanics 7th edition by fowles

Solution Manual for Analytical Mechanics 7th Edition by Fowles CHAPTER FUNDAMENTAL CONCEPTS: VECTORS 1.1 v v (a) A + B = (iˆ + ˆj ) + ( ˆj + kˆ) = iˆ + ˆj + kˆ v v A + B = (1 + + 1) = v v (b) A − B = 3(iˆ + ˆj ) − 2( ˆj + kˆ) = 3iˆ + ˆj − 2kˆ v v (c) A ⋅ B = (1)(0) + (1)(1) + (0)(1) = iˆ ˆj kˆ v v (d) A × B = 1 = iˆ(1 − 0) + ˆj (0 − 1) + kˆ(1 − 0) = iˆ − ˆj + kˆ 1 v v A × B = (1 + + 1) = )( ) v v v 1.2 (a) A ⋅ ( B + C ) = 2iˆ + ˆj ⋅ iˆ + ˆj + kˆ = (2)(1) + (1)(4) + (0)(1) = ( ( A + B ) ⋅ C = ( 3iˆ + ˆj + kˆ ) ⋅ ˆj = (3)(0) + (1)(4) + (1)(0) = v v v v v v (b) A ⋅ ( B × C ) = 1 = −8 v v v v v v ( A × B ) ⋅ C = A ⋅ ( B × C ) = −8 ( ) ( ) v v v v v v v v v (c) A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C = iˆ + kˆ − ˆj = 4iˆ − ˆj + 4kˆ v v v v v v v v v v v v ( A × B ) × C = −C × ( A × B ) = − ⎡⎣( C ⋅ B ) A − ( C ⋅ A) B ⎤⎦ = − ⎡0 ( 2iˆ + ˆj ) − ( iˆ + kˆ ) ⎤ = 4iˆ + 4kˆ ⎣ ⎦ Solution Manual for Analytical Mechanics 7th Edition by Fowles v v A ⋅ B (a )(a ) + (2a )(2a) + (0)(3a ) 5a 1.3 cosθ = = = AB a 14 5a 14a θ = cos −1 ≈ 53° 14 1.4 v (a) A = iˆ + ˆj + kˆ v v A = A⋅ A v (b) B = iˆ + ˆj v v B = B⋅B : body diagonal = iˆ ⋅ iˆ + ˆj ⋅ ˆj + kˆ ⋅ kˆ = : face diagonal = iˆ ˆj kˆ v v v (c) C = A × B = 1 1 v v 1−1 A⋅ B (d) cos θ = = =0 AB 1.5 ∴θ = 90o v v v B = B = A × C = AC sin θ v v A ⋅ C = AC cosθ = u v v A C = Cx + A u v = A+ A 1.6 B A u ∴ Cx = C cos θ = A ∴ C y = C sin θ = v v v v B× A u v B× A⎛ B ⎞ v v Cy = A + ⎜ ⎟ A AB ⎝ A ⎠ B× A v v B× A A2 v dA ˆ d d d = i (α t ) + ˆj ( β t ) + kˆ (γ t ) = iˆα + ˆj β t + kˆ3γ t dt dt dt dt v d2A ˆ = j β + kˆ6γ t dt Solution Manual for Analytical Mechanics 7th Edition by Fowles v v = A ⋅ B = ( q )( q ) + ( 3)( − q ) + (1)( ) = q − 3q + 1.7 ( q − ) ( q − 1) = , q = or v v2 v v v v v v A + B = ( A + B ) ⋅ ( A + B ) = A2 + B + A ⋅ B 1.8 v v ⎡ A + B ⎤ = A2 + B + AB ⎣ ⎦ v v Since A ⋅ B = AB cosθ ≤ AB , v v v v A+ B ≤ A + B v v v v v v A ⋅ B = AB cosθ = A B cosθ ≤ A B B cos θ ≤ B v v v v v v v v v Show A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C 1.9 iˆ v A × Bx Cx or kˆ v v Bz = ( AxCx + Ay C y + Az Cz ) B − ( Ax Bx + Ay By + Az Bz ) C Cz ˆj By Cy = ( Ax Bx Cx + 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 Cz − 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 Solution Manual for Analytical Mechanics 7th Edition by Fowles iˆ Ax By Cz − Bz C y iˆ ( Ay BxC y − Ay By Cx − Az Bz Cx + Az Bx Cz ) kˆ Az = + ˆj ( Az By Cz − Az Bz C y − Ax BxC 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 ⎠ v v Α = A× B 1.11 iˆ v v v v A ⋅ ( B × C ) = A ⋅ Bx Cx 1.12 x r u kˆ Ax Bz = Bx Cz Cx ˆj By Cy Ay By Cy Bx Az Bz = − Ax Cx Cz By Ay Cy Bz v v v Az = − B ⋅ ( A × C ) Cz z r A r C y r B r v v Let A = (Ax,Ay,Az), B = (0,By,0) and C = (0,Cy,Cz) v v r r r Cz is the perpendicular distance between the plane A , B and its opposite u = B x C is r r r r directed along the x-axis since the vectors B , C are in the y,z plane u x = B x C = ByCz r r is the area of the parallelogram formed by the vectors B , C Multiply that area times the v v height of plane A , B = Ax to get the volume of the parallopiped r r r V = Ax u x = Ax By Cz = A • B x C ( ) Solution Manual for Analytical Mechanics 7th Edition by Fowles 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′ = cosθ = kˆ ⋅ kˆ′, iˆ ⋅ kˆ′ = sin θ kˆ ⋅ iˆ′ = − sin θ ⎛ cos θ t ⎜ T =⎜ ⎜ ⎝ sin θ ˆj ⋅ ˆj ′ = − sin θ ⎞ ⎛ cos φ ⎟⎜ ⎟ ⎜ − sin φ cos θ ⎠⎟ ⎝⎜ sin φ cos φ 0 ⎞ ⎛ cosθ cos φ ⎟ ⎜ ⎟ = ⎜ − sin φ ⎠⎟ ⎝⎜ sin θ cos φ cos θ sin φ cos φ sin θ sin φ − sin θ ⎞ ⎟ ⎟ cos θ ⎟⎠ 1.14 iˆ ⋅ iˆ′ = cos 30o = ˆj ⋅ iˆ′ = sin 30o = kˆ ⋅ iˆ′ = iˆ ⋅ ˆj ′ = − sin 30o = − ˆj ⋅ ˆj ′ = cos 30o = kˆ ⋅ ˆj ′ = ˆj ⋅ kˆ′ = iˆ ⋅ kˆ′ = ⎡ ⎤ 3⎤ ⎡ 0⎥ ⎢ 3+ ⎥ ⎢ 2 ⎥⎡ ⎤ ⎢ ⎡ Ax′ ⎤ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢3 ⎢ A ⎥ = ⎢− ⎥ ⎢ y′ ⎥ ⎢ 2 ⎥ ⎢ ⎥ = ⎢ − 1⎥ ⎥ ⎢ −1⎥ ⎢⎣ Az′ ⎥⎦ ⎢ ⎥ ⎣ ⎦ ⎢⎢ −1 ⎥⎥ ⎢ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎣⎢ ⎦⎥ v A = 3.232iˆ′ + 1.598 ˆj′ − kˆ′ kˆ ⋅ kˆ′ = Solution Manual for Analytical Mechanics 7th Edition by Fowles 1.15 Rotate thru φ about z-axis φ = 45o Rφ Rotate thru θ about x’-axis Rθ Rotate thru ψ about z’-axis θ = 45 ψ = 45o ⎛ ⎜ ⎜ ⎜ Rφ = ⎜ − ⎜ ⎜ ⎜ ⎝ 2 ⎛ ⎜1 ⎜ ⎜ Rθ = ⎜ ⎜ ⎜ ⎜0 − ⎝ o ⎞ 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⎞ v ⎜ ⎟ where x′ = ⎜ ⎟ and ⎜0⎟ ⎝ ⎠ 2 we have: ψ + β + α = v v Condition is: x′ = Rx v v Since x ⋅ x = After a lot of algebra: α = 1.16 Rψ ⎞ ⎟ ⎟ ⎛1⎞ ⎛α ⎞ ⎟ ⎜ ⎟ or ⎜ ⎟ = R (ψ ,θ , φ ) ⎜⎜ β ⎟⎟ ⎟ ⎜0⎟ ⎜γ ⎟ ⎟ ⎝ ⎠ ⎝ ⎠ ⎟ ⎟ 2⎠ ⎛α ⎞ v ⎜ ⎟ x =⎜β ⎟ ⎜γ ⎟ ⎝ ⎠ 2 − , β= + , γ = 4 v v = vτˆ = ctτˆ v2 c 2t v a = v&τˆ + nˆ = cτˆ + nˆ b ρ Solution Manual for Analytical Mechanics 7th Edition by Fowles b v , v = τˆ bc and c v v v ⋅a c bc cosθ = = = va bc 2c v a = cτˆ + cnˆ at t = θ = 45o 1.17 v ˆ ω sin (ω t ) + ˆj 2bω cos (ω t ) v ( t ) = −ib 1 v v = ( b 2ω sin ω t + 4b 2ω cos ω t ) = bω (1 + 3cos ω t ) v ˆ ω cos ω t − ˆj 2bω sin ω t a ( t ) = −ib v a = bω (1 + 3sin ω t ) 1.18 v v = 2bω ; t = 0, at t= at π , 2ω v v = bω v ˆ ω cos ω t − ˆjbω sin ω t + kˆ 2ct v ( t ) = ib v ˆ ω sin ω t − ˆjbω cos ω t + kˆ 2c a ( t ) = −ib 1 v a = ( b 2ω sin ω t + b 2ω cos ω t + 4c ) = ( b 2ω + 4c ) 1.19 v & ˆr + rθ&eˆθ = bkekt eˆr + bcekt eˆθ v = re v a = && r − rθ& eˆr + rθ&& + 2r&θ& eˆθ = b ( k − c ) ekt eˆr + 2bckekt eˆθ v v b k ( k − c ) e kt + 2b c ke2 kt v ⋅a = cos φ = 1 va 2 2 2 2 ⎤2 kt kt ⎡ 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 ) v && − Rφ&2 eˆ + R&φ& + Rφ&& eˆ + &&ˆ a= R zez φ R v a = −bω eˆR + 2ceˆz v a = ( b 2ω + 4c ) Solution Manual for Analytical Mechanics 7th Edition by Fowles 1.21 1.22 v r ( t ) = iˆ (1 − e− kt ) + ˆjekt v ˆ − kt + ˆjke kt r ( t ) = ike v ˆ e − kt + ˆjk e kt r ( t ) = −ik v v = eˆr r& + eˆφ rφ sin θ + eˆθ rθ& ⎧π ⎡ π v ⎤⎫ v = eˆφ bω sin ⎨ ⎢1 + cos ( 4ω t ) ⎥ ⎬ − eˆθ b ω sin ( 4ω t ) ⎦⎭ ⎩2 ⎣ π v ⎡π ⎤ v = eˆφ bω cos ⎢ cos ( 4ω t ) ⎥ − eˆθ bω sin ( 4ω t ) ⎣8 ⎦ ⎡ ⎤2 v ⎛π ⎞ π v = bω ⎢cos ⎜ cos 4ω t ⎟ + sin 4ω t ⎥ ⎝8 ⎠ ⎣ ⎦ Path is sinusoidal oscillation about the equator 1.23 v v v ⋅ v = v2 v v dv v v dv ⋅v + v ⋅ = 2vv& dt dt v v 2v ⋅ a = 2vv& v v v ⋅ a = vv& Solution Manual for Analytical Mechanics 7th Edition by Fowles 1.24 1.25 v d v v v dr v v v d v v ⎡⎣ r ⋅ ( v × a ) ⎤⎦ = ⋅(v × a ) + r ⋅ (v × a ) dt dt dt v v v v v v ⎡⎛ dv v ⎞ ⎛ v da ⎞ ⎤ = v ⋅ ( v × a ) + r ⋅ ⎢⎜ × a ⎟ + ⎜ v × ⎟ ⎥ dt ⎠ ⎦ ⎠ ⎝ ⎣⎝ dt v ⎡ v v& ⎤ = + r ⋅ ⎣0 + ( v × a )⎦ d v v v v v v ⎡⎣ r ⋅ ( v × a ) ⎤⎦ = r ⋅ ( v × a& ) dt v v v = vτˆ and a = aτ τˆ + an nˆ v v v ⋅a v v v ⋅ a = vaτ , so aτ = v a = aτ2 + an2 , so an = ( a − aτ2 ) 1.26 For 1.14, aτ = aτ = −b 2ω cos ω t ⋅ sin ω t + b 2ω sin ω t ⋅ cos ω t + 4c 2t ( b2ω cos2 ωt + b2ω sin ωt + 4c2t ) 4c t (b ω 2 + 4c 2t ) ⎛ ⎞2 16c 4t an = ⎜ b 2ω + 4c − 2 ⎟ b ω + 4c t ⎠ ⎝ b k ( k − c ) e2 kt + 2b 2c ke kt kt 2 = b ke k + c For 1.15, aτ = ( ) bekt ( k + c ) 1 2 an = ⎡⎢b e kt ( k + c ) − b k e kt ( k + c ) ⎤⎥ = bce kt ( k + c ) ⎣ ⎦ 1.27 nˆ dθ τˆ dτˆ = − nˆ dθ ρ Solution Manual for Analytical Mechanics 7th Edition by Fowles The figure above shows the unit vectors nˆ and τˆ which are normal and tangent to the curve The vectors are shown at different points along the trajectory Since the particle is moving tangentially to the curve and in a direction perpendicular to ρ , the local radius of curvature, we have … v v = v τˆ v2 v a = v& τˆ + v τˆ& = − vθ& nˆ = − nˆ (since v& = and from the above figure) ρ ⎛ v2 v v v × a = v τˆ × ⎜ − ⎝ ρ ⎞ v v3 nˆ ⎟ = v = ρ ρ ⎠ 1.28 v ˆ sin θ + ˆjb cosθ ro P = ib v ˆ θ& cos θ − ˆjbθ& sin θ vrel = ib v ˆ θ&& cosθ − θ& sin θ − ˆjb θ&&sin θ + θ& cosθ arel = ib v v at the point θ = π , vrel = −v v So, v = bθ& = v ( ) ( ) rel v& ao = b b v2 v2 v Now, arel = v&relτˆ + rel nˆ = aoτˆ + nˆ b ρ θ& = v b θ&& = ⎛ v4 ⎞ v arel = ⎜ ao2 + ⎟ b ⎠ ⎝ v v v v v v vP = v + vrel and aP = ao + arel ⎡ ⎛ ao ⎞ ⎤ ˆ ⎛ ao ⎞ v2 v2 v ˆ aP = i ⎢ ao + b ⎜ cos θ − sin θ ⎟ ⎥ − jb ⎜ sin θ + cos θ ⎟ b b ⎠ ⎝b ⎠⎦ ⎝b ⎣ ⎛ ⎞2 v4 2v v aP = ao ⎜ + cos θ + 2 − sin θ ⎟ ao b aob ⎝ ⎠ v aP is a maximum at θ = , i.e., at the top of the wheel 10 Solution Manual for Analytical Mechanics 7th Edition by Fowles −2sin θ − 2v cosθ = aob ⎛ v2 ⎞ ⎟ ⎝ aob ⎠ Comments : Note that a point on the bottom of the wheel is instantaneously at rest, i.e., there is no relative motion between the ground and the bottom of the wheel assuming no slipping 0⎞ ⎛ x -x ⎞⎛ x x ⎞ ⎛ x ⎜ ⎜ ⎟⎜ ⎟ % = x x − x x = x ⎟ Therefore, x = 1.29 RR ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎜ 0 ⎟⎜ 0 ⎟ ⎜ ⎟ 1⎠ ⎝ ⎠⎝ ⎠ ⎝ The transformation represents a rotation of 45o about the z-axis (see Example 1.8.2) θ = tan −1 ⎜ − 1.30 (a) a θ b φ a = iˆ cos θ + b = iˆ cos ϕ + ˆj sin θ 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 ... z Solution Manual for Analytical Mechanics 7th Edition by Fowles iˆ Ax By Cz − Bz C y iˆ ( Ay BxC y − Ay By Cx − Az Bz Cx + Az Bx Cz ) kˆ Az = + ˆj ( Az By Cz − Az Bz C y − Ax BxC y + Ax By. .. volume of the parallopiped r r r V = Ax u x = Ax By Cz = A • B x C ( ) Solution Manual for Analytical Mechanics 7th Edition by Fowles 1.13 For rotation about the z axis: iˆ ⋅ iˆ′ = cos φ =.. .Solution Manual for Analytical Mechanics 7th Edition by Fowles v v A ⋅ B (a )(a ) + (2a )(2a) + (0)(3a ) 5a 1.3 cosθ = = =