Analytical mechanics solutions manual fowles05

CHAPTER NONINERTIAL REFERENCE SYSTEMS 5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net force G acting on him is zero The scale exerts an upward force, N , whose value is equal to the scale reading - the “weight,” W’, of the observer in the accelerated frame Thus G G G N + mg − mA0 = G N N − mg − mA0 = N − mg − m G A0 g = N − mg = 4 5 mg = W 4 W ′ = 150lb W′ = N = (a) (b) G mg (b) The acceleration is downward, in the same G direction as g W g N − mg + m   = W ′ = W − = W 4 4 W ′ = 90lb G G G G 5.2 (a) Fcent = −mω × (ω × r ′ ) G G G For ω ⊥ r ′ , Fcent = mω r ′ eˆr ′ ω = 500 s −1 = 1000π s −1 G Fcent = 10−6 × (1000π ) × eˆr = 5π dynes outward Fcent mω r ′ (1000π ) = = = 5.04 × 104 980 Fg mg (b) 5.3 G G G mg + T − mAD = (See Figure 5.1.2) g − mg ˆj + T cos θ ˆj + T sin θ iˆ − m   iˆ =  10  mg T cos θ = mg , and T sin θ = 10 tan θ = , θ = 5.71D 10 mg = 1.005mg T= cos θ G 5.4 The non-inertial observer thinks that g ′ points downward in the direction of the hanging plumb bob… Thus g G G G g ′ = g − AD = g ˆj − iˆ 10 For small oscillations of a simple pendulum: T = 2π g′ g g ′ = g +   = 1.005 g  10  T = 2π 5.5 (a) 1 = 1.995π 1.005g g f = − µ mg is the frictional force acting on the G A0 G f box, so G G G f − mA0 = ma′ (b) (a) G ( a′ is the acceleration of the box relative to the truck See G Equation 5.1.4b.) Now, f the only real force acting horizontally, so the acceration relative to the road is f µ mg g a= =− = −µ g = − m m (For + in the direction of the moving truck, the – indicates that friction opposes the forward sliding of the box.) g AD = − (The truck is decelerating.) from above, ma − mA0 = ma′ so g g g a′ = a − AD = − + = K 5.6 (a) r = iˆ ( xD + R cos Ωt ) + ˆjR sin Ωt K r = −iˆΩR sin Ωt + ˆjΩR cos Ωt K K r ⋅ r = v = Ω R ∴ v = ΩR circular motion of radius R K K K K K ˆ ′ + ˆjy′ (b) r′ = r − ω × r ′ where r ′ = ix ˆ ′ + ˆjy′ = −iˆΩR sin Ωt + ˆjΩR cos Ωt − ω kˆ × ix ( ) = −iˆΩR sin Ωt + ˆjΩR cos Ωt − ˆjω x′ + iˆω y′ x ′ = ω y′ − ΩR sin Ωt y ′ = −ω x′ + ΩR cos Ωt here i = −1 ! (c) Let u′ = x′ + iy′ u ′ = x ′ + iy ′ = ω y′ − ΩR sin Ωt − iω x′ + iΩR cos Ωt = −ω y′ iω u ′ = +iω x′ ∴ u ′ + iω u′ = −ΩR sin Ωt + iΩR cos Ωt = iΩ ReiΩt Try a solution of the form u′ = Ae − iω t + BeiΩt u ′ = −iω Ae − iω t + iΩBeiΩt iω u′ = iω Ae− iω t + iω BeiΩt ∴ u ′ + iω u ′ = i (ω + Ω ) Beiω t so B = ΩR ω +Ω Also at t = the coordinate systems coincide so u ′ = A + B = x′ ( ) + iy′ ( ) = xD + R ΩR ω +Ω ΩR iΩt + e ω +Ω ∴ A = xD + R − B = xD + R − ω R  − iω t  Thus, u′ =  xD + e ω + Ω   so, A = xD + ωR ω +Ω 5.7 The x, y frame of reference is attached to the Earth, but the x-axis always points away from the Sun Thus, it rotates once every year relative to the fixed stars The X,Y frame of reference is fixed inertial frame attached to the Sun (a) In the x, y rotating frame of reference x ( t ) = R cos ( Ω − ω ) t − Rε y ( t ) = − R sin ( Ω − ω ) t where R is the radius of the asteroid’s orbit and RE is the radius of the Earth’s orbit Ω is the angular frequency of the Earth’s revolution about the Sun and ω is the angular frequency of the asteroid’s orbit (b) x ( t ) = − ( Ω − ω ) R sin ( Ω − ω ) t → at t = y ( t ) = − ( Ω − ω ) R cos ( Ω − ω ) t → − ( Ω − ω ) R at t = (c) K K K K K K K K K K a = A − Aε − Ω × r − 2Ω × r − Ω × Ω × r K Where a is the acceleration of the asteroid in the x, y frame of reference, K K A, Aε are the accelerations of the asteroid and the Earth in the fixed, inertial frame of reference K K K K K 1st : examine: A − Aε − Ω × Ω × r K K K K K K K K K K = ω × ω × R − Ω × Ω × Rε − Ω × Ω × ( R − Rε ) K K K K K K K K = (ω × ω − Ω × Ω ) × R = − (ω − Ω ) R note: ω = ω kˆ , Ω = Ωkˆ Thus: K K K K a = ( Ω − ω ) R − 2Ω × v Therefore: ˆ + ˆjy = ( Ω − ω ) iR ˆ ˆ ˆ ˆ ix  cos ( Ω − ω ) t − jR sin ( Ω − ω ) t  −2 jΩx + 2i Ωy ( ) Thus:  x = ( Ω − ω ) R cos ( Ω − ω ) t + 2Ωy  y = − ( Ω − ω ) R sin ( Ω − ω ) t − 2Ωx Let  x = ( Ω − ω ) y and Then, we have y = ( Ω − ω ) R cos ( Ω − ω ) t +  y = ( Ω − ω ) x 2Ω y (Ω − ω ) which reduces to y = − ( Ω − ω ) R cos ( Ω − ω ) t Integrating … y = − R sin ( Ω − ω ) t → at t = Also, or − x ( Ω − ω ) = − ( Ω − ω ) R sin ( Ω − ω ) t − 2Ωx x = ( Ω + ω ) R sin ( Ω − ω ) t + 2Ωx x = − ( Ω − ω ) R sin ( Ω − ω ) t Integrating … x = R cos ( Ω − ω ) t + const x = R cos ( Ω − ω ) t − Rε → R − Rε at t = 5.8 Relative to a reference frame fixed to the turntable the cockroach travels at a constant speed v’ in a circle Thus y′ v ′2 G G ′ a eˆr ′ = − ω b Since the center of the turntable is fixed G ′ x b AD = The angular velocity, ω, of the turntable is constant, so G G ω = ω kˆ′ , with ω = G G G G r ′ = beˆr ′ , so ω × (ω × r ′ ) = −bω eˆr ′ G G G v′ = v′eˆθ ′ , so ω × v′ = −ω v′eˆr ′ G G G G G G G From eqn 5.2.14, a = a′ + 2ω × v′ + ω × (ω × r ′) and putting in terms from above v′ − 2ω v′ − bω ar ′ = − b G G For no slipping F ≤ µ s mg , so a ≤ µ s g v′ + 2ω v′ + bω ≤ µ s g b vm′ + 2ω bvm′ + b 2ω − bµ s g = vm′ = −ω b ± ω 2b − b 2ω + bµ s g Since v′ was defined positive, the +square root is used vm′ = −ω b + bµ s g (b) 5.9 G v′ = −v′eˆθ ′ G G ω × v′ = +ω v′eˆr ′ v′ + 2ω v′ − bω ar ′ = − b ′ v − 2ω v′ + bω ≤ µ s g b vm′ = ω b + bµ s g G VD VD ˆ iˆ′ As in Example 5.2.2, ω = j ′ and AD = G ρ ρ For the point at the front of the wheel: G G V  r ′ = D ˆj ′ and v′ = −VD kˆ′ b G ω =0 Vb G G V ω × r ′ = D kˆ′ × −bjˆ′ = D iˆ′ ρ G G G ω × (ω × r ′ ) = G G ω × v′ = ( ) ρ VD ˆ VDb ˆ VD b ˆ k′× i′ = j′ ρ ρ ( ) ρ VD ˆ k ′ × −VD kˆ′ = ρ G VD  VD VD 2b  G G G G G ˆ a = r ′ + ω × (ω × r ′ ) + AD = i′ +  +  ˆj ′ b ρ ρ   5.10 (See Example 5.3.3) mω x′ = mx′ x′ ( t ) = Aeω t + Be −ω t x ′ ( t ) = ω Aeω t − ω Be −ω t Boundary Conditions: l x2 ( 0) = = A + B x ′ ( ) = = ω ( A − B ) ∴A= l B= l x′ ( t ) = cosh ω t l l l x′ (T ) = + = cosh ωT 2 (a) (b) ∴ cosh ωT = l x ′ ( t ) = ω sinh ω t when the bead reaches the end of the rod T= or ω cosh −1 = 1.317 G v′ = 400 ˆj ′ mph = 586.67 ˆj ′ ft ⋅ s −1 G ω = 7.27 ×10−5 cos 41D ˆj ′ + sin 41D kˆ′ s −1 G G ω × v′ = − ( 7.27 ×10−5 ) ( 586.67 ) ( sin 41D ) iˆ′ ft ⋅ s −2 G G −2mω × v′ Fcor = Fgrav mg ( = ) ( 7.27 ×10−5 ) ( 586.67 ) ( sin 41D ) 32 Fcor = 0.0017 Fgrav G G The Coriolis force is in the −ω × v′ direction, i.e., +iˆ′ or east 5.12 ω l x ′ (T ) = ω sinh ωT l l = ω sinh cosh −1  = ω (1.732 ) = 0.866ω l 2 l ωl = 0.866ω l or ω  cosh ωT − 1 = 2 (c) 5.11 l (See Figure 5.4.3) G ω = ω y′ ˆj′ + ω z′ kˆ′ G v′ = vx′iˆ′ + v y′ ˆj ′ G G ω × v′ = −ω z′v y′iˆ′ + ω z′vx′ ˆj′ − ω y′vx′ kˆ′ G G (ω × v′ )horiz = −ω z′v′y′iˆ′ + ω z′vx′ ˆj′ G G = (ω z2′v y2′ + ω z2′vx2′ ) = ω z′ ( vx2′ + v y2′ ) = ω z′v′ G G G Fcor = −2mω × v′ G G G G = 2m (ω × v′ )horiz = 2mω z′ v′ , independent of the direction of v′ Fcor (ω × v′ )horiz ( ) 5.13 horiz From Example 5.4.1 … 1  8h  ′ xh = ω   cos λ  g  and yh′ = 3 −5 −1  × 1250 ft  D ′ xh = ( 7.27 × 10 s )   cos 41 −2  32 ft ⋅ s  xh′ = 0.404 ft to the east 5.14 From Example 5.4.2: ωH ∆≈ sin λ is the deflection of the baseball towards the south since it vD was struck due East at Yankee Stadium at latitude λ = 41D N (problem 5.13) v0 is the initial speed of the baseball whose range is H From eqn 4.3.18b, without air resistance in an inertial reference frame, the horizontal range is … vD2 sin 2α H= g Solving for v0 …  gH  vD =    sin 2α   32 ft ⋅ s −2 × 200 ft  −1 vD =   = 113 ft ⋅ s D sin 30   ∴ ( 7.27 ×10 ∆≈ −5 s −1 )( 2002 ft ) sin 41D = 0.0169 ft = 0.2 in 113 ft ⋅ s −1 A deflection of 0.2 inches should not cause the outfielder any difficulty 5.15 Equation 5.2.10 gives the relationship between the time derivative of any vector in a fixed and rotating frame of reference Thus … G G G  da  G G  da   r =  =   +ω × a  dt  fixed  dt  rot G G G G G G G G G a = r ′ + ω × r ′ + 2ω × r′ + ω × (ω × r ′ ) G G G G G G G G G G  da     = r ′ + ω × r ′ + ω × r ′ + 2ω × r ′ + 2ω × r ′  dt rot G G G G G G G G G +ω × (ω × r ′ ) + ω × ω × r ′ + ω × ω × r′ ( ) ( ) G G G G G G G G G G G G G G ω × a = ω ×  r ′ + ω × (ω × r ′ ) + 2ω × (ω × r′ ) + ω × ω × (ω × r ′ )  G G is ⊥ to ω and r ′ Let this define a direction nˆ : G G G G ω × r ′ = ω × r ′ nˆ G G G G G Since ω ⊥ nˆ , ω × (ω × r ′ ) is in the plane defined by ω and r ′ and G G G G G G G G ω × (ω × r ′) = ω × nˆ ω × r ′ = ω ω × r ′ G G G G Since ω ⊥ ω × (ω × r ′ ) … G G G G G G ω × ω × (ω × r ′ )  = ω ω × r ′ G G G G And ω × ω × (ω × r ′ )  is in the direction − nˆ G G G G G G Thus ω × ω × (ω × r ′ )  = −ω (ω × r ′ ) G G G G G G G G G G G ω × a = ω ×  r ′ + ω × ω × r ′ + 2ω × ω × r′ − ω (ω × r ′ ) Now G G (ω × r ′ ) ( ) ( ) G G G G G G G G G G G  r =  r ′ + ω × r ′ + 3ω × r′ + 3ω ×  r ′ + ω × (ω × r ′ ) G G G G G G G G +2ω × ω × r ′ + 3ω × ω × r′ − ω (ω × r ′ ) ( 5.16 ) ( ) With xD′ = yD′ = zD′ = xD′ = yD′ = , and zD′ = vD′ Equations 5.4.15a – 5.4.15c become: x′ ( t ) = ω gt cos λ − ω t vD′ cos λ y′ ( t ) = z ′ ( t ) = − gt + vD′t When the bullet hits the ground z ′ ( t ) = , so 2v′ t= D g  8v′   4v ′  x′ = ω g  D3  cos λ − ω  D2  vD′ cos λ  g   g  4ω vD′ x′ = − cos λ 3g x′ is negative and therefore is the distance the bullet lands to the west 5.17 With xD′ = yD′ = zD′ = and xD′ = vD cos α yD′ = zD′ = vD sin α we can solve equation 5.4.15c to find the time it takes the projectile to strike the ground … z ′ ( t ) = − gt + vD′ t sin α + ω vDt cos α cos λ = 2vD′ sin α 2v′ sin α ≈ D t= or g − 2ω vD′ cos α cos λ g We have ignored the second term in the denominator—since vD′ would have to be impossibly large for the value of that term to approach the magnitude g For example, for λ = 41o and α = 45o g − 2ω vD′ cos α cos λ ≈ g − ω vD′ or vD′ ≈ g ω ≈ 144 km ! s Substituting t into equation 5.4.15b to find the lateral deflection gives 4ω vD′ y ( t ) = − [ω vD cos α sin λ ] t = − sin λ sin α cos α g2 5.18 Let … G r y x G R0 G aD = acceleration of object relative to Earth G ωD = ωD kˆ = its angular speed G AD = acceleration of satellite G ω = ω kˆ = its angular speed G G G G G G G G aD = a + 2ω × v + ω × (ω × r ) + AD (Equation 5.2.14) G G G G G G G G a = aD − AD − 2ω × v − ω × ω × r G R As in problem 5.7 Evaluate the term … G G G G G G G G G G G G G G G G ∆ a = aD − AD − ω × ω × r = ω D × ω D × RD − ω × ω × R − ω × ω × RD − R G G ∆ a = (ω D2 − ω ) RD ( … given the condition that but ωD2 RD3 = ω R G G  R3  ∆ a = −ω RD  −   RD  G G G G G G RD ⋅ RD = R + r ⋅ R + r = R + r + 2rR cos θ ( Letting x = cos θ )( ) G G  2x  RD ⋅ RD = R + r + Rx ≈ R  +  for small r R  − R3  x  or and = 1 +  RD3  R G −   G G R x   ∆ a = −ω RD 1 −  +   ≈ −3ω x D ≈ −3ω x iˆ for small r   R  R    2x 2 RD3 ≈ R 1 +  R  ) G G G G ˆ + ˆjy Hence: a = ∆ a − 2ω × v = −3ω x iˆ − 2ω kˆ × ix G  ˆ − 2ω xˆj a = iˆ x + ˆjy = −3ω xiˆ + 2ω yi  So x − 2ω y − 3ω x =  y + 2ω x = ( 5.19 G G G G mr = qE + q v × B ( Equation 5.2.14 ) ) G G G G G G G G G  r =  r ′ + ω × r ′ + 2ω × v′ + ω × (ω × r ′ ) G G G G v = v′ + ω × r ′ Equation 5.2.13 q G G G ω =− B so ω = 2m G G G G q G G G G G G G mr′ − q B × v′ − B × (ω × r ′ ) = qE + q ( v′ + ω × r ′ ) × B  G G G G G G q G G G G G G mr ′ + q v′ × B + (ω × r ′ ) × B = qE + q v′ × B + q (ω × r ′ ) × B G q G G G G mr ′ = qE + (ω × r ′ ) × B G q qB q G G (ω × r ′) × B =   ( r ′)( sin θ )( B ) ∝ B 2  2m  G G Neglecting terms in B , mr′ = qE 5.20 ( ) ( ) ( ) For x′ = x cos ω ′t + y sin ω ′t y′ = − x sin ω ′t + y cos ω ′t x ′ = x cos ω ′t − xω ′ sin ω ′t + y sin ω ′t + yω ′ cos ω ′t y ′ = − x sin ω ′t − xω ′ cos ω ′t + y cos ω ′t − yω ′ sin ω ′t x ′ = x cos ω ′t + y sin ω ′t + ω ′ y′ y ′ = − x sin ω ′t + y cos ω ′t − ω ′x′  x′ =  x cos ω ′t − xω ′ sin ω ′t +  y sin ω ′t + yω ′ cos ω ′t + ω ′ y ′  y′ = −  x sin ω ′t − xω ′ cos ω ′t +  y cos ω ′t − yω ′ sin ω ′t − ω ′x ′  x′ =  x cos ω ′t +  y sin ω ′t + 2ω ′ y ′ + ω ′2 x′  y′ = −  x sin ω ′t +  y cos ω ′t − 2ω ′x ′ + ω ′2 y′ Substituting into Eqns 5.6.3:  x cos ω ′t +  y sin ω ′t + 2ω ′ y ′ + ω ′2 x′ g g = − x cos ω ′t − y sin ω ′t + 2ω ′ y ′ l l ′ ′ ′   x sin ω t + y cos ω t − 2ω x ′ + ω ′2 y′ g g = + x sin ω ′t − y cos ω ′t − 2ω ′x ′ l l 10 Collecting terms and neglecting terms in ω ′2 : g  g    x + x  cos ω ′t +   y + y  sin ω ′t =   l  l    g  g    x + x  sin ω ′t −   y + y  cos ω ′t =   l  l    24 hours sin λ 24 T= = 73.7 hours sin19D 5.21 T= 5.22 Choose a coordinate system with the origin at the center of the wheel, the x′ and y′ axes pointing toward fixed points on the rim of the wheel, and the z ′ axis pointing toward the center of curvature of the track Take the initial position of G the x′ axis to be horizontal in the −VD direction, so the initial position of the y′ axis is vertical V The bicycle wheel is rotating with angular velocity D about its axis, so … b V G ωl = kˆ′ D b A unit vector in the vertical direction is: Vt Vt nˆ = iˆ′ sin D + ˆj ′ cos D b b At the instant a point on the rim of the wheel reaches its highest point: Vt Vt G  r ′ = bnˆ = b  iˆ′ sin D + ˆj ′ cos D  b b   Since the coordinate system is moving with the wheel, every point on the rim is fixed in that coordinate system G G r′ = and  r′ = The x′y′z ′ coordinate system also rotates as the bicycle wheel completes a circle around the track: V V  Vt Vt G ω = nˆ D = D  iˆ′ sin D + ˆj ′ cos D  b b  ρ ρ The total rotation of the coordinate axes is represented by: Vt Vt V G G G V  ω = ω1 + ω = D  iˆ′ sin D + ˆj ′ cos D  + kˆ′ D b b  b ρ Vt Vt G V  ω = D  iˆ′ cos D − ˆj ′ sin D  ρb  b b  Vt Vt  V2 G G V  ω × r ′ = D  kˆ′ cos D + kˆ′ sin D  = D kˆ′ b b  ρ ρ  11 G G ω × r′ = VDb  ˆ VDt VDt ˆ Vt Vt Vt Vt  − k ′ sin D cos D  + VD  ˆj ′ sin D − iˆ′ cos D   k ′ sin cos b b b b  b b  ρ   2 V  Vt Vt V  Vt Vt G G G ω × (ω × r ′ ) = D  kˆ′ sin D + kˆ′ cos D  + D  −iˆ′ sin D − ˆj ′ cos D  b b  b  b b  ρ  2 V V G G G ω × (ω × r ′ ) = kˆ′ D − nˆ D ρ b Since the origin of the coordinate system is traveling in a circle of radius ρ : G V2 AD = kˆ′ D G G ω × r′ = ρ G G G G G G G G G G  r =  r ′ + ω × r ′ + 2ω × r′ + +ω × (ω × r ′ ) + AD V2 V2 V2 V2 G  r = kˆ′ D + kˆ′ D − nˆ D + kˆ′ D ρ ρ ρ b V2 V2 G  r = D kˆ′ − D nˆ b ρ With appropriate change in coordinate notation, this is the same result as obtained in Example 5.2.2 12