MEN 201 Problems For Chapter 21 with Solutions, Spring 2012

Engineering Mechanics - Dynamics Chapter 21 Problem 21-1 Show that the sum of the moments of inertia of a body, Ixx+Iyy+Izz , is independent of the orientation of the x, y, z axes and thus depends only on the location of the origin. Solution: Ixx + Iyy + Izz m y2 + z2 ⌠⎮ ⎮⌡ d m x2 + z2 ⌠⎮ ⎮⌡ d + m x2 + y2 ⌠⎮ ⎮⌡ d + = m m m Ixx + Iyy + Izz 2 m x2 + y2 + z2 ⌠⎮ ⎮⌡ d = m However, x2 + y2 + z2 = r2 , where r is the distance from the origin O to dm. The magnitude | r | does not depend on the orientation of the x, y, z axes. Consequently, Ixx + Iyy + Izz is also independent of the x, y, z axes. Problem 21-2 Determine the moment of inertia of the cylinder with respect to the a-a axis of the cylinder. The cylinder has a mass m. Solution: m 0 h y ρ πa2 ⌠ ⎮⌡ d = = hρ πa2 ρ m hπa2 = Iaa m hπa2 0 h y a2 4 + y2 ⎛⎜ ⎝ ⎞⎟ ⎠ πa2 ⌠ ⎮ ⎮ ⌡ d ⎡⎢ ⎢ ⎢⎣ ⎤⎥ ⎥ ⎥⎦ = m hπa2 1 4 hπa4 1 + 3 π a2 h3 ⎛⎜⎝ ⎞⎟⎠ = 714 Engineering Mechanics - Dynamics Chapter 21 Iaa m a2 4 h2 + 3 ⎛⎜ ⎝ ⎞⎟ ⎠ = Problem 21-3 Determine the moments of inerta Ix and Iy of the paraboloid of revolution. The mass of the paraboloid is M. Given: M 20 slug = r 2 ft = h 2 ft = Solution: V 0 h y πr2 y h ⌠⎮ ⎮⌡ d = ρ M V = Ix 0 h y ρ 1 2 πr2 y h ⎛⎜⎝ ⎞⎟⎠ r2 y h ⎛⎜⎝ ⎞⎟⎠ ⌠⎮ ⎮⌡ d = Ix 26.7 slug ft2 ⋅ = Iy 0 h y ρ πr2 y h ⎛⎜⎝ ⎞⎟⎠ 1 4 r2 y h ⎛⎜⎝ ⎞⎟⎠ + y2 ⎡⎢⎣ ⎤⎥⎦ ⌠⎮ ⎮⌡ d = Iy 53.3 slug ft2 ⋅ = *Problem 21-4 Determine the product of inertia Ixy of the body formed by revolving the shaded area about the line x = a + b. Express your answer in terms of the density ρ. Given: a 3 ft = b 2 ft = c 3 ft = 715 Engineering Mechanics - Dynamics Chapter 21 Solution: k 0 a x 2πc x a a + b − x ( ) ⎡⎢⎣ ⎤⎥⎦ a + b ( ) c 2 x a ⌠⎮ ⎮⌡ d = Ixy = kρ k 636ft5 = Problem 21-5 Determine the moment of inertia Iy of the body formed by revolving the shaded area about the line x = a + b. Express your answer in terms of the density ρ. Given: a 3 ft = b 2 ft = c 3 ft = Solution: k 0 a x 2πc x a a + b − x ( ) ⎡⎢⎣ ⎤⎥⎦ a + b − x ( )2 a + b ( )2 + ⎡⎣ ⎤⎦ ⌠⎮ ⎮⌡ d = Iy = kρ k 4481ft5 = Problem 21-6 Determine by direct integration the product of inertia Iyz for the homogeneous prism. The density of the material is ρ. Express the result in terms of the mass m of the prism. Solution: ρ 2m a2h = 716 Engineering Mechanics - Dynamics Chapter 21 Iyz 0 a y 2m a2h ⎛⎜ ⎝ ⎞⎟ ⎠ h a − y ( ) h 2 a − y 2 ⎛⎜⎝ ⎞⎟⎠ ⌠ ⎮ ⎮ ⌡ d = 1 6 mah = Iyz 1 6 mah = Problem 21-7 Determine by direct integration the product of inertia Ixy for the homogeneous prism. The density of the material is ρ. Express the result in terms of the mass m of the prism. Solution: ρ 2m a2h = 717 Engineering Mechanics - Dynamics Chapter 21 Ixy 0 a y 2m a2h ⎛⎜ ⎝ ⎞⎟ ⎠ h a − y ( ) y a − y 2 ⎛⎜⎝ ⎞⎟⎠ ⌠ ⎮ ⎮ ⌡ d = 1 12 a4 m a2 ⎛⎜ ⎝ ⎞⎟ ⎠ = Ixy 1 12 a2m = *Problem 21-8 Determine the radii of gyration kx and ky for the solid formed by revolving the shaded area about the y axis. The density of the material is ρ. Given: a 4 ft = b 0.25 ft = ρ 12 slug ft3 = Solution: M 0 b y ρ πa2 ⌠ ⎮⌡ d b a y ρ π a2 b2 y2 ⌠⎮ ⎮ ⎮⌡ d + = M 292.17 slug = Ix 0 b y ρ a2 4 + y2 ⎛⎜ ⎝ ⎞⎟ ⎠ πa2 ⌠ ⎮ ⎮ ⌡ d b a y ρ a2 b2 4y2 + y2 ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ π a2 b2 y2 ⌠⎮ ⎮ ⎮⌡ d + = Ix 948.71 slug ft2 ⋅ = Iy 0 b y ρ a2 2 ⎛⎜ ⎝ ⎞⎟ ⎠ πa2 ⌠ ⎮ ⎮ ⌡ d b a y ρ a2 b2 2y2 ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ π a2 b2 y2 ⌠⎮ ⎮ ⎮⌡ d + = Iy 1608.40 slug ft2 ⋅ = kx Ix M = kx 1.80 ft = 718 Engineering Mechanics - Dynamics Chapter 21 ky Iy M = ky 2.35 ft = Problem 21-9 Determine the mass moment of inertia of the homogeneous block with respect to its centroidal x' axis. The mass of the block is m. Solution: m = ρabh ρ m abh = Ix' m abh h − 2 h 2 z 1 12a2 + z2 ⎛⎜⎝ ⎞⎟⎠ ab ⌠⎮ ⎮ ⎮⌡ d = m abh 1 12 a3 bh 1 + 12 abh3 ⎛⎜⎝ ⎞⎟⎠ = Ix' m 12 a2 + h2 ( ) = Problem 21-10 Determine the elements of the inertia tensor for the cube with respect to the x, y, z coordinate system. The mass of the cube is m. 719 Engineering Mechanics - Dynamics Chapter 21 Solution: Ixx = Iyy = Izz 2 3ma2 = Ixy = Ixz −m a 2 ⎛⎜⎝ ⎞⎟⎠ 2 = −m a2 4 = Iyz m a 2 ⎛⎜⎝ ⎞⎟⎠ 2 = m a2 4 = I ma2 12 8 3 3 3 8 3 − 3 3 − 8 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Remember to change the signs of the products of inertia to put them in the inertia tensor Problem 21-11 Compute the moment of inertia of the rod-and-thin-ring assembly about the z axis. The rods and ring have a mass density ρ. Given: ρ 2 kg m = l 500 mm = h 400 mm = θ 120 deg = Solution: r l2 − h2 = φ acos h l ⎛⎜⎝ ⎞⎟⎠ = Iz 3 ρl l2 3 sin φ( )2 ⎛⎜ ⎝ ⎞⎟ ⎠ ρ 2πr ( ) r2 + = Iz 0.43 kg m2 ⋅ = 720 Engineering Mechanics - Dynamics Chapter 21 *Problem 21-12 Determine the moment of inertia of the cone about the z' axis. The weight of the cone is W, the height is h, and the radius is r. Given: W 15 lb = h 1.5 ft = r 0.5 ft = g 32.2 ft s2 = Solution: θ atan r h ⎛⎜⎝ ⎞⎟⎠ = Ix 3 80W 4r2 + h2 ( ) W 3h 4 ⎛⎜⎝ ⎞⎟⎠ 2 + = Iy = Ix Iz 3 10W r2 = Iz' Ix sin θ( )2 + Izcos θ( )2 = Iz' 0.0962 slug ft2 ⋅ = Problem 21-13 The bent rod has weight density γ. Locate the center of gravity G(x', y') and determine the principal moments of inertia Ix', Iy', and Iz' of the rod with respect to the x', y', z' axes. Given: γ 1.5 lb ft = a 1 ft = b 1 ft = g 32.2 ft s2 = 721 Engineering Mechanics - Dynamics Chapter 21 Solution: x' 2a a 2 ba + 2a + b = x' 0.667 ft = y' ab b b 2 + 2a + b = y' 0.50 ft = Ix' γa y'2 γa b − y' ( )2 + 1 + 12γbb2 γb b 2 − y' ⎛⎜⎝ ⎞⎟⎠ 2 + = Ix' 0.0272 slug ft2 ⋅ = Iy' 2 12γaa2 2γa a 2 − x' ⎛⎜⎝ ⎞⎟⎠ 2 + γb a − x' ( )2 + = Iy' 0.0155 slug ft2 ⋅ = Iz' Ix' + Iy' = Iz' 0.0427 slug ft2 ⋅ = Problem 21-14 The assembly consists of two square plates A and B which have a mass MA each and a rectangular plate C which has a mass MC. Determine the moments of inertia Ix, Iy and Iz. Given: MA 3 kg = MC 4.5 kg = θ 60 deg = θ1 90 deg = θ2 30 deg = a 0.3 m = b 0.2 m = c 0.4 m = Solution: ρA MA c 2b ( ) = 722 Engineering Mechanics - Dynamics Chapter 21 Ix 1 12MC 2a ( )2 2 0 c ξ ρA 2b ( ) a ( + ξ cos θ( ) )2 (ξ sin θ( ) )2 + ⎡⎣ ⎤⎦ ⌠ ⎮ ⌡ d + = Iy 1 12MC 2b ( )2 2 b − b x 0 c ξ ρA x2 + ξ2 sin θ( )2 ( ) ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d + = Iz 1 12MC (2b )2 (2a )2 + ⎡⎣ ⎤⎦ 2 b − b x 0 c ξ ρA x2 a ( + ξ cos θ( ) )2 + ⎡⎣ ⎤⎦ ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d + = Ix 1.36 kg m2 ⋅ = Iy 0.380 kg m2 ⋅ = Iz 1.25 kg m2 ⋅ = Problem 21-15 Determine the moment of inertia Ix of the composite plate assembly. The plates have a specific weight γ. Given: γ 6 lb ft2 = a 0.5 ft = b 0.5 ft = c 0.25 ft = g 32.2 ft s2 = Solution: θ atan a b ⎛⎜⎝ ⎞⎟⎠ = I1 γc2 a2 + b2 c2 3 (2a )2 (2b )2 + 12 + ⎡⎢ ⎣ ⎤⎥ ⎦ = 723 Engineering Mechanics - Dynamics Chapter 21 I2 γc2 a2 + b2 c2 3 = Ix 2 I1 sin θ( )2 + I2 cos θ( )2 ( ) γ 2a ( ) 2b ( ) (2a )2 12 + = Ix 0.0293 slug ft2 ⋅ = *Problem 21-16 Determine the product of inertia Iyz of the composite plate assembly. The plates have a specific weight γ. Solution: Due to symmetry, Iyz 0 = Problem 21-17 Determine the moment of inertia of the composite body about the aa axis. The cylinder has weight Wc and each hemisphere has weight Wh. Given: Wc 20 lb = Wh 10 lb = b 2 ft = c 2 ft = g 32.2 ft s2 = Solution: θ atan c b ⎛⎜⎝ ⎞⎟⎠ = Iz 2 2 5Wh c 2 ⎛⎜⎝ ⎞⎟⎠ 2 1 2Wc c 2 ⎛⎜⎝ ⎞⎟⎠ 2 + = Iz 0.56 slug ft2 ⋅ = 724 Engineering Mechanics - Dynamics Chapter 21 Iy 2 83 320Wh c 2 ⎛⎜⎝ ⎞⎟⎠ 2 2Wh b 2 3 8 c 2 + ⎛⎜⎝ ⎞⎟⎠ 2 + Wc b2 12 c 2 ⎛⎜⎝ ⎞⎟⎠ 2 1 4 + ⎡⎢ ⎣ ⎤⎥ ⎦ + = Iy 1.70 slug ft2 ⋅ = Iaa Izcos θ( )2 + Iy sin θ( )2 = Iaa 1.13 slug ft2 ⋅ = Problem 21-18 Determine the moment of inertia about the z axis of the assembly which consists of the rod CD of mass MR and disk of mass MD. Given: MR 1.5 kg = MD 7 kg = r 100 mm = l 200 mm = Solution: θ atan r l ⎛⎜⎝ ⎞⎟⎠ = I1 1 3MRl2 1 + 4MDr2 + MDl2 = I2 I1 = I3 1 2MDr2 = Imat 1 0 0 0 cos θ( ) sin θ( ) − 0 sin θ( ) cos θ( ) ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ I1 0 0 0 I2 0 0 0 I3 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 1 0 0 0 cos θ( ) sin θ( ) 0 sin θ( ) − cos θ( ) ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡⎢ ⎢ ⎢⎣ ⎤⎥ ⎥ ⎥⎦ = Iz Imat2 2 , = Iz 0.0915 kg m2 ⋅ = 725 Engineering Mechanics - Dynamics Chapter 21 Problem 21-19 The assembly consists of a plate A of weight WA, plate B of weight WB, and four rods each of weight Wr. Determine the moments of inertia of the assembly with respect to the principal x, y, z axes. Given: WA 15 lb = WB 40 lb = Wr 7 lb = rA 1 ft = rB 4 ft = h 4 ft = Solution: L rB − rA ( )2 + h2 = L 5.00 ft = θ asin h L ⎛⎜⎝ ⎞⎟⎠ = θ 53.13 deg = Ix 2Wr L2 3 ⎛⎜ ⎝ ⎞⎟ ⎠ sin θ( )2 2 Wr L2 12 ⎛⎜ ⎝ ⎞⎟ ⎠ Wr h 2 ⎛⎜⎝ ⎞⎟⎠ 2 rA rB + 2 ⎛⎜⎝ ⎞⎟⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + WB rB 2 4 + WA rA 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + WAh2 + ... = Iy = Ix by symmetry Iz 4 Wr L2 12 cos θ( )2 Wr rB rA + 2 ⎛⎜⎝ ⎞⎟⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ WA rA 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + WB rB 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + = Ix Iy Iz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 20.2 20.2 16.3 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ slug ft2 ⋅ = *Problem 21-20 The thin plate has a weight Wp and each of the four rods has weight Wr. Determine the moment of 726 Engineering Mechanics - Dynamics Chapter 21 inertia of the assembly about the z axis. Given: Wp 5 lb = Wr 3 lb = h 1.5 ft = a 0.5 ft = Solution: L h2 + a2 + a2 = θ acos h L ⎛⎜⎝ ⎞⎟⎠ = Iz 4 1 3WrL2 sin θ( )2 1 12Wp (2a )2 (2a )2 + ⎡⎣ ⎤⎦ + = Iz 0.0881 slug ft2 ⋅ = Problem 21-21 If a body contains no planes of symmetry, the principal moments of inertia can be determined mathematically. To show how this is done, consider the rigid body which is spinning with an angular velocity ω, directed along one of its principal axes of inertia. If the principal moment of inertia about this axis is I, the angular momentum can be expressed as H = Iω = Iωxi + Iωyj + Iωzk. The components of H may also be expressed by Eqs. 21-10, where the inertia tensor is assumed to be known. Equate the i, j, and k components of both expressions for H and consider ωx, ωy, and ωz to be unknown. The solution of these three equations is obtained provided the determinant of the coefficients is zero. Show that this determinant, when expanded, yields the cubic equation I3 -(Ixx+ Iyy + Izz)I2 + (IxxIyy + IyyIzz + IzzIxx- I2 xy - I2 yz-I2 zx)I -(IxxIyyIzz - 2IxyIyzIzx - IxxI2 yz - IyyI2 zx - IzzI2 xy) = 0. The three positive roots of I, obtained from the solution of this equation, represent the principal moments of inertia Ix, Iy, and Iz. Solution: H = Iω Iωxi + Iωyj + Iωzk = Equating the i, j, and k components to the scalar (Eq. 21 - 10) yields Ixx − I ( )ωx − Ixyωy − Ixzωz 0 = −Iyx ωx Iyy − I ( )ωy + − Iyzωz 0 = 727 Engineering Mechanics - Dynamics Chapter 21 −Izx ωx − Izyωy Izz − I ( )ωz + 0 = Solution for nontrivial ωx, ωy, and ωz requires Ixx − I ( ) Iyx − Izx − Ixy − Iyy − I ( ) Izy − Ixz − Iyz − Izz − I ( ) ⎡⎢ ⎢ ⎢⎣ ⎤⎥ ⎥ ⎥⎦ 0 = Expanding the determinant produces the required equation QED Problem 21-22 Show that if the angular momentum of a body is determined with respect to an arbitrary point A, then HA can be expressed by Eq. 21-9. This requires substituting ρA = ρG + ρGA into Eq. 21-6 and expanding, noting that ∫ρGdm = 0 by definition of the mass center and vG = vA+ ω × ρGA. Solution: HA ρA m ⌠⎮ ⎮⌡ d × vA m ρA ω ( × ρA ) × ⌠⎮ ⎮⌡ d + = HA m ρG + ρGA ⌠⎮ ⎮⌡ d × vA m ρG + ρGA ( ) ω ρG + ρGA ( ) ⎡⎣ × ⎤⎦ × ⌠⎮ ⎮⌡ d + = HA ρG m ⌠⎮ ⎮⌡ d × vA ρGA × m vA + m ρG ω ( × ρG ) × ⌠⎮ ⎮⌡ d + ρG m ⌠⎮ ⎮⌡ d ω ( × ρGA ) × + ρGA ω ρG m ⌠⎮ ⎮⌡ d ⎛⎜ × ⎜⎝ ⎞⎟ ⎟⎠ × ρGA ω ( × ρGA )m × + + ... = 728 Engineering Mechanics - Dynamics Chapter 21 Since ρG m ⌠⎮ ⎮⌡ d = 0 and HG m ρG ω ( × ρG ) × ⌠⎮ ⎮⌡ d = HA ρGA × m vA + HG ρGA × ω (ρGA ) × m + = ρGA × m vA ω × ρGA + ( ) + HG = HA ρG × m vG + HG = Q.E.D Problem 21-23 The thin plate of mass M is suspended at O using a ball-and-socket joint. It is rotating with a constant angular velocity ω = ω1k when the corner A strikes the hook at S, which provides a permanent connection. Determine the angular velocity of the plate immediately after impact. Given: M 5 kg = ω1 2 rad s = a 300 mm = b 400 mm = Solution: Angular Momentum is conserved about the line OA. OA 0 a b − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = oa OA OA = I2 1 = 3M b2 I3 1 12M 2a ( )2 = I1 I2 + I3 = Imat I1 0 0 0 I2 0 0 0 I3 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = Ioa oaTImat oa = Guess ω2 1 rad s = 729 Engineering Mechanics - Dynamics Chapter 21 Given Imat 0 0 ω1 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ oa Ioaω2 = ω2 Find ω2 ( ) = ω2oa 0.00 0.75 − 1.00 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = *Problem 21-24 Rod AB has weight W and is attached to two smooth collars at its end points by ball-and-socket joints. If collar A is moving downward at speed vA, determine the kinetic energy of the rod at the instant shown. Assume that at this instant the angular velocity of the rod is directed perpendicular to the rod’s axis. Given: W 6 lb = vA 8 ft s = a 3 ft = b 6 ft = c 2 ft = Solution: L a2 + b2 + c2 = Guesses vB 1 ft s = ωx 1 rad s = ωy 1 rad s = ωz 1 rad s = Given 0 0 vA − ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ c b a − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × + vB 0 0 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ c b a − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 ft s = 730 Engineering Mechanics - Dynamics Chapter 21 vB ωx ωy ωz ⎛⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟⎠ Find vB ωx , , ωy , ωz ( ) = ω ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ω 0.98 1.06 − 1.47 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = vB 12.00 ft s = vG 0 0 vA − ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ ω c 2 b 2 −a 2 ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ × + = T 1 2 W g ⎛⎜⎝ ⎞⎟⎠ (vGvG ) 1 2 W g ⎛⎜⎝ ⎞⎟⎠ L2 12 ω ω ( ) + = T 6.46 lb ft ⋅ = Problem 21-25 At the instant shown the collar at A on rod AB of weight W has velocity vA. Determine the kinetic energy of the rod after the collar has descended a distance d. Neglect friction and the thickness of the rod. Neglect the mass of the collar and the collar is attached to the rod using ball-and-socket joints. Given: W 6 lb = vA 8 ft s = a 3 ft = b 6 ft = c 2 ft = d 3 ft = Solution: L a2 + b2 + c2 = Guesses vB 1 ft s = ωx 1 rad s = ωy 1 rad s = ωz 1 rad s = 731 Engineering Mechanics - Dynamics Chapter 21 Given 0 0 vA − ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ c b a − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × + vB 0 0 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ c b a − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 ft s = vB ωx ωy ωz ⎛⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟⎠ Find vB ωx , , ωy , ωz ( ) = ω ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ω 0.98 1.06 − 1.47 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = vB 12.00 ft s = vG 0 0 vA − ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ ω c 2 b 2 −a 2 ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ × + = T1 1 2 W g ⎛⎜⎝ ⎞⎟⎠ vG vG ⋅ ( ) 1 2 W g ⎛⎜⎝ ⎞⎟⎠ L2 12 ⎛⎜ ⎝ ⎞⎟ ⎠ ω ω ( ⋅ ) + = T1 6.46 lb ft ⋅ = In position 2 the center of mass has fallen a distance d/2 T1 + 0 T2 W d 2 ⎛⎜⎝ ⎞⎟⎠ − = T2 T1 W d 2 ⎛⎜⎝ ⎞⎟⎠ + = T2 15.5 lb ft ⋅ = Problem 21-26 The rod AB of mass MAB is attached to the collar of mass MA at A and a link BC of mass MBC using ball-and-socket joints. If the rod is released from rest in the position shown, determine the angular velocity of the link after it has rotated 180°. Given: MAB 4 kg = MA 1 kg = 732 Engineering Mechanics - Dynamics Chapter 21 MBC 2 kg = a 1.2 m = b 0.5 m = c a2 + b2 = Solution: θ atan b a ⎛⎜⎝ ⎞⎟⎠ = I 1 3MAB c2 sin θ( )2 1 + 3MBC b2 = Guess ω 1 rad s = Given MAB + MBC ( )gb 1 2Iω2 = ω Find ω ( ) = ω 10.85 rad s = Problem 21-27 The rod has weight density γ and is suspended from parallel cords at A and B. If the rod has angular velocity ω about the z axis at the instant shown, determine how high the center of the rod rises at the instant the rod momentarily stops swinging. Given: γ 3 lb ft = ω 2 rad s = a 3 ft = g 32.2 ft s2 = Solution: T1 + V1 T2 + V2 = 733 Engineering Mechanics - Dynamics Chapter 21 1 2 1 12 γ 2a ( ) g ⎡⎢⎣ ⎤⎥⎦ (2a )2 ω2 γ2ah = h 1 6 a2 ω2 g ⎛⎜ ⎝ ⎞⎟ ⎠ = h 2.24 in = *Problem 21-28 The assembly consists of a rod AB of mass mAB which is connected to link OA and the collar at B by ball-and-socket joints. When θ = 0 and y = y1, the system is at rest, the spring is unstretched, and a couple moment M, is applied to the link at O. Determine the angular velocity of the link at the instant θ = 90°. Neglect the mass of the link. Units Used: kN 103 N = Given: mAB 4 kg = M 7 N m = a 200 mm = y1 600 mm = k 2 kN m = Solution: L a2 + a2 y1 2 + = I 1 3mAB L2 = δ L − y1 = Guess ω 1 rad s = Given mAB g a 2 M 90 deg ( ) + 1 2Iω2 1 + 2kδ2 = 734 Engineering Mechanics - Dynamics Chapter 21 ω Find ω ( ) = ω 6.10 rad s = ωOA ω L a = ωOA 20.2 rad s = Problem 21-29 The assembly consists of a rod AB of mass mAB which is connected to link OA and the collar at B by ball-and-socket joints. When θ = 0 and y = y1, the system is at rest, the spring is unstretched, and a couple moment M = M0(bθ + c), is applied to the link at O. Determine the angular velocity of the link at the instant θ = 90°. Neglect the mass of the link. Units Used: kN 103 N = Given: mAB 4 kg = M0 1 N m = y1 600 mm = a 200 mm = b 4 = c 2 = k 2 kN m = Solution: L a2 + a2 y1 2 + = I 1 3mAB L2 = δ L − y1 = Guess ω 1 rad s = Given mAB g a 2 0 90 deg θ M0 bθ + c ( ) ⌠ ⎮⌡ d + 1 2Iω2 1 + 2kδ2 = ω Find ω ( ) = 735 Engineering Mechanics - Dynamics Chapter 21 ω 5.22 rad s = ωOA ω L a = ωOA 17.3 rad s = Problem 21-30 The circular plate has weight W and diameter d. If it is released from rest and falls horizontally a distance h onto the hook at S, which provides a permanent connection, determine the velocity of the mass center of the plate just after the connection with the hook is made. Given: W 19 lb = d 1.5 ft = h 2.5 ft = g 32.2 ft s2 = Solution: vG1 2gh = vG1 12.69 ft s = W g ⎛⎜⎝ ⎞⎟⎠ vG1 d 2 ⎛⎜⎝ ⎞⎟⎠ 5 4 W g ⎛⎜⎝ ⎞⎟⎠ d 2 ⎛⎜⎝ ⎞⎟⎠ 2 ω2 = ω2 8vG1 5d = ω2 13.53 rad s = vG2 ω2 d 2 = vG2 10.2 ft s = Problem 21-31 A thin plate, having mass M, is suspended from one of its corners by a ball-and-socket joint O. If a stone strikes the plate perpendicular to its surface at an adjacent corner A with an impulse Is, determine the instantaneous axis of rotation for the plate and the impulse created at O. Given: M 4 kg = a 200 mm = 736 Engineering Mechanics - Dynamics Chapter 21 θ −45 deg = Is 60 − 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N s⋅ = Solution: I1 2 = 3M a2 I2 1 3M a2 = I3 = I2 I23 M a2 4 = Cmat 1 0 0 0 cos θ( ) sin θ( ) − 0 sin θ( ) cos θ( ) ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Imat Cmat I1 0 0 0 I2 I23 − 0 I23 − I3 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Cmat T = Guesses IOx IOy IOz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 1 1 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N s = vx vy vz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 1 1 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m s = ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 1 1 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = Given Is IOx IOy IOz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ + M vx vy vz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = a 2 0 1 − 1 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × Is Imat ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 0 0 a − 2 ⎛⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟⎠ × vx vy vz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = 737 Engineering Mechanics - Dynamics Chapter 21 IOx IOy IOz ωx ωy ωz vx vy vz ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ Find IOx IOy , , IOz , ωx , ωy , ωz , vx , vy , vz ( ) = Impulse IOx IOy IOz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ω ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = axis ω ω = Impulse 8.57 0.00 0.00 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N s⋅ = axis 0.00 0.14 0.99 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = *Problem 21-32 Rod AB has weight W and is attached to two smooth collars at its ends by ball-and-socket joints. If collar A is moving downward with speed vA when z = a, determine the speed of A at the instant z = 0. The spring has unstretched length c. Neglect the mass of the collars. Assume the angular velocity of rod AB is perpendicular to its axis. Given: W 6 lb = vA 8 ft s = a 3 ft = b 6 ft = c 2 ft = k 4 lb ft = δ 2 ft = 738 Engineering Mechanics - Dynamics Chapter 21 g 32.2 ft s2 = Solution: L a2 + b2 + c2 = First Position Guesses vB1 1 ft s = ωx1 1 rad s = ωy1 1 rad s = ωz1 1 rad s = Given 0 0 vA − ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ ωx1 ωy1 ωz1 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ c b a − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × + vB1 0 0 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ωx1 ωy1 ωz1 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ c b a − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 ft s = ωx1 ωy1 ωz1 vB1 ⎛⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟⎠ Find ωx1 ωy1 , , ωz1 , vB1 ( ) = ω1 ωx1 ωy1 ωz1 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ω1 0.98 1.06 − 1.47 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = vG1 0 0 vA − ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ ω1 1 2 c b a − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ × + = vG1 6.00 0.00 4.00 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ft s = T1 1 2 W g vG1 vG1 ⋅ ( ) 1 2 W g L2 12 ω1 ω1 ⋅ ( ) + = T1 6.46 lb ft ⋅ = Work - Energy T1 + V1 T2 + V2 = 739 Engineering Mechanics - Dynamics Chapter 21 T2 T1 W a 2 + 1 2k L2 − b2 − c ( ) 2 − = T2 10.30 lb ft ⋅ = Second Position Note that B becomes the instantaneous center Guesses vA2 1 ft s = vB2 1 ft s = ωx2 1 rad s = ωy2 1 rad s = ωz2 1 rad s = Given 0 0 vA2 − ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ ωx2 ωy2 ωz2 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ L2 b2 − b 0 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ × + vB2 0 0 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ωx2 ωy2 ωz2 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ L2 b2 − b 0 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 0 ft s = T2 1 2 W g L2 3 ωx2 2 ωy2 2 + ωz2 2 + ( ) = vA2 vB2 ωx2 ωy2 ωz2 ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find vA2 vB2 , , ωx2 , ωy2 , ωz2 ( ) = ωx2 ωy2 ωz2 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 2.23 1.34 − 0.00 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = vB2 0.00 ft s = vA2 18.2 ft s = Problem 21-33 The circular disk has weight W and is mounted on the shaft AB at angle θ with the horizontal. Determine the angular velocity of the shaft when t = t1 if a constant torque M is applied to the shaft. The shaft is originally spinning with angular velocity ω1 when the torque is applied. 740 Engineering Mechanics - Dynamics Chapter 21 Given: W 15 lb = θ 45 deg = t1 3 s = M 2 lb ft ⋅ = ω1 8 rad s = r 0.8ft = g 32.2 ft s2 = Solution: IAB W g ⎛⎜⎝ ⎞⎟⎠ r2 4 ⎛⎜ ⎝ ⎞⎟ ⎠ cos θ( )2 W g ⎛⎜⎝ ⎞⎟⎠ r2 2 ⎛⎜ ⎝ ⎞⎟ ⎠ sin θ( )2 + = IAB 0.11 lb ft ⋅ ⋅s2 = α M IAB = ω2 ω1 + αt1 = ω2 61.7 rad s = Problem 21-34 The circular disk has weight W and is mounted on the shaft AB at angle of θ with the horizontal. Determine the angular velocity of the shaft when t = t1 if a torque M = M0 ebt applied to the shaft. The shaft is originally spinning at ω1 when the torque is applied. Given: W 15 lb = θ 45 deg = t1 2 s = M0 4 lb ft ⋅ = ω1 8 rad s = r 0.8 ft = g 32.2 ft s2 = 741 Engineering Mechanics - Dynamics Chapter 21 b 0.1 s 1 − = Solution: IAB W g ⎛⎜⎝ ⎞⎟⎠ r2 4 ⎛⎜ ⎝ ⎞⎟ ⎠ cos θ( )2 W g ⎛⎜⎝ ⎞⎟⎠ r2 2 ⎛⎜ ⎝ ⎞⎟ ⎠ sin θ( )2 + = IAB 0.11 lb ft ⋅ ⋅s2 = ω2 ω1 1 IAB 0 t1 t M0 eb t ⌠ ⎮⌡ d ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ + = ω2 87.2 rad s = Problem 21-35 The rectangular plate of mass mp is free to rotate about the y axis because of the bearing supports at A and B. When the plate is balanced in the vertical plane, a bullet of mass mb is fired into it, perpendicular to its surface, with a velocity v. Compute the angular velocity of the plate at the instant it has rotated 180°. If the bullet strikes corner D with the same velocity v, instead of at C, does the angular velocity remain the same? Why or why not? Given: mp 15 kg = mb 0.003 kg = v 2000 m s = a 150 mm = b 150 mm = g 9.81 m s2 = Solution: Guesses ω2 1 rad s = ω3 1 rad s = Given mb va 1 3mp a2ω2 = 1 2mp a2 3 ω2 2 mp g a 2 + 1 2mp a2 3 ω3 2 mp g a 2 − = 742 Engineering Mechanics - Dynamics Chapter 21 ω2 ω3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find ω2 ω3 , ( ) = ω2 8.00 rad s = ω3 21.4 rad s = If the bullet strikes at D, the result will be the same. *Problem 21-36 The rod assembly has a mass density ρ and is rotating with a constant angular velocity ω = ω1k when the loop end at C encounters a hook at S, which provides a permanent connection. Determine the angular velocity of the assembly immediately after impact. Given: ρ 2.5 kg m = a 0.5 m = ω1 2 rad s = h 0.5 m = Solution: OC 0 a h − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = oc OC OC = I1 1 3ρh3 1 12ρ 2a ( )3 + + ρ2ah2 = I2 1 3ρh3 + ρ2ah2 = I3 I1 − I2 = Imat I1 0 0 0 I2 0 0 0 I3 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = IOC ocTImat oc = Guess ω2 1 rad s = 743 Engineering Mechanics - Dynamics Chapter 21 Given Imat 0 0 ω1 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ oc = IOCω2 ω2 Find ω2 ( ) = ω2 −0.63 rad s = ω2oc 0.00 0.44 − 0.44 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = Problem 21-37 The plate of weight W is subjected to force F which is always directed perpendicular to the face of the plate. If the plate is originally at rest, determine its angular velocity after it has rotated one revolution (360°). The plate is supported by ball-and-socket joints at A and B. Given: W 15 lb = F 8 lb = a 0.4 ft = b 1.2 ft = g 32.2 ft s2 = Solution: θ atan a b ⎛⎜⎝ ⎞⎟⎠ = θ 18.43 deg = IAB W g ⎛⎜⎝ ⎞⎟⎠ a2 12 ⎛⎜ ⎝ ⎞⎟ ⎠ cos θ( )2 W g ⎛⎜⎝ ⎞⎟⎠ b2 12 ⎛⎜ ⎝ ⎞⎟ ⎠ sin θ( )2 + = IAB 0.0112 lb ft ⋅ ⋅s2 = Guess ω 1 rad s = Given F acos θ( ) 2π ( ) 1 2IABω2 = ω Find ω ( ) = ω 58.4 rad s = 744 Engineering Mechanics - Dynamics Chapter 21 The space capsule has mass mc and the radii of gyration are kx = kz and ky. If it is traveling with a velocity vG, compute its angular velocity just after it is struck by a meteoroid having mass mm and a velocity vm = ( vxi +vyj +vzk ). Assume that the meteoroid embeds itself into the capsule at point A and that the capsule initially has no angular velocity. Units Used: Mg 1000 kg = Given: mc = 3.5 Mg mm 0.60 kg = kx = 0.8 m vx −200 m s = ky = 0.5 m vy −400 m s = vG 600 m s = vz 200 m s = a = 1 m b = 1 m c 3 m = Solution: Guesses ωx 1 rad s = ωy 1 rad s = ωz 1 rad s = Given a c b − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ mm vx vy vG − vz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ ⎡⎢ ⎢ ⎢⎣ ⎤⎥ ⎥ ⎥⎦ × mc kx 2 0 0 0 ky 2 0 0 0 kx 2 ⎛⎜ ⎜ ⎜ ⎜ ⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎠ ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ = ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find ωx ωy , , ωz ( ) = ωx ωy ωz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 0.107 − 0.000 0.107 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = Problem 21-39 Derive the scalar form of the rotational equation of motion along the x axis when Ω ≠ ω and the moments and products of inertia of the body are not constant with respect to time. 745 Problem 21-38 Engineering Mechanics - Dynamics Chapter 21 Solution: In general M t Hxi + Hyj + Hzk ( ) d d = M H'xi + H'yj + H'zk ( ) Ω Hxi + Hyj + Hzk ( ) × + = Substitute Ω Ωxi + Ωyj + Ωzk = and expanding the cross product yields M H'x − ΩzHy + Ωy Hz ( )i H'y − Ωx Hz + ΩzHx ( )j + H'z − Ωy Hx + Ωx Hy ( )k + ... = Substitute Hx, Hy, and Hz using Eq. 21 - 10. For the i component ΣMx t Ixωx − Ixyωy − Ixzωz ( ) Ωz Iyωy − Iyzωz − Iyxωx ( ) − Ωy Izωz − Izxωx − Izyωy ( ) + ... ⎡⎢⎣ ⎤⎥⎦ d d = One can obtain y and z components in a similar manner. *Problem 21-40 Derive the scalar form of the rotational equation of motion along the x axis when Ω ≠ ω and the moments and products of inertia of the body are constant with respect to time. Solution: In general M t Hxi + Hyj + Hzk ( ) d d = M H'xi + H'yj + H'zk ( ) Ω Hxi + Hyj + Hzk ( ) × + = Substitute Ω Ωxi + Ωyj + Ωzk = and expanding the cross product yields M H'x − ΩzHy + Ωy Hz ( )i H'y − Ωx Hz + ΩzHx ( )j + H'z − Ωy Hx + Ωx Hy ( )k + ... = Substitute Hx, Hy, and Hz using Eq. 21 - 10. For the i component ΣMx t Ixωx − Ixyωy − Ixzωz ( ) Ωz Iyωy − Iyzωz − Iyxωx ( ) − Ωy Izωz − Izxωx − Izyωy ( ) + ... ⎡⎢⎣ ⎤⎥⎦ d d = 746 Engineering Mechanics - Dynamics Chapter 21 For constant inertia, expanding the time derivative of the above equation yields ΣMx Ixω'x − Ixyω'y − Ixzω'z ( ) Ωz Iyωy − Iyzωz − Iyxωx ( ) − Ωy Izωz − Izxωx − Izyωy ( ) + ... = One can obtain y and z components in a similar manner. Problem 21-41 Derive the Euler equations of motion for Ω ≠ ω i.e., Eqs. 21-26. Solution: In general M t Hxi + Hyj + Hzk ( ) d d = M H'xi + H'yj + H'zk ( ) Ω Hxi + Hyj + Hzk ( ) × + = Substitute Ω Ωxi + Ωyj + Ωzk = and expanding the cross product yields M H'x − ΩzHy + Ωy Hz ( )i H'y − Ωx Hz + ΩzHx ( )j + H'z − Ωy Hx + Ωx Hy ( )k + ... = Substitute Hx, Hy, and Hz using Eq. 21 - 10. For the i component ΣMx t Ixωx − Ixyωy − Ixzωz ( ) Ωz Iyωy − Iyzωz − Iyxωx ( ) − Ωy Izωz − Izxωx − Izyωy ( ) + ... ⎡⎢⎣ ⎤⎥⎦ d d = Set Ixy = Iyz = Izx = 0 and require Ix Iy , , Iz to be constant. This yields ΣMx Ixω'x − IyΩzωy + IzΩyωz = One can obtain y and z components in a similar manner. Problem 21-42 The flywheel (disk of mass M) is mounted a distance d off its true center at G. If the shaft is rotating at constant speed ω, determine the maximum reactions exerted on the journal bearings at A and B. 747 Engineering Mechanics - Dynamics Chapter 21 Given: M = 40 kg a 0.75 m = d = 20 mm b 1.25 m = ω 8 rad s = g 9.81 m s2 = Solution: Check both up and down positions Guesses Aup = 1 N Bup 1 N = Given Aup + Bup − M g −M dω2 = −Aup a + Bupb 0 = Aup Bup ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find Aup Bup , ( ) = Aup Bup ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 213.25 ⎛⎜⎝127.95 ⎞⎟⎠ N = Guesses Adown = 1 N Bdown 1 N = Given Adown + Bdown − M g M dω2 = −Adown a + Bdownb 0 = Adown Bdown ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find Adown Bdown , ( ) = Adown Bdown ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 277.25 ⎛⎜⎝166.35 ⎞⎟⎠ N = Thus Amax max Aup Adown , ( ) = Amax 277 N = Bmax max Bup Bdown , ( ) = Bmax 166 N = Problem 21-43 The flywheel (disk of mass M) is mounted a distance d off its true center at G. If the shaft is rotating at constant speed ω, determine the minimum reactions exerted on the journal bearings at A and B during the motion. 748 Engineering Mechanics - Dynamics Chapter 21 Given: M = 40 kg a 0.75 m = d = 20 mm b 1.25 m = ω 8 rad s = g 9.81 m s2 = Solution: Check both up and down positions Guesses Aup = 1 N Bup 1 N = Given Aup + Bup − M g −M dω2 = −Aup a + Bupb 0 = Aup Bup ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find Aup Bup , ( ) = Aup Bup ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 213.25 ⎛⎜⎝127.95 ⎞⎟⎠ N = Guesses Adown = 1 N Bdown 1 N = Given Adown + Bdown − M g M dω2 = −Adown a + Bdownb 0 = Adown Bdown ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find Adown Bdown , ( ) = Adown Bdown ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 277.25 ⎛⎜⎝166.35 ⎞⎟⎠ N = Thus Amin min Aup Adown , ( ) = Amin 213 N = Bmin min Bup Bdown , ( ) = Bmin 128 N = 749 Engineering Mechanics - Dynamics Chapter 21 *Problem 21-44 The bar of weight W rests along the smooth corners of an open box. At the instant shown, the box has a velocity v = v1k and an acceleration a = a1k. Determine the x, y, z components of force which the corners exert on the bar. Given: W = 4 lb a 2 ft = v1 5 ft s = b 1 ft = c 2 ft = a1 2 ft s2 = Solution: Guesses Ax = 1 lb Bx 1 lb = Ay = 1 lb By 1 lb = Bz 1 lb = Given Ax + Bx 0 = Ay + By 0 = Bz − W W g a1 = 1 2 c − b a − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Bx By Bz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ × 1 2 c b − a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Ax Ay 0 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ × + 0 = Ax Ay Bx By Bz ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ Find Ax Ay , , Bx , By , Bz ( ) = Ax Ay ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.12 − ⎛⎜⎝ 1.06 ⎞⎟⎠ lb = Bx By Bz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 2.12 1.06 − 4.25 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb = 750 Engineering Mechanics - Dynamics Chapter 21 Problem 21-45 The bar of weight W rests along the smooth corners of an open box. At the instant shown, the box has a velocity v = v1j and an acceleration a = a1j. Determine the x, y, z components of force which the corners exert on the bar. Given: W = 4 lb a 2 ft = v1 3 ft s = b 1 ft = c 2 ft = a1 −6 ft s2 = Solution: Guesses Ax = 1 lb Bx 1 lb = Ay = 1 lb By 1 lb = Bz 1 lb = Given Ax + Bx 0 = Ay + By W g a1 = Bz − W 0 = 1 2 c − b a − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Bx By Bz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ × 1 2 c b − a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Ax Ay 0 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ × + 0 = Ax Ay Bx By Bz ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ Find Ax Ay , , Bx , By , Bz ( ) = Ax Ay ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.00 − ⎛⎜⎝ 0.63 ⎞⎟⎠ lb = Bx By Bz ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 2.00 1.37 − 4.00 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb = 751 Engineering Mechanics - Dynamics Chapter 21 Problem 21-46 The conical pendulum consists of a bar of mass m and le