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 s 2 = Solution: θ atan r h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I x 3 80 W 4r 2 h 2 + () W 3h 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += I y I x = I z 3 10 Wr 2 = I z' I x sin θ() 2 I z cos θ() 2 += I z' 0.0962 slug ft 2 ⋅= 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 I x' , I y' , and I z' of the rod with respect to the x', y', z' axes. Given: γ 1.5 lb ft = a 1ft= b 1ft= g 32.2 ft s 2 = 721 Engineering Mechanics - Dynamics Chapter 21 Solution: x' 2a a 2 ba+ 2ab+ = x' 0.667 ft= y' ab b b 2 + 2ab+ = y' 0.50 ft= I x' γay' 2 γab y'−() 2 + 1 12 γbb 2 + γb b 2 y'− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += I x' 0.0272 slug ft 2 ⋅= I y' 2 12 γaa 2 2γa a 2 x'− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + γba x'−() 2 += I y' 0.0155 slug ft 2 ⋅= I z' I x' I y' += I z' 0.0427 slug ft 2 ⋅= Problem 21-14 The assembly consists of two square plates A and B which have a mass M A each and a rectangular plate C which has a mass M C . Determine the moments of inertia I x , I y and I z . Given: M A 3kg= M C 4.5 kg= θ 60 deg= θ 1 90 deg= θ 2 30 deg= a 0.3 m= b 0.2 m= c 0.4 m= Solution: ρ A M A c 2b() = 722 Engineering Mechanics - Dynamics Chapter 21 I x 1 12 M C 2a() 2 2 0 c ξρ A 2b()a ξ cos θ()+() 2 ξ sin θ()() 2 + ⎡ ⎣ ⎤ ⎦ ⌠ ⎮ ⌡ d+= I y 1 12 M C 2b() 2 2 b− b x 0 c ξρ A x 2 ξ 2 sin θ() 2 + () ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d+= I z 1 12 M C 2b() 2 2a() 2 + ⎡ ⎣ ⎤ ⎦ 2 b− b x 0 c ξρ A x 2 a ξ cos θ()+() 2 + ⎡ ⎣ ⎤ ⎦ ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d+= I x 1.36 kg m 2 ⋅= I y 0.380 kg m 2 ⋅= I z 1.25 kg m 2 ⋅= Problem 21-15 Determine the moment of inertia I x of the composite plate assembly. The plates have a specific weight γ. Given: γ 6 lb ft 2 = a 0.5 ft= b 0.5 ft= c 0.25 ft= g 32.2 ft s 2 = Solution: θ atan a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I 1 γc2 a 2 b 2 + c 2 3 2a() 2 2b() 2 + 12 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = 723 Engineering Mechanics - Dynamics Chapter 21 I 2 γc2 a 2 b 2 + c 2 3 = I x 2 I 1 sin θ() 2 I 2 cos θ() 2 + γ 2a()2b() 2a() 2 12 += I x 0.0293 slug ft 2 ⋅= *Problem 21-16 Determine the product of inertia I yz of the composite plate assembly. The plates have a specific weight γ. Solution: Due to symmetry, I yz 0= Problem 21-17 Determine the moment of inertia of the composite body about the aa axis. The cylinder has weight W c and each hemisphere has weight W h . Given: W c 20 lb= W h 10 lb= b 2ft= c 2ft= g 32.2 ft s 2 = Solution: θ atan c b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I z 2 2 5 W h c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 1 2 W c c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += I z 0.56 slug ft 2 ⋅= 724 Engineering Mechanics - Dynamics Chapter 21 I y 2 83 320 W h c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 2W h b 2 3 8 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + W c b 2 12 c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 1 4 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ += I y 1.70 slug ft 2 ⋅= I aa I z cos θ() 2 I y sin θ() 2 += I aa 1.13 slug ft 2 ⋅= Problem 21-18 Determine the moment of inertia about the z axis of the assembly which consists of the rod CD of mass M R and disk of mass M D . Given: M R 1.5 kg= M D 7kg= r 100 mm= l 200 mm= Solution: θ atan r l ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I 1 1 3 M R l 2 1 4 M D r 2 + M D l 2 += I 2 I 1 = I 3 1 2 M D r 2 = I mat 1 0 0 0 cos θ() sin θ()− 0 sin θ() cos θ() ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ I 1 0 0 0 I 2 0 0 0 I 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1 0 0 0 cos θ() sin θ() 0 sin θ()− cos θ() ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ = I z I mat 22, = I z 0.0915 kg m 2 ⋅= 725 Engineering Mechanics - Dynamics Chapter 21 Problem 21-19 The assembly consists of a plate A of weight W A , plate B of weight W B , and four rods each of weight W r . Determine the moments of inertia of the assembly with respect to the principal x, y, z axes. Given: W A 15 lb= W B 40 lb= W r 7lb= r A 1ft= r B 4ft= h 4ft= Solution: Lr B r A −() 2 h 2 += L 5.00 ft= θ asin h L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 53.13 deg= I x 2W r L 2 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ() 2 2 W r L 2 12 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ W r h 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 r A r B + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + W B r B 2 4 + W A r A 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ W A h 2 ++ ...= I y I x = by symmetry I z 4 W r L 2 12 cos θ() 2 W r r B r A + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ W A r A 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + W B r B 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += I x I y I z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 20.2 20.2 16.3 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ slug ft 2 ⋅= *Problem 21-20 The thin plate has a weight W p and each of the four rods has weight W r . Determine the moment of 726 Engineering Mechanics - Dynamics Chapter 21 inertia of the assembly about the z axis. Given: W p 5lb= W r 3lb= h 1.5 ft= a 0.5 ft= Solution: Lh 2 a 2 + a 2 += θ acos h L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I z 4 1 3 W r L 2 sin θ() 2 1 12 W p 2a() 2 2a() 2 + ⎡ ⎣ ⎤ ⎦ += I z 0.0881 slug ft 2 ⋅= 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 = Iω x i + Iω y j + Iω z k. 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 ω x , ω y , and ω z 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 I 3 -(I xx + I yy + I zz )I 2 + (I xx I yy + I yy I zz + I zz I xx - I 2 xy - I 2 yz -I 2 zx )I -(I xx I yy I zz - 2I xy I yz I zx - I xx I 2 yz - I yy I 2 zx - I zz I 2 xy ) = 0. The three positive roots of I, obtained from the solution of this equation, represent the principal moments of inertia I x , I y , and I z . Solution: H Iω= Iω x i Iω y j+ Iω z k+= Equating the i, j, and k components to the scalar (Eq. 21 - 10) yields I xx I−()ω x I xy ω y − I xz ω z − 0= I yx − ω x I yy I−()ω y + I yz ω z − 0= 727 Engineering Mechanics - Dynamics Chapter 21 I zx − ω x I zy ω y − I zz I−()ω z + 0= Solution for nontrivial ω x , ω y , and ω z requires I xx I−() Iyx− I zx − I xy − I yy I−() I zy − I xz − I yz − I zz 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 H A can be expressed by Eq. 21-9. This requires substituting ρ A = ρ G + ρ GA into Eq. 21-6 and expanding, noting that ∫ρ G dm = 0 by definition of the mass center and v G = v A + ω × ρ GA . Solution: H A mρ A ⌠ ⎮ ⎮ ⌡ d v A × mρ A ωρ A ×()× ⌠ ⎮ ⎮ ⌡ d+= H A mρ G ρ GA + ⌠ ⎮ ⎮ ⌡ d v A × mρ G ρ GA +()ωρ G ρ GA +()× ⎡ ⎣ ⎤ ⎦ × ⌠ ⎮ ⎮ ⌡ d+= H A mρ G ⌠ ⎮ ⎮ ⌡ d v A × ρ GA m× v A + mρ G ωρ G ×()× ⌠ ⎮ ⎮ ⌡ d+ mρ G ⌠ ⎮ ⎮ ⌡ d ωρ GA ×()×+ ρ GA ω mρ G ⌠ ⎮ ⎮ ⌡ d× ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × ρ GA ωρ GA ×()m×++ ...= 728 Engineering Mechanics - Dynamics Chapter 21 Since mρ G ⌠ ⎮ ⎮ ⌡ d0= and H G mρ G ωρ G ×()× ⌠ ⎮ ⎮ ⌡ d= H A ρ GA m× v A H G + ρ GA ω× ρ GA ()× m+= ρ GA m× v A ωρ GA ×+()H G += H A ρ G m× v G H G += 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 = ω 1 k 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 5kg= ω 1 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 = I 2 1 3 Mb 2 = I 3 1 12 M 2a() 2 = I 1 I 2 I 3 += I mat I 1 0 0 0 I 2 0 0 0 I 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = I oa oa T I mat oa= Guess ω 2 1 rad s = 729 Engineering Mechanics - Dynamics Chapter 21 Given I mat 0 0 ω 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ oa I oa ω 2 = ω 2 Find ω 2 ()= ω 2 oa 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 v A , 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 6lb= v A 8 ft s = a 3ft= b 6ft= c 2ft= Solution: La 2 b 2 + c 2 += Guesses v B 1 ft s = ω x 1 rad s = ω y 1 rad s = ω z 1 rad s = Given 0 0 v A − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ c b a− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×+ v B 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ c b a− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 ft s = 730 Engineering Mechanics - Dynamics Chapter 21 v B ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v B ω x , ω y , ω z ,()= ω ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω 0.98 1.06− 1.47− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = v B 12.00 ft s = v G 0 0 v A − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ω c 2 b 2 a− 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ×+= T 1 2 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v G v G () 1 2 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L 2 12 ωω()+= T 6.46 lb ft⋅= Problem 21-25 At the instant shown the collar at A on rod AB of weight W has velocity v A . 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 6lb= v A 8 ft s = a 3ft= b 6ft= c 2ft= d 3ft= Solution: La 2 b 2 + c 2 += Guesses v B 1 ft s = ω x 1 rad s = ω y 1 rad s = ω z 1 rad s = 731 Engineering Mechanics - Dynamics Chapter 21 Given 0 0 v A − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ c b a− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×+ v B 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ c b a− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 ft s = v B ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v B ω x , ω y , ω z ,()= ω ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω 0.98 1.06− 1.47− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = v B 12.00 ft s = v G 0 0 v A − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ω c 2 b 2 a− 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ×+= T 1 1 2 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v G v G ⋅() 1 2 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L 2 12 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ωω⋅()+= T 1 6.46 lb ft⋅= In position 2 the center of mass has fallen a distance d/2 T 1 0+ T 2 W d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= T 2 T 1 W d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += T 2 15.5 lb ft⋅= Problem 21-26 The rod AB of mass M AB is attached to the collar of mass M A at A and a link BC of mass M BC 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: M AB 4kg= M A 1kg= 732 Engineering Mechanics - Dynamics Chapter 21 M BC 2kg= a 1.2 m= b 0.5 m= ca 2 b 2 += Solution: θ atan b a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I 1 3 M AB c 2 sin θ() 2 1 3 M BC b 2 += Guess ω 1 rad s = Given M AB M BC +()gb 1 2 Iω 2 = 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 3ft= g 32.2 ft s 2 = Solution: T 1 V 1 + T 2 V 2 += 733 Engineering Mechanics - Dynamics Chapter 21 1 2 1 12 γ 2a() g ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2a() 2 ω 2 γ2ah= h 1 6 a 2 ω 2 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = h 2.24 in= *Problem 21-28 The assembly consists of a rod AB of mass m AB which is connected to link OA and the collar at B by ball-and-socket joints. When θ = 0 and y = y 1 , 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 10 3 N= Given: m AB 4kg= M 7Nm= a 200 mm= y 1 600 mm= k 2 kN m = Solution: La 2 a 2 + y 1 2 += I 1 3 m AB L 2 = δ Ly 1 −= Guess ω 1 rad s = Given m AB g a 2 M 90 deg()+ 1 2 Iω 2 1 2 kδ 2 += 734 Engineering Mechanics - Dynamics Chapter 21 ω Find ω()= ω 6.10 rad s = ω OA ω L a = ω OA 20.2 rad s = Problem 21-29 The assembly consists of a rod AB of mass m AB which is connected to link OA and the collar at B by ball-and-socket joints. When θ = 0 and y = y 1 , the system is at rest, the spring is unstretched, and a couple moment M = M 0 (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 10 3 N= Given: m AB 4kg= M 0 1Nm= y 1 600 mm= a 200 mm= b 4= c 2= k 2 kN m = Solution: La 2 a 2 + y 1 2 += I 1 3 m AB L 2 = δ Ly 1 −= Guess ω 1 rad s = Given m AB g a 2 0 90 deg θM 0 bθ c+() ⌠ ⎮ ⌡ d+ 1 2 Iω 2 1 2 kδ 2 += ω Find ω()= 735 Engineering Mechanics - Dynamics Chapter 21 ω 5.22 rad s = ω OA ω L a = ω OA 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 s 2 = Solution: v G1 2gh= v G1 12.69 ft s = W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v G1 d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 5 4 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ω 2 = ω 2 8v G1 5d = ω 2 13.53 rad s = v G2 ω 2 d 2 = v G2 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 I s , determine the instantaneous axis of rotation for the plate and the impulse created at O. Given: M 4kg= a 200 mm= 736 Engineering Mechanics - Dynamics Chapter 21 θ 45− deg= I s 60− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Ns⋅= Solution: I 1 2 3 Ma 2 = I 2 1 3 Ma 2 = I 3 I 2 = I 23 M a 2 4 = C mat 1 0 0 0 cos θ() sin θ()− 0 sin θ() cos θ() ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = I mat C mat I 1 0 0 0 I 2 I 23 − 0 I 23 − I 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ C mat T = Guesses I Ox I Oy I Oz ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1 1 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Ns= v x v y v z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1 1 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m s = ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1 1 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = Given I s I Ox I Oy I Oz ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + M v x v y v z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = a 2 0 1− 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ I s × I mat ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 0 a− 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ × v x v y v z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = 737 Engineering Mechanics - Dynamics Chapter 21 I Ox I Oy I Oz ω x ω y ω z v x v y v z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find I Ox I Oy , I Oz , ω x , ω y , ω z , v x , v y , v z ,()= Impulse I Ox I Oy I Oz ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = axis ω ω = Impulse 8.57 0.00 0.00 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Ns⋅= 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 v A 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 6lb= v A 8 ft s = a 3ft= b 6ft= c 2ft= k 4 lb ft = δ 2ft= 738 Engineering Mechanics - Dynamics Chapter 21 g 32.2 ft s 2 = Solution: La 2 b 2 + c 2 += First Position Guesses v B1 1 ft s = ω x1 1 rad s = ω y1 1 rad s = ω z1 1 rad s = Given 0 0 v A − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ω x1 ω y1 ω z1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ c b a− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×+ v B1 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω x1 ω y1 ω z1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ c b a− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 ft s = ω x1 ω y1 ω z1 v B1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find ω x1 ω y1 , ω z1 , v B1 ,()= ω 1 ω x1 ω y1 ω z1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω 1 0.98 1.06− 1.47− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = v G1 0 0 v A − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ω 1 1 2 c b a− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= v G1 6.00 0.00 4.00− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ft s = T 1 1 2 W g v G1 v G1 ⋅() 1 2 W g L 2 12 ω 1 ω 1 ⋅()+= T 1 6.46 lb ft⋅= Work - Energy T 1 V 1 + T 2 V 2 += 739 Engineering Mechanics - Dynamics Chapter 21 T 2 T 1 W a 2 + 1 2 kL 2 b 2 − c− () 2 −= T 2 10.30 lb ft⋅= Second Position Note that B becomes the instantaneous center Guesses v A2 1 ft s = v B2 1 ft s = ω x2 1 rad s = ω y2 1 rad s = ω z2 1 rad s = Given 0 0 v A2 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ω x2 ω y2 ω z2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ L 2 b 2 − b 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ v B2 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω x2 ω y2 ω z2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ L 2 b 2 − b 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 ft s = T 2 1 2 W g L 2 3 ω x2 2 ω y2 2 + ω z2 2 +()= v A2 v B2 ω x2 ω y2 ω z2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A2 v B2 , ω x2 , ω y2 , ω z2 ,()= ω x2 ω y2 ω z2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 2.23 1.34− 0.00 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ rad s = v B2 0.00 ft s = v A2 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 = t 1 if a constant torque M is applied to the shaft. The shaft is originally spinning with angular velocity ω 1 when the torque is applied. 740 Engineering Mechanics - Dynamics Chapter 21 Given: W 15 lb= θ 45 deg= t 1 3s= M 2lbft⋅= ω 1 8 rad s = r 0.8ft= g 32.2 ft s 2 = Solution: I AB W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ() 2 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ() 2 += I AB 0.11 lb ft⋅ s 2 ⋅= α M I AB = ω 2 ω 1 αt 1 += ω 2 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 = t 1 if a torque M = M 0 e bt applied to the shaft. The shaft is originally spinning at ω 1 when the torque is applied. Given: W 15 lb= θ 45 deg= t 1 2s= M 0 4lbft⋅= ω 1 8 rad s = r 0.8 ft= g 32.2 ft s 2 = 741 Engineering Mechanics - Dynamics Chapter 21 b 0.1 s 1− = Solution: I AB W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ() 2 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ() 2 += I AB 0.11 lb ft⋅ s 2 ⋅= ω 2 ω 1 1 I AB 0 t 1 tM 0 e bt ⌠ ⎮ ⌡ d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += ω 2 87.2 rad s = Problem 21-35 The rectangular plate of mass m p 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 m b 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: m p 15 kg= m b 0.003 kg= v 2000 m s = a 150 mm= b 150 mm= g 9.81 m s 2 = Solution: Guesses ω 2 1 rad s = ω 3 1 rad s = Given m b va 1 3 m p a 2 ω 2 = 1 2 m p a 2 3 ω 2 2 m p g a 2 + 1 2 m p a 2 3 ω 3 2 m p g a 2 −= 742 Engineering Mechanics - Dynamics Chapter 21 ω 2 ω 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find ω 2 ω 3 ,()= ω 2 8.00 rad s = ω 3 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 = ω 1 k 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= ω 1 2 rad s = h 0.5 m= Solution: OC 0 a h− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = oc OC OC = I 1 1 3 ρh 3 1 12 ρ 2a() 3 + ρ2ah 2 += I 2 1 3 ρh 3 ρ2ah 2 += I 3 I 1 I 2 −= I mat I 1 0 0 0 I 2 0 0 0 I 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = I OC oc T I mat oc= Guess ω 2 1 rad s = 743 Engineering Mechanics - Dynamics Chapter 21 Given I mat 0 0 ω 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ oc I OC ω 2 = ω 2 Find ω 2 ()= ω 2 0.63− rad s = ω 2 oc 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 8lb= a 0.4 ft= b 1.2 ft= g 32.2 ft s 2 = Solution: θ atan a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 18.43 deg= I AB W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 2 12 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ() 2 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 12 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ() 2 += I AB 0.0112 lb ft⋅ s 2 ⋅= Guess ω 1 rad s = Given Facos θ()2π() 1 2 I AB ω 2 = ω Find ω()= ω 58.4 rad s = 744 Engineering Mechanics - Dynamics Chapter 21 The space capsule has mass m c and the radii of gyration are k x = k z and k y . If it is traveling with a velocity v G , compute its angular velocity just after it is struck by a meteoroid having mass m m and a velocity v m = ( v x i +v y j +v z k ). 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: m c 3.5 Mg= m m 0.60 kg= k x 0.8 m= v x 200− m s = k y 0.5 m= v y 400− m s = v G 600 m s = v z 200 m s = a 1m= b 1m= c 3m= Solution: Guesses ω x 1 rad s = ω y 1 rad s = ω z 1 rad s = Given a c b− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m m v x v y v G − v z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ × m c k x 2 0 0 0 k y 2 0 0 0 k x 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find ω x ω y , ω z ,()= ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 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 H x i H y j+ H z k+() d d = M H' x i H' y j+ H' z k+()Ω H x i H y j+ H z k+()×+= Substitute Ω Ω x i Ω y j+ Ω z k+= and expanding the cross product yields M H' x Ω z H y − Ω y H z +()i H' y Ω x H z − Ω z H x +()j+ H' z Ω y H x − Ω x H y + k+ ...= Substitute H x , H y , and H z using Eq. 21 - 10. For the i component ΣM x t I x ω x I xy ω y − I xz ω z −()Ω z I y ω y I yz ω z − I yx ω x −()− Ω y I z ω z I zx ω x − I zy ω y −+ ... ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 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 H x i H y j+ H z k+() d d = M H' x i H' y j+ H' z k+()Ω H x i H y j+ H z k+()×+= Substitute Ω Ω x i Ω y j+ Ω z k+= and expanding the cross product yields M H' x Ω z H y − Ω y H z +()i H' y Ω x H z − Ω z H x +()j+ H' z Ω y H x − Ω x H y + k+ ...= Substitute H x , H y , and H z using Eq. 21 - 10. For the i component ΣM x t I x ω x I xy ω y − I xz ω z −()Ω z I y ω y I yz ω z − I yx ω x −()− Ω y I z ω z I zx ω x − I zy ω y −+ ... ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ d d = 746 Engineering Mechanics - Dynamics Chapter 21 For constant inertia, expanding the time derivative of the above equation yields ΣM x I x ω' x I xy ω' y − I xz ω' z −()Ω z I y ω y I yz ω z − I yx ω x −()− Ω y I z ω z I zx ω x − I zy ω y −+ ...= 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 H x i H y j+ H z k+() d d = M H' x i H' y j+ H' z k+()Ω H x i H y j+ H z k+()×+= Substitute Ω Ω x i Ω y j+ Ω z k+= and expanding the cross product yields M H' x Ω z H y − Ω y H z +()i H' y Ω x H z − Ω z H x +()j+ H' z Ω y H x − Ω x H y + k+ ...= Substitute H x , H y , and H z using Eq. 21 - 10. For the i component ΣM x t I x ω x I xy ω y − I xz ω z −()Ω z I y ω y I yz ω z − I yx ω x −()− Ω y I z ω z I zx ω x − I zy ω y −+ ... ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ d d = Set I xy I yz = I zx = 0= and require I x I y , I z , to be constant. This yields ΣM x I x ω' x I y Ω z ω y − I z Ω y ω z += 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 s 2 = Solution: Check both up and down positions Guesses A up 1N= B up 1N= Given A up B up + Mg− M− dω 2 = A up − aB up b+ 0= A up B up ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find A up B up ,()= A up B up ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 213.25 127.95 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= Guesses A down 1N= B down 1N= Given A down B down + Mg− Mdω 2 = A down − aB down b+ 0= A down B down ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find A down B down ,()= A down B down ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 277.25 166.35 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= Thus A max max A up A down ,()= A max 277 N= B max max B up B down ,()= B max 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 s 2 = Solution: Check both up and down positions Guesses A up 1N= B up 1N= Given A up B up + Mg− M− dω 2 = A up − aB up b+ 0= A up B up ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find A up B up ,()= A up B up ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 213.25 127.95 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= Guesses A down 1N= B down 1N= Given A down B down + Mg− Mdω 2 = A down − aB down b+ 0= A down B down ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find A down B down ,()= A down B down ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 277.25 166.35 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= Thus A min min A up A down ,()= A min 213 N= B min min B up B down ,()= B min 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 = v 1 k and an acceleration a = a 1 k. Determine the x, y, z components of force which the corners exert on the bar. Given: W 4lb= a 2ft= v 1 5 ft s = b 1ft= c 2ft= a 1 2 ft s 2 = Solution: Guesses A x 1lb= B x 1lb= A y 1lb= B y 1lb= B z 1lb= Given A x B x + 0= A y B y + 0= B z W− W g a 1 = 1 2 c− b a− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ B x B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ × 1 2 c b− a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ A x A y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0= A x A y B x B y B z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , B x , B y , B z ,()= A x A y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.12− 1.06 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= B x B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 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 = v 1 j and an acceleration a = a 1 j. Determine the x, y, z components of force which the corners exert on the bar. Given: W 4lb= a 2ft= v 1 3 ft s = b 1ft= c 2ft= a 1 6− ft s 2 = Solution: Guesses A x 1lb= B x 1lb= A y 1lb= B y 1lb= B z 1lb= Given A x B x + 0= A y B y + W g a 1 = B z W− 0= 1 2 c− b a− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ B x B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ × 1 2 c b− a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ A x A y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0= A x A y B x B y B z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , B x , B y , B z ,()= A x A y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.00− 0.63 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= B x B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 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 length L that is supported by the pin at its end A. If the pin is subjected to a rotation ω, determine the angle θ that the bar makes with the vertical as it rotates. Solution: I x I z = 1 3 mL 2 = I y 0= ω x 0= ω y ω− cos θ()= ω z ωsin θ()= ω' x 0= ω' y 0= ω' z 0= ΣM x I x ω' x I y I z −()ω y ω x −= m− g L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ() 00 1 3 mL 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ω− cos θ()()ωsin θ()()−= g 2 1 3 Lω 2 cos θ()= θ acos 3g 2Lω 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Problem 21-47 The plate of weight W is mounted on the shaft AB so that the plane of the plate makes an angle θ with the vertical. If the shaft is turning in the direction shown with angular velocity ω, determine the vertical reactions at the bearing supports A and B when the plate is in the position shown. 752 Engineering Mechanics - Dynamics Chapter 21 W 20 lb= θ 30 deg= ω 25 rad s = a 18 in= b 18 in= c 6in= Solution: I x W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c 2 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I z I x 2 = I y I z = ω x ωsin θ()= ω y ω− cos θ()= ω z 0 rad s = Guesses F A 1lb= F B 1lb= Given F A F B + W− 0= 0 0 F B bF A a− ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ I x 0 0 0 I y 0 0 0 I z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ω x ω y ω z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ ×= F A F B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find F A F B ,()= F A F B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 8.83 11.17 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= *Problem 21-48 The car is traveling around the curved road of radius ρ such that its mass center has a constant speed v G . Write the equations of rotational motion with respect to the x, y, z axes. Assume that the car’s six moments and products of inertia with respect to these axes are known. 753 Given: Engineering Mechanics - Dynamics Chapter 21 Solution: Applying Eq. 21-24 with ω x 0= ω y 0= ω z v G ρ = ω' x ω' y = ω' z = 0= ΣM z 0= ΣM x I yz v G ρ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = ΣM y I zx v G ρ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = Note: This result indicates the normal reactions of the tires on the ground are not all necessarily equal. Instead, they depend upon the speed of the car, radius of curvature, and the products of inertia, I yz and I zx . (See Example 13-6.) Problem 21-49 The rod assembly is supported by journal bearings at A and B, which develops only x and z force reactions on the shaft. If the shaft AB is rotating in the direction shown with angular velocity ω, determine the reactions at the bearings when the assembly is in the position shown. Also, what is the shaft’s angular acceleration? The mass density of each rod is ρ. Given: ω 5− rad s = ρ 1.5 kg m = a 500 mm= b 300 mm= c 500 mm= d 400 mm= e 300 mm= g 9.81 m s 2 = 754 Engineering Mechanics - Dynamics Chapter 21 I xx ρ ab+ c+() ab+ c+() 2 3 ρda 2 + ρe e 2 12 + ρea b+() 2 e 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ += I zz ρ ab+ c+() ab+ c+() 2 3 ρd d 2 12 + ρda 2 d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + ρea b+() 2 += I yy ρd d 2 3 ρe e 2 3 += I xy ρda d 2 = I yz ρea b+() e 2 = I mat I xx I xy − 0 I xy − I yy I yz − 0 I yz − I zz ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = I mat 1.5500 0.0600− 0.0000 0.0600− 0.0455 0.0540− 0.0000 0.0540− 1.5685 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kg m 2 ⋅= Guesses A x 1N= A z 1N= B x 1N= B z 1N= ω' y 1 rad s 2 = Given A x 0 A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ B x 0 B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 0 ρ ab+ c+ d+ e+()g ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ − ρ− d d 2 ω 2 ρe e 2 ω' y + 0 ρ− d d 2 ω' y ρe e 2 ω 2 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = 0 ab+ c+ 2 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 0 0 ρ− ab+ c+()g ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ × d 2 a 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 0 0 ρ− dg ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×+ 0 ab+ e 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 0 0 ρ− eg ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × 0 ab+ c+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ B x 0 B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×++ ... I mat 0 ω' y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 ω− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ I mat 0 ω− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= A x A z B x B z ω' y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A z , B x , B z , ω' y ,()= A x A z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.17− 12.33 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= B x B z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.0791− 12.3126 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= ω' y 25.9 rad s 2 = 755 Solution: Engineering Mechanics - Dynamics Chapter 21 Problem 21-50 The rod assembly is supported by journal bearings at A and B, which develops only x and z force reactions on the shaft. If the shaft AB is subjected to a couple moment M 0 j and at the instant shown the shaft has an angular velocity ω j, determine the reactions at the bearings when the assembly is in the position shown. Also, what is the shaft's angular acceleration? The mass density of each rod is ρ. Given: ω 5− rad s = c 500 mm= ρ 1.5 kg m = d 400 mm= a 500 mm= e 300 mm= b 300 mm= g 9.81 m s 2 = M 0 8Nm⋅= Solution: I xx ρ ab+ c+() ab+ c+() 2 3 ρda 2 + ρe e 2 12 + ρea b+() 2 e 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ += I zz ρ ab+ c+() ab+ c+() 2 3 ρd d 2 12 + ρda 2 d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + ρea b+() 2 += I yy ρd d 2 3 ρe e 2 3 += I xy ρda d 2 = I yz ρea b+() e 2 = I mat I xx I xy − 0 I xy − I yy I yz − 0 I yz − I zz ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = I mat 1.5500 0.0600− 0.0000 0.0600− 0.0455 0.0540− 0.0000 0.0540− 1.5685 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kg m 2 ⋅= Guesses A x 1N= A z 1N= B x 1N= B z 1N= ω' y 1 rad s 2 = 756 Engineering Mechanics - Dynamics Chapter 21 A x 0 A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ B x 0 B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 0 ρ ab+ c+ d+ e+()g ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ − ρ− d d 2 ω 2 ρe e 2 ω' y + 0 ρ− d d 2 ω' y ρe e 2 ω 2 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = 0 ab+ c+ 2 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 0 0 ρ− ab+ c+()g ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ × d 2 a 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 0 0 ρ− dg ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×+ 0 ab+ e 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 0 0 ρ− eg ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × 0 ab+ c+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ B x 0 B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0 M 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ++ ... I mat 0 ω' y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 ω− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ I mat 0 ω− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= A x A z B x B z ω' y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A z , B x , B z , ω' y ,()= A x A z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3.39 0.66− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= B x B z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 7.2243 4.1977 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= ω' y 201.7 rad s 2 = Problem 21-51 The rod assembly has a weight density r. It is supported at B by a smooth journal bearing, which develops x and y force reactions, and at A by a smooth thrust bearing, which develops x, y, and z force reactions. If torque M is applied along rod AB, determine the components of reaction at the bearings when the assembly has angular velocity ω at the instant shown. 757 Given Engineering Mechanics - Dynamics Chapter 21 Given: γ 5 lb ft = a 4ft= d 2ft= M 50 lb ft⋅= b 2ft= g 32.2 ft s 2 = ω 10 rad s = c 2ft= Solution: ρ γ g = I yz ρcb c 2 ρdc b d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += I zz ρc c 2 3 ρdc 2 += I xx ρ ab+() ab+() 2 3 ρc c 2 12 + ρcb 2 c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + ρd d 2 12 + ρdc 2 b d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ += I yy ρ ab+() ab+() 2 3 ρcb 2 + ρd d 2 12 + ρdb d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += I mat I xx 0 0 0 I yy I yz − 0 I yz − I zz ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = I mat 16.98 0.00 0.00 0.00 15.32 2.48− 0.00 2.48− 1.66 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb ft⋅ s 2 ⋅= Guesses A x 1lb= A y 1lb= A z 1lb= B x 1lb= B y 1lb= α 1 rad s 2 = Given A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ B x B y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 0 ρ ab+ c+ d+()g ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ − ρ− c c 2 αρdcα− ρ− c c 2 ω 2 ρdcω 2 − 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = 758 dhanesh_h Mathcad - CombinedMathcads