ProblemsH. Okubo A a B r=0.25 m O 15 kg 30 Figure 1: Problem 1 1. For what acceleration a of the frame will the uniform slender rod maintain the orientation shown in the figure? Neglect the friction and mass√of the small rollers at A and B. (a = g 3) Solution. Applying the three equations of motion gives ma = RA (1) 0 = RB − mg (2) 0 = RB l cos 30◦ − RA l sin 30◦ 8 kg 10 kg Figure 2: Problem 2 vectors, i, j, k, are attached to point O, with the y-axis vertical and the zxplane horizontal. The moment of inertia of the drum is 1 I = Mr 2 (5) 2 where M is the drum mass. The equations of motion are (3) Substituting Eqs (1) and (2) into Eq. (3) gives √ 0 = mg( 3/2) − ma/2 √ 0 = g 3−a √ a = g 3 (4) r(T10 − T8 ) = Iα T10 − m10 g = m10 a10 T8 − m8 g = m8 a8 2. Calculate the downward acceleration a of the 10-kg cylinder. The drum is a uniform cylinder, and friction at the pivot is negligible. Compare your answer with that obtained by ignoring the inertia effects of the drum. (a = 0.769 m/s2 , a = 1.090 m/s2 ) Solution. The axes x-y-z with unit (6) (7) (8) where T10 is the tension of the cable attached the 10-kg cylinder, T8 is the tension of the cable for the 8-kg cylinder, m10 and m8 are the masses of 10kg and 8-kg cylinders. The accelerations of the cylinders are a10 and a8 , 1 861 b ) Solution. ˙ With M = 0. T10 . aB is −−→ aB = aG + ω˙ × GB P 3P b b = j+ × − i+ j m mb 2 2 P (−3i − j) (14) = 2m maG = P j P aG = j (10) m where m is the mass of the plate. (a = (P/2m)(−3i − j)) Solution. and aG is the acceleration vector of the center of mass. a = 1. a can be expressed as α = a/r.y O B b b b 2b A x Figure 4: Problem 4 where ω˙ is the angular acceleration of the plate. ω˙ = 3. Friction is negligible.09 m/s2 . (6)-(8). The moment of inertia 2 . (0. and I is the moment of inertia which is Z b/2 Z b/2 m I = dxdy(x2 + y 2 ) 2 b −b/2 −b/2 Z Z m b/2 b/2 2 x dxdy = 2 b −b/2 −b/2 Z Z m b/2 2 b/2 dxdy y + 2 b −b/2 −b/2 P Figure 3: Problem 3 and the angular acceleration of drum is α. The uniform square plate of mass m is lying motionless on the horizontal surface when the force P is applied at A as shown. and T8 from Eqs. Determine the resulting initial acceleration of point B. (11). Eliminating α. The uniform rectangular plate is released from rest in the position shown. Determine the maximum angular velocity ω during the ensuring motion. we have r 2 (−m10 a + m10 g − m8 a − m8 g) 1 = Mr 2 a 2 m10 − m8 g a= M/2 + m10 + m8 = 0. −→ I ω˙ = GA × P j (11) 4. Friction p g at the pivot is negligible. we calculate ω.769 m/s2 (9) = mb2 6 (12) Using the expressionn of I and Eq. We apply the equations of motion 3P k mb (13) The acceleration of B. The relationship between a10 and a8 is a = −a10 = a8 . cos θ (α = 3g 2b ) Solution. The 75-kg flywheel has a radius of gyration about its shaft axis of k = 0. determine its angular velocity ω at t = 3 s. (ω = 1. If the bars are released from rest in the position shown. Substituting into the energy equation gives 5 2 2 1 √ mb ω − ( 5 − 1)mgb 6√ 2 3( 5 − 1) g ω2 = 5r b g (18) ω = 0.50 m and is subjected to the torque M = 10(1 − e−t Nm. determine their initial angular acceleration α as they collapse in the vertical plane. where t is in seconds.861 b 0 = Substitution into the work-energy equation for virtual change gives b 1 0 = mb2 αdθ − mg cos θdθ 3 2 (22) from which the angular acceleration is 5.093 rad/s) Solution.about O is Z IO = = A 2b 0 Z b m 2 (x + y 2)dydx 2 2b 0 5 2 mb 3 b θ θ b O (15) Figure 5: Problem 5 The change in kinetic energy is 1 5 ∆T = IO ω 2 = mb2 ω 2 2 6 B The virtual change in kinetic energy is (16) b b dT = m( α)( dθ) + Iαdθ 2 2 m 2 m = b αdθ + b2 αdθ 4 12 1 2 mb αdθ (20) = 3 The change in gravitational potential energy is √ ! 5 b ∆Vg = mg − b − mg − 2 2 1 √ = − ( 5 − 1)mgb (17) 2 The virtual change in potential energies of the system becomes b b dVg = d(−mg sin θ + mg ) 2 2 b = d{mg (1 − sin θ)} 2 b = −mg cos θdθ (21) 2 when the center of gravity is located at the lowest position. The moment of inertia of the slender bar is Z b/2 m m 2 x dx = b2 (19) I= 12 −b/2 b α= 3g cos θ 2b (23) 6. If the flywheel is at rest at time t = 0. Application of the angular impulse-momentum equation over the 3 . The two uniform slender bars are hinged at O and supported on the horizontal surface by their end rollers of negligible mass. the angular velocity vector becomes (24) 10(3 + e−3 − 1) 75(0.8i − 1. The collar at O and attached shaft OC rotate about the fixed x0 -axis at the constant rate Ω = 4 rad/s.62 m/s (27) α = Ωi × pk = −Ωpj = −40j rad/s2 4 (30) .2j + 1. Find the velocity v A of point A on the rim if its position vector at this instant is r = 0.52 )ω = 10 t + e−t 0 8. The velocity v A becomes ω = Ωi + pk = 4i + 10k rad/s The magnitude of ω is p ω = Ω2 + p2 √ 42 + 102 = √ = 116 = 10. (ω = 10. α = −40 rad/s2 ) Solution.4 m B A A θ Ω z C y O Ω x0 x p x Figure 7: Problem 8 Figure 6: Problem 7 interval gives Z 3 2 Mdt = mk ω 0 3 75(0.093 rad/s (25) ω = 7. The rotor and shaft are mounted in a clevis which can rotate about the z-axis with an angular velocity Ω.5j + 2k (26) (28) (29) The angular acceleration vector is The rim speed vB equal the magnitude of velocity of A. Thus.25) = 1.8 rad/s vA = ω0 × r = (−4j − 3k) ×(0. Determine the magnitude of the total angular velocity ω of the disk and find its angular acceleration α.8)2 + (−1.8 rad/s.5)2 + 22 = 2. We use x-y-z components with unit vectors i. Simultaneously. With Ω = 0 and θ constant. The system has two components of angular velocity: Ω about x0 -axis and p about zaxis. the circular disk rotates about OC at the constant rate p = 10 rad/s.ω0 z v y 0.62 m/s) Solution. The x0 -axis coincides with x-axis.1k m. Thus p vB = (−0.2j + 1. j. What is the rim speed vB of any point B? (vB = 2.1k) = −0.5i + 1. the rotor has an angular velocity ω 0 = −4j − 3k rad/s. k.5i + 1. the angular velocity vector becomes ω = Ωi + pk = 4i + 10k rad/s b b O b z ω x b Figure 8: Problem 10 velocity ω as shown.3j +(4i + 10k) ×{(4i + 10k) × 0. k for x-y-z components. j.6j + 1.y 9. For the conditions of Prob. Thus. The x0 -axis coincides with xaxis. The three small spheres. H O = 3mb2 ω( 31 i + 32 j − 32 k)) Solution. j. The linear momentum is (31) The velocity of A is given by −→ v A = v C + ω × CA = 4i × 0. each of mass m. (v A = −3i − 1.4k + (4i + 10k) × 0.4k m/s2 ) Solution.3j} = −34. determine the velocity v A and acceleration αA of point A on the disk as it passes the position shown. 8. We use unit vectors i. are rigidly mounted to the horizontal shaft which rotates with the angular 5 . (G = mbω 2( √12 i − √12 j).3j = −3i − 1.4k) +(−40j) × 0.2k (32) −→ The angular acceleration vector of CA is α = Ωi × pk = −40j rad/s2 b G = Σmω × r = m{(−ωk) × (bj + 2bk) +(−ωk) × (bi + bk) +(−ωk) × (−bk)} = mbω(i − j) √ 1 1 = mbω 2( √ i − √ j) (35) 2 2 (33) The acceleration of A becomes −→ aA = Ωi × (Ωi × OC) −→ −→ +α × CA + ω × (ω × CA) = 4i × (4i × 0.8j − 6.6j + 1. Neglect the radius of each sphere compared with the other dimensions and write expressions for the magnitudes of their linear momentum G and their angular momentum H O about the origin √ O of the coordinates. We use x-y-z components with unit vectors i. aA = −34. k. The vector −→ CA has two components of angular velocity: Ω about x0 -axis and p about z-axis.8j − 6. Reference axes xy-z are attached to the collar at O and its shaft OC.2k m/s.4k m/s2 (34) The angular momentum is H O = Σr × m(ω × r) = m(bj + 2bk) × bωi +m(bi + bk) × (−bωj) = mb2 ω(i + 2j − 2k) 1 2 2 = 3mb2 ω( i + j − k)(36) 3 3 3 10. Determine the bending moment M acting on each rod at its base. locity vmax which result if the cylinThe general moment releation is exder is displaced 0. (M = 21 mblω 2 ) Solution. Hy . ωx = ωy = 0. The z-x plane includes the two rods. the systhe components of angular velocity vectem period τ . the moment is Figure 9: Problem 11 11. Each of the two rods of mass m is welded to the face of the disk which rotates about the vertical axis with a constant angular velocity ω. about the respective axes. and the maximum vetor ω. and ωz = ω. Hx . and ωx . and Ixy (= so that 1 Iyz ). General angular momentum H can be described as M = −Hy ωi + Hx ωj = Iyz ω 2i − Ixz ω 2 j (40) Iyz and Ixz are calculated as Iyz = 0 Ixz = Z xzdm l m bz dz l 0 Z l mb = zdz l 0 1 = mbl 2 = H = (Ixx ωx − Ixy ωy − Ixz ωz )i +(−Iyx ωx + Iyy ωy − Iyz ωz )j +(−Izx ωx − Izy ωy + Izz ωz )k = Hx i + Hy j + Hz k (37) Z (41) (42) The moment equation gives 1 M = − mblω 2 j 2 (43) where the quantities Ixx . The rotating reference axes x-y-z with unit vectors i. and Izz are the moment of inertia of the body The bending moment M B equals −M . For the spring-mass system shown. Iyy . ωy . and Hz are 12. Iyz (= Izy ). ωz are termine the static deflection δst .1 m downward from 6 . decomponents of H. j and k are attached to the disk. The x and y axes lie in the horizontal plane of the disk and the z axis is vertical.b b pressed by X ˙ M = H dH = + ω × H (38) dt xyz = (H˙ x − Hy ωz + Hz ωy )i l +(H˙ y − Hz ωx + Hx ωz )j +(H˙ z − Hx ωy + Hy ωx )k ω (39) When H˙ x = H˙ y = H˙ z = 0. and Izx (= Ixz ) are the M B = mblω 2 (44) products of inertia with respect to the 2 coordinates axes. 0 = Bωn cos 0 steady-state motion of the 10-kg mass if (a) c = 500 Ns/m and (b) c = Solving the two equations for A and B 0. ((a)X = 1.81) δst = = = 0. Determine the value of the damping ra6 3 ωn tio ζ for the simple spring-mass-dashpot = = Hz fn = 2π 2π π system shown.27(10−2) m) the velocity is Solution.1 m and B = 0.75) 1 π Solution. (δst = 0. We compute the damping τ= = s ratio ζ.6 m/s m 7 . 2.75 (47) 2mωn 2(2)(14) Evaluating the displacement and velocity at time t = 0 gives 14. Determine the amplitude X of the 0. vmax = 0.1 = A cos 0.1(6) = 0.1ωn sin ωn t m¨ x + cx˙ + kx = f cos ωt from which the maximum velocity is f x¨ + 2ζωn x˙ + ωn2 = cos ωt(48) vmax = 0.273 m F=1000 cos 120t N k 144 (45) Figure 12: Problem 14 r r k 144 = = 6 rad/s ωn = m 4 13. (ζ = 0.6 m/s) Solution. The equation of motion is y˙ = −0. k=100 kN/m c m=10 kg 0 = mg − kδst mg 4(9. τ 3 The motion of the mass is p p ωn = k/m = 392/2 = 14 rad/s y = A cos ωn t + B sin ωn t (46) y˙ = −Aωn sin ωn t + Bωn cos ωn t c 42 ζ= = = 0.x k=144 N/m k=392 N/m Equilibrium position 2 kg m=4 kg c=42 Ns/m y Figure 11: Problem 13 Figure 10: Problem 12 its equilibrium position and released. we see that at equilibrium. Therefore. τ = π3 s.344(10−2) m (b)X = yields A = 0. From the spring relationship.273 m. 344(10−2) m 7440. (τ = 2ζωnωf /m (55) X2 = 6π m/5k) (ωn2 − ω 2)2 + (2ζωnω)2 Solution. ζ= 3 m 2(10)(100) The period τ is p r (ωn2 − ω 2)2 + (2ζωnω)2 2π m p = 6π τ= = 44002 + (2(0. The light rod and attached sphere of Therefore.43 − ω )X1 + 2ζωnωX2 8 . (48) for the steadystate can be expressed as k x = X1 cos ωt + X2 sin ωt (49) x˙ = ω(−X1 sin ωt + X2 cos ωt) (50) 2 x¨ = −ω (X1 cos ωt + X2 sin ωt) (51) b 2 b Figure 13: Problem 15 Substituting Eqs (49)-(51) into Eq. Determine the period (ωn − ω )f /m X1 = (54) τ for small oscillations in the verti(ωn2 − ω 2)2 + (2ζωnω)2 calpplane about the pivot O.where ωn = r k c . The equation of motion is p The amplitude X = X12 + X22 is m(3b)2 θ¨ = −k(2b)2 θ − kb2 θ − mg(3b) 3mg f /m 2 = −5kb θ + (57) (56) X=p 5kb (ωn2 − ω 2 )2 + (2ζωnω)2 p 3mg k 5 θ+ (58) θ¨ = − ωn = 100(103)/10 = 100 9m 5kb (ωn2 − ω 2 )2 = (1002 − 1202)2 = 44002 The angular velocity ω is r (a) 1 5k 500 ω= = 0. mass m are at rest in the horizontal 2 2 position shown.ζ = m 2mωn k O m The solution of Eq.43 (b) f = m (52) = 0 (53) ζ=0 p (ωn2 − ω 2 )2 + (2ζωn ω)2 = 4400 1000/10 −2ζωn ωX1 + (ωn2 − ω 2 )X2 = 2.27(10−2 ) m X= 4400 15. (48) gives (ωn2 b X= 1000/10 = 1.25)(100)(120))2 ω 5k = 7440.25.