1 1. Show that the sum of the moments of inertia of a bod, I + I + I, is independent of the orientation of the,, aes and thus depends onl on the location of its origin. I + I + I = Lm ( + )dm + Lm ( + )dm + Lm ( + )dm = Lm ( + + )dm However, + + = r,where r is the distance from the origin O to dm. Since ƒ r ƒ is constant, it does not depend on the orientation of the,, ais. Consequentl, I + I + I is also indepenent of the orientation of the,, ais. Q.E.D. 1109
1. Determine the moment of inertia of the cone with respect to a vertical ais passing through the cone s center of mass. What is the moment of inertia about a parallel ais that passes through the diameter of the base of the cone? The cone has a mass m. a The mass of the differential element is dm = rdv = r(p ) d = rpa. h d h di = 1 4 dm + dm However, I = L di = rpa 4h 4 = 1 4 B rpa h dr a a h b + rpa h d = rpa 4h 4 (4h + a ) 4 d (4h + a ) L 0 h 4 d = rpa h 0 (4h + a ) m = Lm dm = rpa h L0 h d = rpa h 3 Hence, Using the parallel ais theorem: I = 3m 0 (4h + a ) I = I + md 3m 0 (4h + a ) = I + ma 3h 4 b I = 3m 80 (h + 4a ) I ' = I + md = 3m 80 (h + 4a ) + ma h 4 b = m 0 (h + 3a ) I = 3m 80 (h + 4a ) I = m 0 (h + 3a ) 1110
1 3. Determine moment of inertia I of the solid formed b revolving the shaded area around the ais. The densit of the material is r = 1 slug/ft 3. = ft The mass of the differential element is di = 1 dm + dm dm = rdv = rap B d = rpd. 4ft = 1 4 [rpd]() + (rpd) = rp( 1 4 + 3 ) d 4 I = di = rp ( 1 L L 4 + 3 ) d = 69.33 pr 0 = 69.33(p)(1) = 614 slug # ft I = 614 slug # ft 1111
*1 4. Determine the moments of inertia I and I of the paraboloid of revolution. The mass of the paraboloid is 0 slug. ft The mass of the differential element is dm = rdv = rap B d = rpd. m = 0 = dm = rpd Lm L 0 = 4rp r = 5 p slug/ft3 di = 1 4 dm + dma B = 1 4 [rpd]() + [rpd] = A5 + 10 3 B d 0 ft I = di = A5 + 10 3 B d = 53.3 slug # ft L L 0 di = 1 dm = rp d = 10 d I = di = 10 d = 6.7 slug # ft L L 0 I = 53.3 slug # ft I = 6.7 slug # ft 111
1 5. Determine b direct integration the product of inertia I for the homogeneous prism. The densit of the material is r. Epress the result in terms of the total mass m of the prism. a a h The mass of the differential element is dm = rdv = rhd = rh(a - )d. Using the parallel ais theorem: m = Lm dm = rh L a 0 (a - )d = ra h di = (di ) G + dm G G = 0 + (rhd) () a h b = rh d = rh (a - ) d a I = rh (a - ) d = ra3 h L0 1 = 1 6 a ra h b(ah) = m 6 ah I = m 6 ah 1113
1 6. Determine b direct integration the product of inertia I for the homogeneous prism. The densit of the material is r. Epress the result in terms of the total mass m of the prism. a a h The mass of the differential element is dm = rdv = rhd = rh(a - )d. Using the parallel ais theorem: m = Lm dm = rh L a 0 (a - )d = ra h di = (di ) G + dm G G = 0 + (rhd)a b() = rh d = rh (3 - a + a ) d I = rh = ra 4 h 4 a ( 3 - a + a ) d L0 = 1 1 a ra h ba = m 1 a I = m 1 a 1114
1 7. Determine the product of inertia I of the object formed b revolving the shaded area about the line = 5 ft. Epress the result in terms of the densit of the material, r. 3ft ft 3 3 3 3 dm = rp (5 - ) d = r p (5 - )3 d = 38.4rp L0 L 0 L 0 3 ~ 3 dm = rp (5 - ) d L0 L 0 3 = rp (5 - )(3) d L Thus, The solid is smmetric about, thus I = 0 = 40.5rp = 40.5rp = 1.055 ft 38.4rp I = I + m 0 = 0 + 5(1.055)(38.4rp) I = 636r I = 636r 1115
*1 8. Determine the moment of inertia I of the object formed b revolving the shaded area about the line = 5ft. Epress the result in terms of the densit of the material, r. 3ft ft 3 3 1 I = L dm r - 1 (m )() 3 0 1 L0 dm r = 1 rp(5 - ) 4 d L0 = 1 rp a5-4 L0 3 b d = 490.9 rp m =rp() (3) = 1 rp 3 3 I = 490.9 rp - 1 (1 r p)() = 466.9 r p Mass of bod; 3 m = rp(5 - ) d - m L 0 3 = rp(5 - L 3 ) d - 1 rp 0 = 38.4 rp I = 466.9 rp + (38.4 rp)(5) = 146.9 rp I = 4.48(10 3 ) r Also, 3 I = r dm L 3 = (5 - ) r (p)(5 - ) d L 3 = rp (5 - ) 3 (3) 1/ d L = 466.9 rp 3 m = dm L 0 0 0 3 = rp (5 - ) d L 0 3 = rp (5 - )(3) 1/ d L 0 0 = 38.4 rp I = 466.9 rp + 38.4 rp(5) = 4.48(10 3 )r I = 4.48(10 3 ) r I = 4.48(10 3 ) r 1116
1 9. Determine the moment of inertia of the cone about the ais. The weight of the cone is 15 lb, the height is h = 1.5 ft, and the radius is r = 0.5 ft. h r u = tan -1 ( 0.5 1.5 ) = 18.43 I = I = [ 3 80 m{4(0.5) + (1.5) }] + m[1.5 - ( 1.5 4 )] I = I = 1.3875 m I = 3 10 m(0.5) = 0.075 m I = I = I = 0 Using Eq. 1 5. I = u I + u I + u I = 0 + [cos(108.43 )] (1.3875m) + [cos(18.43 )] (0.075m) = 0.06m I = 0.06( 15 3. ) = 0.0961 slug # ft I = 0.0961 slug # ft 1117
1 10. Determine the radii of gration k and k for the solid formed b revolving the shaded area about the ais.the densit of the material is r. 0.5 ft For :The mass of the differential element is dm = rdv = r(p ) d = rp d. k 4ft 1 0.5 ft di = 1 dm = 1 C rpd DA 1 B = 1 rpd 4 4ft 4 d I = di = 1 rp + 1 L L 4 0.5 Cr(p)(4) (0.5)D(4) = 134.03r However, Hence, m = Lm dm = rp L 4 0.5 d I k = A m = 134.03r A 4.35r + rcp(4) (0.5)D = 4.35r =.35 ft For : 0.5 ft 6 4ft k di = 1 4 dm + dm = 1 4 crpd I = L di = rp L 4 = rpa 1 4 4 + 1B d 0.5 I = 1 4 Crp(4) (0.5)D(4) + Crp(4) (0.5)D(0.15) = 50.46r da I b + arpd b A 1 4 4 + 1B d = 8.53r I = I + I = 8.53r + 50.46r = 78.99r Hence, k = I m = 78.99r = 1.80 ft 4.35r k =.35 ft k = 1.80 ft 1118
1 11. Determine the moment of inertia of the clinder with respect to the a a ais of the clinder. The clinder has a mass m. a a The mass of the differential element is dm = rdv = r Apa B d. di aa = 1 4 dma + dma B = 1 4 CrApa B dda + CrApa B dd = A 1 4 rpa4 + rpa B d h I aa = di aa = A 1 4 rpa4 + rpa B d L L 0 a h However, Hence, = rpa h 1 (3a + 4h ) h m = dm = rapa B d = rpa h Lm L I aa = m 1 A3a + 4h B 0 I aa = m 1 (3a + 4h ) 1119
*1 1. Determine the moment of inertia I of the composite plate assembl.the plates have a specific weight of 6lb>ft. 0.5 ft 0.5 ft Horiontial plate: 0.5 ft 0.5 ft 0.5 ft I = 1 1 (6(1)(1) 3. )(1) = 0.0155 Vertical plates: l = 0.707, l = 0.707, l = 0 I = 1 3 (6(1 4 )(1) )( 1 3. 4 ) = 0.00137 I = ( 6(1 4 )(1) 3. )( 1 1 )[(1 4 ) + (1) ] + ( 6(1 4 )(1) )( 1 3. 8 ) = 0.0135 Using Eq. 1 5, I = (0.707) (0.00137) + (0.707) (0.0135) = 0.00686 Thus, I = 0.0155 + (0.00686) = 0.09 slug # ft I = 0.09 slug # ft 110
1 13. Determine the product of inertia I of the composite plate assembl. The plates have a weight of 6 lb>ft. 0.5 ft 0.5 ft Due to smmetr, 0.5 ft 0.5 ft 0.5 ft I = 0 I = 0 111
1 14. Determine the products of inertia I, I, and I, of the thin plate. The material has a densit per unit area of 50 kg>m. 400 mm 00 mm The masses of segments 1 and shown in Fig. a are m 1 = 50(0.4)(0.4) = 8 kg and m = 50(0.4)(0.) = 4 kg. Due to smmetr I = I = I = 0 for segment 1 and I = I = I = 0 for segment. 400 mm I = I + m G G = C0 + 8(0.)(0.)D + C0 + 4(0)(0.)D = 0.3 kg # m I = I + m G G = C0 + 8(0.)(0)D + C0 + 4(0.)(0.1)D = 0.08 kg # m I = I + m G G = C0 + 8(0.)(0)D + C0 + 4(0)(0.1)D = 0 I = 0.3 kg # m I = 0.08 kg # m I = 0 11
1 15. Determine the moment of inertia of both the 1.5-kg rod and 4-kg disk about the ais. 300 mm 100 mm ' Due to smmetr I = I + I = 0 I = I = B 1 4 (4)(0.1) + 4(0.3) R + 1 3 (1.5)(0.3) = 0.415 kg # m I = 1 (4)(0.1) = 0.0 kg # m u = cos (18.43 ) = 0.9487, u = cos 90 = 0, u = cos (90 + 18.43 ) = -0.316 I = I u + I u + I u - I u u - I u u - I u u = 0.415(-0.316) + 0 + 0.0(0.9487) - 0-0 - 0 = 0.0595 kg # m I = 0.0595 kg # m 113
*1 16. The bent rod has a mass of 3 kg>m. Determine the moment of inertia of the rod about the O a ais. a 0.3 m 1 m O Solution The bent rod is subdivided into three segments and the location of center of mass for each segment is indicated in Fig. a. The mass of each segments is m 1 = 3(1) = 3 kg, m = 3(0.5) = 1.5 kg and m 3 = 3(0.3) = 0.9 kg. 0.5 m I = c 1 1 (3)(1 ) + 3(0.5 ) d + 30 + 1.5(1 ) 4 + c 1 1 (0.9)(0.3 ) + 0.9(0.15 + 1 ) d = 3.47 kg # m I = 0 + c 1 1 (1.5)(0.5 ) + 1.5(0.5 ) d + c 1 1 (0.9)(0.3 ) + 0.9(0.15 + 0.5 ) d = 0.377 kg # m I = c 1 1 (3)(1 ) + 3(0.5 ) d + c 1 1 (1.5)(0.5 ) + 1.5(1 + 0.5 ) d + 30 + 0.9(1 + 0.5 ) 4 = 3.75 kg # m I = [0 + 0] + [0 + 1.5(0.5)(-1)] + [0 + 0.9(0.5)(-1)] = -0.85 kg # m I = [0 + 0] + [0 + 0] + [0 + 0.9(-1)(0.15)] = -0.135 kg # m I = [0 + 0] + [0 + 0] + [0 + 0.9(0.15)(0.5)] = 0.0675 kg # m The unit vector that defines the direction of the O a ais is U Oa = 0.5i - 1j + 0.3k 0.5 + (-1) + 0.3 = 0.5 1.34 i - 1 1.34 j + 0.3 1.34 k Thus, u = 0.5 u = - 1 u = 0.3 1.34 1.34 1.34 114
*1 16. Continued Then I Oa = I u + I u + I u - I u u - I u u - I u u = 3.47a 0.5 1.34 b 1 + 0.377a - 1.34 b + 3.75a 0.3 1.34 b - (-0.85)a 0.5 1.34 ba - 1 1.34 b 1 0.3 0.3 0.5 -(-0.135)a - ba b - (0.0675)a ba 1.34 1.34 1.34 1.34 b = 0.4813 kg # m = 0.481 kg # m I Oa = 0.481 kg # m 115
1 17. The bent rod has a weight of 1.5 lb>ft. Locate the center of gravit G(, ) and determine the principal moments of inertia I, I, and I of the rod with respect to the,, aes. 1ft Due to smmetr = 0.5 ft 1ft A _ G _ = W w = (-1)(1.5)(1) + C(-0.5)(1.5)(1)D 3C1.5(1) D = -0.667 ft I = ca 1.5 3. b(0.5) d + 1 1 a 1.5 3. b(1) = 0.07 slug # ft I = c 1 1 a 1.5 3. b(1) + a 1.5 3. b(0.667-0.5) d + a 1.5 b(1-0.667) 3. = 0.0155 slug # ft I = c 1 1 a 1.5 3. b(1) + a 1.5 3. b(0.5 + 0.1667 ) d + 1 1 a 1.5 3. b(1) + a 1.5 3. b(0.3333) = 0.047 slug # ft = 0.5 ft = -0.667 ft I = 0.07 slug # ft I = 0.0155 slug # ft I = 0.047 slug # ft 116
1 18. Determine the moment of inertia of the rod-and-disk assembl about the ais.the disks each have a weight of 1 lb. The two rods each have a weight of 4 lb, and their ends etend to the rims of the disks. 1ft ft 1ft For a rod: u = tan -1 a 1 1 b = 45 u = cos 90 = 0, u = cos 45 = 0.7071, u = cos (90 + 45 ) = -0.7071 I = I = a 1 1 ba 4 3. b C() + () D = 0.088 slug # ft I = 0 I = I = I = 0 I = 0 + 0 + (0.088)(-0.7071) = 0.04141 slug # ft For a disk: I = a 1 1 ba 3. b(1) = 0.1863 slug # ft Thus. I = (0.04141) + (0.1863) = 0.455 slug # ft I = 0.455 slug # ft 117
1 19. Determine the moment of inertia of the composite bod about the aa ais. The clinder weighs 0 lb, and each hemisphere weighs 10 lb. a ft u a = 0.707 u a = 0 u a = 0.707 I = 1 ( 0 3. )(1) + [ 5 ( 10 3. )(1) ] = 0.5590 slug # ft I = I = 1 1 ( 0 3. )[3(1) + () ] + [0.59( 10 3. )(1) + 10 3. (11 8 ) ] I = I = 1.6975 slug # ft I aa = 0 + (0.707) (1.6975) + (0.707) (0.559) a ft I aa = 1.13 slug # ft I aa = 1.13 slug # ft 118
*1 0. Determine the moment of inertia of the disk about the ais of shaft AB. The disk has a mass of 15 kg. 30 B A 150 mm Due to smmetr I = I = I = 0 I = I = 1 4 (15)(0.15) = 0.084375 kg # m I = 1 (15)(0.15) = 0.16875 kg # m u = cos 90 = 0, u = cos 30 = 0.8660 u = cos (30 + 90 ) = -0.5 I = I u + I u + I u - I u u - I u u - I u u = 0 + 0.16875(0.8660) + 0.084375(-0.5) - 0-0 - 0 = 0.148 kg # m I = 0.148 kg # m 119
1 1. The thin plate has a weight of 5 lb and each of the four rods weighs 3 lb. Determine the moment of inertia of the assembl about the ais. 1.5 ft For the rod: 0.5 ft 0.5 ft I = 1 1 a 3 3. ba(0.5 + (0.5) b = 0.00388 slug # ft 0.5 ft 0.5 ft For the composite assembl of rods and disks: I = 4B0.00388 + a 3 3. b 0.5 + 0.5 R + 1 1 a 5 3. b(1 + 1 ) = 0.0880 slug # ft I = 0.0880 slug # ft 1130
1. If a bod contains no planes of smmetr, the principal moments of inertia can be determined mathematicall. To show how this is done, consider the rigid bod which is spinning with an angular velocit V, directed along one of its principal aes of inertia. If the principal moment of inertia about this ais is I, the angular momentum can be epressed as H = IV = Iv i + Iv j + Iv k.the components of H ma also be epressed b Eqs. 1 10, where the inertia tensor is assumed to be known. Equate the i, j, and k components of both epressions for H and consider v, v, and v to be unknown. The solution of these three equations is obtained provided the determinant of the coefficients is ero. Show that this determinant, when epanded, ields the cubic equation O V I 3 - (I + I + I )I + (I I + I I + I I - I - I - I )I - (I I I - I I I - I I - I I - I I ) = 0 The three positive roots of I, obtained from the solution of this equation, represent the principal moments of inertia,,and. I I I H = Iv = Iv i + Iv j + Iv k Equating the i, j, k components to the scalar equations (Eq. 1 10) ields (I - I) v - I v - I v = 0 -I v + (I - I) v - I v = 0 -I v - I v + (I - I) v = 0 Solution for,,and requires v v v (I - I) -I -I 3 -I (I - I) -I 3 = 0 -I -I (I - I) Epanding I 3 - (I + I + I )I + AI I + I I + I I - I - I - I BI - AI I I - I I I - I I - I I - I I B = 0 Q.E.D. I 3 - (I + I + I )I + (I I + I I + I I - I - I - I )I - (I I I - I I I - I I - I I - I I ) = 0 Q.E.D. 1131
1 3. Show that if the angular momentum of a bod is determined with respect to an arbitrar point A, then H A can be epressed b Eq. 1 9. This requires substituting R A = R G + R G>A into Eq. 1 6 and epanding, noting that 1 R G dm = 0 b definition of the mass center and v G = v A + V : R G>A. Z A R G/A G R G R A P Y H A = a Lm r A dmb * v A + Lm r A * (v * r A )dm X = a Lm (r G + r G>A ) dmb * v A + Lm (r G + r G>A ) * Cv * r G + r G>A )Ddm = a Lm r G dmb * v A + (r G>A * v A ) Lm dm + Lm r G * (v * r G ) dm + a Lm r G dmb * (v * r G>A ) + r G>A * av * Lm r G dmb + r G>A * (v * r G>A ) Lm dm Since r G dm = 0 and from Eq. 1 8 H G = r G * (v * r G )dm Lm Lm H A = (r G>A * v A )m + H G + r G>A * (v * r G>A )m = r G>A * (v A + (v * r G>A ))m + H G = (r G>A * mv G ) + H G Q.E.D. H A = (r G>A * mv G ) + H G 113
*1 4. The 15-kg circular disk spins about its ale with a constant angular velocit of v 1 = 10 rad>s. Simultaneousl, the oke is rotating with a constant angular velocit of v = 5 rad>s. Determine the angular momentum of the disk about its center of mass O, and its kinetic energ. O 150 mm The mass moments of inertia of the disk about the,, and aes are v 1 10 rad/s I = I = 1 4 mr = 1 4 (15)(0.15 ) = 0.084375 kg # m I = 1 mr = 1 (15)(0.15 ) = 0.16875 kg # m Due to smmetr, I = I = I = 0 Here, the angular velocit of the disk can be determined from the vector addition of and.thus, v 1 v v = v 1 + v = [-10j + 5k] rad>s so that v 5 rad/s v = 0 v = -10 rad>s v = 5 rad>s Since the disk rotates about a fied point O, we can appl H = I v = 0.084375(0) = 0 H = I v = 0.16875(-10) = -1.6875 kg # m >s H = I v = 0.084375(5) = 0.41875 kg # m >s Thus, H O = [-1.69j + 0.4k] kg # m >s The kinetic energ of the disk can be determined from T = 1 I v + 1 I v + 1 I v = 1 (0.084375)(0 ) + 1 (0.16875)(-10) + 1 (0.084375)(5 ) = 9.49 J H O = [ - 1.69j + 0.4k] kg # m >s T = 9.49 J 1133
1 5. The large gear has a mass of 5 kg and a radius of gration of k = 75 mm. Gears B and C each have a mass of 00 g and a radius of gration about the ais of their connecting shaft of 15 mm. If the gears are in mesh and C has an angular velocit of ωc = {15j} rad>s, determine the total angular momentum for the sstem of three gears about point A. B 100 mm 40 mm 40 mm A 45 vc ={15j} rad/s C I A = 5(0.075) = 8.15A10-3 B kg # m I B = I C = 0.(0.015) = 45A10-6 B kg # m Kinematics: v C = v B = 15 rad>s v = (0.04)(15) = 0.6 m>s v A = a 0.6 0.1 b = 6 rad>s H B = I B v B = A45A10-6 BB(15) = 675A10-6 B H B = -675A10-6 B sin 45 i - 675A10-6 B cos 45 j H B = -477.3A10-6 Bi - 477.3A10-6 B j H C = I C v C = A45A10-6 BB(15) = 675A10-6 B H C = 675A10-6 B j H A = I A v A = 8.15A10-3 B(6) = 0.16875 H A = 0.16875k The total angular momentum is therefore, H = H B + H C + H A = -477 10-6 i + 198 10-6 j + 0.169k kg # m s H = {-477(10-6 )i + 198(10-6 )j + 0.169k} kg # m >s 1134
1 6. The circular disk has a weight of 15 lb and is mounted on the shaft AB at an angle of 45 with the horiontal. Determine the angular velocit of the shaft when t = 3sif a constant torque M = lb # ft is applied to the shaft. The shaft is originall spinning at v 1 = 8 rad>s when the torque is applied. A M 0.8 ft 45 v 1 8 rad/s B Due to smmetr I = I = I = 0 I = I = 1 A 15 3. B (0.8) = 0.07453 slug # ft I = 1 A 15 3. B (0.8) = 0.1491 slug # ft For ' ais u = cos 45 = 0.7071 u = cos 45 = 0.7071 u = cos 90 = 0 I = I u + I u + I u - I u u - I u u - I u u = 0.1491(0.7071) + 0.07453(0.7071) + 0-0 - 0-0 = 0.1118 slug # ft Principle of impulse and momentum: (H ) 1 + L M dt = (H ) 0.1118(8) + (3) = 0.1118 v v = 61.7 rad/s v = 61.7 rad>s 1135
1 7. The circular disk has a weight of 15 lb and is mounted on 0.8 ft the shaft AB at an angle of 45 with the horiontal. 45 Determine the angular velocit of the shaft when t = sif v 1 8 rad/s a torque M = 14e 0.1t lb # ft, where t is in seconds, is applied A B to the shaft. The shaft is originall spinning at v 1 = 8 rad>s when the torque is applied. M Due to smmetr I = I = I = 0 I = I = 1 4 A 15 3. B (0.8) = 0.07453 slug # ft I = 1 A 15 3. B (0.8) = 0.1491 slug # ft For ' ais u = cos 45 = 0.7071 u = cos 45 = 0.7071 u = cos 90 = 0 I = I u + I u + I u - I u u - I u u - I u u = 0.1491(0.7071) + 0.07453(0.7071) + 0-0 - 0-0 = 0.1118 slug # ft Principle of impulse and momentum: (H ) 1 + L M dt = (H ) 0.1118(8) + 4e 0.1 t dt = 0.1118v L 0 v = 87. rad/s v = 87. rad>s 1136
*1 8. The rod assembl is supported at G b a ball-and-socket joint. Each segment has a mass of 0.5 kg/m. If the assembl is originall at rest and an impulse of I = 5-8k6 N # s is applied at D, determine the angular velocit of the assembl just after the impact. 0.5 m A 1m B 1m G C D I = { 8k} N s 0.5 m Moments and products of inertia: I = 1 1 C(0.5)D() + C0.5(0.5)D(1) = 0.8333 kg # m I = 1 1 C1(0.5)D(1) = 0.04166 kg # m I = 1 1 C(0.5)D() + c 1 1 C0.5(0.5)D(0.5) + C0.5(0.5)D(1 + 0.5 ) d = 0.875 kg # m I = C0.5(0.5)D(-0.5)(1) + C0.5(0.5)D(0.5)(-1) = -0.15 kg # m I = I = 0 From Eq. 1 10 H = 0.8333v + 0.15v H = 0.15v + 0.04166v H = 0.875v (H G ) 1 + L t t 1 M G dt = (H G ) 0 + (-0.5i + 1j) * (-8k) = (0.8333v + 0.15v )i + (0.15v + 0.04166v )j + 0.875v k Equating i, j and k components -8 = 0.8333v + 0.15v -4 = 0.15v + 0.04166v 0 = 0.875v (1) () (3) Solving Eqs. (1) to (3) ields: v = 8.73 rad s v -1 - rad s v = 0 Then v = {8.73i - 1j} rad s v = {8.73i - 1j} rad>s 1137
1 9. The 4-lb rod AB is attached to the 1-lb collar at A and a -lb link BC using ball-and-socket joints. If the rod is released from rest in the position shown, determine the angular velocit of the link after it has rotated 180. A 1.3 m 1. m B C 0.5 m T 1 + V 1 = T + V 0 + 4(0.5) + (0.5) = T - 4(0.5) - (0.5) T = 3 v AB = 0.5v 1.3 = 0.3846v T = 1 [1 3 ( 3. )(0.5) ]v + 1 [ 1 1 ( 4 3. )(1.3) ](0.3846v ) + 1 ( 4 3. )(0.5v ) T = 0.007764v = 3 v = 19.7 rad/s v = 19.7 rad>s 1138
1 30. The rod weighs 3lb>ft and is suspended from parallel cords at A and B. If the rod has an angular velocit of rad>s about the ais at the instant shown, determine how high the center of the rod rises at the instant the rod momentaril stops swinging. 3 ft 3 ft A v rad/s T 1 + V 1 = T + V 1 c 1 W 1 g l dv + 0 = 0 + Wh h = 1 l v 4 g = 1 (6) () 4 (3.) h = 0.1863 ft =.4 in. h =.4 in. 1139
1 31. The 4-lb rod AB is attached to the rod BC and collar A using ball-and-socket joints. If BC has a constant angular velocit of rad>s, determine the kinetic energ of AB when it is in the position shown. Assume the angular velocit of AB is directed perpendicular to the ais of AB. 3ft 1ft B rad/s 1ft A C v A = A i v B + {-j} ft>s v AB = v i + v j + v k r B/A = {-3i + 1j + 1k} ft r G/B = {1.5i - 0.5j - 0.5k} Equating i, j and k components v B = v A + v AB * r B/A i j k -j = A i + 3 v v v 3-3 1 1 v - v + A = 0 v + 3v = v + 3v = 0 (1) () (3) Since v AB is perpendicular to the ais of the rod, v AB r B/A = (v i + v j + v k) (-3i + 1j + 1k) = 0-3v + 1v + 1v = 0 (4) Solving Eqs. (1) to (4) ields: v = 0.1818 rad>s v = -0.06061 rad>s v = 0.6061 rad>s A = 0.6667 ft>s Hence v AB = {0.1818i - 0.06061j + 0.6061k} rad>s v A = {0.6667i} ft>s v G = v B + v AB * r G/B i j k = -j + 3 0.1818-0.06061 0.6061 3 1.5-0.5-0.5 = {0.3333i - 1.0j} ft>s v AB = 0.1818 + (-0.06061) + 0.6061 = 0.4040 G = (0.3333) + (-1.0) = 1.111 T = 1 m G + 1 I G v AB = 1 a 4 3. b(1.111) + 1 c 1 1 a 4 3. b A 3 + 1 + 1 B d(0.4040) = 0.090 ft # lb T = 0.090 ft # lb 1140
*1 3. The -kg thin disk is connected to the slender rod which is fied to the ball-and-socket joint at A. If it is released from rest in the position shown, determine the spin of the disk about the rod when the disk reaches its lowest position. Neglect the mass of the rod.the disk rolls without slipping. A 0.5 m 30 B 0.1 m C I = I = 1 4 ()(0.1) + (0.5) - = 0.505 kg # m I = 1 ()(0.1) = 0.01 kg # m v = v + v = -v v j + v # sin 11.31 j + v # cos 11.31 k = (0.1961v - v )j + (0.98058v # )k Since v A = v C = 0, then v C = v A + v * r C>A 0 = 0 + C(0.1961v # - v )j + (0.98058v # )kd * (0.5j - 0.1k) 0 = -0.01961v + 0.1v - 0.4909v # v = 0.1961v Thus, v = -0.96154v j, 0.1931v k h 1 = 0.5 sin 41.31 = 0.3301 m, h = 0.5 sin 18.69 = 0.160 m T 1 + V 1 = T + V 0 + (9.81)(0.3301) = c0 + 1 (0.01)(-0.96154v ) + 1 (0.505)(0.1931v ) d - (9.81)(0.160 v = 6. rad s v = 6. rad>s 1141
1 33. The 0-kg sphere rotates about the ale with a constant angular velocit of v s = 60 rad>s. If shaft AB is subjected to a torque of M = 50 N # m, causing it to rotate, determine the value of v p after the shaft has turned 90 from the position shown. Initiall, v p = 0.Neglect the mass of arm CDE. D 0.4 m v s 60 rad/s 0.3 m E C B 0.1 m The mass moments of inertia of the sphere about the,,and aes are v p I = I = I = 5 mr = 5 (0)(0.1 ) = 0.08 kg # m A M 50 N m When the sphere is at position 1,Fig. a, v p = 0.Thus, the velocit of its mass center is ero and its angular velocit is v 1 = [60k] rad>s.thus, its kinetic energ at this position is T = 1 m(v G) 1 + 1 I (v 1 ) + 1 I (v 1 ) + 1 I (v 1 ) = 0 + 0 + 0 + 1 (0.08) (60 ) = 144 J When the sphere is at position, Fig. a, v p = v p i. Then the velocit of its mass center is (v G ) = v p * v G>C = (v p i) * (-0.3j + 0.4k) = -0.4v p j - 0.3v p k.then (v G ) = (-0.4v p ) + (-0.3v p ) = 0.5v p. Also, its angular velocit at this position is v = v p i - 60j.Thus,its kinetic energ at this position is T = 1 m(v G) + 1 I (v ) + 1 I (v ) + 1 I (v ) = 1 (0)A0.5v p B + 1 (0.08)Av p B + 1 (0.08)(-60) =.54v p + 144 When the sphere moves from position 1 to position, its center of gravit raises verticall = 0.1 m.thus,its weight W does negative work. U W = -W = -0(9.81)(0.1) = -19.6 J Here, the couple moment M does positive work. U W = Mu = 50 a p b = 5pJ Appling the principle of work and energ, T 1 + U 1 - = T 144 + 5p + (-19.6) =.54v p + 144 v p = 4.8 rad>s v p = 4.8 rad>s 114
1 34. The 00-kg satellite has its center of mass at point G. Its radii of gration about the,, aes are k = 300 mm, k = k = 500 mm, respectivel. At the instant shown, the satellite rotates about the,, and aes with the angular velocit shown, and its center of mass G has a velocit of v G = 5 50i + 00j + 10k6 m>s. Determine the angular momentum of the satellite about point A at this instant. The mass moments of inertia of the satellite about the,,and aes are Due to smmetr, the products of inertia of the satellite with respect to the,, and coordinate sstem are equal to ero. The angular velocit of the satellite is Thus, Then, the components of the angular momentum of the satellite about its mass center G are Thus, I = I = 00A0.5 B = 50 kg # m I = 00A0.3 B = 18 kg # m I = I = I = 0 v = [600i + 300j + 150k] rad>s v = 600 rad>s v = -300 rad>s v = 150 rad>s (H G ) = I v = 50(600) = 30 000 kg # m >s (H G ) = I v = 50(-300) =-15 000 kg # m >s (H G ) = I v = 18(150) = 500 kg # m >s H G = [30 000i - 15 000j + 500k] kg # m >s The angular momentum of the satellite about point A can be determined from V 600 rad/s 800 mm, V G A 150 rad/s v G V 300 rad/s H A = r G>A * mv G + H G = (0.8k) * 00(-50i + 00j + 10k) + (30 000i - 15 000j + 500k) = [-000i - 55 000j + 500k] kg # m /s H A = {-000i - 55 000j + 500k6kg # m >s 1143
1 35. The 00-kg satellite has its center of mass at point G. Its radii of gration about the,, aes are k = 300 mm, k = k = 500 mm, respectivel. At the instant shown, the satellite rotates about the,,and aes with the angular velocit shown, and its center of mass G has a velocit of v G = 5 50i + 00j + 10k6 m>s. Determine the kinetic energ of the satellite at this instant., V 150 rad/s The mass moments of inertia of the satellite about the,,and aes are I = I = 00A0.5 B = 50 kg # m I = 00A0.3 B = 18 kg # m V 600 rad/s 800 mm G A v G V 300 rad/s Due to smmetr, the products of inertia of the satellite with respect to the,, and coordinate sstem are equal to ero. I = I = I = 0 The angular velocit of the satellite is v = [600i - 300j + 150k] rad>s Thus, v = 600 rad>s v = -300 rad>s v = 150 rad>s Since v G = (-50) + 00 + 10 = 116 900 m >s, the kinetic energ of the satellite can be determined from T = 1 mv G + 1 I v + 1 I v + 1 I v = 1 (00)(116 900) + 1 (50)A600 B + 1 (50)(-300) + 1 (18)A150 B = 37.005A10 6 BJ = 37.0 MJ T = 37.0 MJ 1144
*1 36. The 15-kg rectangular plate is free to rotate about the ais because of the bearing supports at A and B.When the plate is balanced in the vertical plane, a 3-g bullet is fired into it, perpendicular to its surface, with a velocit v = 5-000i6 m>s. Compute the angular velocit of the plate at the instant it has rotated 180. If the bullet strikes corner D with the same velocit v, instead of at C, does the angular velocit remain the same? Wh or wh not? A D 150 mm C 150 mm 150 mm Consider the projectile and plate as an entire sstem. v B Angular momentum is conserved about the AB ais. (H AB ) 1 = -(0.003)(000)(0.15)j = {-0.9j} (H AB ) 1 = (H AB ) -0.9j = I v i + I v j + I v k Equating components, v = 0 v = 0 v = -0.9 c 1 1 (15)(0.15) + 15(0.075) d = -8 rad/s T 1 + V 1 = T + V 1 c 1 1 (15)(0.15) + 15(0.075) d(8) + 15(9.81)(0.15) = 1 c 1 1 (15)(0.15) + 15(0.075) dv AB v AB = 1.4 rad/s If the projectile strikes the plate at D, the angular velocit is the same, onl the impulsive reactions at the bearing supports A and B will be different. v AB = 1.4 rad>s 1145
1 37. The 5-kg thin plate is suspended at O using a ball-andsocket joint. It is rotating with a constant angular velocit V = 5k6 rad>s when the corner A strikes the hook at S, which provides a permanent connection. Determine the angular velocit of the plate immediatel after impact. V {k} rad/s 300 mm 300 mm O Solution 400 mm Angular momentum is conserved about the OA ais. (H O ) 1 = I v k = c 1 1 (5)(0.6) d ()k = 0.30k A S u OA = 50.6j - 0.8k6 (H OA ) 1 = (H O ) 1 # uoa = (0.30)(-0.8) = -0.4 v = 0.6vj - 0.8vk (1) (H O ) = I v j + I v k = 1 3 (5)(0.4) v j + 1 1 (5)(0.6) v k = 0.667v j + 0.150v k From Eq. (1), v = 0.6v v = -0.8v (H O ) = 0.16vj - 0.10vk (H OA ) = (H O ) - u OA = 0.16v(0.6) + (0.1v)(0.8) = 0.19v Thus, (H OA ) 1 = (H OA ) -0.4 = 0.19v v = -1.5 rad>s v = -1.5u OA v = 5-0.750j + 1.00k6rad>s V = 5-0.750j + 1.00k6 rad>s 1146
1 38. Determine the kinetic energ of the 7-kg disk and 1.5-kg rod when the assembl is rotating about the ais at v = 5 rad>s. B v 5 rad/s Solution Due to smmetr I = I = I = 0 I = I = c 1 4 (7)(0.1) + 7(0.) d + 1 3 (1.5)(0.) = 0.3175 kg # m C A D 100 mm 00 mm I = 1 (7)(0.1) = 0.035 kg # m For ais u = cos 6.56 = 0.8944 u = cos 116.57 = -0.447 u = cos 90 = 0 I = I u + I u + I u - I u u - I u u - I u u = 0 + 0.3175(-0.447) + 0.035(0.8944) - 0-0 - 0 = 0.0915 kg # m T = 1 I v + 1 I v + 1 I v = 0 + 0 + 1 (0.0915)(5) = 1.14 J T = 1.14 J 1147
1 39. Determine the angular momentum H of the 7-kg disk and 1.5-kg rod when the assembl is rotating about the ais at v = 5 rad>s. B v 5 rad/s Solution Due to smmetr I = I = I = 0 I = I = c 1 4 (7)(0.1) + 7(0.) d + 1 3 (1.5)(0.) = 0.3175 kg # m I = 1 (7)(0.1) = 0.035 kg # m C A D 100 mm 00 mm For ais u = cos 6.56 = 0.8944 u = cos 116.57 = -0.447 u = cos 90 = 0 I = I u + I u + I u - I u u - I u u - I u u = 0 + 0.3175(-0.447) + 0.035(0.8944) - 0-0 - 0 = 0.0915 kg # m v = v = 0 H = -I v - I v + I v = -0-0 + 0.0915(5) = 0.4575 kg # m >s H = 0.4575 kg # m >s 1148
*1 40. Derive the scalar form of the rotational equation of motion about the ais if æzvand the moments and products of inertia of the bod are not constant with respect to time. In general M = d dt (H i + H j + H k) = AH # i + H # j + H # kb +Æ*(H i + H j + H k) Substitute Æ=Æ i +Æ j +Æ k and epanding the cross product ields M = a AH # B -Æ H +Æ H bi + a AH # B -Æ H +Æ H bj Subsitute H, H and H using Eq. 1 10. For the i component, + a AH # B -Æ H +Æ H bk M = d dt (I v - I v - I v ) -Æ (I v - I v - I v ) +Æ (I v - I v - I v ) One can obtain and components in a similar manner. ΣM = d dt (I v - I v - I v ) - Ω (I v - I v - I v ) + Ω (I v - I v - I v ) 1149
1 41. Derive the scalar form of the rotational equation of motion about the ais if æzvand the moments and products of inertia of the bod are constant with respect to time. In general M = d dt (H i + H j + H k) = AH # i + H # j + H # kb +Æ*(H i + H j + H k) Substitute Æ=Æ i +Æ j +Æ k and epanding the cross product ields M = a AH # B -Æ H +Æ H bi + a AH # B -Æ H +Æ H bj Substitute H, H and H using Eq. 1 10. For the i component + a AH # B -Æ H +Æ H bk M = d dt (I v - I v - I v ) -Æ (I v - I v - I v ) For constant inertia, epanding the time derivative of the above equation ields M = (I v # - I v # - I v # ) -Æ (I v - I v - I v ) +Æ (I v - I v - I v ) +Æ (I v - I v - I v ) One can obtain and components in a similar manner. ΣM = (I v # - I v # - I v # ) - Ω (I v - I v - I v ) + Ω (I v - I v - I v ) Similarl for ΣM and ΣM. 1150
1 4. Derive the Euler equations of motion for Eqs. 1 6. æzv, i.e., In general M = d dt (H i + H j + H k) = AH # i + H # j + H # kb +Æ*(H i + H j + H k) Substitute Æ=Æ i +Æ j +Æ k and epanding the cross product ields M = a AH # B -Æ H +Æ H bi + a AH # B -Æ H +Æ H bj Substitute H, H and H using Eq. 1 10. For the i component + a AH # B -Æ H +Æ H bk M = d dt (I v - I v - I v ) -Æ (I v - I v - I v ) +Æ (I v - I v - I v ) Set I = I = I = 0 and require I, I, I to be constant. This ields M = I v # - I Æ v + I Æ v One can obtain and components in a similar manner. ΣM = I. v - I Ω v + I Ω v 1151
1 43. The 4-lb bar rests along the smooth corners of an open bo. At the instant shown, the bo has a velocit v = 53j6 ft>s and an acceleration a = 5-6j6 ft>s. Determine the,, components of force which the corners eert on the bar. A F = m(a G ) ; A + B = 0 [1] ft B F = m(a G ) ; A + B = 4 3. (-6) [] 1ft ft F = m(a G ) ; B - 4 = 0 B = 4lb Appling Eq. 1 5 with v = v = v = 0 v # = v # = v # = 0 (M G ) = I v # - (I - I )v v ; B (1) - A (1) + 4(0.5) = 0 [3] (M G ) = I v # - (I - I )v v ; A (1) - B (1) + 4(1) = 0 [4] Solving Eqs. [1] to [4] ields: A = -.00 lb A = 0.67 lb B =.00 lb B = -1.37 lb (M G ) = I v # - (I - I ) v v ; (-.00)(0.5) - (.00)(0.5) - (-1.37)(1) + (0.67)(1) = 0 (O.K!) B = 4 lb A = -.00 lb A = 0.67 lb B =.00 lb B = -1.37 lb 115
*1 44. The uniform rectangular plate has a mass of m = kg and is given a rotation of v = 4 rad>s about its bearings at A and B. If a = 0. m and c = 0.3 m, determine the vertical reactions at the instant shown. Use the,, aes shown and note that I = - a mac - a 1 bac c + a b. A c V a B v = 0, v = 0, v = -4 v # = 0, v # = 0, v # = 0 M = I v # - (I - I )v v - I av # - v v b - I Av - v B - I av # + v v b B B a a b 1 + a c R b - A B a a b 1 + a c R b = -I (v) B - A = a mac 6 b c - a v Ca + c D 3 F = m(a G ) ; A + B - mg = 0 Substitute the data, Solving: B - A = (0.)(0.3) 6 A + B = (9.81) C (0.3) - (0.) S(-4) = 0.34135 C(0.3) + (0.) D 3 A = 9.64 N B = 9.98 N A = 9.64 N B = 9.98 N 1153
1 45. If the shaft AB is rotating with a constant angular velocit of v = 30 rad>s, determine the X, Y, Z components of reaction at the thrust bearing A and journal bearing B at the instant shown. The disk has a weight of 15 lb. Neglect the weight of the shaft AB. A 1.5 ft 30 1 ft The rotating frame is set with its origin at the plate s mass center, Fig. a.this frame will be fied to the disk so that its angular velocit is Æ=vand the,, and aes will alwas be the principle aes of inertia of the disk. Referring to Fig. b, 0.5 ft v 30 rad/s B v = [30 cos 30 j-30 sin 30 k] rad>s = [5.98j-15k] rad>s Thus, v = 0 v = 5.98 rad>s v = -15 rad>s Since v is alwas directed towards the + Y ais and has a constant magnitude, v # = 0.Also, since Æ=v, Av # B = v # = 0.Thus, v # = v # = v # = 0 The mass moments of intertia of the disk about the,, aes are I = I = 1 4 a 15 3. b A0.5 B = 0.0911 slug # ft I = 1 a 15 3. b A0.5 B = 0.0583 slug # ft Appling the equations of motion, M = I v # - (I - I )v v ; B Z (1) - A Z (1.5) = 0 - (0.0583-0.0911)(5.98)(-15) B Z - 1.5A Z = 11.35 (1) M = I v # - (I - I )v v ; B X (1 sin 30 ) - A X (1.5 sin 30 ) = 0-0 B X - 1.5A X = 0 () M = I v # - (I - I )v v ; B X (1 cos 30 ) - A X (1.5 cos 30 ) = 0-0 B X - 1.5A X = 0 F X = m(a G ) X ; F Y = m(a G ) Y ; F Z = m(a G ) Z ; A X + B X = 0 A Y = 0 A Z + B Z - 15 = 0 (3) (4) Solving Eqs. (1) through (4), A Z = 1.461 lb A X = B X = 0 B Z = 13.54 lb = 13.5 lb A Z = 1.46 lb B Z = 13.5 lb A X = A Y = B X = 0 1154
1 46. The assembl is supported b journal bearings at A and B, which develop onl and force reactions on the shaft. If the shaft is rotating in the direction shown at V = 5i6 rad>s, determine the reactions at the bearings when the assembl is in the position shown. Also, what is the shaft s angular acceleration? The mass per unit length of each rod is 5 kg>m. Solution B 1 m m v A 1 m Equations of Motion. The inertia properties of the assembl are I = 1 3 [5(1)](1 ) = 1.6667 kg # m I = 1 3 [5(3)](3 ) + 30 + 5(1)(1 ) 4 = 50.0 kg # m I = 1 3 [5(3)](3 ) + e 1 1 [5(1)](1 ) + 5(1)(1 + 0.5 ) f = 51.67 kg # m I = 0 + [5(1)](1)(0.5) =.50 kg # m I = I = 0 with v # = rad>s, v = v = 0 and v # = v # = 0 b referring to the FBD of the assembl, Fig. a, ΣM = I v # ; -[5(1)](9.81)(0.5) = 1.6667 v # v # = -14.715 rad>s = -14.7 rad>s ΣM = -I v # ; [5(1)](9.81)(1) + [5()](9.81)(1.5)-B (3) = -.50(-14.715) B = 77.665 N = 77.7 N ΣM = -I v ; -B (3) = -.50( ) B = 3.3333 N = 3.33 N ΣF = M(a G ) ; A = 0 ΣF = M(a G ) ; -A - 3.333 = [5(1)]3 - (0.5)4 A = 6.667 N = 6.67 N ΣF = M(a G ) ; A + 77.665 - [5(3)](9.81) - [5(1)](9.81) = [5(1)][-14.715(0.5)] A = 81.75 N v # = -14.7 rad>s B = 77.7 N B = 3.33 N A = 0 A = 6.67 N A = 81.75 N 1155
1 47. The assembl is supported b journal bearings at A and B, which develop onl and force reactions on the shaft. If the shaft A is subjected to a couple moment M = 540i6 N # m, and at the instant shown the shaft has an angular velocit of V = 5i6 rad>s, determine the reactions at the bearings of the assembl at this instant. Also, what is the shaft s angular acceleration? The mass per unit length of each rod is 5 kg>m. Solution B 1 m m v A 1 m Equations of Motions. The inertia properties of the assembl are I = 1 3 [5(1)](1 ) = 1.6667 kg # m I = 1 3 [5(3)](3 ) + 30 + [5(1)](1 ) 4 = 50.0 kg # m I = 1 3 [5(3)](3 ) + e 1 1 [5(1)](1 ) + [5(1)](1 + 0.5 ) f = 51.67 kg # m I = 0 + [5(1)](1)(0.5) =.50 kg # m I = I = 0 with v = rad>s, v = v = 0 and v # = v # = 0 b referring to the FBD of the assembl, Fig. a, ΣM = I v # ; 40 - [5(1)](9.81)(0.5) = 1.6667 v # # = 9.85 rad>s v ΣM = -I v # ; [5(1)](9.81)(1) + [5(3)](9.81)(1.5) - B (3) = -.50(9.85) B = 97.665 N = 97.7 N ΣM = -I v ; -B (3) = -.50( ) B = 3.3333 N = 3.33 N ΣF = M(a G ) ; A = 0 ΣF = M(a G ) ; -A - 3.3333 = [5(1)]3 - (0.5)4 A = 6.6667 N = 6.67 N ΣF = M(a G ) ; A + 97.665 - [5(3)](9.81) - [5(1)](9.81) = [5(1)][9.85(0.5)] A = 11.75 N = 1 N # v = 9.85 rad>s B = 97.7 N B = 3.33 N A = 0 A = 6.67 N A = 1 N 1156
*1 48. The man sits on a swivel chair which is rotating with a constant angular velocit of 3 rad>s. He holds the uniform 5-lb rod AB horiontal. He suddenl gives it an angular acceleration of rad>s, measured relative to him, as shown. Determine the required force and moment components at the grip, A, necessar to do this. Establish aes at the rod s center of mass G, with + upward, and + directed along the ais of the rod towards A. B 3 ft ft rad/s A 3 rad/s I = I = 1 1 A 5 3. B(3) = 0.1165 ft 4 I = 0 Æ=v = 3k v = v = 0 v = 3 rad>s Æ # = (v # ) +Æ*v = -i + 0 v # = - rad>s v # = v # = 0 (a G ) = 0 (a G ) = (3.5)(3) = 31.5 ft>s (a G ) = (1.5) = 3ft>s F = m(a G ) ; A = 0 F = m(a G ) ; A = 5 (31.5) = 4.89 lb 3. F = m(a G ) ; -5 + A = 5 3. (3) A = 5.47 lb M = I v # - (I - I )v v ; M + 5.47(1.5) = 0.1165(-) - 0 M = -8.43 lb # ft M = I v # - (I - I )v v ; 0 + M = 0-0 M = 0 M = I v # - (I - I )v v ; M = 0-0 M = 0 A = 0 A = 4.89 Ib A = 5.47 Ib M = -8.43 Ib # ft M = 0 M = 0 1157
1 49. The rod assembl is supported b a ball-and-socket joint at C and a journal bearing at D, which develops onl and force reactions. The rods have a mass of 0.75 kg/m. Determine the angular acceleration of the rods and the components of reaction at the supports at the instant v = 8 rad>s as shown. Æ=v = 8k v = v = 0, v # = v # = 0, I = I = 0 v = 8 rad>s v # = v # I = 0.75(1)()(0.5) = 0.75 kg # m m m D ω =8rad/s 1m A B 50 N m C I = 1 3 (0.75)(1)(1) = 0.5 kg # m Eqs. 1 4 become M = I v M = -I v # M = I v # Thus, -D (4) - 7.3575(0.5) = 0.75(8) D = -1.9 N D (4) = -0.75v 50 = 0.5 v # v # = 00 rad>s D = -37.5 N F = m(a G ) ; C - 37.5 = -1(0.75)(00)(0.5) C = -37.5 N F = m(a G ) ; C - 1.9 = -(1)(0.75)(8) (0.5) C = -11.1 N F = m(a G ) ; C - 7.3575-9.43 = 0 C = 36.8 N v # = 00 rad>s D = -1.9 N D = -37.5 N C = -37.5 N C = -11.1 N C = 36.8 N 1158
1 50. The bent uniform rod ACD has a weight of 5 lb>ft and is supported at A b a pin and at B b a cord. If the vertical shaft rotates with a constant angular velocit v = 0 rad>s, determine the,, components of force and moment developed at A and the tension in the cord. 1 ft A 0.5 ft C w = w = 0 B 1 ft w = 0 rad>s w # = w # = w # = 0 The center of mass is located at ω D = 5(1)a 1 b 5() = 0.5 ft = 0.5 ft (smmetr) I = 5 3. (-0.5)(1) = -0.0776 slug # ft I = 0 Eqs. 1 4 reduce to ΣM = I (w ) ; -T B (0.5) - 10(0.75) = -0.0776(0) T B = 47.1 lb ΣM = 0; M = 0 ΣM = 0; M = 0 ΣF = ma ; A = 0 ΣF = ma ; A = -a 10 3. b(0) (1-0.5) A = -93. lb ΣF = ma ; A - 47.1-10 = 0 A = 57.1 lb T B = 47.1 lb M = 0 M = 0 A = 0 A = -93. lb A = 57.1 lb 1159
1 51. The uniform hatch door, having a mass of 15 kg and a mass center at G,is supported in the horiontal plane b bearings at A and B. If a vertical force F = 300 N is applied to the door as shown, determine the components of reaction at the bearings and the angular acceleration of the door.the bearing at A will resist a component of force in the direction, whereas the bearing at B will not. For the calculation, assume the door to be a thin plate and neglect the sie of each bearing.the door is originall at rest. Eqs. 1 5 reduce to B - A = -440 v = v = v = 0 v # = v # = 0 M = 0; 300(0.5-0.03) + B (0.15) - A (0.15) = 0 00 mm 00 mm (1) A 30 mm 100 mm G F 150 mm 150 mm B 100 mm 30 mm M = I v # ; 15(9.81)(0.) - (300)(0.4-0.03) = [ 1 1 (15)(0.4) + 15(0.) ]v # v # = -10 rad>s M = 0; -B (0.15) + A (0.15) = 0 F = m(a G ) ; -A + B = 0 A = B = 0 F = m(a G ) ; A = 0 F = m(a G ) ; 300-15(9.81) + B + A = 15(101.96)(0.) B + A = 153.03 () Solving Eqs. (1) and () ields A = 97 N B = -143 N v # = -10 rad>s A = B = 0 A = 0 A = 97 N B = -143 N 1160
*1 5. The 5-kg circular disk is mounted off center on a shaft which is supported b bearings at A and B. If the shaft is rotating at a constant rate of v = 10 rad>s, determine the vertical reactions at the bearings when the disk is in the position shown. A 100 mm ω 0 mm 100 mm G 100mm B v = 0, v = -10 rad/s, v = 0 v # = 0, v # = 0, v # = 0 a + M = I v # - (I - I )v v -(0.)(F A ) + (0.)(F B ) = 0 F A = F B +T F = ma ; F A + F B - 5(9.81) = -5(10) (0.0) F A = F B = 19.5 N F A = F B = 19.5 N 1161
1 53. Two uniform rods, each having a weight of 10 lb, are pin connected to the edge of a rotating disk. If the disk has a constant angular velocit v D = 4 rad>s, determine the angle u made b each rod during the motion, and the components of the force and moment developed at the pin A. Suggestion: Use the,, aes oriented as shown. 1.75 ft ω D = 4 rad/s A ft θ θ G ft I = 1 1 a 10 3. b(4) = 0.4141 slug # ft I = 0 B Appling Eq. 1 5 with v = 4 sin u v = 4 cos u v = 0 v # = v # = v # = 0 ΣM = I v # - (I - I )v v ; -A () = 0 - (0.4141-0)(4 sin u)(4 cos u) (1) ΣM = I v # - (I - I )v v ; M + A () = 0 () ΣM = I v # - (I - I )v v ; M = 0 Also, ΣF = m(a G ) ; A = 0 From Eq. () M = 0 ΣF = m(a G ) ; A - 10 sin u = -a 10 3. b(1.75 + sin u)(4) cos u (3) ΣF = m(a G ) ; A - 10 cos u = a 10 3. b(1.75 + sin u)(4) sin u (4) Solving Eqs. (1), (3) and (4) ields: u = 64.1 A = 1.30 lb A = 0. lb M = 0 A = 0 M = 0 u = 64.1 A = 1.30 lb A = 0. lb 116
1 54. The 10-kg disk turns around the shaft AB, while the shaft rotates about BC at a constant rate of v = 5 rad>s. If the disk does not slip, determine the normal and frictional force it eerts on the ground. Neglect the mass of shaft AB. C v 5 rad/s B Solution Kinematics. The instantaneous ais of ero velocit (IA) is indicated in Fig. a. Here the resultant angular velocit is alwas directed along IA. The fied reference frame is set to coincide with the rotating frame using the similar triangle, m A 0.4 m v = v 0.4 ; v = (5) = 5.0 rad>s 0.4 Thus, V = V + V = 5-5i + 5.0k6 rad>s Here, (v # ) = (v # ) = 0 since v is constant. The direction of V will not change that alwas along ais when Ω = v. Then V # = (V # ) + V * V = 0 The direction of V b does not change with reference to the rotating frame if this frame rotates with Ω = v = 5-5i 6 rad>s. Then Finall V # = (v # ) + V * v = 0 + (-5i) * (5.0k) = 515j6 rad>s V # = V # + V # = 0 + 15j = 515j6 rad>s 1163
1 54. Continued Equations of Motion. The mass moments of inertia of the disk about the, and aes are I = I = 1 4 (10)(0.4 ) + 10( ) = 40.4 kg # m I = 1 (10)(0.4 ) = 0.800 kg # m B referring to the FBD of the disk, Fig. b, ΣM = I v # - (I - I )v v ; N() - 10(9.81)() = 40.4(15) - (0.8-40.4)(5.0)(-5) N = 148.1 N = 148 N ΣM = I v # - (I - I )v v ; F f (0.4) = 0-0 F f = 0 N = 148 N F f = 0 1164
1 55. The 0-kg disk is spinning on its ale at v s = 30 rad>s, while the forked rod is turning at v 1 = 6 rad>s. Determine the and moment components the ale eerts on the disk during the motion. 00 mm O v s 30 rad/s Solution Solution I Kinematics. The fied reference XYZ frame is set coincident with the rotating frame. Here, this rotating frame is set to rotate with Ω = V p = 5-6k6 rad>s. The angular velocit of the disk with respect to the XYZ frame is v 1 6 rad/s A V = V p + V s = 530j - 6k6 rad>s Then v = 0 v = 30 rad>s v = -6 rad>s Since V p and V s does not change with respect to frame, V # with respect to this frame is V # = 0. Then v # = 0 v # = 0 v # = 0 Equation of Motion. Although the disk spins about the ais, but the mass moment of inertia of the disk remain constant with respect to the frame. I = I = 1 4 (0)(0. ) = 0. kg # m I = 1 (0)(0. ) = 0.4 kg # m 1165
1 55. Continued With Ω V with Ω = 0, Ω = 0 and Ω = -6 rad>s ΣM = I v # - I Ω v + I Ω v ; (M 0 ) = 0-0.4(-6)(30) + 0 (M 0 ) = 7.0 N # m ΣM = I v # - I Ω v + I Ω v 0 = 0 (Satisfied!) ΣM = I v # - I Ω v + I Ω v ; Solution II (M 0 ) = 0-0 + 0 = 0 Here, the frame is set to rotate with Ω = V = 530j - 6k6 rad>s. Setting another frame coincide with and XYZ frame to have an angular velocit of Ω = V p = 5-6k6 rad>s, Thus with Ω = v, v # = (v # ) = (v # ) + Ω * v = 0 + (-6k) * (30j - 6k) = 5180i6 rad>s v # = 180 rad>s v # = 0 v # = 0 ΣM = I v # - (I - I )v v ; (M 0 ) = 0.(180) - (0.4-0.)(30)(-6) = 7.0 N # m ΣM = I v # - (I - I )v v ; 0 = 0-0 (Satisfied) ΣM = I v # - (I - I )v v (M 0 ) = 0-0 = 0 (M 0 ) = 7.0 N # m (M 0 ) = 0 1166
*1 56. The 4-kg slender rod AB is pinned at A and held at B b a cord.the ale CD is supported at its ends b ball-and-socket joints and is rotating with a constant angular velocit of rad>s. Determine the tension developed in the cord and ω the magnitude of force developed at the pin A. C B 40 m I = 1 3 (4)() = 5.3333 kg # m I = 0 A D Appling the third of Eq. 1 5 with v = cos 40 = 1.531 rad>s v = sin 40 = 1.856 rad>s M = I v # - (I - I ) v v ; T( cos 40 ) - 4(9.81)(1 sin 40 ) = 0 - (0-5.3333)(1.531)(1.856) T = 3.3 N Also, F = m(a G ) ; A = 0 F = m(a G ) ; A - 3.3 = -4() (1 sin 40 ) A = 13.03 N F = m(a G ) ; A - 4(9.81) = 0 A = 39.4 N F A = A + A + A = 0 + 13.03 + 39.4 = 41.3 N T = 3.3 N F A = 41.3 N 1167
1 57. The blades of a wind turbine spin about the shaft S with a constant angular speed of v s, while the frame precesses about the vertical ais with a constant angular speed of v p. Determine the,, and components of moment that the shaft eerts on the blades as a function of u. Consider each blade as a slender rod of mass m and length l. v p u The rotating frame shown in Fig. a will be attached to the blade so that it rotates with an angular velocit of Æ=v, where v = v s + v p. Referring to Fig. b v p = v p sin ui + v p cos uk.thus, v = v p sin ui + v s j + v p cos uk.then u S v s v = v p sin u v = v s v = v p cos u The angular acceleration of the blade v # with respect to the XYZ frame can be obtained b setting another frame having an angular velocit of Æ = v p = v p sin ui + v p cos uk.thus, v # = Av # B +Æ *v = (v # 1) + (v # ) +Æ *v S +Æ *v P = 0 + 0 + Av p sin ui + v p cos ukb * (v s j) + 0 = -v s v p cos ui + v s v p sin uk Since Æ=v, v # = v #.Thus, v # = -v s v p cos u v # = 0 v # = v s v p sin u Also, the,, and aes will remain as principle aes of inertia for the blade.thus, I = I = 1 1 (m)(l) = 3 ml I = 0 Appling the moment equations of motion and referring to the free-bod diagram shown in Fig. a, M = I v # - AI - I Bv v ; M = 3 ml (-v s v p cos u) - a 3 ml - 0b(v s )(v p cos u) = - 4 3 ml v s v p cos u M = I v # - AI - I Bv v ; M = 0 - a0-3 ml b(v p cos u)(v p sin u) = 1 3 ml v p sin u M = I v # - (I - I )v v ; M = 0-0 = 0 M = - 4 3 ml v s v p cos u M = 1 3 ml v p sin u M = 0 1168
1 58. The 15-lb clinder is rotating about shaft AB with a constant angular speed v = 4 rad>s. If the supporting shaft at C, initiall at rest, is given an angular acceleration a C = 1 rad>s,determine the components of reaction at the bearings A and B. The bearing at A cannot support a force component along the ais, whereas the bearing at B does. 0.5 ft 1ft A G 1ft ω B v = {-4i} rad>s α C v # = v # +Æ*v = 1k + 0 * (-4i) = {1k} rad>s C Hence v = -4, v = v = 0, v # = v # = 0, v # = 1 rad>s M = I v # - (I - I ) v v, 0 = 0 M = I v # - (I - I ) v v, B (1) - A (1) = 0 M = I v # - (I - I ) v v, A (1) - B (1) = c 1 1 a 15 3. b A3(0.5) + () B d(1) - 0 F = m(a G ) ; B = 0 F = m(a G ) ; A + B = -a 15 3. b(1)(1) F = m(a G ) ; A + B - 15 = 0 Solving, A = -1.69 lb B = -3.90 lb A = B = 7.5 lb B = 0 B = -3.90 lb A = -1.69 lb A = B = 7.5 lb 1169
1 59. The thin rod has a mass of 0.8 kg and a total length of 150 mm. It is rotating about its midpoint at a constant rate u. = 6 rad>s, while the table to which its ale A is fastened is rotating at fastened is rotating at rad>s. Determine the,, moment components which the ale eerts on the rod when the rod is in an position u. rad/s A u u. The,, aes are fied as shown. v = sin u v = cos u v = u # = 6 v # = u # cos u = 1 cos u v # = -u # sin u = -1 sin u v # = 0 I = 0 I = I = 1 1 (0.8)(0.15) = 1.5(10-3 ) Using Eqs. 1 5: M = 0-0 = 0 M = 1.5(10-3 )(-1 sin u) - [1.5(10-3 )-0](6)( sin u) M = (-0.036 sin u) N# m M = 0 - [0-1.5(10-3 )]( sin u)( cos u) M = 0.006 sin u cos u = (0.003 sin u) N# m ΣM = 0 ΣM = (-0.036 sin u) N # m ΣM = (0.003 sin u) N # m 1170
*1 60. Show that the angular velocit of a bod, in terms of Euler angles,, and, can be epressed as v (f # = (f # f sin u cos u + c # sin c + u # u c cos c)i + (f # sin u cos c - u # sin c)j + )k, where i, j, and k are directed along the,, aes as shown in Fig. 1 15d. From Fig. 1 15b. due to rotation f, the,, components of f # are simpl f # along ais. From Fig 1 15c, due to rotation, the,, components of and are in the direction, f # u in the direction, and u # f # u # f # sin u cos u in the direction. Lastl, rotation c.fig. 1 15d, produces the final components which ields v = Af # sin u sin c + u # cos cbi + Af # sin u cos c - u # sin cbj + Af # cos u + c # Bk Q.E.D. v = (ḟ sin u sin c + u. cos c)i + (ḟ sin u cos c - u. sin c)j + (ḟ cos u + ċ )k 1171
1 61. Athin rod is initiall coincident with the Z ais when it is given three rotations defined b the Euler angles f = 30, u = 45, and c = 60. If these rotations are given in the order stated, determine the coordinate direction angles a, b, g of the ais of the rod with respect to the X, Y, and Z aes. Are these directions the same for an order of the rotations? Wh? u = (1 sin 45 ) sin 30 i - (1 sin 45 ) cos 30 j + 1cos 45 k u = 0.3536i - 0.614j + 0.7071k a = cos - 1 0.3536 = 69.3 b = cos - 1 (-0.614) = 18 g = cos - 1 (0.7071) = 45 No, the orientation of the rod will not be the same for an order of rotation, because finite rotations are not vectors. a = 69.3 b = 18 g = 45 No, the orientation will not be the same for an order. Finite rotations are not vectors. 117
1 6. The groscope consists of a uniform 450-g disk D which is attached to the ale AB of negligible mass. The supporting frame has a mass of 180 g and a center of mass at G. If the disk is rotating about the ale at v D =90rad>s, determine the constant angular velocit v p at which the frame precesses about the pivot point O. The frame moves in the horiontal plane. M = I Æ v 5 mm 0 mm 80 mm 5 mm 35 mm A G B D ω D ω p O (0.450)(9.81)(0.15) + (0.180)(9.81)(0.080) = 1 (0.450)(0.035) v P (90) v P = 7.9 rad>s v P = 7.9 rad>s 1173
1 63. The to groscope consists of a rotor R which is attached to the frame of negligible mass. If it is observed that the frame is precessing about the pivot point O at v p = rad>s, determine the angular velocit v R of the rotor. The stem OA moves in the horiontal plane. The rotor has a mass of 00 g and a radius of gration k OA = 0 mm about OA. ω p O 30 mm ω R ΣM = I Ω w (0.)(9.81)(0.03) = 30.(0.0) 4()(w R ) R A vr = 368 rad>s v R = 368 rad>s 1174
*1 64. The top consists of a thin disk that has a weight of 8 lb and a radius of 0.3 ft. The rod has a negligible mass and a length of 0.5 ft. If the top is spinning with an angular velocit v s = 300 rad>s, determine the stead-state precessional angular velocit v p of the rod when u = 40. 0.5 ft ω p 0.3 ft ω s θ ΣM = -I f # sin u cos u + I f # sin u( # f cos u + c # ) 8(0.5 sin 40 ) = - c 1 4 a 8 3. b(0.3) + a 8 3. b(0.5) d v p sin 40 cos 40 + c 1 a 8 3. b(0.3) d v p sin 40 (v p cos 40 + 300) 0.0783v p -.1559v p +.571 = 0 v p = 1.1 rad>s (Low precession) v p = 76.3 rad>s (High precession) w p = 1.1 rad>s w p = 76.3 rad>s 1175