I xx + I yy + I zz = (y 2 + z 2 )dm + (x 2 + y 2 )dm. (x 2 + z 2 )dm + (x 2 + y 2 + z 2 )dm = 2

Transcription

1 9196_1_s1_p /8/09 1:09 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 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. 95

2 9196_1_s1_p /8/09 1:09 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1. Determine the moment of inertia of the cone with respect to a vertical ais that passes 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 v = 1 4 dm + dm = 1 4 rpa h dr a a h b + r pa h d = r pa 4h 4 (4h + a ) 4 d I v = di v = rpa L 4h 4 (4h + a ) L 0 h 4 d = r pa h 0 (4h + a ) However, m = dm = r pa Lm h L0 h d = r pa h 3 Hence, Using the parallel ais theorem: I = 3m 0 (4h + a ) I v = I v + md 3m 0 (4h + a ) = I v + ma 3h 4 b I v = 3m 80 (h + 4a ) I v = I + md = 3m 80 (h + 4a ) + ma h 4 b = m 0 (h + 3a ) 96

3 9196_1_s1_p /8/09 1:09 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 3. Determine the moments of inertia I and I of the paraboloid of revolution. The mass of the paraboloid is m. a a m = r p d = rp L L 0 0 a r r bd = rra a ba r a r Thus, 1 I v = Lm dm = 1 rp 4 d = 1 r4 rpa L0 a b L0 a a v d = rpa r4 6 ba a I = 1 3 mr I = a 1 Lm 4 dm + dm b = 1 a a 4 rp 4 d + r p d L0 L 0 a = 1 r4 r pa 4 a b d + rpa r L0 a b L0 a 3 d = rpr 4 a 1 + rpr a 3 4 = 1 6 mr + 1 ma I = m 6 (r + 3a ) *1 4. 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 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 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 97

4 9196_1_s1_p /8/09 1:09 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 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 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 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 h = 1 1 a ra ba = m 1 a 1 6. Determine the product of inertia I for the homogeneous tetrahedron. The densit of the material is r. Epress the result in terms of the total mass m of the solid. Suggestion: Use a triangular element of thickness d and then epress di in terms of the sie and mass of the element using the result of Prob a dm = r dv = rc 1 (a - )(a - ) dd = r (a - ) d a m = r L0 a (a - a + )d = ra3 6 a From Prob. 1 5 the product of inertia of a triangular prism with respect to the and planes is I. For the element above, di = rd = ra 4 h (a - )4. Hence, 4 4 a I = r (a 4-4a a - 4a )d 4 L0 I = ra 5 10 or, I = ma 0 98

5 9196_1_s1_p /8/09 1:10 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 7. Determine the moments of inertia for the homogeneous clinder of mass m about the,, aes. r r Due to smmetr I = I = I = 0 I = I = 1 1 m(3r + r ) + ma r b = 7mr 1 I = 1 mr For, u = cos 135 = - 1, u = cos 90 = 0, u = cos 135 = - 1 I = I u + I u + I u - I u u - I u u - I u u = 7mr 1 = 13 4 mr - 1 b mr 1 a - b For, I = I = 7mr 1 For, u = cos 135 = - 1, u = cos 90 = 0, u = cos 45 = 1 I = I u + I u + I u - I u u - I u u - I u u = 7mr 1 = 13 4 mr - 1 b mr 1 a - b

6 9196_1_s1_p /8/09 1:10 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 8. Determine the product of inertia I of the homogeneous triangular block. The material has a densit of r. Epress the result in terms of the total mass m of the block. b h The mass of the differential rectangular volume element shown in Fig. a is dm = rdv = rbd. Using the parallel - plane theorem, a di = di + dm G G = 0 + [rbd]a b b = rb d However, = h (a - ). Then a di = rb c h a (a - ) dd = rb h a a - d Thus, I = L di = rb h a L0 a a - d = rb h a a - 3 a 3 b 0 = 1 1 ra b h However, m = rv = a 1. Then ahbb = 1 rabh I = 1 1 ra b m h = 1 1 rabh 6 mab 930

7 9196_1_s1_p /8/09 1:10 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 9. The slender rod has a mass per unit length of 6 kg>m. Determine its moments and products of inertia with respect to the,, aes. m O The mass of segments (1), (), and (3) shown in Fig. a is m 1 = m = m 3 = 6() = 1 kg. The mass moments of inertia of the bent rod about the,, and aes are m m I = I + m G + G = c 1 1 (1) d + c 1 1 (1) + 1C + -1 D d = 80 kg # m I = I + m G + G = c 1 1 (1) d + c d + c 1 1 (1) + 1C + -1 D d = 18 kg # m I = I + m G + G = c 1 1 (1) d + c 1 (1) d + c d = 176 kg # m Due to smmetr, the products of inertia of segments (1), (), and (3) with respect to their centroidal planes are equal to ero. Thus, I = I + m G G = C0 + 1(1)(0)D + C0 + 1()(1)D + C0 + 1()()D = 7 kg # m I = I + m G G = C0 + 1(0)(0)D + C0 + 1(1)(0)D + C0 + 1()(-1)D = -4 kg # m I = I + m G G = C0 + 1(1)(0)D + C0 + 1()(0)D + C0 + 1()(-1)D = -4 kg # m 931

8 9196_1_s1_p /8/09 1:11 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl Determine the products of inertia I, I, and I of the homogeneous solid. The material has a densit of 7.85 Mg>m mm 00 mm 100 mm The masses of segments (1) and () shown in Fig. a are m 1 = r V 1 = 7850(0.4)(0.4)(0.1) = 15.6 kg and m = r V = 7850(0.)(0.)(0.1) = 31.4 kg. Due to smmetr I = I = I = 0 for segment (1) and I = I = I = 0 for segment (). Since segment () is a hole, it should be considered as a negative segment. Thus 00 mm 00 mm I = I + m G G = C (0.)(0.) D - C (0.3)(0.1) D = 4.08 kg # m I = I + m G G = C (0.)(0.05) D - C (0.1)(0.05) D = 1.10 kg # m I = I + m G G = C (0.)(0.05) D - C (0.3)(0.05) D = kg # m 93

9 9196_1_s1_p /8/09 1:11 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The assembl consists of two thin plates and which have a mass of 3 kg each and a thin plate C which has a mass of 4.5 kg. Determine the moments of inertia I, I and I. I = I = 1 1 (3)(0.4) = 0.04 kg # m 0.4 m m 0.3 m C 0.4 m m 0. m I = 1 1 (3)[(0.4) + (0.4) ] = 0.08 kg # m I = I = I = 0 For G, u = 0 u = cos 60 = 0.50 u = cos 30 = I G = (0.5) (0.866) = 0.05 kg # m I G = I = 0.04 kg # m For G, u = 0 u = cos 30 = u = cos 10 = I G = (0.866) (-0.5) = 0.07 kg # m I = 1 1 (4.5)(0.6) + [ {( ) + (0.173) }] I = 1.36 kg # m I = 1 1 (4.5)(0.4) + [ (0.173) ] I = kg # m I = 1 1 (4.5)[(0.6) + (0.4) ] + [ ( ) ] I = 1.6 kg # m 933

10 9196_1_s1_p /8/09 1:1 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 1. 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. 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 (). I = I + m G G = C0 + 8(0.)(0.)D + C0 + 4(0)(0.)D 400 mm 400 mm 00 mm = 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 = 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. 1 ft Due to smmetr = 0.5 ft 1 ft _ G _ = W w = (-1)(1.5)(1) + C(-0.5)(1.5)(1)D 3C1.5(1) D = ft I = ca b(0.5) d a b(1) = 0.07 slug # ft I = c 1 1 a b(1) + a b( ) d + a 1.5 b( ) 3. = slug # ft I = c 1 1 a b(1) + a b( ) d a b(1) + a b(0.3333) = slug # ft 934

11 9196_1_s1_p /8/09 1:1 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The assembl consists of a 10-lb slender rod and a 30-lb thin circular disk. Determine its moment of inertia about the ais. The mass of moment inertia of the assembl about the,, and aes are I = I = c 1 4 a b d + c 1 1 a b d = slug # ft ft 1 ft I = 1 a b = slug # ft Due to smmetr, I = I = I = 0. From the geometr shown in Fig. a, u = tan -1 a 1. Thus, the direction of the ais is defined b the unit b = 6.57 vector u = cos 6.57 j - sin 6.57 k = j k Thus, u = 0 u = u = Then I = I u + I u + I u - I u u - I u u - I u u = (0) (0.8944) (-0.447) = 1.5 slug # ft The top consists of a cone having a mass of 0.7 kg and a hemisphere of mass 0. kg. Determine the moment of inertia when the top is in the position shown. I 30 mm I = I = 3 80 (0.7)C(4)(0.3) + (0.1) D + (0.7)c 3 4 (0.1) d + a b(0.)(0.03) + (0.)c 3 8 (0.03) + (0.1) d = kg # m I = a 3 10 b(0.7)(0.03) + a 5 b(0.)(0.03) I = kg # m 100 mm 30 mm 45 u = cos 90 = 0, u = cos 45 = , u = cos 45 = I = I u + I u + I u - I u u - I u u - I u u = (0.7071) + (0.61)10-3 (0.7071) I = kg # m 935

12 9196_1_s1_p /8/09 1:13 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 16. Determine the products of inertia I, I, and I of the thin plate. The material has a mass per unit area of 50 kg>m. 00 mm 00 mm 00 mm The masses of segments (1), (), and (3) shown in Fig. a are m 1 = m = 50(0.4)(0.4) = 8 kg and m 3 = 50cp(0.1) d = 0.5p kg. 00 mm Due to smmetr I = I = I = 0 for segment (1), I = I = I = 0 for segment (), and I = I = I = 0 for segment (3). Since segment (3) is a hole, it should be considered as a negative segment. Thus 400 mm 100 mm I = I + m G G = C0 + 8(0.)(0.) D + C0 + 8(0)(0.) D - C p(0)(0.) D 400 mm = 0.3 kg # m I = I + m G G = C0 + 8(0.)(0) D + C0 + 8(0.)(0.) D - C p(0.)(0.) D = 0.57 kg # m I = I + m G G = C0 + 8(0.)(0)D + C0 + 8(0)(0.)D - C p(0)(0.)D = 0 kg # m Determine the product of inertia I for the bent rod. The rod has a mass per unit length of kg>m. Product of Inertia: ppling Eq we have I = I G + m G G = [ () (0) (0.5)] + [ () (0.3) (0.5)] + [ () (0.6) (0.5)] = kg # m 600 mm 400 mm 500 mm 936

13 9196_1_s1_p /8/09 1:13 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl Determine the moments of inertia I, I, I for the bent rod. The rod has a mass per unit length of kg>m. 400 mm 600 mm Moments of Inertia: ppling Eq. 1 3, we have I = (I ) G + m( G + G) = c 1 1 (0.4) () () d = 0.66 kg # m I = (I ) G + m( G + G) = c 1 1 (0.4) () () d = kg # m I = (I ) G + m( G + G) = C () D = 1.09 kg # m + C () D + c 1 1 (0.5) () () d + c 1 1 (0.6) () () d + C () D + c 1 1 (0.6) () () d + c 1 1 (0.5) () () d 500 mm 937

14 9196_1_s1_p /8/09 1:13 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl Determine the moment of inertia of the rod-andthin-ring assembl about the ais. The rods and ring have a mass per unit length of kg>m. O For the rod, 500 mm D 400 mm u = 0.6, u = 0, u = 0.8 I = I = 1 3 [(0.5)()](0.5) = kg # m C 10 I = 0 I = I = I = 0 From Eq. 1 5, I = (0.6) I = 0.03 kg # m For the ring, The radius is r = 0.3 m Thus, I = mr = [ (p)(0.3)](0.3) = kg # m Thus the moment of inertia of the assembl is I = 3(0.03) = 0.49 kg # m 938

15 9196_1_s1_p /8/09 1:14 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 0. 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 + 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. 939

16 9196_1_s1_p /8/09 1:14 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 1. Show that if the angular momentum of a bod is determined with respect to an arbitrar point, then H can be epressed b Eq This requires substituting R = R G + R G> into Eq. 1 6 and epanding, noting that 1 R G dm = 0 b definition of the mass center and v G = v + V : R G>. Z R G/ G R G R P Y H = a Lm r dmb * v + Lm r * (v * r )dm X = a Lm (r G + r G> ) dmb * v + Lm (r G + r G> ) * Cv * r G + r G> )Ddm = a Lm r G dmb * v + (r G> * v ) Lm dm + Lm r G * (v * r G ) dm + a Lm r G dmb * (v * r G> ) + r G> * av * Lm r G dmb + r G> * (v * r G> ) Lm dm Since r G dm = 0 and from Eq. 1 8 H G = r G * (v * r G )dm Lm Lm H = (r G> * v )m + H G + r G> * (v * r G> )m = r G> * (v + (v * r G> ))m + H G = (r G> * mv G ) + H G Q.E.D. 940

17 9196_1_s1_p /8/09 1:14 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1. The 4-lb rod is attached to the disk and collar using ball-and-socket joints. If the disk has a constant angular velocit of rad>s, determine the kinetic energ of the rod when it is in the position shown.ssume the angular velocit of the rod is directed perpendicular to the ais of the rod. v = v + v * r > 3 ft 1 ft rad/s i j k v i = -(1)()j + 3 v v v Epand and equate components: v = -v + v = v + 3 v 0 = -v - 3 v (1) () (3) lso: v # r> = 0 3v - v - v = 0 (4) Solving Eqs. (1) (4): v is perpendicular to the rod. v = rad>s v = rad>s v = rad>s v = ft>s v = rad>s, v = rad>s, v = rad>s v = {-j} ft>s r > = {3i - 1j - 1k} ft v G = v + v * r > v G = -j + 1 i j k v G = {0.333i - 1j} ft>s v G = (0.333) + (-1) = ft>s v = (0.1818) + ( ) + (0.6061) = rad>s T = a 1 ba 4 3. b(1.054) + a 1 bc 1 1 a 4 3. b(3.3166) d(0.6356) T = ft # lb 941

18 9196_1_s1_p /8/09 1:15 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 3. Determine the angular momentum of rod in Prob. 1 about its mass center at the instant shown. ssume the angular velocit of the rod is directed perpendicular to the ais of the rod. rad/s 1 ft 3 ft v = v + v * r > i j k v i = -(1)()j + 3 v v v Epand and equate components: v = -v + v = v + 3 v 0 = -v - 3 v (1) () (3) lso: v # r> = 0 3v - v - v = 0 (4) Solving Eqs. (1) (4): v = rad>s v = rad>s v = rad>s v = ft>s v is perpendicular to the rod. r > = (3) + (-1) + (-1) = ft I G = a 1 1 ba 4 3. b(3.3166) = slug # ft H G = I G v = (0.1818i j k) H G = {0.007i j k) slug # ft >s 94

19 9196_1_s1_p /8/09 1:16 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 4. The uniform thin plate has a mass of 15 kg. Just before its corner strikes the hook, it is falling with a velocit of v G = 5-5k6 m>s with no rotational motion. Determine its angular velocit immediatel after corner strikes the hook without rebounding. Referring to Fig. a, the mass moments of inertia of the plate about the,, and aes are I = I + m G + G = 1 1 (15) = 0.8 kg # m I = I + m G + G = 1 1 (15) c(-0.3) + 0 d = 1.8 kg # m I = I + m G + G = 1 1 (15) c(-0.3) + 0. d =.6 kg # m Due to smmetr, I = I = I = 0. Thus, Since the plate falls without rotational motion just before the impact, its angular momentum about point is Since the plate rotates about point just after impact, the components of its angular momentum at this instant can be determined from Thus, I = I + m G G = (-0.3)(0.) = -0.9 kg # m I = I + m G G = (0.)(0) = 0 I = I + m G G = (-0.3)(0) = 0 (H ) 1 = r G> * mv G = (-0.3i + 0.j) * 15(-5k) = [-15i -.5j] kg # m >s C(H ) D = I v - I v - I v = 0.8v - (-0.9)v - 0(v ) = 0.8v + 0.9v C(H ) D = -I v + I v - I v = -(-0.9)v + 1.8v - 0(v ) = 0.9v + 1.8v C(H ) D = -I v + I v - I v = 0(v ) - 0(v ) +.6v =.6v (H ) = (0.8v + 0.9v )i + (0.9v + 1.8v )j +.6v k Referring to the free-bod diagram of the plate shown in Fig. b, the weight W is a nonimpulsive force and the impulsive force F acts through point. Therefore, angular momentum of the plate is conserved about point. Thus, 00 mm 00 mm G v G 300 mm 300 mm (H ) 1 = (H ) -15i -.5j = (0.8v + 0.9v )i + (0.9v + 1.8v )j +.6v k Equating the i, j, and k components, -15 = 0.8v + 0.9v -.5 = 0.9v + 1.8v 0 =.6v Solving Eqs. (1) through (3), v = rad>s v = rad>s v = 0 Thus, v = [-10.7i j] rad>s (1) () (3) 943

20 9196_1_s1_p /8/09 1:16 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 5. The 5-kg disk is connected to the 3-kg slender rod. If the assembl is attached to a ball-and-socket joint at and the 5-N # m couple moment is applied, determine the angular velocit of the rod about the ais after the assembl has made two revolutions about the ais starting from rest. The disk rolls without slipping. M 5 N m 1.5 m 0. m I = I = 1 4 (5)(0.) + 5(1.5) (3)(1.5) = I v = 1 (5)(0.) = v = -v j +v k = -v j +v sin j +v cos k = (0.1316v - v )j v k Since points and C have ero velocit, v C = v + v * r C> 0 = 0 + C( v - v )j v k D * (1.5j -0.k ) 0 = v v + 0. v v = v Thus, v = v j v k T 1 + U 1 - = T 0 + 5(p) () = (0.100)( v ) + 1 (13.55)( v ) v = 58 rad>s 944

21 9196_1_s1_p /8/09 1:17 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 6. The 5-kg disk is connected to the 3-kg slender rod. If the assembl is attached to a ball-and-socket joint at and the 5-N # m couple moment gives it an angular velocit about the ais of v = rad>s, determine the magnitude of the angular momentum of the assembl about. M 5 N m 1.5 m 0. m I = I = 1 4 (5)(0.) + 5(1.5) (3)(1.5) = I = 1 (5)(0.) = v = -v j +v k = -v j +v sin j +v cos k = (0.1316v - v )j v k Since points and C have ero velocit, v C = v + v * r C> 0 = 0 + C( v - v ) j v k D * (1.5j -0.k ) 0 = v v + 0.v v = v Thus, v = v j v k Since v = rad>s v = j k So that, H = I v i +I v j + I v k = ( )j (1.985) k = j +6.86k H = ( ) + (6.86) = 6.9 kg # m >s 945

22 9196_1_s1_p /8/09 1:17 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 7. The space capsule has a mass of 5 Mg and the radii of gration are k = k = 1.30 m and k = 0.45 m. If it travels with a velocit v G = 5400j + 00k6 m>s, compute its angular velocit just after it is struck b a meteoroid having a mass of 0.80 kg and a velocit v m = 5-300i + 00j - 150k6 m>s. ssume that the meteoroid embeds itself into the capsule at point and that the capsule initiall has no angular velocit. v m G v G (0.8 m, 3. m, 0.9 m) Conservation of ngular Momentum: The angular momentum is conserved about the center of mass of the space capsule G. Neglect the mass of the meteroid after the impact. (H G ) 1 = (H G ) r G * m m v m = I G v (0.8i + 3.j + 0.9k) * 0.8(-300i + 00j - 150k) = v i v j v k -58i - 10j + 896k = 8450v i v j v k Equating i, j and k components, we have -58 = 8450v v = rad>s -10 = 101.5v v = rad>s 896 = 8450v v = rad>s Thus, v = {-0.065i j k} rad>s 946

23 9196_1_s1_p /8/09 1:17 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 8. Each of the two disks has a weight of 10 lb. The ale weighs 3 lb. If the assembl rotates about the ais at v = 6 rad>s, determine its angular momentum about the ais and its kinetic energ. The disks roll without slipping. ft ft 1 ft 1 ft v 6 rad/s 6 v = 1 v = 1 rad>s v = {-1i} rad>s v = {1i} rad>s H = c 1 a b(1) d(1i) + c 1 a b(1) d(-1i) b c 1 4 a b(1) () d(6) a 3 3. b(4) (6) rk H = {16.6k} slug # ft >s T = 1 I v + 1 I v + 1 I v = 1 ca 1 a b(1) bd(1) b c 1 4 a b(1) () d a 3 3. b(4) r(6) = 7.1 lb # ft 947

24 9196_1_s1_p /8/09 1:18 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 9. The 10-kg circular disk spins about its ale with a constant angular velocit of v 1 = 15 rad>s. Simultaneousl, arm O and shaft O rotate about their aes with constant angular velocities of v = 0 and v 3 = 6 rad>s, respectivel. Determine the angular momentum of the disk about point O, and its kinetic energ. O 600 mm The mass moments of inertia of the disk about the centroidal,, and aes, Fig. a, are V3 V 150 mm I = I = 1 4 mr = 1 4 (10)0.15 = kg # m V 1 I = 1 mr = 1 (10)0.15 = kg # m Due to smmetr, the products of inertia of the disk with respect to its centroidal planes are equal to ero. I = I = I = 0 Here, the angular velocit of the disk can be determined from the vector addition of and. Thus, v 1 v 3 v = v 1 + v = [6i + 15k] rad>s The angular momentum of the disk about its mass center G can be obtained b appling H = I v = (6) = kg # m >s H = I v = (0) = 0 H = I v = 0.115(15) = kg # m >s Thus, H G = [0.3375i k] kg # m >s Since the mass center G rotates about the ais with a constant angular velocit of v 3 = [6i] rad>s, its velocit is v G = v 3 * r C>O = (6i) * (0.6j) = [3.6k] m>s Since the disk does not rotate about a fied point O, its angular momentum must be determined from H O = r C>O * mv C + H G = (0.6j) * 10(3.6k) + (0.3375i k) = [1.9375i k] kg # m >s = [1.9i k] kg # m >s The kinetic energ of the disk is therefore T = 1 v # HO = 1 (6i + 15k) # (1.9375i k) = 78.5 J 948

25 9196_1_s1_p /8/09 1:19 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The 10-kg circular disk spins about its ale with a constant angular velocit of v 1 = 15 rad>s. Simultaneousl, arm O and shaft O rotate about their aes with constant angular velocities of v = 10 rad>s and v 3 = 6 rad>s, respectivel. Determine the angular momentum of the disk about point O, and its kinetic energ. O 600 mm The mass moments of inertia of the disk about the centroidal,, and aes. Fig. a, are V3 V 150 mm I = I = 1 4 mr = 1 4 (10)0.15 = kg # m V 1 I = 1 mr = 1 (10)0.15 = kg # m Due to smmetr, the products of inertia of the disk with respect to its centroidal planes are equal to ero. 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 3 v = v 1 + v + v 3 = [6i + 10j + 15k] rad>s The angular momentum of the disk about its mass center G can be obtained b appling H = I v = (6) = kg # m H = I v = (10) = kg # m H = I v = 0.115(15) = kg # m Thus, H G = [0.3375i j k] kg # m Since the mass center G rotates about the fied point O with an angular velocit of Æ=v + v 3 = [6i + 10j], its velocit is v G =Æ*r G>O = (6i + 10j) * (0.6j) = [3.6k] m>s Since the disk does not rotate about a fied point O, its angular momentum must be determined from H O = r C>O * mv G + H G = (0.6j) * 10(3.6k) + (0.3375i j k) = [1.9375i j k] kg # m >s = [1.9i j k] kg # m >s The kinetic energ of the disk is therefore T = 1 v # HO = 1 (6i + 10j + 15k) # (1.9375i j k) = 81.3J 949

26 9196_1_s1_p /8/09 1:19 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 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. t 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 at this instant., V 150 rad/s G v G The mass moments of inertia of the satellite about the,, and aes are V 600 rad/s 800 mm V 300 rad/s I = I = = 50 kg # m I = = 18 kg # m 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 Then, the components of the angular momentum of the satellite about its mass center G are (H G ) = I v = 50(600) = kg # m >s (H G ) = I v = 50(-300) = kg # m >s (H G ) = I v = 18(150) = 500 kg # m >s Thus, H G = [30 000i j + 500k] kg # m >s The angular momentum of the satellite about point can be determined from H = r G> * mv G + H G = (0.8k) * 00(-50i + 00j + 10k) + (30 000i j + 500k) = [-000i j + 500k] kg # m >s 950

27 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 3. 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. t 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 G v G The mass moments of inertia of the satellite about the,, and aes are V 600 rad/s 800 mm V 300 rad/s I = I = = 50 kg # m I = = 18 kg # m 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) = 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)( ) + 1 (50) (50)(-300) + 1 (18)150 = J = 37.0MJ 951

28 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The 5-lb thin plate is suspended from a ball-andsocket joint at O. 0.-lb projectile is fired with a velocit of v = 5-300i - 50j + 300k6 ft>s into the plate and becomes embedded in the plate at point. Determine the angular velocit of the plate just after impact and the ais about which it begins to rotate. Neglect the mass of the projectile after it embeds into the plate. 0.5 ft O 0.5 ft ngular momentum about point O is conserved. (H O ) = (H O ) 1 = r O * m p v p (H O ) 1 = (0.5j k) * a b(-300i - 50j + 300k) = { i j k} lb # ft # s I = a 1 5 ba 1 3. b C(1) + (1) D + a 5 3. b(0.5) = slug # ft I = a 1 5 ba 1 3. b(1) + a 5 3. b(0.5) = slug # ft I = a 1 5 ba 1 3. b(1) = slug # ft 0.5 ft v 0.75 ft 0.5 ft (H O ) 1 = (H O ) i j k = 0.335v i v j v k v = = rad>s v = = rad>s v = = 7.00 rad>s v = {-.16i j + 7.0k} rad>s is of rotation line is along v: -.160i j k u O = (-.160) + (5.400) + (7.00) = -0.33i j k 95

29 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl Solve Prob if the projectile emerges from the plate with a velocit of 75 ft>s in the same direction. 0.5 ft O 0.5 ft u v = a bi - a bj + a bk = i j k I = a 1 5 ba 1 3. b C(1) + (1) D + a 5 3. b(0.5) = slug # ft I = a 1 5 ba 1 3. b(1) + a 5 3. b(0.5) = slug # ft I = a 1 5 ba 1 3. b(1) = slug # ft 0.5 ft v 0.75 ft 0.5 ft H 1 + L M O dt = H (0.5j k) * a b(-300i - 50j + 300k) + 0 = v i v j v k +(0.5j k) * a 0. b(75)(-0.609i j k) 3. Epanding, the i, j, k, components are: = v = v = v v = , v =.3844, v = v = {-0.954i +.38j k} rad>s is of rotation is along v: i +.38j k u = (-0.954) + (.38) + (3.18) u = -0.33i j k 953

30 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl thin plate, having a mass of 4 kg, is suspended from one of its corners b a ball-and-socket joint O. If a stone strikes the plate perpendicular to its surface at an adjacent corner with an impulse of I s = 5-60i6 N # s, determine the instantaneous ais of rotation for the plate and the impulse created at O. O 00 mm (H O ) 1 + L M O dt = (H O ) I s { 60i} N s 0 + r >O * I S = (H O ) 00 mm 0 + (-0.(0.7071)j - 0.(0.7071)k) * (-60i) = (I O ) v i + (I O ) v j + (I O ) v k Epand and equate components: 0 = (I O ) v = (I O ) v = (I O ) v (1) () (3) I = 0, I = 0, I = 0 I = a 1 1 b(4)(0.) = , I = a 1 1 b(4)(0.) = u = cos 90 = 0, u = cos 135 = , u = cos 45 = (I G ) = I u + I u + I u - I u u - I u u - I u u = 0 + ( )( ) + ( )(0.7071) (I G ) = (I O ) = For (I O ), use the parallel ais theorem. (I O ) = C0.7071(0.)D, (I O ) = Hence, from Eqs. (1) and (): v = 0, v = , v = The instantaneous ais of rotation is thus, u l = j k = 0.141j k (90.914) + ( ) The velocit of G just after the plate is hit is v G = v * r G>O v G = (90.914j k) * (-0.(0.7071)k) = i m(v G ) 1 + L F dt = m(v G ) 0-60i + L F O dt = -4(1.857)i L F O dt = {8.57i} N # s 954

31 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 36. The 15-lb plate is subjected to a force F = 8 lb which is alwas directed perpendicular to the face of the plate. If the plate is originall at rest, determine its angular velocit after it has rotated one revolution (360 ). The plate is supported b ball-and-socket joints at and. Due to smmetr 1. ft F 8 lb I = I = I = 0 I = 1 1 a b(1.) = slug # ft I = 1 1 a b( ) = slug # ft I = 1 1 a b(0.4) = slug # ft 0.4 ft For ais u = cos = u = cos 90 = 0 u = cos = I = I u + I u + I u - I u u - I u u - I u u = (0.316) (0.9487) = slug # ft Principle of work and energ: T 1 + U 1 - = T 0 + 8(1. sin )(p) = 1 ( )v v = 58.4 rad>s The plate has a mass of 10 kg and is suspended from parallel cords. If the plate has an angular velocit of 1.5 rad>s about the ais at the instant shown, determine how high the center of the plate rises at the instant the plate momentaril stops swinging. Consevation Energ: Datum is set at the initial position of the plate. When the plate is at its final position and its mass center is located h above the datum. Thus, its gravitational potential energ at this position is 10(9.81)h = 98.1h. Since the plate momentaril stops swinging, its final kinetic energ T = 0. Its initial kinetic energ i T 1 = 1 I G v = 1 c 1 (10)0.5 d 1.5 = J 50 mm rad/s T 1 + V 1 = T + V = h h = m = 3.58 mm 955

32 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The satellite has a mass of 00 kg and radii of gration of k = k = 400 mm and k = 50 mm. When it is not rotating, the two small jets and are ignited simultaneousl, and each jet provides an impulse of I = 1000 N # s on the satellite. Determine the satellite s angular velocit immediatel after the ignition. 500 mm 500 mm The mass moments of inertia of the satellite about the,, and aes are I = I = = 3 kg # m I = = 1.5 kg # m 500 mm I G I 400 mm Due to smmetr, I = I = I = 0 Thus, the angular momentum of the satellite about its mass center G is H = I v = 3v H = I v = 3v H = I v = 1.5v ppling the principle of angular impulse and momentum about the,, and aes, t (H ) 1 + M dt = (H ) L t = 3v v = 0 t H 1 + M dt = H L t (0.4) (0.5) = 3v v = rad>s t (H ) 1 + M dt = (H ) L t (0.5) (0.5) = 1.5v v = 80 rad>s Thus v = {-8.1j + 80k} rad>s 956

33 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The bent rod has a mass per unit length of 6 kg>m, and its moments and products of inertia have been calculated in Prob If shaft rotates with a constant angular velocit of v = 6 rad>s, determine the angular momentum of the rod about point O, and the kinetic energ of the rod. m V O Here, the angular velocit of the rod is m v = [6k] rad>s m Thus, v = v = 0 v = 6 rad>s The rod rotates about a fied point O. Using the results of Prob H = I v - I v - I v = 80(0) - 7(0) - (-4)(6) = 144 kg # m >s H = -I v + I v - I v = -7(0) + 18(0) - (-4)(6) = 144 kg # m >s Thus, H = -I v - I v + I v = -(-4)(0) - (-4)(0) + 176(6) = 1056 kg # m >s H O = [144i + 144j k] kg # m >s The kinetic energ of the rod can be determined from T = 1 v # HO = 1 (6k) # (144i + 144j k) = 3168 J = 3.17 kj 957

34 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *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) = H # i + H # j + H # k +Æ*(H i + H j + H k) Substitute Æ=Æ i +Æ j +Æ k and epanding the cross product ields M = a H # -Æ H +Æ H bi + a H # -Æ H +Æ H bj Subsitute H, H and H using Eq For the i component + a H # -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 Derive the scalar form of the rotational equation of motion about the ais if æzv and 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) = H # i + H # j + H # k +Æ*(H i + H j + H k) Substitute Æ=Æ i +Æ j +Æ k and epanding the cross product ields M = a H # -Æ H +Æ H bi + a H # -Æ H +Æ H bj Substitute H, H and H using Eq For the i component + a H # -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. 958

35 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl 1 4. Derive the Euler equations of motion for æzv, i.e., Eqs In general M = d dt (H i + H j + H k) = H # i + H # j + H # k +Æ*(H i + H j + H k) Substitute Æ=Æ i +Æ j +Æ k and epanding the cross product ields M = a H # -Æ H +Æ H bi + a H # -Æ H +Æ H bj Substitute H, H and H using Eq For the i component + a H # -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. 959

36 9196_1_s1_p /8/09 1:57 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The uniform rectangular plate has a mass of m = kg and is given a rotation of v = 4 rad>s about its bearings at and. If a = 0. m and c = 0.3 m, determine the vertical reactions at and at the instant the plate is vertical as shown. Use the,, aes shown and note that I = - a mac - a 1 bac c + a b. c V a 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 v - v - I av # + v v b a a b + a c 1 R b - a a b + a c 1 R b = -I (v) - = a mac 6 b c - a v Ca + c D 3 F = m(a G ) ; + - mg = 0 Substitute the data, Solving: - = (0.)(0.3) 6 + = (9.81) C (0.3) - (0.) S(-4) = C(0.3) + (0.) D 3 = 9.64 N = 9.98 N 960

37 9196_1_s1_p /8/09 1:58 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 44. The disk, having a mass of 3 kg, is mounted eccentricall on shaft. If the shaft is rotating at a constant rate of 9 rad>s, determine the reactions at the journal bearing supports when the disk is in the position shown. 1 m 1.5 m v 9 rad/s 50 mm 75 mm v = 0, v = -9, v = 0 M = I v # - (I - I )v v (1.5) - (1) = 0-0 M = I v # - (I - I )v v (1) - (1.5) = 0-0 F = ma ; + = 0 F = ma ; + - 3(9.81) = 3(9) (0.05) Solving, = = 0 = 3.1 N = 18.5 N 961

38 9196_1_s1_p /8/09 1:58 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The slender rod has a mass m and it is connected to the bracket b a smooth pin at. The bracket is rigidl attached to the shaft. Determine the required constant angular velocit of V of the shaft, in order for the rod to make an angle of u with the vertical. V L 3 The rotating frame is set with its origin at the rod s mass center, Fig. a. This frame will be attached to the rod so that its angular velocit is Æ=vand the,, aes will alwas be the principal aes of inertia. Referring to Fig. b, v = -v cos uj + v sin uk u L Thus, v = 0 v = -v cos u v = v sin u Since both the direction and the magnitude is constant. lso, since, v # = v # v # = 0 Æ=v = 0. Thus, v # = v # = v # = 0 The mass moments of inertia of the rod about the,, aes are I = I = 1 1 ml I = 0 ppling the equation of motion and referring to the free-bod diagram of the rod, Fig. a, M = I v # - I - I v v ; - a L b = ml (-v cos u)(v sin u) = mv L 6 sin u cos u (1) The acceleration of the mass center of the rod can be determined from a G = v r = v a L and is directed as shown in sin u + L 3 b = v L (3 sin u + ) 6 Fig. c. Thus, F = m(a G ) ; - mg sin u = - mv L 6 (3 sin u + ) cos u = mg sin u = - mv L 6 (3 sin u + ) cos u () Equating Eqs. (1) and (), mv L 6 sin u cos u = mg sin u - mv L 6 (3 sin u + ) cos u 3g tan u v = L( sin u + 1) 96

39 9196_1_s1_p /8/09 1:58 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The 5-kg rod is supported b a rotating arm.the support at is a journal bearing, which develops reactions normal to the rod.the support at is a thrust bearing, which develops reactions both normal to the rod and along the ais of the rod. Neglecting friction, determine the,, components of reaction at these supports when the frame rotates with a constant angular velocit of v = 10 rad>s. I = I = 1 1 (5)(1) = kg # m I = m G 0.5 m ω = 10 rad/s ppling Eq. 1 5 with v = v = 0 v = 10 rad>s v # = v # = v # = 0 M = I v # - (I - I )v v ; 0 = 0 M = I v # - (I - I )v v ; (0.5) - (0.5) = 0 M = I v # - (I - I )v v ; (0.5) - (0.5) = 0 (1) () lso, F = m(a G ) ; = -5(10) (0.5) = -50N F = m(a G ) ; + = 0 F = m(a G ) ; + - 5(9.81) = 0 (3) (4) Solving Eqs. (1) to (4) ields: = = 0 = = 4.5 N The car travels around the curved road of radius r such that its mass center has a constant speed v G. Write the equations of rotational motion with respect to the,, aes. ssume that the car s si moments and products of inertia with respect to these aes are known. ppling Eq. 1 4 with v = 0, v = 0, v = v G, r v = v = v = 0 G r M = -I 0 - a v G r b R = I r v G M = -I a v G r b - 0R = - I r v G M = 0 Note:This result indicates the normal reactions of the tires on the ground are not all necessaril equal. Instead, the depend upon the speed of the car, radius of curvature, and the products of inertia I and I. (See Eample 13 6.) 963

40 9196_1_s1_p /8/09 1:58 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 48. The shaft is constructed from a rod which has a mass per unit length of kg>m. Determine the,, components of reaction at the bearings and if at the instant shown the shaft spins freel and has an angular velocit of v = 30 rad>s. What is the angular acceleration of the shaft at this instant? earing can support a component of force in the direction, whereas bearing cannot. v 30 rad/s O 0.6 m 0. m 0.1 m 0. m 0.6 m W = C3(0.) + 1.D()(9.81) = N W = 0C1.()(9.81) D + 0.1C0.4()(9.81) D + 0.C0.()(9.81) D = N # m = W W = = m I = c 1 3 C0.()D(0.) d + C0.()D(0.) = kg # m ppling Eq. 1 5 with v = v = 0 v = 30 rad>s v # = v # = 0 M = I v # - (I - I )v v ; (0.7) - (0.7) = 0 (1) M = I v # - (I - I )v v ; ( ) = v # v # = 58.9 rad>s M = I v # - (I - I )v v ; (0.7) - (0.7) = 0 () lso, F = m(a G ) ; - - = -1.8()( )(30) F = m(a G ) ; = 0 F = m(a G ) ; = -1.8()( )(58.9) (3) (4) Solving Eqs. (1) to (4) ields: = = 7.0 N = = 1.9 N 964

41 9196_1_s1_p /8/09 1:59 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl Four spheres are connected to shaft. If m C = 1 kg and m E = kg, determine the mass of spheres D and F and the angles of the rods, u D and u F, so that the shaft is dnamicall balanced, that is, so that the bearings at and eert onl vertical reactions on the shaft as it rotates. Neglect the mass of the rods. V 0.1 m C 0.1 m u D 0.1 m 0.1 m D 30 E 0. m u F 0.1 m 0.1 m F 0.1 m For = 0; 1 m 1 = 0 (0.1 cos 30 )() - (0.1 sin u F )m F - (0. sin u D )m D = 0 (1) For = 0; 1 m 1 = 0 (0.1)(1) - (0.1 sin 30 )() + (0. cos u D )m D + (0.1 cos u F )m F = 0 () For I = 0; 1 1 m 1 = 0 -(0.)(0. sin u D )m D + (0.3)(0.1 cos 30 )() - (0.4)(0.1 sin u F )m F = 0 (3) For I = 0; 1 1 m 1 = 0 (0.1)(0.1)(1) + (0.)(0. cos u D )m D - (0.3)(0.1 sin 30 )() + (0.1 cos u F )(0.4)(m F ) = 0 (4) Solving, u D = 139 m D = kg u F = 40.9 m F = 1.3 kg 965

42 9196_1_s1_p /8/09 1:59 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl man stands on a turntable that rotates about a vertical ais with a constant angular velocit of v p = 10 rad>s. If the wheel that he holds spins with a constant angular speed of v s = 30 rad>s, determine the magnitude of moment that he must eert on the wheel to hold it in the position shown. Consider the wheel as a thin circular hoop (ring) having a mass of 3 kg and a mean radius of 300 mm. 500 mm 300 mm v s 30 rad/s The rotating frame will be set with an angular velocit of Æ=v P = [10k] rad>s. Since the wheel is smmetric about its spinning ais, the,,and aes will remain as the principle aes of inertia. Thus, I = I = 1 mr = 1 (3)0.3 = kg # m v p 10 rad/s I = mr = 30.3 = 0.7 kg # m The angular velocit of the wheel is v = v s + v P = [-30i + 10k] rad>s. Thus, v = -30 rad>s v = 0 v = 10 rad>s Since the directions of v s and v do not change with respect to the frame and their magnitudes are constant, v # p = 0. Thus, v # = v # = v # = 0 ppling the equations of motion and referring to the free-bod diagram shown in Fig. a, M = I v # - I Æ v + I Æ v ; M = 0 M = I v # - I Æ v + I Æ v ; M = (10)(-30) = N # m M = I v # - I Æ v + I Æ v ; M = 0 Thus, M = M + M + M = 0 + (-81.0) + 0 = 81.0 N # m 966

43 9196_1_s1_p /8/09 1:59 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl The 50-lb disk spins with a constant angular rate of v 1 = 50 rad>s about its ale. Simultaneousl, the shaft rotates with a constant angular rate of v = 10 rad>s. Determine the,, components of the moment developed in the arm at at the instant shown. Neglect the weight of arm. ft The rotating frame is established as shown in Fig. a. This frame will be set to have an angular velocit of Æ=v = [10i] rad>s. Since the disk is smmetric about its spinning ais, the,, and aes will remain as the principle aes of inertia.thus, v 10 rad/s v 1 50 rad/s 0.75 ft I = I = 1 4 a b 0.75 = slug # ft I = 1 a b 0.75 = slug # ft The angular velocit of the disk is v = v s + v p = [10i + 50k] rad>s. Thus, v = 10 rad>s v = 0 v = 50 rad>s Since the directions of v 1 and v do not change with respect to the frame and their magnitudes are constant, v # = 0. Thus, v # = v # = v # = 0 ppling the equations of motion and referring to the free-bod diagram shown in Fig. a, M = I v # - I Æ v + I Æ v ; M X - Z () = 0 (1) M = I v # - I Æ v + I Æ v ; M Y = (10)(50) + 0 M Y = lb # ft = -18 lb # ft M = I v # - I Æ v + I Æ v ; M Z = M Z = 0 Since the mass center of the disk rotates about the X ais with a constant angular velocit of v, its acceleration is a G = v # = 0-10 (j) = [-00j] ft>s * r G - v 1 = [10i] rad>s r G. Thus, F Z = m(a G ) Z ; Z - 50 = (0) Z = 50 lb Substituting this result into Eq. (1), we have M X = 100 lb # ft 967

44 9196_1_s1_p /8/09 1:59 PM Page Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the currentl *1 5. The man stands on a turntable that rotates about a vertical ais with a constant angular velocit of v 1 = 6 rad>s. If he tilts his head forward at a constant angular velocit of v = 1.5 rad>s about point O, determine the magnitude of the moment that must be resisted b his neck at O at the instant u = 30. ssume that his head can be considered as a uniform 10-lb sphere, having a radius of 4.5 in. and center of gravit located at G, and point O is on the surface of the sphere. G 4.5 in. O u V 1.5 rad/s The rotating frame shown in Fig. a will be attached to the head so that it rotates with an angular velocit of Æ=v, where v = v 1 + v. Referring to Fig. b, v 1 = [6 cos 30 j + 6 sin 30 k] rad>s = [5.196j + 3k] rad>s. Thus, v = [-1.5i j + 3k] rad>s. Then v = -1.5 rad>s v = rad>s v = 3 rad>s The angular acceleration of the head v # with respect to the XYZ frame can be obtained b setting another frame having an angular velocit of Æ = v 1 = [5.196j + 3k] rad>s. Thus v # = v # +Æ *v V 1 6 rad/s Since Æ=v, v # = v # = [-4.5j k] rad>s. Thus, v # = 0 v # = -4.5 rad>s v # = rad>s lso, the,, aes will remain as principal aes of inertia. Thus, I = I = 5 = (v # 1) + (v # ) +Æ *v 1 +Æ *v = (5.196j + 3k) * (-1.5i) = [-4.5j k] rad>s a b = slug # ft I = 5 a b = slug # ft ppling the moment equations of motion and referring to the free-bod diagram shown in Fig. a, M = I v # - I - I v v ; M - 10 sin 30 (0.375) = 0 - ( )(5.196)(3) M =.556 lb # ft M = I v # - (I - I )v v ; M = (-4.5) - 0 M = lb # ft M = I v # - (I - I )v v ; M = (7.794) - ( )(-1.5)(5.196) = lb # ft Thus, M = M + M + M = ( ) =.68 lb # ft 968