Dynamics_CH19_HW

Dynamics_CH19_HW - *19-4. The space capsule has a mass of...

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Unformatted text preview: *19-4. The space capsule has a mass of 1200 kg and a moment of inertia [G = 900 kg - m2 about an axis passing through G and directed perpendicular to the page. If it is traveling forward with a speed 125 = 800 m/s and executes a turn by means of two jets, which provide a constant thrust of 400 N for 0.3 s, determine the capsule‘s | 5 - "‘ 15' angular velocity just after the jets are turned off. Ls... won t+ (HG), + ZIMGd = (HG), fit] “00M " 35' 0 + 2[400¢0515°(0.3)(1.5)] = 900:», m, = 0.386 rad/s Ans 19-5. Solve Prob. 17-55 using the principle of impulse and momentum. <(+> ("(2)1 +£IMo dz=(Ha)z 0 (“lo 3(l-r‘°"’)a!:= (o. 13).» 3(r+52‘°'“)|3 = 0.13.» «D: 20.8 rad/s A... _'___________—_————_-——————-—-o -6. Solve Prob. 17—54 using the principle of impulse d momentum; (5.4)] + zJ’MAfi 8' (HA): 0 + I’sm = [10(o.2)’]m 0 mu 5 1 _ — 0.411) a) = 56.2 rad/s An: “’- m(v¢;)l + EIFdr = m(vG), M (3+) 0+A,(3) = o A, =0 Ans vvt~v‘-f’2“>'fi<1;gw immanwa'swgn‘vv‘m") (+1) 0 + A,(3) — 98.1(3) = o A, = 98.1 N Am 509 19-11. The pilot of a crippled F-lS fighter was able to control his plane by throttling the two engines. If the plane has a weight of 17 000 lb and a radius of gyration of k(; = 4.7 ft about the mass center G, determine the angular velocity of the plane and the velocity of its mass center Gin I: 5 s if the thrust in each engine is altered to T, = 5000 lb and T2 = 800 lb as shown. Originally the plane is flying straight at 1200 ft/s. Neglect the effects of drag and the loss of fuel. ((+l (HG)1+2JMGdl=(HG)z 0+ 500015)” 25>— 800(5)(l.25) = [(%0%0)(4.7)1]m a): 225 rad/s Ans "10101): +ZJdel=m(Vax)z 17000 17000 (322 )(1200)+5800(5)‘(32,2 )(Vch (VG). =1.25(10’)n/s Am *19-12. The spool has a mass of 30 kg and a radius of gyration k0 = 0.25 m. Block A has a mass of 25 kg, and block 8 has a mass of 10 kg. If they are released from rest, determine the time required for block A to attain a speed of 2 m/s. Neglect the mass of the ropes. v‘ - 2m/s 2 m- —6.667nd/a M 0.18m v. 1: 6.667(0.18) = 1.20 m/s u (He), + that: = (H0). 0 + 25(9.81)(0.3)r — 10(9.81)(0.18)(!) = 25(2)(o.3) + 30(o.25)‘(6.667) + toumxo. 19-13. The man pulls the rope off the reel with a constant force of 8 lb in the direction shown. If the reel has a weight of 250 lb and radius of gyration kg = 0.8 ft about the trunnion (pin) at A, determine the angular velocity of the reel in 3 5 starting from rest. Neglect friction and the weight of rope that is removed. H (11.). + tIMAdI = (HA). 0 + 30.7.90) = [32ng(0.8)2]¢» a) = 6.04 rad/a Am 512 19-17. The drum has a mass of 70 kg, a radius of 300 mm, and radius of gyration k0 = 125 mm. If the coefficients of static and kinetic friction at A are [1,, = 0.4 and pk = 0.3, respectively, determine the drum’s angular y 2 s after it is released from rest. Take 0 = 30° velocit L+ (HA)l + EIMA dz = (HA), 0 + 70(9.8l)(sin30°)(0.3)(2) = (70(o.125)z + 70(03):)(0 m = 27.863 rad/s = 27.9 rad/s Ans mm). + :15 d: = mm), x +/ 0 + 70(9.81)sin30°(2) — F(2) = 70(27.863)(0.3) F = 50.79 N +'\ 0 + ND(2) - 70(9.81)cos30°(2) = 0 ND = 594.7 N If", - 0.4[(594.7] = 237.37 N > 50.79 N —___—_______________________ 19-18. The double pulley consists of two wheels which are attached to one another and turn at the same rate. The pulley has a mass of 15 kg and a radius of gyration k0 = 110mm. If the block at A has a mass of 40 kg determine the speed of the block in 3 s after a constant force F = 2 kN is applied to the rope wrapped around the Inner hub of the pulley.The block is originally at rest. Neglect the mass of the rope. «4) (Ho), + zluod: = (Ho), 0 + momma) — 40(9.81)(0.2)(3) = manor.» + 40(0-2m)(0.2) a, = 120.4 ml: V‘ = 02020.4) = 24.111115 An- 19-30. The square plate has a mass m and is suspended at its corner A by a cord. lfit receives a horizontal impulse I at corner B, determine the location y of the point P about which the plate appears to rotate during the impact. (to (Hal + EIMad: = (HG): T a "I z z 5 0+[(__) = -—-(a +a)a) fi 12 K —)v (3*) “(Va-x): '9' 2154‘ = “(Va-1): 0+I=mva 6! a):— fzan Puke Which the car exerts on the ole whil ' . e AC 15 mntrally vertical. P 4r“ l G = 2.25 m. Determine the horizontal l l *) (HG)| +2IMG d; = (flab 0+UFdr](3.5)= i75(2.25)2<60) IFdf=15.2k.N-s Ans 57.1 19-43. A thin disk of mass m has an angular velocity w, while rotating on a smooth surface. Determine its new angular velocity just after the hook at its edge strikes the peg P and the disk starts to rotate about P without rebounding. HI :3], I 1 (Enr’m. = [En-r’ + mrzlmz ml = "'0’; Ann * a::-:4.10_1I;ie pendulum consists of a 5-lb slender rod AB fired. t h wooden block. A projectile weighing 0.2 lb is in o t 6 center of the block with a velocity of 1000 ft/ If the pendulum is initially at rest, and the projectile embedss itself into the block determin I , e the an ul ' pendulum just after the impact. g m velOClty Of the Mavs Moment of Inertia: The mass moment inertia of the pendulum and the embeded bullet about point A is 5 >m+3:_.<12> = 2.239 slug ['11 fans Prvati’an of Angular Momentum : Since force que to the impact is inlrmai tn the system consisting of the pendulum and the bullet. it will cancel out Thus. angular momentum is conserved about point/4. Applying Eq l9— 17, we have (HAM = (HA )2 (mi, Uh)(’b) = (IA)! 0’; 0.2 (3:5)(1000H25) =1.239m2 (u2 = 6.94 rad/s Ans 528 19-45. A thin ring having a mass of 15 kg strikes the 20-mm-high step. Determine the largest angular velocity u“ the ring can have so that it will not rebound off the step at A when it strikes it. The weight is non — impulsive “’ 1’; (HA)1=(HA)z ) 1,. z z ‘ /”""m 15a», )(0.18)(0.18—0.02)+[15(0. IS)’](aa)=[15(o.1s) +15(o.rs) ]ah 9’ an =0.9444m [gnomm 3.00"" +\zr_ = m(aa),.: (15)(9.81)cos O—NA = lSaflOJB) [gunk] 160 When hoop is abouuorcbound. NA -0. Also. c059: 1—86. and so 3 ya» =6.9602rad/s k 9 FA 4’ =T4=737 rad/s Ann (M3M A)‘ 529 19-46. Determine the height h at which a billiard ball of mass m must be struck so that no frictional force develops between it and the table at A. Assume that the cue C only exerts a horizontal force P on the ball. For lheball (L) M. + ZIFd: = mv, 0 + HA!) = luv, (1) £+ "1401*sz 4’: (HA): 2 o + (nun) = [311”) + "Mint; (2) WV; = tulr (3) Solving qu.(l)—(3) {all yield: h=~r Au 19-47. The disk has a mass of 15 kg. If it is released from- rest when 0 = 30°, determine the maximum angle 0 of rebound after it collides with the wall.The coefficient of $1 ‘ restitution between the disk and the wall is e = 0.6. When t9 = 0°, the disk hangs such that it just touches the wall, ‘1 Neglect friction at the pin C. Datum at lower position of G. n+w=n+w o l 3 2 o 0+(15)(9_81)(0.15)(l —cos30 ) = i[5(15>(0‘15) 10;" a): 3,4l8 rad/s ,' o-(-0.15m’) i”) e =0’6Z3.41t3(015)-0 a} = 2,0508 rad/s T: + ‘6 = T3 + ‘5 l[2(‘5)(0>15)l](2.0503)l + 0 = 0+ 15(9.81)(0. lS)(l ~0050) 2 2 9 2 17 9° Ans 530 ...
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This note was uploaded on 07/09/2011 for the course EGN 3311 taught by Professor Hudyma during the Fall '08 term at UNF.

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Dynamics_CH19_HW - *19-4. The space capsule has a mass of...

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