Dynamics Homework 10 - ENGINEERING MECHANICS Dynamics W.F...

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Unformatted text preview: ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 20—1* The radial—arm saw that has an operating speed of 1500 rev/min. The blade and motor armature have a combined weight of 3 lb and a centroidal radius of gyration of k6 = 1 in. When the saw is turned off, bearing friction and a magnetic brake exert a constant braking torque T on the blade and motor. Determine the length of time that the blade rotates before coming to rest if T = 0.015 lb°in. (bearing friction only). Determine the torque T necessary to stop the blade in just 0.25 s. Solution The axle of the saw is a fixed axis of rotation. The moment of inertia of the blade and armature relative to this axis is I; [—32%] [—l-z—JZ -6 2 646.998(10 ) slug-ft a. If the only torque about the axis of rotation is bearing friction, then the angular momentum principle (Eq. 20—9) gives (1500 rev/min = SON rad/s) —6 . 646.998(10 )(sou) — (3—2:; t H O t = 81.3 s . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . .. Ans. b. If the blade stops rotating in just 0.25 s, then the angular momentum principle gives -6 646.998(10 )(SOfl) - T(0.25) = 0 T = 0.40652 lb‘ft = 4.88 lb'in. ... . . . . . . ........ . . . . . . .. Ans. ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 20-18 The uniform slender rod AB (m = 3 kg, 3 = 800 mm) is resting on a frictionless horizontal surface when it is struck with an impulse P At = 20 N's as shown. If b = 650 mm, determine a. The angular velocity w of the rod immediately after the impact. b. The velocity VA of end A immediately after the impact. Solution The moment of inertia of the rod relative to an axis through (a 1its mass center is 1 2 IG— 12 (3)(0.8> u-S 2 —-. = 0.16000 kg'm The principles of linear and angular momentum give '4") + = 0 0 3VGX +T + = 0 20 3VGY C+ o + 0.25(20) = 0.160000 VGx = 0 m/s = T VGy 20/3 m/s w = 31.3 rad/s C . . . . . . . . . . . . . . . . . . . . . . . . . ............. Ans. h. Then, the relative velocity equation gives _ = _ + - vA VG vA/G = mm 3 + (0.4)<31.2s><—3) = —5.83 3 m/s = 5.83 m/s ‘L . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 20-24* A 3-kg uniform slender rod AB 800-mm long hangs in a vertical plane from a frictionless pivot when a 0.03—kg bullet strikes the rod and becomes embedded in it. If the initial velocity of the bullet is v0 = 350 m/s, determine a. The rotation rate of the rod and bullet immediately after the impact. b. The average force of contact between the rod and the bullet for an impact duration of At = 0.001 s. c. The average magnitude of the force exerted on the rod by the frictionless pin at A for an impact duration of At = 0.001 s. d. The total system energy lost in the impact. e. The maximum angle through which the rod will swing after the impact. Solution a. When the bullet and rod are considered together, there is no angular impulse on the system about an axis through the fixed point A. Therefore, the final angular momentum about A is equal to the initial angular momentum about A C+ HA1.- 0 + 0.500(o.o3)(350) = 5.2500 n-m-s C+ HAf [—:'(3)(0.800)2+(0.03)(0.500)2]w f wf = 8.11 rad/s C . . . . . . . . . . . . . . . .... b. For the bullet alone, the linear F: momentum equation gives «— +-> 0.03(350) — F(0.001) = 0.03(0.500)(8.108) 0,03% F = 10,380 N = 10.38 kN .......... . . . . . . ....... . . . . . . .. Ans. (Problem 20-24 continues ...) ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges (Problem 20—24 — cont.) c. For the rod alone the linear momentum equations +9 0 + (Ax + F)(0.00l) 3(O.400)(8.108) O+A—W0.00l o ( y )( ) -648 N = 648 N 6— w= 3(9.8l) 29.43 N T V 6482 + 29.432 = 649 N initial and final energy of the system are 2 %(o.03)(350) = 1837.5 J 2 2 2 T i[—-l—(3)(o.800) + 0.03(o.500) ](8.108) = 21.283 J f 2 3 and the percent loss of energy is Ti - Tf is loss —-—T(100) i 1837.5 — 21.283 1837.5 e. Finally, use work-energy to determine the angle of swing of the rod. 21.283 + 3(9.8l)(—0.400) + 0.03(9.81)(-O.500) + O = O + 3(9.81)(—0.4OO cos 9) + 0.03(9.81)(-0.500 cos 6) ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 20-29* The uniform rod falls horizontally and strikes the rigid corner with its right end. If the velocity of the rod just before the impact is v0 and the impact is perfectly elastic (e = 1), determine the angular velocity u and the velocity of the mass center VG immediately after the impact. Solution Just prior to the impact, the velocity of the right end of the rod is the same as that of the mass center, v0 downward. The line of impact is the y-direction and the definition of the coefficient of restitution gives Also, the linear and angular momentum equations across the impact give +¢ . _ A = . V mvo B t vaf (b) I; 1 2 + ——— A = ———— C+ C [2 (B t) 12 m6 (of (c) Finally, using the relative velocity equation a = - - w ‘ VB VG + VB/G to relate va, va, and f gives _ _ i w d VBf ‘ vs: 2 f ( ) Combining Eqs. a through d gives mv — [imc‘Jw “(Lu -VO] 0 6 f 2 f (o = I, . . . . . . . f 3vo/ C Ans = ~L . . . . . . VGf Vo/2 Ans ENGINEERING MECHANICS — Dynamics W.F. Ri16y & L.D. Sturges 20-37 A slender uniform rod (C = 30 in., WAR = 4 lb) is released 0 from rest at an angle of 6 = 70 to the horizontal and strikes a hard horizontal surface as shown. If the initial height of the rod is h = 60 in. and the coefficient of restitution is e = 0.7, determine a. The angular velocity of the rod immediately after the impact. b. The velocity of the mass center of the rod immediately after the impact. If end B of the rod will strike the surface as the rod rotates immediately after the impact. Solution 3. Use work—energy to determine the speed of the bar just before it strikes the horizontal surface 60 1 4 2 0+4[12]+°‘ 2[32.2]VGi+0 = , i vGi 17 94436 ft/s The line of impact is the yhdirection. Then the definition of coefficient of restitution gives (uAf) - 0 e = 0.7 = - ——Y——-——— (-17.94436) — o A G (vAf)y = 12.56105 ft/s T “3A6 = (VGf)y — (15/12)(.‘.)ABf cos 70 A \709 = (va)y — 0.42753wABf and the linear and angular momentum equations across the impact give 4 T: [W](—17.94436) + AM __;i__ 32.2 (va)y 2 15 o 1 4 30 Ca' 0 [12 C°S 70] Am: 12 [32.2 12] wABf (Problem 20-37 continues ..J EMHNEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges (Problem 20—37 - cont.) Rewriting these last three equations (ng)y - 0.427536)“f = 12.56105 0.776f0wABf + 5.13030 AAt = 0 (va)y — 8.050 AAt = —17.94436 and solving gives (vgf)y = 4.64 ft/s T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. wABf = -18.54 rad/s = 18.54 rad/s D . . . . . . . . . . . . . . . . . .. Ans. c. After the impact, the rod rotates at a constant rate of 18.535 rad/s. Therefore, the rod will be vertical with B on the bottom A when 18.535t = 2.79253 rad (= 90° + 70°) t = 0.15066 5 Also, after the impact the motion of the mass center is given by - A 2 a = -32.2 J ft/s G VG = (4.637 — 32.2t) j ft/s —° 2 A r6 = (1.25 sin 700 + 4.637t - 16.1t ) J ft When t = 0.15066 5, yG = 1.50773 ft = 18.09 in. Therefore, end B will be 3.09 in. above the surface and it Will not hit the surface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 20—39* Bar AB of Fig. a is attached to a frictionless pin at A; bar CD is attached to a frictionless pin at E B and rests on a frictionless support at F. Both AB and CD are uniform slender bars 36 in. long and weighing 5 lb. . 12 mar—12 final E F Both bars are initially at rest when a slight disturbance causes bar AB to fall to the right and strike bar CD as shown in Fig. b. If the coefficient of restitution is e = 0.6, determine The angular velocities of both bars immediately after the impact. The maximum angle of rebound of bar AB after the impact. The average magnitude of the support reaction at E for an impact duration of 0.005 5. Solution Both bars move in fixed axis rotation. The moments of inertia of the bars about these fixed axes are 2 1 5 36 2 IA 3 [ 32.2 ][ l2 ] — 0.46584 slug ft 2 2 1 5 36 5 6 2 IE 12 [ 32.2 ][ 12 ] + [ 32.2 ][ 12 J ’ 0‘15528 Slug'ft a. Use work—energy to determine the angular velocity of bar AB before the impact 1 2 o + + = "‘—' . (O + ' 0 5(18/12) 0 2 (0 46584) ABi 5(18/12) Sin 30 (.0 = . A81 4 01248 rad/s 3 Bar AB rotates about a fixed axis through A = (18/12)(4.01248) = 6.01872 “ GABi en CABi = (27/12)(4.01248) en = 9.02808 : (,0 - . (1.) (18/12) ABf( ) 1 5000 AB ) ( GABf <1 <1 <1 <1 —§ —§ 11 = — . to (27/12) H 2 2500 A (.0 CABf ABf Bar CD rotates about a fixed axis through E w = 0 CD1 (Problem 20—39 continues ...) ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges (Problem 20-39 — cont.) VGCDJ'. = Vccm' = o - = w A vGCDf (6/12) CDf J ft/s I = w _A VCCDf (12/12) CDf( J) 00 A o A = w — w ' CDf cos 3 en CDf 511'! 30 et ft/s The line of impact is the n-direction. Then the definition of coefficient of restitution gives - . w — w ( 2 2500 ABf) ( CDf 9.02808 — 0 while the angular momentum equations give CA: 0.46584(-4.01248) + (27/12)Fnt = 0.465840 ABf O : + A = . to CE 0 (12/12)F t cos 3o 0 15528 CDf Solving these three equations gives wABf = 0.432 rad/s c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. w = . . . . . . . . . . . . . . . . . . . . . . . .; . . . . . . . . . . .. . CDf 5 13 rad/s C Ans FAt = 0.92021 lb°s b. Using work—energy again on bar AB after the impact gives 2 —:~(0.46584)(O.43211) + 5(18/12) sin 300 = 0 + 5(18/12) sin 6 9 = 30.380 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. c. Finally, the linear momentum equations for bar CD give —+: o + (F sin 30° + EX)At = o o 5 6 = — A=——-————. T o + (Ey F cos 3o 1 t [ 32.2 ][ 12 ](5 13219) EX = —92.02 lb = 92.02 lb e— E = 239.1 lb T y D E = 256 lb 5 68.95 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ...
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This note was uploaded on 07/13/2009 for the course ME a taught by Professor Dr. during the Spring '09 term at University of Nevada, Las Vegas.

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Dynamics Homework 10 - ENGINEERING MECHANICS Dynamics W.F...

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