69 110 Vibrações 9th

PROBLEM 19.69\nDetermine the period of small oscillations of a small particle which moves without friction inside a cylindrical surface of radius R.\n\nSOLUTION\n\nDatum at:\nPosition\nT_i = 0\nV_i = WR(1 - cos θ_i)\n\nSmall oscillations:\n(1 - cos θ_i) ≈ 2 sin^2(θ_i/2)\nV_i = WRθ_i^2 / 2\nV_f = WRθ_o^2 / 2\n\nPosition\nV_f = Rθ_o^2 / 2\nT_f = 1/2 mR^2(ω_i^2 + ω_f^2)\nV_f = 0\n\nConservation of energy:\n0 + WRθ_o^2 / 2 - 1/2 mR^2ω_i^2 + 0 = mR^2(α_o^2 + α_f^2)\nW = mg\n\nmR^2α_o^2 / 2 - 1/2 mR^2ω_i^2 = W_s = √(g/R)\n\nPeriod of oscillations.\nT_s = 2π √(R/g)\n\nT_s = 2π / ω_s = 2π √(R/g) ←\n\nPROPRIETARY MATERIAL © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this material may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this material, you are using it without permission.\n2208 PROBLEM 19.70\nA 14-oz sphere A and a 10-oz sphere C are attached to the ends of a rod AC of negligible weight which can rotate in a vertical plane about an axis at B. Determine the period of small oscillations of the rod.\n\nSOLUTION\n\nDatum at:\nPosition\nT_i = 0\nV_i = W_c * h_C - W_A * h_A\nh_c = B*C(1 - cos θ_i)\nh_a = A*A(1 - cos θ_i)\n\nSmall angles:\n1 - cos θ ≈ θ^2 / 2\nV_i = (W_C * B * C)(1/12) - (W_A * A * B_A)(1/2)θ^2 / 2\n\nV_i = [(10 lb)(12) / 16 * (8/12) + (14 lb)(16 / 12)]θ^2 / 2\n\nV_i = (0.4167 - 0.3646)θ^2 / 2 = 0.05208θ^2 / 2\n\nPosition\nV_f = 0\nT_f = T_2 = (1/2 m_c(v_c)^2 + 1/2 m_A(v_A)^2) θ^2\n\nh = 8 / 12\nθ = 5θ^2 \nT_f = 1/2 m_c(θ_f)^2 + (....)W*C\n\nT_f = 1/2(8/12)(5)θ_f^2 + W = W*Aθ_a^2\n\nPROPRIETARY MATERIAL © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this material may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this material, you are using it without permission.\n2209 PROBLEM 19.70 (Continued)\n\nConservation of energy.\nT_i + V_i = T_2 + V_2; 0 + 0.05208 = 0.4297θ^2 / 2\n\nW_s^2 = (32.2)(0.05208) / (0.4297)\n\nPeriod of oscillations.\nT_s = 2π / W_s = 2π / √3.902\n\nT_s = 3.18 s ←\n\nPROPRIETARY MATERIAL © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this material may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this material, you are using it without permission.\n2210 PROBLEM 19.71\n\nA 1.8-kg collar A is attached to a spring of constant 800 N/m and can slide without friction on a horizontal rod. If the collar is moved 70 mm to the left from its equilibrium position and released, determine the maximum velocity and maximum acceleration of the collar during the resulting motion.\n\nSOLUTION\n\nDatum at O:\n\nPosition 1\nT1 = 0 V1 = 1/2 kx2\n\nxk = axk\n\nT2 = 1/m v2 = 0 v2 = xk\n\nxk = 0.070 m\nT1 + V1 = T2 + V2 0 + 0 = kx2/2\n\n1/2 kx2 = 1/2 kx2/m2 v2 = 800 N/m m = 1.8 kg\n\n\\( \\alpha_{c}^{2} = 4444.4 s^{-2}, \\alpha_{2} = 21.08 \\text{ rad/s} \\)\n\nxk = \\alpha_{c} xk = (21.08 s^{-1})(0.070 m) = 1.476 m/s\n\nxk = \\alpha_{c} xk = (21.08 s^{-1})^{2}(0.070 m) = 31.1 m/s²\n\nPROPRIETARY MATERIAL © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or beyond the limited distribution to teachers and students permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.\n\n2211 PROBLEM 19.72\n\nA 3-lb collar A is attached to a spring of constant 5 lb/in. and can slide without friction on a horizontal rod. The collar is at rest when it is struck with a mallet and given an initial velocity of 35 in/s. Determine the amplitude of the resulting motion and the maximum acceleration of the collar.\n\nSOLUTION\n\nm = W/g = 3 / 32.2 = 0.093168 lb·s²/ft k = 5 lb/in. = 60 lb/ft\n\nPosition 1 Immediately after collar is struck:\n\nx = 0, v = 35 in/s = 2.9167 ft/s\nVi = 0, T1 = 1/m v1² = (0.093167)(2.9167)² = 0.39628 ft·lb\n\nPosition 2, v = 0, x is maximum, i.e., x = xk\n\nVi = 0 T2 = 1/2 kx2 = (60)(xk) = 30v2 T2 = 0\n\n0.39628 + 0 = 0 + 30v2\n\nxk = 0.11493 ft\n\nxk = 1.379 in. \n\nMaximum force:\n\nPmax = kx = (60)(0.11493) = 6.8959 lb\n\nMaximum acceleration:\n\naa = Pmax/m = 6.8959/0.093168 = 74.016 lb/ft² aa = 888 in/s²\n\nPROPRIETARY MATERIAL © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or beyond the limited distribution to teachers and students permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.\n\n2212 PROBLEM 19.73\n\nA uniform rod AB can rotate in a vertical plane about a horizontal axis at C located at a distance c above the mass center G of the rod. For small oscillations, determine the value of c for which the frequency of the motion will be maximum.\n\nSOLUTION\n\nFind \\( \\omega_{n} \\) as a function of c.\n\nDatum at O:\n\nT1 = 0 V1 = mgh\nVi = 0 T1 = mgc(1 - cos(θn))\n\n1 - cos(θn) = 2sin²(θn/2)\n\nV1 = mgc(θn²/2)\n\nT2 = 1/2 cω²θn\n\nV2 = 0\n\nT1 + V1 = T2 + V2 + m(gc)(θn²/2) = 1/2 [(r²/(12+c))ω²]θn² = 0\n\nT1 + V1 = T2 + 0\n\nConservation of energy:\n\n\\( \\omega_{n}^{2} = \\frac{gc}{(\\frac{1}{12}+c)} \\)\n\nMaximum c, when:\n\ndoz/dc = 0: g(2(c²+t²) - 2c²g) = 0\n\nr²/12 - c² = 0\n\nc = 1/√2\n\nPROPRIETARY MATERIAL © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or beyond the limited distribution to teachers and students permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.\n\n2213 PROBLEM 19.74\nA homogeneous wire of length 2l is bent as shown and allowed to oscillate about a frictionless pin at B. Denoting by T0 the period of small oscillations when β = 0, determine the angle β for which the period of small oscillations is 2T0.\n\nSOLUTION\nWe denote by m the mass of half the wire.\n\nPosition 1 Maximum deflections:\nT1 = 0; V1 = -mg l/2 cos(θ - β) - mg l/2 cos(θ + β) =\n\n= -mg l/2 (cos θ cos β + sin θ sin β) - mg l/2 cos β - sin θ - β;\nV1 = -mg l/ cos β cos θ;\n\nFor small oscillations,\ncos θ = 1 - 1/2 θ²;\n...\n\nThus,\nT2 = 1/2 l^2 / I0; but I0 = 2/3 m l²;\n\nI0 = 2/3 m l²;\n\nPROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this material may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or beyond the limited distribution to teachers and instructors permitted by McGraw-Hill for their authorized course preparation. If you are a student using this material, you are using it without permission.\n2214 PROBLEM 19.74 (Continued)\n\nConservation of energy:\nT1 + V1 = T2 + V2\n-mg l cos β + 1/2 mg l/cos β = 1/2 I0 θ˙2 - mg cos β\n\nSetting θ˙2 = ω², I0 = mg /3 λ γ cos β\n\nBut for β = 0,\nT0 = 2π √(2l /3g);\nFor T = 2T0,\n2π = 2l /3g;\n\nSquaring and reducing,\n1/cos β = 4 cos β/4\nβ = 75.5°\n\nPROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this material may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or beyond the limited distribution to teachers and instructors permitted by McGraw-Hill for their authorized course preparation. If you are a student using this material, you are using it without permission.\n2215 PROBLEM 19.75\nThe inner rim of an 85-lb flywheel is placed on a knife edge, and the period of its small oscillations is found to be 1.26 s. Determine the centroidal moment of inertia of the flywheel.\n\nSOLUTION\nDatum at 0:\nPosition 1\nT1 = 1/2 I0 θ˙2; V1 = 0\n\nPosition 2\nT2 = 0; V2 = mg h\nh = r(1 - cos θ) = r 2 sin^2 θ/2\nv2 = (θ˙2)/2\n\nConservation of energy:\nT1 + V1 = T2 + V2:\n1/2 I0 θ˙2 + 0 = mg h + 1/2 m v2^2\n\nFor simple harmonic motion,\n\nI0 θ˙2 = mg h; ω2 = m g/ I0;\n\nT = 2π √(I0 /mg)\n\nMoment of inertia,\nI0 = T + m r^2; T = I0( g)/(4r);\n\nT = (8/4) (4{\'4} (θ˙2)/(m r))\n\nT = 1.096 lb. ft. s²\n\nPROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this material may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or beyond the limited distribution to teachers and instructors permitted by McGraw-Hill for their authorized course preparation. If you are a student using this material, you are using it without permission.\n2216 PROBLEM 19.76\nA connecting rod is supported by a knife edge at Point A; the period of its small oscillations is observed to be 1.03 s. Knowing that the distance r_a is 150 mm, determine the centroidal radius of gyration of the connecting rod.\n\nSOLUTION\nPosition (1) Displacement is maximum.\nT_1 = 0, V_1 = mgr(1 - cos \\theta_1) = \\frac{1}{2} mg r_g \\dot{\\theta}^2\n\nPosition (2) Velocity is maximum.\n(v_c)_{h} = r\\dot{\\theta}\nT_2 = \\frac{1}{2} m l^2 \\dot{\\theta}^2 = \\frac{1}{2} m r_g^2 \\dot{\\theta}^2\nV_c = 0\nFor simple harmonic motion,\nConservation of energy:\nT_1 + V_1 = T_2 + V_2\n0 + \\frac{1}{2} mg r_g^2 = \\frac{1}{2} m (l^2 + r_g^2) \\dot{\\theta}^2 + 0\n\nData:\nr_a = 1.03 s\\, \\omega_a = \\frac{2 \\pi}{T} = \\frac{2 \\pi}{1.03} = 6.1002 rad/s\nr_a = 150 mm = 0.15 m \\Rightarrow g = 9.81 m/s^2\nr_c^2 = (9.81)(0.15)(0.15) = 0.039543 = 0.017043 m^2\\, (6.1002)^2 = 37.252\n\\k = 0.13055 m\n\\k = 130.6 mm ▲\n\nPROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be deployed, reproduced or distributed in any form by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this manual, you are using it without permission.\n2277