Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual

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Solutions Manual to AccompanyMECHANICAL VIBRATIONS: THEORY AND APPLICATIONS,1STEDITIONS. GRAHAM KELLY

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ContentsPrefaceviiChapter 11Chapter 255Chapter 3134Chapter 4220Chapter 5369Chapter 6424Chapter 7526Chapter 8660Chapter 9727Chapter 10765Chapter 11853Chapter 12921Chapter 13993

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1CHAPTER 1: INTRODUCTIONShort Answer Problems1.1True: The earth is taken to be non-accelerating for purposes of modeling systems onthe surface of the earth.1.2False: Systems undergoing mechanical vibrations are not subject to nuclear reactions isan example of an implicit assumption.1.3True: Basic laws of nature can only be observed and postulated.1.4False: The point of application of surface forces is on the surface of the body.1.5False: The number of degree of freedom necessary to model a mechanical system isunique.1.6False: Distributed parameter systems are another name for continuous systems.1.7True: The Buckingham Pi theorem states that the number of dimensionless variablesrequired in the formulation of a dimensional relationship is the number of dimensionalvariables, including the dependent variable, minus the number of dimensions involved inthe dimensional variables.1.8True: The displacement of its mass center (x and y coordinates) and the rotation aboutan axis perpendicular to the mass center are degrees of freedom the motion of anunconstrained rigid body undergoing planar motion.1.9False: A particle traveling in a circular path has a velocity which is tangent to thecircle.1.10False: The principle of work and energy is derived from Newton’s second law byintegrating the dot product of the law with a differential displacement vector as the particlemoves from one location to another.1.11The continuum assumption treats all matter as a continuous material and implies thatproperties are continuous functions of the coordinates used in modeling the system.1.12An explicit assumption must be stated every time it is used, whereas an implicitassumption is taken for granted.1.13Constitutive equations are used to model the stress-strain relationships in materials.They are used in vibrations to model the force-displacement relationships in materials thatbehave as a spring.1.14A FBD is a diagram of a body abstracted from its surroundings and showing theeffects of the surroundings as forces. They are drawn at an arbitrary time.

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Chapter 1: Introduction21.15The equation represents simple harmonic motion1.16(a) X is the amplitude of motion; (b)߱ is the frequency at which the motion occurs(c)߶is the phase between the motion and a pure sinusoid.1.17The phase angle is positive for simply harmonic motion. Thus the response lags apure sinusoid.1.18A particle has mass that is concentrated at a point. A rigid body has a distribution ofmass about the mass center.1.19A rigid body undergoes planar motion if (1) the path of its mass center lies in a planeand (2) rotation occurs only about an axis perpendicular to the plane of motion of the masscenter.1.20The acceleration of a particle traveling in a circular path has a tangential componentthat is the radius of the circle times the angular acceleration of the particle and a centripetalacceleration which is directed toward the center of the circle which is the radius time thesquare of the angular velocity.1.21An observer fixed at A observes, instantaneously that particle B is moving in acircular path of radiusܚห஻/஺หabout A.1.22It is applied to the FBD of the particle.1.23The effective forces for a rigid body undergoing planar motion are a force applied atthe mass center equal to݉ ത ܉and a moment equal toߙҧܫ.1.24The two terms of the kinetic energy of a rigid body undergoing planar motion areଵ݉ଶݒҧଶ, the translational kinetic energy, andଵଶ߱ҧܫଶ, the rotational kinetic energy.1.25The principle of impulse and momentum states that a body’s momentum (linear orangular) momentum atݐଵplus the external impulses applied to the body (linear or angular)betweenݐଵandݐଶis equal to the system’s momentum (linear or angular) atݐଶ.1.26One, letߠbe the angular rotation of the bar, measured positive counterclockwise,from the system’s equilibrium position.1.27Four, letݔଵbe the absolute displacement of the cart,ݔଶthe displacement of theleftmost block relative to the cart,ݔଷthe displacement of the rightmost block away fromthe cart andߠthe counterclockwise angular rotation of the bar.1.28Four, letݔଵrepresent the displacement of the center of the disk to the right,ݔଶthedownward displacement of the hanging mass,ݔଷthe displacement of the sliding mass tothe left andߠthe counterclockwise angular rotation of the rightmost pulley.

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Chapter 1: Introduction31.29Two, letߠbe clockwise the angular displacement of the bar and x the downwarddisplacement of the hanging mass.1.30Three, let x be the downward displacement of the middle of the upper bar,ߠitsclockwise angular rotation and߶the clockwise angular rotation of the lower bar.1.31Three, letߠrepresent the clockwise angular rotation of the leftmost disk,߶theclockwise angular rotation of the rightmost disk and x the upward displacement of theleftmost hanging mass.1.32Infinite, let x be a coordinate measured along the neutral axis of the beam measuredfor the fixed support. Then the displacement is a continuous function of x and t, w(x,t).1.33Three,letݔଵbethedownwarddisplacementofthehand,ݔଶthedownwarddisplacement of the palm andݔଷthe displacement of the fingers.1.34Given: Uniform acceleration, a=2 m/s. (a)ݒ ൅ ݐ ܽൌ ሻݐሺݒ଴ ฺݒሺ5ሻ ൌ ቀ2௠௦మቁ ሺ5 sሻ ൅ቀ0୫ୱቁ ൌ 10 m(b)ܽൌ ሻݐሺݔ௧మଶݒ ൅଴ݔ ൅ ݐ଴ ฺݔሺ5ሻ ൌଵଶቀ2୫ୱమቁ ሺ5 sሻଶൌ 25 m1.35Given: ܞൌ 2 cos 2 ܑݐ൅ 3 sin 2 ܒ ݐ൅ 0.4m/s. (a)ൌ ܉ܞௗௗ௧ൌ െ4 sin 2ݐ ܑ൅6 cos 2 ܒ ݐm/sଶฺܽ ሺߨሻ ൌ െ4 sin 2ߨ ܑ൅ 6 cos 2ߨ ܒ୫ୱమൌ 6 ܒm/sଶ(b)ൌ ݐ݀ܞ ׬ ൌ ܚቂሺsin 2 ݐ൅ ܥଵሻ ܑ൅ ቀെଷଶcos 2 ݐ൅ ܥଶቁ ܒ൅ ሺ0.4 ݐ൅ ܥଷቃܓሻm. The particle starts at the originat t = 0. Application of this condition leads to)ܚሺtሻ ൌ ቂሺsin 2 ݐሻ ܑ൅ ቀെଷଶcos 2ݐ ൅ଷଶቁ ܒ ൅ܓݐ0.4m. Evaluation atߨleads toሺܚπሻൌsin൅ ܑ ߨ2−32cosܒ൅32ߨ2ܓߨ൅0.4m =ܓߨ0.4m.1.36Given: v=2 m/s, r=3 m,ߠሺ0ሻ ൌ 0(a)ൌ ݒௗ௦ௗ௧ฺൌ ݏݐ2 ൌ ݐ݀ݒ ׬at t=2 s the particlehas traveled 4 m. Butߠݎ ൌ ݏthusൌ ߠସ ୫ଷ ୫ൌ 1.33 rad ൌ 76.2°. (b) The acceleration of aparticle traveling on a circular path has two components. One isௗ௩ௗ௧which is tangent to thecircle and is zero for this problem. The other component is௩మ௥ൌሺଶ ୫/ୱሻమଷ ୫ൌ 1.33 mdirectedtoward the center of the circle from the position of the particle.1.37Given:m=2kg,ܫҧ ൌ 0.5 kg · mଶ܉ ,ത ൌ ሺ5 ܑ൅ 3 ܒሻ m/sଶ, ߙൌ 10 rad/sଶ.Effectiveforces are݉ ܉ത ൌ ሺ2 kgሻ ቂሺ5 ܑ൅ 3ܒ ሻ୫ୱమቃ ൌ ሺ10 ܑ൅ 15ܒ ሻ୫ୱమapplied at the mass center and acoupleܫҧ ߙൌ ሺ0.5 kg · mଶሻሺ10 rad/sଶሻ ൌ 5 N · m.1.38Given: m = 0.1 kg, ܞൌ ሺ9 ܑ൅ 11j ሻ m/s. The kinetic energy of the particle isܶ ൌଵ݉ଶ|ܞ|ଶൌଵଶሺ0.1 kgሻ൫√9ଶ൅ 11݉ ଶ൯ݏ/ଶൌ 0.711 J.

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Chapter 1: Introduction41.39Given: m=3 kg,ܞത ൌ ሺ3 ܑ൅ 4 ܒሻ m/s, d=0.2 m The angular velocity is calculated fromܞ|ฺ߱߱݀ൌ |തൌܞ|ത|ௗ=ହ ୫/ୱ଴.ଶ ୫=20 rad/s.1.40Given:ൌ 100 J, ܫൌ 0.03 kg · mଶThe kinetic energy of a rigid body which rotatesabout its centroidal axis isܶ ൌଵଶ߱ܫଶ. Thus100 J ൌଵଶሺ0.03 kg · mଶ߱ሻଶwhich leads to߱ ൌ 81.65୰ୟୢୱୣୡ.1.41Given: m = 5 kg,ݒҧ ൌ 4 m/s,߱ ൌ 20 rad/s,ܫҧ ൌ 0.08 kg · mଶ. The kinetic energy ofarigidbodyundergoingplanarmotionisܶ ൌଵ݉ଶݒҧଶ൅ଵଶ߱ҧܫଶൌଵଶሺ5 kgሻሺ4 m/sሻଶ൅ଵଶሺ0.08 kg · mଶሻሺ20 rad/sሻଶൌ 56 J.1.42Given: F=12,000 N,∆ ݐൌ 0.03 s. The impulse applied to the system isൌ ݐ∆ܨൌ ܫሺ12,000 Nሻሺ0.03 sሻ ൌ 360 N · s.1.43Given: m = 3 kg,ݒଵൌ 0 m/s, force as given in Figure(a) The impulse imparted totheparticleis׬ ൌ ܫݐ݀ܨଷ ୱ଴ൌଵଶሺ1ሻሺ100ሻ ൅ 2ሺ100ሻ ൅ଵଶሺ1ሻሺ100ሻ ൌ 300 N · s(b)Thevelocity at t=2 s is given by the principle of impulse and momentum݉ ൌ ݒ׬ݐ݀ܨଶ ୱ଴ฺൌ ݒ׬ிௗ௧మ ౩బ௠ൌଶହ଴ N·ୱଷ ୩୥ൌ 83.3 m/s. (c) The velocity after 5 s isൌ ݒ׬ிௗ௧ఱ ౩బ௠ൌଷ଴଴ N·ୱଷ ୩୥ൌ100 m/s.1.44Given: m = 2 kg, F=6 N, t=10 s,ݒଵൌ 4 m/s. The principle of work and energy isused to calculate how far the particle travelsܶ ଵܷ൅ଵ՜ଶܶൌଶafter the velocityݒଶiscalculatedfromtheprincipleofimpulseandmomentum݉ ݒଵݒ ݉ൌ ܫ ൅ଶฺݒଶൌ௠௩భାூ௠ൌሺଶ ୩୥ሻሺସ ୫/ୱሻାሺ଺ Nሻሺଵ଴ ୱሻଶ ୩୥ൌ 34 m/s.Thenlettingxbethedistancetraveledapplicationofworkandenergygivesଵଶሺ2 kgሻሺ4 m/sሻଶ൅ ሺ6 Nሻݔ ൌଵଶሺ2 kgሻሺ34 m/s2which is solved to yield x=190 m.1.45(a) -(ii) (b)-(iv) (c)-(i) (d)-(v) (e)-(i) (f)-(v) (g)-(vi) (h)-(iii) (i)-(ix)

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Chapter 1: Introduction5Chapter Problems1.1The one-dimensional displacement of a particle isݔሺݐሻ ൌ 0.5݁ି଴.ଶ௧sin 5ݐ m(a) What is the maximum displacement of the particle? (b) What is the maximum velocityof the particle? (c) What is the maximum acceleration of the particle?Given: x(t)Find:ݔሻ ܽሺ௠௔௫ݒሻ ܾሺ௠௔௫ܽሻ ܿሺ௠௔௫Solution: (a) The maximum displacement occurs when the velocity is zero. Thusݔሶሺݐሻ ൌ 0.5݁ି଴.ଶ௧ሺെ0.2 sin 5 ݐ൅ 5 cos 5ݐሻSetting the velocity to zero leads toെ0.2 sin 5 ݐ൅ 5 cos 5 ݐൌ 0ortan 5 ݐൌ 25. The first time that the solution is zero is t=0.3062. Substituting this valueof t into the expression for x(t) leads toݔ௠௔௫ൌ 0.4699 m(b) The maximum velocity occurs when the acceleration is zeroݔሷሺݐሻ ൌ 0.5݁ି଴.ଶ௧ሾെ0.2ሺെ0.2 sin 5 ݐ൅ 5 cos 5ݐሻ െ cos 5 ݐെ 25 sin 5ݐሿൌ 0.5݁ି଴.ଶ௧ሺെ24.96 sin 5 ݐെ 6 cos 5ݐሻThe acceleration is zero when24.96 sin 5 ݐെ 6 cos 5 ݐൌ 0ฺ tan 5 ݐൌ െ0.240.The first time that this is zero is t=0.5812 which leads to a velocity ofݒ௠௜௡ൌ െ2.185 m/s(c) The maximum acceleration occurs whenݔഺ ൌ 0,ݔഺ ൌ 0.5݁ି଴.ଶ௧ሾെ0.2ሺെ24.96 sin 5 ݐെ 6 cos 5ݐሻ െ ሺ24.96ሻሺ5ሻ cos 5 ݐ൅ 30 sin 5ݐሿൌ 0.5݁ି଴.ଶ௧ሺ34.992 sin 5 ݐെ 123.6 cos 5ݐሻThemaximumaccelerationoccurswhen34.992 sin 5 ݐെ 123.6 cos 5 ݐൌ 0ฺtan 5 ݐൌ 3.53. The time at which the maximum acceleration occurs is t=0.2589 swhich leads toܽ௠௔௫ൌ െ12.18 m/sଶProblem 1.1 illustrates the relationships between displacement, velocity and acceleration.

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Chapter 1: Introduction61.2The one-dimensional displacement of a particle ismsin.00.24)+(5te0.5=x(t)2t-(1)(a) What is the maximum displacement of the particle? (b) What is the maximum velocity ofthe particle? (c) What is the maximum acceleration of the particle?Given: x(t)Find:ݔሻ ܽሺ௠௔௫ݒሻ ܾሺ௠௔௫ܽሻ ܿሺ௠௔௫Solution: (a) The maximum displacement occurs when the velocity is zero. Thusݔሶሺݐሻ ൌ 0.5݁ି଴.ଶ௧ሾെ0.2 sinሺ5 ݐ൅ 0.24ሻ ൅ 5 cosሺ5 ݐ൅ 0.24ሻሿSetting the velocity to zero leads toെ0.2 sinሺ5 ݐ൅ 0.24ሻ ൅ 5 cosሺ5 ݐ൅ 0.24ሻ ൌ 0ortanሺ5 ݐ൅ 0.24ሻ ൌ 0.2582.Thefirsttimethatthesolutioniszeroist=0.3062.Substituting this value of t into the expression for x(t) leads toݔ௠௔௫ൌ 0.4745 m(b) The maximum velocity occurs when the acceleration is zeroݔሷሺݐሻ ൌ 0.5݁ି଴.ଶ௧ሼെ0.2ሾሺെ0.2 sinሺ5 ݐ൅ 0.24ሻ ൅ 5 cosሺ5 ݐ൅ 0.24ሻሻሿെ cosሺ5 ݐ൅ 0.24ሻ െ 25 sinሺ5 ݐ൅ 0.24ሻሽൌ 0.5݁ି଴.ଶ௧ሾെ24.96 sinሺ5 ݐ൅ 0.24ሻ െ 6 cosሺ5 ݐ൅ 0.24ሻሿThe acceleration is zero whenെ24.96 sinሺ5 ݐ൅ 0.24ሻ െ 6 cosሺ5 ݐ൅ 0.24ሻ ൌ 0ฺ tanሺ5 ݐ൅ 0.24ሻ ൌ െ0.240.The first time that this is zero is t = 0.5332 which leads to a velocity ofݒ௠௜௡ൌ െ2.0188 m/s(c) The maximum acceleration occurs whenݔഺ ൌ 0,ݔഺ ൌ 0.5݁ି଴.ଶ௧ሼെ0.2ሾെ24.96 sinሺ5 ݐ൅ 0.24ሻ െ 6 cosሺ5 ݐ൅ 0.24ሻሿെ ሺ24.96ሻሺ5ሻ cosሺ5 ݐ൅ 0.24ሻ ൅ 30 sinሺ5 ݐ൅ 0.24ሻሽൌ 0.5݁ି଴.ଶ௧ሾ34.992 sinሺ5 ݐ൅ 0.24ሻ െ 123.6 cosሺ5 ݐ൅ 0.24ሻሿThe maximum acceleration occurs when34.992 sinሺ5 ݐ൅ 0.24ሻ െ 123.6 cosሺ5 ݐ൅ 0.24ሻ ൌ 0ฺ tanሺ5 ݐ൅ 0.24ሻ ൌ 3.53.The time at which the maximum acceleration occurs is t=0.2109 s which leads toܽ௠௔௫ൌ െ12.30 m/sଶProblem 1.2 illustrates the relationships between displacement, velocity and acceleration.

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Chapter 1: Introduction742m/sec75m/sec221.3At the instant shown in Figure P1.3, the slenderrod has a clockwise angular velocity of 5 rad/sec and acounterclockwise angular acceleration of 14 rad/sec2.At the instant shown, determine (a) the velocity ofpointPand (b) the acceleration of pointP.Given:ω= 5 rad/sec,α= 14 rad/sec2,θ= 10°Find:ݒ௣,aPSolution: The particle at the pin support, call it O, is fixed. Hence its velocity and accelerationare zero. Using the relative velocity and acceleration equations between two particles on arigid bodyܞ௉ܞ ൌܗ൅ ૑ ൈ ܚ௉/ைൌ െ5 ܓൈ ሺ3 cos 10°ܑെ 3 sin 10°ܒሻ ൌ െ15 sin 10°ܑെ 15 cos 10°ܒൌ െ2.604 ܑെ 14.772 ܒandOPOPOP+)(+=//αxrωxrxωaasm9.85sm22=PP)54.3+(-66.5=ajiaAlternate solution: The bar is rotating about a fixed point. Thus any point on the bar moves ona circular arc about the point of support. The particle P has two components of acceleration,one directed between P and O (the normal acceleration), and one tangent to the path of Pwhose direction is determined using the right hand rule (the tangential component).The component normal to the path of P issmsrad5m275=)(3=a2nand is directed between P and O. The tangential acceleration issmsradm2242=))(14(3=atThe normal and tangential components of acceleration are illustrated on the diagram below.

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Chapter 1: Introduction8Problem 1.3 illustrates the use of the relative acceleration equation of rigid body kinetics.1.4At t = 0, a particle of mass 1.2 kg is traveling with a speed of 10 m/s that is increasingat a rate of 0.5 m/s2. The local radius of curvature at this instant is 50 m. After the particletravels 100 m, the radius of curvature of the particle's path is 50 m.(a) What is the speed of the particle after it travels 100 m?(b) What is the magnitude of the particle’s acceleration after it travels 100 m?(c) How long does it take the particle to travel 100 m?(d) What is the external force acting on the particle after it travels 100 m?Given: m = 1.2 kg, v(t=0) = 10 m/s, dv/dt= 0.5 m/s2, and r = 25 m when s = 100 mFind: (a) v when s = 100 m, (b) a when s = 3 m, (c) t when s = 3 mSolution: Let s(t) be the displacement of the particle, measured from t = 0. The particle’svelocity is∫∫+==+=tttdtvdtdtdvtv00105.05.0)0()(By definition v=ds/dt. Thus the displacement of the particle is obtained as()∫∫+=+=+=ttttdttsdtvts0021025.0105.0)0()(When s = 100 m,s28.81025.0m1002=⇒+=ttt(a)The velocity when s = 100 m ism/s14.1410)28.8(5.0=+=v(b)Since the particle is traveling along a curved path, its acceleration has two components:a tangential component equal to the rate of change of the velocity2m/s5.0==dtdvatand a normal component directed toward the center of curvature222m/s00.4m50)m/s14.14(===rvanThe magnitude of the acceleration at this instant is

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Chapter 1: Introduction92222222m/s03.4a)m/s00.4()m/s5.0(a=+=+=ntaa(c)The time for the particle to travel 100 m is previously calculated as t = 8.28 s(d)The external force equation written in terms of magnitudes is෍|۴| ൌ݉| ܽ|which upon application to the particle gives෍|۴| ൌ ሺ1.2 kgሻ ቀ4.03 msଶቁ ൌ 4.84 NProblem 1.4 illustrates the kinematics of a particle traveling along a curved path.1.5The machine of Figure P1.15 has a vertical displacement,x(t). The machine has component which rotates with a constantangular speed,ω. The center of mass of the rotating componentis a distanceefrom its axis of rotation. The center of mass of therotating component is as shown att= 0. Determine the verticalcomponent of the acceleration of the rotating component.Given: e,ω, x (t)Find: aySolution: The particle of interest is on a component that movesrelative to the machine. From the relative acceleration equation,MGMGaaa+=where()jatxM&&−=and()jiaθθωsincos2−−=eMGSince the angular velocity of the rotating component is constant andθ= 0 when t = 0,tωθ=Hence the vertical acceleration of the center of mass of the rotating component is

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Chapter 1: Introduction10()tetxa2yωωsin−−=&&Problem 1.5 illustrates application of the relative acceleration equation. Vibrations ofmachines subject to a rotating unbalance are considered in Chapter 4.1.6The rotor of Figure P1.6 consists of a disk mountedon a shaft. Unfortunately, the disk is unbalanced, andthe center of mass is a distanceefrom the center of theshaft. As the disk rotates, this causes a phenomenacalled “whirl”, where the disk bows. Letrbe theinstantaneous distance from the center of the shaft tothe original axis of the shaft andߠbe the angle madeby a given radius with the horizontal. Determine theacceleration of the mass center of the disk.Given: e, rFind:܉തSolution: The position vector from the origin to the center of the disk isܑݎ௥where r varieswith time. The mass center moves in a circular path about the center of the disk. Therelative acceleration equation gives܉௖܉ ൌ௢൅ હ ൈ ܑݎ௥൅ ૑ ൈ ሺ૑ ൈ ܑݎ௥ሻ ൅ ݎሷܑ௥൅ 2૑ ൈ ݎሶܑ௥܉௖ൌ ൫ݎሷ െ ߠݎሶଶܑ൯௥ܑሶ൯ߠሶݎ2 ൅ ሷߠݎ൫ ൅ఏThe acceleration of the mass center is then܉ത ൌ ൫ݎሷ െ ߠݎሶଶܑ൯௥ܑሶ൯ߠሶݎ2 ൅ ሷߠݎ൫ ൅ఏ߱݁െଶሾcosሺ ݐ߱െ ߠܑሻ௥൅ sinሺ ݐ߱െ ߠܑሻఏሿProblem 1.6 illustrates application of the relative acceleration equation.1.7A 2 ton truck is traveling down anicy, 10º hill at 50 mph when the driversees a car stalled at the bottom of thehill 250 ft away. As soon as he sees thestalled car, the driver applies his brakes,but due to the icy conditions, a brakingforce of only 2000 N is generated. Doesthe truck stop before hitting the car?Given: W = 4000 lb.,θ= 10o, d = 250 ft., Fb= 2000 N = 449.6 lb,vo= 50 mph = 73.33 ft/sec10º250 ft

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Chapter 1: Introduction11N=WWgaFbEXTERNAL FORCESEFFECTIVE FORCESFind: v = 0 before x = 250 ft.Solution: Application of Newton’s Law to the free body diagram of the truck at anarbitrary instant()()202..secft973.110sinlb4000lb449.6secft2.32sinsin=⎟⎟⎠⎞⎜⎜⎝⎛+−=⎟⎠⎞⎜⎝⎛+−==+−=∑∑aaWFgaagWWFFFbbeffxextxθθSince the acceleration is constant, the velocity and displacement of the truck aretttvtaxtvatv33.73986.0233.73973.12020+=+=+=+=The acceleration is positive thus the vehicle speeds up as it travels down the incline. Thetruck does not stop before hitting the car.Problem 1.7 illustrates application of Newton’s Law to a particle and kinematics ofconstant acceleration.1.8The contour of a bumpy road is approximated byy(x) = 0.03 sin(0.125x) m. What is theamplitude of the vertical acceleration of the wheels of an automobile as it travels over thisroad at a constant horizontal speed of 40 m/s?

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Chapter 1: Introduction12Given:y(x) = 0.03sin(0.125x) m,v= 40 m/sFind: ASolution: Since the vehicle is traveling at a constant horizontal speed its horizontal distancetraveled in a time t isx=vt. Thus the vertical displacement of the vehicle is[]m)5sin(03.0)40(125.0sin03.0)(ttty==The vertical velocity and acceleration of the vehicle are calculated as2m/s)5sin(75.0)5sin()5(15.0)(m/s)5cos(15.0)5cos()5(03.0)(tttatttv−=−===Thus the amplitude of acceleration is A=0.75 m/s2.Problem 1.8 illustrates the relationship between displacement, velocity, and accelerationfor the motion of a particle.1.9The helicopter of Figure P1.9 has a horizontal speed of 110 ft/s and a horizontalacceleration of 3.1 ft/s2. The main blades rotate at a constant speed of 135 rpm. At theinstant shown, determine the velocity and acceleration of particleA.Given: vh= 110 ft/s, ah=3ft/s2,ω= 135 rpm = 14.1 rad/s, r = 2.1 ftFind:vA,aASolution: Construct a x-y coordinate system in the horizontal planeAs illustrated. Using this coordinate system2ft/s3ft/s,110iaiv−=−=The position vector of A relative to the helicopter at this instant is[]jijirhA48.148.1)4/sin()4/cos(/−=−=ππrThe relative velocity equation is used to determine the velocity of particle A asft/s9.201.89)48.148.1(1.14110/jivjikivrkvvAAhAhA+−=−×+−=×+=ωThe relative acceleration equation is used to determine the acceleration of particle A as

Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual - Page 16 preview image

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Chapter 1: Introduction132//ft/s6.2944.297)87.2087.20(1.141.3)(jiajikiarkkrkaaAAhAhAhA+−=+×+−=××+×+=ωωαProblem 1.9 illustrates the use of the relative velocity and relative acceleration equations.1.10For the system shown in Figure P1.10,the angular displacement of the thin disk isgiven byߠሺݐሻ ൌ 0.03 sin ቀ30ݐ ൅గସቁ rad. Thediskrollswithoutslippingonthesurface.Determine the following as functions of time.(a) The acceleration of the center of the disk.(b) The acceleration of the point of contactbetween the disk and the surface. (c) Theangularaccelerationofthebar.(d)Theverticaldisplacementoftheblock.(Hint:Assume small angular oscillations߶of thebar. Thensin ߶ൎ ߶.)Given:ሻݐሺߠ,ݎௗൌ 0.1m,ℓൌ 0.3m,݀ ൌ 0.2mFind: (a)ܽ തௗ(b)ܽ ത௖(c)ߙ(d) xSolution: (a) The angular acceleration of the disk isߠሷሺݐሻ ൌ െሺ30ሻଶ0.03 sin ቀ30 ݐ൅ ߨ4ቁ ൌ െ27 sin ቀ30 ݐ൅ ߨ4ቁ radsଶSince the disk rolls without slip the acceleration of the mass center isܽ ௗൌ ߙݎൌ ሺ0.1 mሻ ൬െ27 sin ቀ30 ݐ൅ ߨ4ቁ radsଶ൰ ൌ െ2.7 sin ቀ30 ݐ൅ ߨ4ቁ msଶ(b) Since the disk rolls without slip the horizontal acceleration of the point of contact iszero. The vertical acceleration isሶߠݎଶtowards the center܉௖ൌ ሺ0.1 mሻ ൤ሺ30ሻሺ0.03ሻ cos ቀ30 ݐ൅ ߨ4ቁ rads ൨ଶܒ

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