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Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual

Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual is your ultimate textbook solutions guide, providing answers to the most difficult questions.

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    Solutions Manual to Accompany MECHANICAL VIBRATIONS: THEORY AND APPLICATIONS, 1 ST EDITION S. GRAHAM KELLY

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    Contents Preface vii Chapter 1 1 Chapter 2 55 Chapter 3 134 Chapter 4 220 Chapter 5 369 Chapter 6 424 Chapter 7 526 Chapter 8 660 Chapter 9 727 Chapter 10 765 Chapter 11 853 Chapter 12 921 Chapter 13 993

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    1 CHAPTER 1: INTRODUCTION Short Answer Problems 1.1 True: The earth is taken to be non-accelerating for purposes of modeling systems on the surface of the earth. 1.2 False: Systems undergoing mechanical vibrations are not subject to nuclear reactions is an example of an implicit assumption. 1.3 True: Basic laws of nature can only be observed and postulated. 1.4 False: The point of application of surface forces is on the surface of the body. 1.5 False: The number of degree of freedom necessary to model a mechanical system is unique. 1.6 False: Distributed parameter systems are another name for continuous systems. 1.7 True: The Buckingham Pi theorem states that the number of dimensionless variables required in the formulation of a dimensional relationship is the number of dimensional variables, including the dependent variable, minus the number of dimensions involved in the dimensional variables. 1.8 True: The displacement of its mass center (x and y coordinates) and the rotation about an axis perpendicular to the mass center are degrees of freedom the motion of an unconstrained rigid body undergoing planar motion. 1.9 False: A particle traveling in a circular path has a velocity which is tangent to the circle. 1.10 False: The principle of work and energy is derived from Newton’s second law by integrating the dot product of the law with a differential displacement vector as the particle moves from one location to another. 1.11 The continuum assumption treats all matter as a continuous material and implies that properties are continuous functions of the coordinates used in modeling the system. 1.12 An explicit assumption must be stated every time it is used, whereas an implicit assumption is taken for granted. 1.13 Constitutive equations are used to model the stress-strain relationships in materials. They are used in vibrations to model the force-displacement relationships in materials that behave as a spring. 1.14 A FBD is a diagram of a body abstracted from its surroundings and showing the effects of the surroundings as forces. They are drawn at an arbitrary time.

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    Chapter 1: Introduction 2 1.15 The equation represents simple harmonic motion 1.16 (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. 1.17 The phase angle is positive for simply harmonic motion. Thus the response lags a pure sinusoid. 1.18 A particle has mass that is concentrated at a point. A rigid body has a distribution of mass about the mass center. 1.19 A rigid body undergoes planar motion if (1) the path of its mass center lies in a plane and (2) rotation occurs only about an axis perpendicular to the plane of motion of the mass center. 1.20 The acceleration of a particle traveling in a circular path has a tangential component that is the radius of the circle times the angular acceleration of the particle and a centripetal acceleration which is directed toward the center of the circle which is the radius time the square of the angular velocity. 1.21 An observer fixed at A observes, instantaneously that particle B is moving in a circular path of radius ܚห ஻/஺ ห about A. 1.22 It is applied to the FBD of the particle. 1.23 The effective forces for a rigid body undergoing planar motion are a force applied at the mass center equal to ݉ ത ܉ and a moment equal to ߙҧܫ . 1.24 The two terms of the kinetic energy of a rigid body undergoing planar motion are ଵ ݉ଶ ݒ ҧ ଶ , the translational kinetic energy, and ଵ ଶ ߱ҧܫ ଶ , the rotational kinetic energy. 1.25 The principle of impulse and momentum states that a body’s momentum (linear or angular) momentum at ݐ ଵ plus the external impulses applied to the body (linear or angular) between ݐ ଵ and ݐ ଶ is equal to the system’s momentum (linear or angular) at ݐ ଶ . 1.26 One, let ߠ be the angular rotation of the bar, measured positive counterclockwise, from the system’s equilibrium position. 1.27 Four, let ݔ ଵ be the absolute displacement of the cart, ݔ ଶ the displacement of the leftmost block relative to the cart, ݔ ଷ the displacement of the rightmost block away from the cart and ߠ the counterclockwise angular rotation of the bar. 1.28 Four, let ݔ ଵ represent the displacement of the center of the disk to the right, ݔ ଶ the downward displacement of the hanging mass, ݔ ଷ the displacement of the sliding mass to the left and ߠ the counterclockwise angular rotation of the rightmost pulley.

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    Chapter 1: Introduction 3 1.29 Two, let ߠ be clockwise the angular displacement of the bar and x the downward displacement of the hanging mass. 1.30 Three, let x be the downward displacement of the middle of the upper bar, ߠ its clockwise angular rotation and ߶ the clockwise angular rotation of the lower bar. 1.31 Three, let ߠ represent the clockwise angular rotation of the leftmost disk, ߶ the clockwise angular rotation of the rightmost disk and x the upward displacement of the leftmost hanging mass. 1.32 Infinite, let x be a coordinate measured along the neutral axis of the beam measured for the fixed support. Then the displacement is a continuous function of x and t, w(x,t). 1.33 Three, let ݔ ଵ be the downward displacement of the hand, ݔ ଶ the downward displacement of the palm and ݔ ଷ the displacement of the fingers. 1.34 Given: Uniform acceleration, a=2 m/s. (a) ݒ ൅ ݐ ܽൌ ሻݐሺݒ ଴ ฺ ݒ ሺ5ሻ ൌ ቀ2 ௠ ௦ మ ቁ ሺ5 sሻ ൅ ቀ0 ୫ ୱ ቁ ൌ 10 m (b) ܽൌ ሻݐሺݔ ௧ మ ଶ ݒ ൅ ଴ ݔ ൅ ݐ ଴ ฺ ݔ ሺ5ሻ ൌ ଵ ଶ ቀ2 ୫ ୱ మ ቁ ሺ5 sሻ ଶ ൌ 25 m 1.35 Given: ܞൌ 2 cos 2 ܑݐ൅ 3 sin 2 ܒ ݐ൅ 0.4 m/s. (a) ൌ ܉ ܞௗ ௗ௧ ൌ െ4 sin 2ݐ ܑ ൅ 6 cos 2 ܒ ݐm/s ଶ ฺܽ ሺߨሻ ൌ െ4 sin 2ߨ ܑ ൅ 6 cos 2ߨ ܒ ୫ ୱ మ ൌ 6 ܒm/s ଶ (b) ൌ ݐ݀ܞ ׬ ൌ ܚ ቂሺsin 2 ݐ൅ ܥ ଵ ሻ ܑ൅ ቀെ ଷ ଶ cos 2 ݐ൅ ܥ ଶ ቁ ܒ൅ ሺ0.4 ݐ൅ ܥ ଷ ቃܓሻ m. The particle starts at the origin at t = 0. Application of this condition leads to) ܚሺtሻ ൌ ቂሺsin 2 ݐሻ ܑ൅ ቀെ ଷ ଶ cos 2ݐ ൅ ଷ ଶ ቁ ܒ ൅ ܓݐ0.4 m. Evaluation at ߨ leads to ሺܚ π ሻ ൌsin ൅ ܑ ߨ2 − 32cos ܒ൅32ߨ2 ܓߨ൅0.4 m = ܓߨ0.4 m. 1.36 Given: v=2 m/s, r=3 m, ߠሺ0ሻ ൌ 0 (a) ൌ ݒ ௗ௦ ௗ௧ฺ ൌ ݏ ݐ2 ൌ ݐ݀ݒ ׬ at t=2 s the particle has traveled 4 m. But ߠݎ ൌ ݏ thus ൌ ߠ ସ ୫ ଷ ୫ ൌ 1.33 rad ൌ 76.2° . (b) The acceleration of a particle traveling on a circular path has two components. One is ௗ௩ ௗ௧ which is tangent to the circle and is zero for this problem. The other component is ௩ మ ௥ ൌ ሺଶ ୫/ୱሻ మ ଷ ୫ ൌ 1.33 m directed toward the center of the circle from the position of the particle. 1.37 Given: m=2 kg, ܫҧ ൌ 0.5 kg · m ଶ ܉ , ത ൌ ሺ5 ܑ൅ 3 ܒሻ m/s ଶ , ߙൌ 10 rad/s ଶ . Effective forces are ݉ ܉ത ൌ ሺ2 kgሻ ቂሺ5 ܑ൅ 3ܒ ሻ ୫ ୱ మ ቃ ൌ ሺ10 ܑ൅ 15ܒ ሻ ୫ ୱ మ applied at the mass center and a couple ܫҧ ߙൌ ሺ0.5 kg · m ଶ ሻሺ10 rad/s ଶ ሻ ൌ 5 N · m . 1.38 Given: m = 0.1 kg, ܞൌ ሺ9 ܑ൅ 11j ሻ m/s . The kinetic energy of the particle is ܶ ൌ ଵ ݉ଶ |ܞ| ଶ ൌ ଵ ଶ ሺ0.1 kgሻ൫√9 ଶ ൅ 11 ݉ ଶ ൯ݏ/ ଶ ൌ 0.711 J .

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    Chapter 1: Introduction 4 1.39 Given: m=3 kg, ܞ ത ൌ ሺ3 ܑ൅ 4 ܒሻ m/s , d=0.2 m The angular velocity is calculated from ܞ| ฺ߱߱݀ൌ |ത ൌ ܞ| ത| ௗ = ହ ୫/ୱ ଴.ଶ ୫ =20 rad/s. 1.40 Given: ൌ 100 J , ܫൌ 0.03 kg · m ଶ The kinetic energy of a rigid body which rotates about its centroidal axis is ܶ ൌ ଵ ଶ ߱ܫ ଶ . Thus 100 J ൌ ଵ ଶ ሺ0.03 kg · m ଶ ߱ሻ ଶ which leads to ߱ ൌ 81.65 ୰ୟୢ ୱୣୡ . 1.41 Given: m = 5 kg, ݒҧ ൌ 4 m/s , ߱ ൌ 20 rad/s , ܫҧ ൌ 0.08 kg · m ଶ . The kinetic energy of a rigid body undergoing planar motion is ܶ ൌ ଵ ݉ଶ ݒ ҧ ଶ ൅ ଵ ଶ ߱ҧܫ ଶ ൌ ଵ ଶ ሺ5 kgሻሺ4 m/sሻ ଶ ൅ ଵ ଶ ሺ0.08 kg · m ଶ ሻሺ20 rad/sሻ ଶ ൌ 56 J . 1.42 Given: F=12,000 N, ∆ ݐൌ 0.03 s . The impulse applied to the system is ൌ ݐ∆ܨൌ ܫ ሺ12,000 Nሻሺ0.03 sሻ ൌ 360 N · s . 1.43 Given: m = 3 kg, ݒ ଵ ൌ 0 m/s , force as given in Figure (a) The impulse imparted to the particle is ׬ ൌ ܫ ݐ݀ܨ ଷ ୱ ଴ ൌ ଵ ଶ ሺ1ሻሺ100ሻ ൅ 2ሺ100ሻ ൅ ଵ ଶ ሺ1ሻሺ100ሻ ൌ 300 N · s (b) The velocity at t=2 s is given by the principle of impulse and momentum ݉ ൌ ݒ ׬ ݐ݀ܨ ଶ ୱ ଴ฺ ൌ ݒ ׬ ிௗ௧ మ ౩ బ ௠ ൌ ଶହ଴ N·ୱ ଷ ୩୥ ൌ 83.3 m/s . (c) The velocity after 5 s is ൌ ݒ ׬ ிௗ௧ ఱ ౩ బ ௠ ൌ ଷ଴଴ N·ୱ ଷ ୩୥ ൌ 100 m/s . 1.44 Given: m = 2 kg, F=6 N, t=10 s, ݒ ଵ ൌ 4 m/s . The principle of work and energy is used to calculate how far the particle travels ܶ ଵ ܷ൅ ଵ՜ଶ ܶൌ ଶ after the velocity ݒ ଶ is calculated from the principle of impulse and momentum ݉ ݒ ଵ ݒ ݉ൌ ܫ ൅ ଶฺ ݒ ଶ ൌ ௠௩ భ ାூ ௠ ൌ ሺଶ ୩୥ሻሺସ ୫/ୱሻାሺ଺ Nሻሺଵ଴ ୱሻ ଶ ୩୥ ൌ 34 m/s . Then letting x be the distance traveled application of work and energy gives ଵ ଶ ሺ2 kgሻሺ4 m/sሻ ଶ ൅ ሺ6 Nሻݔ ൌ ଵ ଶ ሺ2 kgሻሺ34 m/ s2 which 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: Introduction 5 Chapter Problems 1.1 The one-dimensional displacement of a particle is ݔሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ sin 5ݐ m (a) What is the maximum displacement of the particle? (b) What is the maximum velocity of the particle? (c) What is the maximum acceleration of the particle? Given: x(t) Find: ݔሻ ܽሺ ௠௔௫ ݒሻ ܾሺ ௠௔௫ ܽሻ ܿሺ ௠௔௫ Solution: (a) The maximum displacement occurs when the velocity is zero. Thus ݔሶሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ ሺെ0.2 sin 5 ݐ൅ 5 cos 5ݐሻ Setting the velocity to zero leads to െ0.2 sin 5 ݐ൅ 5 cos 5 ݐൌ 0 or tan 5 ݐൌ 25 . The first time that the solution is zero is t=0.3062. Substituting this value of t into the expression for x(t) leads to ݔ ௠௔௫ ൌ 0.4699 m (b) The maximum velocity occurs when the acceleration is zero ݔሷሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ ሾെ0.2ሺെ0.2 sin 5 ݐ൅ 5 cos 5ݐሻ െ cos 5 ݐെ 25 sin 5ݐሿ ൌ 0.5݁ି ଴.ଶ௧ ሺെ24.96 sin 5 ݐെ 6 cos 5ݐሻ The acceleration is zero when 24.96 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 of ݒ ௠௜௡ ൌ െ2.185 m/s (c) The maximum acceleration occurs when ݔ ഺ ൌ 0 , ݔ ഺ ൌ 0.5݁ି ଴.ଶ௧ ሾെ0.2ሺെ24.96 sin 5 ݐെ 6 cos 5ݐሻ െ ሺ24.96ሻሺ5ሻ cos 5 ݐ൅ 30 sin 5ݐሿ ൌ 0.5݁ି ଴.ଶ௧ ሺ34.992 sin 5 ݐെ 123.6 cos 5ݐሻ The maximum acceleration occurs when 34.992 sin 5 ݐെ 123.6 cos 5 ݐൌ 0ฺ tan 5 ݐൌ 3.53 . The time at which the maximum acceleration occurs is t=0.2589 s which leads to ܽ ௠௔௫ ൌ െ12.18 m/s ଶ Problem 1.1 illustrates the relationships between displacement, velocity and acceleration.

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    Chapter 1: Introduction 6 1.2 The one-dimensional displacement of a particle is m sin . 0 0.24) + (5t e 0.5 = x(t) 2t - (1) (a) What is the maximum displacement of the particle? (b) What is the maximum velocity of the particle? (c) What is the maximum acceleration of the particle? Given: x(t) Find: ݔሻ ܽሺ ௠௔௫ ݒሻ ܾሺ ௠௔௫ ܽሻ ܿሺ ௠௔௫ Solution: (a) The maximum displacement occurs when the velocity is zero. Thus ݔሶሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ ሾെ0.2 sinሺ5 ݐ൅ 0.24ሻ ൅ 5 cosሺ5 ݐ൅ 0.24ሻሿ Setting the velocity to zero leads to െ0.2 sinሺ5 ݐ൅ 0.24ሻ ൅ 5 cosሺ5 ݐ൅ 0.24ሻ ൌ 0 or tanሺ5 ݐ൅ 0.24ሻ ൌ 0.2582 . The first time that the solution is zero is t=0.3062. Substituting this value of t into the expression for x(t) leads to ݔ ௠௔௫ ൌ 0.4745 m (b) The maximum velocity occurs when the acceleration is zero ݔሷሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ ሼെ0.2ሾሺെ0.2 sinሺ5 ݐ൅ 0.24ሻ ൅ 5 cosሺ5 ݐ൅ 0.24ሻሻሿ െ cosሺ5 ݐ൅ 0.24ሻ െ 25 sinሺ5 ݐ൅ 0.24ሻሽ ൌ 0.5݁ି ଴.ଶ௧ ሾെ24.96 sinሺ5 ݐ൅ 0.24ሻ െ 6 cosሺ5 ݐ൅ 0.24ሻሿ The acceleration is zero when െ24.96 sinሺ5 ݐ൅ 0.24ሻ െ 6 cosሺ5 ݐ൅ 0.24ሻ ൌ 0ฺ tan ሺ5 ݐ൅ 0.24ሻ ൌ െ0.240 . The first time that this is zero is t = 0.5332 which leads to a velocity of ݒ ௠௜௡ ൌ െ2.0188 m/s (c) The maximum acceleration occurs when ݔ ഺ ൌ 0 , ݔ ഺ ൌ 0.5݁ି ଴.ଶ௧ ሼെ0.2ሾെ24.96 sinሺ5 ݐ൅ 0.24ሻ െ 6 cosሺ5 ݐ൅ 0.24ሻሿ െ ሺ24.96ሻሺ5ሻ cosሺ5 ݐ൅ 0.24ሻ ൅ 30 sinሺ5 ݐ൅ 0.24ሻሽ ൌ 0.5݁ି ଴.ଶ௧ ሾ34.992 sinሺ5 ݐ൅ 0.24ሻ െ 123.6 cosሺ5 ݐ൅ 0.24ሻሿ The maximum acceleration occurs when 34.992 sinሺ5 ݐ൅ 0.24ሻ െ 123.6 cosሺ5 ݐ൅ 0.24ሻ ൌ 0ฺ tan ሺ5 ݐ൅ 0.24ሻ ൌ 3.53 . The time at which the maximum acceleration occurs is t=0.2109 s which leads to ܽ ௠௔௫ ൌ െ12.30 m/s ଶ Problem 1.2 illustrates the relationships between displacement, velocity and acceleration.

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    Chapter 1: Introduction 7 42m/sec 75m/sec 2 2 1.3 At the instant shown in Figure P1.3, the slender rod has a clockwise angular velocity of 5 rad/sec and a counterclockwise angular acceleration of 14 rad/sec 2 . At the instant shown, determine (a) the velocity of point P and (b) the acceleration of point P . Given: ω = 5 rad/sec, α = 14 rad/sec 2 , θ = 10 ° Find: ݒ ௣ , a P Solution: The particle at the pin support, call it O, is fixed. Hence its velocity and acceleration are zero. Using the relative velocity and acceleration equations between two particles on a rigid body ܞ ௉ ܞ ൌ ܗ ൅ ૑ ൈ ܚ ௉/ை ൌ െ5 ܓൈ ሺ3 cos 10 ° ܑ െ 3 sin 10 ° ܒሻ ൌ െ15 sin 10 ° ܑ െ 15 cos 10 ° ܒ ൌ െ2.604 ܑെ 14.772 ܒ and O P O P O P + ) ( + = / / α xr ω xr x ω a a s m 9 . 85 s m 2 2 = P P ) 54.3 + (-66.5 = a j i a Alternate solution: The bar is rotating about a fixed point. Thus any point on the bar moves on a 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 P whose direction is determined using the right hand rule (the tangential component). The component normal to the path of P is s m s rad 5 m 2 75 = ) ( 3 = a 2 n and is directed between P and O. The tangential acceleration is s m s rad m 2 2 42 = ) )(14 (3 = a t The normal and tangential components of acceleration are illustrated on the diagram below.

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    Chapter 1: Introduction 8 Problem 1.3 illustrates the use of the relative acceleration equation of rigid body kinetics. 1.4 At t = 0, a particle of mass 1.2 kg is traveling with a speed of 10 m/s that is increasing at a rate of 0.5 m/s 2 . The local radius of curvature at this instant is 50 m. After the particle travels 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/s 2 , and r = 25 m when s = 100 m Find: (a) v when s = 100 m, (b) a when s = 3 m, (c) t when s = 3 m Solution: Let s(t) be the displacement of the particle, measured from t = 0. The particle’s velocity is ∫ ∫ + = = + = t t t dt v dt dt dv t v 0 0 10 5 . 0 5 . 0 ) 0 ( ) ( By definition v=ds/dt. Thus the displacement of the particle is obtained as ( ) ∫ ∫ + = + = + = t t t t dt t s dt v t s 0 0 2 10 25 . 0 10 5 . 0 ) 0 ( ) ( When s = 100 m, s 28 . 8 10 25 . 0 m 100 2 = ⇒ + = t t t (a) The velocity when s = 100 m is m/s 14 . 14 10 ) 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 velocity 2 m/s 5 . 0 = = dt dv a t and a normal component directed toward the center of curvature 2 2 2 m/s 00 . 4 m 50 ) m/s 14 . 14 ( = = = r v a n The magnitude of the acceleration at this instant is

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    Chapter 1: Introduction 9 2 2 2 2 2 2 2 m/s 03 . 4 a ) m/s 00 . 4 ( ) m/s 5 . 0 ( a = + = + = n t a a (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 m s ଶ ቁ ൌ 4.84 N Problem 1.4 illustrates the kinematics of a particle traveling along a curved path. 1.5 The machine of Figure P1.15 has a vertical displacement, x ( t ). The machine has component which rotates with a constant angular speed, ω . The center of mass of the rotating component is a distance e from its axis of rotation. The center of mass of the rotating component is as shown at t = 0. Determine the vertical component of the acceleration of the rotating component. Given: e, ω , x (t) Find: a y Solution: The particle of interest is on a component that moves relative to the machine. From the relative acceleration equation, M G M G a a a + = where ( ) j a t x M & & − = and ( ) j i a θ θ ω sin cos 2 − − = e M G Since the angular velocity of the rotating component is constant and θ = 0 when t = 0, t ω θ = Hence the vertical acceleration of the center of mass of the rotating component is

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    Chapter 1: Introduction 10 ( ) t e t x a 2 y ω ω sin − − = & & Problem 1.5 illustrates application of the relative acceleration equation. Vibrations of machines subject to a rotating unbalance are considered in Chapter 4. 1.6 The rotor of Figure P1.6 consists of a disk mounted on a shaft. Unfortunately, the disk is unbalanced, and the center of mass is a distance e from the center of the shaft. As the disk rotates, this causes a phenomena called “whirl”, where the disk bows. Let r be the instantaneous distance from the center of the shaft to the original axis of the shaft and ߠ be the angle made by a given radius with the horizontal. Determine the acceleration of the mass center of the disk. Given: e, r Find: ܉ ത Solution: The position vector from the origin to the center of the disk is ܑݎ ௥ where r varies with time. The mass center moves in a circular path about the center of the disk. The relative acceleration equation gives ܉ ௖ ܉ ൌ ௢ ൅ હ ൈ ܑݎ ௥ ൅ ૑ ൈ ሺ૑ ൈ ܑݎ ௥ ሻ ൅ ݎሷܑ ௥ ൅ 2૑ ൈ ݎሶܑ ௥ ܉ ௖ ൌ ൫ݎሷ െ ߠݎሶ ଶ ܑ൯ ௥ ܑሶ൯ߠሶݎ2 ൅ ሷߠݎ൫ ൅ ఏ The acceleration of the mass center is then ܉ ത ൌ ൫ݎሷ െ ߠݎሶ ଶ ܑ൯ ௥ ܑሶ൯ߠሶݎ2 ൅ ሷߠݎ൫ ൅ ఏ ߱݁െ ଶ ሾcosሺ ݐ߱െ ߠ ܑሻ ௥ ൅ sinሺ ݐ߱െ ߠ ܑሻ ఏ ሿ Problem 1.6 illustrates application of the relative acceleration equation. 1.7 A 2 ton truck is traveling down an icy, 10º hill at 50 mph when the driver sees a car stalled at the bottom of the hill 250 ft away. As soon as he sees the stalled car, the driver applies his brakes, but due to the icy conditions, a braking force of only 2000 N is generated. Does the truck stop before hitting the car? Given: W = 4000 lb., θ = 10 o , d = 250 ft., F b = 2000 N = 449.6 lb, v o = 50 mph = 73.33 ft/sec 10º 250 ft

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    Chapter 1: Introduction 11 N = W W g a F b EXTERNAL FORCES EFFECTIVE FORCES Find: v = 0 before x = 250 ft. Solution: Application of Newton’s Law to the free body diagram of the truck at an arbitrary instant ( ) ( ) 2 0 2 . . sec ft 973 . 1 10 sin lb 4000 lb 449.6 sec ft 2 . 32 sin sin = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = = + − = ∑ ∑ a a W F g a a g W W F F F b b eff x ext x θ θ Since the acceleration is constant, the velocity and displacement of the truck are t t t v t a x t v at v 33 . 73 986 . 0 2 33 . 73 973 . 1 2 0 2 0 + = + = + = + = The acceleration is positive thus the vehicle speeds up as it travels down the incline. The truck does not stop before hitting the car. Problem 1.7 illustrates application of Newton’s Law to a particle and kinematics of constant acceleration. 1.8 The contour of a bumpy road is approximated by y ( x ) = 0.03 sin(0.125 x ) m. What is the amplitude of the vertical acceleration of the wheels of an automobile as it travels over this road at a constant horizontal speed of 40 m/s?

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    Chapter 1: Introduction 12 Given: y ( x ) = 0.03sin(0.125 x ) m, v = 40 m/s Find: A Solution: Since the vehicle is traveling at a constant horizontal speed its horizontal distance traveled in a time t is x = vt . Thus the vertical displacement of the vehicle is [ ] m ) 5 sin( 03 . 0 ) 40 ( 125 . 0 sin 03 . 0 ) ( t t t y = = The vertical velocity and acceleration of the vehicle are calculated as 2 m/s ) 5 sin( 75 . 0 ) 5 sin( ) 5 ( 15 . 0 ) ( m/s ) 5 cos( 15 . 0 ) 5 cos( ) 5 ( 03 . 0 ) ( t t t a t t t v − = − = = = Thus the amplitude of acceleration is A=0.75 m/s 2 . Problem 1.8 illustrates the relationship between displacement, velocity, and acceleration for the motion of a particle. 1.9 The helicopter of Figure P1.9 has a horizontal speed of 110 ft/s and a horizontal acceleration of 3.1 ft/s 2 . The main blades rotate at a constant speed of 135 rpm. At the instant shown, determine the velocity and acceleration of particle A . Given: v h = 110 ft/s, a h =3ft/s 2 , ω = 135 rpm = 14.1 rad/s, r = 2.1 ft Find: v A , a A Solution: Construct a x-y coordinate system in the horizontal plane As illustrated. Using this coordinate system 2 ft/s 3 ft/s, 110 i a i v − = − = The position vector of A relative to the helicopter at this instant is [ ] j i j i r h A 48 . 1 48 . 1 ) 4 / sin( ) 4 / cos( / − = − = π π r The relative velocity equation is used to determine the velocity of particle A as ft/s 9 . 20 1 . 89 ) 48 . 1 48 . 1 ( 1 . 14 110 / j i v j i k i v r k v v A A h A h A + − = − × + − = × + = ω The relative acceleration equation is used to determine the acceleration of particle A as

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    Chapter 1: Introduction 13 2 / / ft/s 6 . 294 4 . 297 ) 87 . 20 87 . 20 ( 1 . 14 1 . 3 ) ( j i a j i k i a r k k r k a a A A h A h A h A + − = + × + − = × × + × + = ω ω α Problem 1.9 illustrates the use of the relative velocity and relative acceleration equations. 1.10 For the system shown in Figure P1.10, the angular displacement of the thin disk is given by ߠሺݐሻ ൌ 0.03 sin ቀ30ݐ ൅ గ ସ ቁ rad . The disk rolls without slipping on the surface. Determine the following as functions of time. (a) The acceleration of the center of the disk. (b) The acceleration of the point of contact between the disk and the surface. (c) The angular acceleration of the bar. (d) The vertical displacement of the block. (Hint: Assume small angular oscillations ߶ of the bar. Then sin ߶ൎ ߶ .) Given: ሻݐሺߠ , ݎ ௗ ൌ 0.1 m, ℓ ൌ 0.3 m, ݀ ൌ 0.2 m Find: (a) ܽ ത ௗ (b) ܽ ത ௖ (c) ߙ (d) x Solution: (a) The angular acceleration of the disk is ߠሷሺݐሻ ൌ െሺ30ሻ ଶ 0.03 sin ቀ30 ݐ൅ ߨ 4 ቁ ൌ െ27 sin ቀ30 ݐ൅ ߨ 4 ቁ rad s ଶ Since the disk rolls without slip the acceleration of the mass center is ܽ ௗ ൌ ߙݎൌ ሺ0.1 mሻ ൬െ27 sin ቀ30 ݐ൅ ߨ 4 ቁ rad s ଶ ൰ ൌ െ2.7 sin ቀ30 ݐ൅ ߨ 4 ቁ m s ଶ (b) Since the disk rolls without slip the horizontal acceleration of the point of contact is zero. The vertical acceleration is ሶߠݎ ଶ towards the center ܉ ௖ ൌ ሺ0.1 mሻ ൤ሺ30ሻሺ0.03ሻ cos ቀ30 ݐ൅ ߨ 4 ቁ rad s ൨ ଶ ܒ

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