Dynamics of Particles and Rigid Bodies: A Systematic Approach
Language: English
Pages: 528
ISBN: 0521187907
Format: PDF / Kindle (mobi) / ePub
Dynamics of Particles and Rigid Bodies: A Systematic Approach is intended for undergraduate courses in dynamics. This work is a unique blend of conceptual, theoretical, and practical aspects of dynamics generally not found in dynamics books at the undergraduate level. In particular, in this book the concepts are developed in a highly rigorous manner and are applied to examples using a step-by-step approach that is completely consistent with the theory. In addition, for clarity, the notation used to develop the theory is identical to that used to solve example problems. The result of this approach is that a student is able to see clearly the connection between the theory and the application of theory to example problems. While the material is not new, instructors and their students will appreciate the highly pedagogical approach that aids in the mastery and retention of concepts. The approach used in this book teaches a student to develop a systematic approach to problem-solving. The work is supported by a great range of examples and reinforced by numerous problems for student solution. An Instructor's Solutions Manual is available.
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of Velocity Suppose we are given acceleration as a function of velocity, i.e., a = a(v). Then, from the chain rule (Kreyszig, 1988) we have a = a(v) = dv dv dx dv = =v dt dx dt dx (2–49) dv = a(v) dx (2–50) Furthermore, from Eq. (2–49) we have v Separating variables in Eq. (2–50) gives dx = vdv a(v) (2–51) Then, integrating both sides of Eq. (2–51) we obtain v x − x0 = v0 ηdη a(η) (2–52) where η is a dummy variable of integration. Finally, solving Eq. (2–52) for x, we obtain v ηdη
Eqs. (2–297), (2–301), and (2–300), respectively, as A det dt A den dt A deb dt = A = −A vκet + A vτeb (2–304) = −A vτen (2–305) vκen (2–303) We see from Eqs. (2–303)–(2–305) that the curvature and torsion are given as κ = 1 Av τ = 1 Av A det dt (2–306) A deb dt (2–307) Finally, substituting the expression for A det /dt from Eq. (2–297) into (2–283), we obtain the acceleration of the particle at point P as A a= d dt A v et + κ A v 2 en (2–308) While for many
perspective of bases {e1 , e2 , e3 } and {u1 , u2 , u3 } for Using Fig. 2–33, we have e1 e3 = = cos φu1 + sin φu3 − sin φu1 + cos φu3 (2–457) The next step in solving this problem is to determine the angular velocity of each reference frame. Because the vertical shaft rotates with angular velocity Ω relative to the supports and the supports are fixed in reference frame F , the angular velocity of reference frame A in reference frame F is given as F ωA = Ω = Ωe1 (2–458) Next, it is seen
˙ θ + r Ω sin θ ez − r Ω sin θ ez = r θe ˙ θ vrel = r θe (2–510) It is seen that the result of Eq. (2–510) is identical to that of Eq. (2–503), i.e., the quantity vrel is the same regardless of whether the relative velocity is computed in reference frame A or reference frame F . Eliminating the dependence in Eq. (2–510) on the reference frame, we obtain vrel as ˙ θ vrel = r θe (2–511) 104 2.15 Chapter 2. Kinematics Kinematics of Rigid Bodies A rigid body, denoted R, is a collection of
r¨ = d˙ r d˙ r dr d˙ r = = r˙ dt dr dt dr (3–106) The expression for r¨ from Eq. (3–106) can then be substituted into Eq. (3–105). This gives d˙ r r (3–107) + r (˙ r 2 + Rg) = 0 (r 2 + R 2 )˙ dr 168 Chapter 3. Kinetics of Particles Rearranging Eq. (3–107), we obtain r (r 2 + R 2 )˙ d˙ r = −r (˙ r 2 + Rg) dr (3–108) Separating r and r˙ in Eq. (3–108), we have r r˙ d˙ r =− 2 dr r + R2 r˙2 + Rg (3–109) Integrating both sides, we obtain r˙ r˙0 ν dν = 2 ν + Rg r r0 − η2 η dη + R2
