DOCSLIB.ORG

1 Graded Problems

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
1 Graded Problems PHY 5246: Theoretical Dynamics, Fall 2015 Assignment # 9, Solutions 1 Graded Problems Problem 1 (1.a) In order to find the equation of motion of the triangle, we need to write the Lagrangian, with generalized coor- dinate θ . The potential energy is going to be based on y { } the location of the center of mass, and we find that via l/2 l/2 ρ l/2 0 − x yCM (θ =0) = 2 dx dyy m 0 √3(l/2 x) Z Z− − 2 m l/2 1 l 2 = dx 3 x CM 1 √3 −m 0 2 2 − ! 2 l 2 l Z 2 8 3 1 l l/2 θ = x √3l2 2 3 2 − ! 0 3 4 l l √3l = = , − 2 −√3l2 8 −2√3 where the density is ρ. This is as expected by symmetry considerations and by the fact that the CM has to be at the geometrical center of the triangle, i.e. at one third of it’s height, 1 √3 l l = . −3 2 −2√3 Note the negative sign is because the triangle lies below the x-axis. So, for generalized θ, ρ yCM = cos θ, −2√3 and the potential energy is l U = mgyCM = mg cos θ. − 2√3 To calculate the kinetic energy, 1 T = T (about 0) = I θ˙2, rot 2 3 we need first to find the principal moment of inertia about the axis of rotation, which is an axis perpendicular to the plane of the triangle, through 0. This is the only moment of inertia we need, since ˙ ~ω = θˆe3 and therefore, 1 1 1 T = ~ωT ˆI ~ω = θ˙2I = θ˙2I . 2 · · 2 33 2 3 l/2 0 2 2 I3 = 2ρ dy(x + y ) 0 √3(l/2 x) Z Z− − l/2 0 l/2 0 = 2ρ dxx2 dy +2ρ dx dyy2 0 √3(l/2 x) 0 √3(l/2 x) Z Z− − Z Z− − l/2 l l/2 1 l 3 = 2ρ dxx2√3 x +2ρ 3√3 x Z0 2 − ! Z0 3 2 − ! 1 l l3 1 l4 1 l4 = 2ρ√3 + 3 2 8 − 3 16 4 16! m 1 l4 ml2 = 2√3 = . 1 √3 2 3 16 6 2 2 l Therefore 1 ml2 T = θ˙2, 2 6 ! and the Lagrangian is 1 ml2 l L = T V = θ˙2 + mg cos θ. − 2 6 ! 2√3 The equation of motion for our generalized coordinate is d ∂L ∂L ml2 l =0 θ¨ + mg sin θ =0 dt ∂θ˙ − ∂θ → 6 2√3 g θ¨ + √3 sin θ =0. ⇒ l Under small oscillations about θ = 0 this is the equation for a simple harmonic oscillator with frequency g ω = √3 . r l (1.b) For this situation, the θ = 0 position of the CM is y (θ =0)= l . Therefore CM − √3 y l l yCM = cos θ U = mg cos θ. −√3 ⇒ − √3 We need to calculate the moment of inertia I3 with re- l/2 l/2 spect to the new center of rotation (and axis through − x it). l/2 √3x − 2 2 I3 = 2ρ dx dx(x + y ) 0 √3/2l Z Z− l/2 √3x l/2 √3x 2 − − 2 CM = 2ρ dxx 3 dy +2ρ dx 3 dyy θ 0 √ l 0 √ l Z Z− 2 Z Z− 2 l/2 l l/2 1 l3 = 2ρ√3 dxx2 x + + dx 3 x3 + √3 0 2 0 3 8 l "Z − ! Z − !# − 2 1 l4 1 1 l4 l4 = 2ρ√3 + l4 + −4 16 3 − 4 16 16! m l4 3+2+6 5 = 2√3 − = ml2. 1 √3 2 16 6 12 2 2 l Thus the kinetic energy and Lagrangian are 1 5 T = ml2 θ˙2 2 12 1 5 l L = T V = ml2 θ˙2 + mg cos θ, − 2 12 √3 and the equation of motion is d ∂L ∂L 5 l =0 ml2θ¨ + mg sin θ =0 dt ∂θ˙ − ∂θ → 12 √3 g 12 1 θ¨ + sin θ =0. ⇒ l 5 √3 Thus the frequency of the small oscillation is 12 g ω = . s5√3 l Problem 2 (a) y y y = l cos θ CM √2 θ l The cube rotates about the side perpendicular to the plane of the figure, through 0. We are therefore referring to the case in which the frame has axes parallel to the side of the cube and origin at one corner. In figure (1), the cube is rotating about the back side, along the y-axis. The 2 2 corresponding moment of inertia is I22 = 3 ml (also see p. 409-410 in Goldstein). We can find the angular velocity of the cube when one face strikes the plane, by imposing conservation of energy: l Ei = Ui = mg √2 l 1 2 2 2 Ef = Uf + Tf = mg + ml ωf 2 2 3 l l 1 2 2 Ei = Ef mg = mg + ml ω . ⇒ √2 2 3 f 1 1 1 2 3 g g = lωf ωf = (√2 1) . ⇒ √2 − 2! 3 ⇒ s2 − l z O y x Figure 1: (b) If the cube can slide, there is no more fixed point. So, we have to calculate the kinetic energy as CM CM T = Ttrans + Trot 1 2 1 1 2 2 = mvCM + ml ωf , 2 2 6 where: l l vCM =y ˙CM = sin θθ˙ = θ,˙ −√2 −2 circ and when θ = 45 , the cube hits the plane. With ω(θ = 45◦)= ωf , we have 1 l2 1 5 T = m ω2 + ml2ω2 = lω2, 2 4 f 12 f 24 f Or 1 1 5 2 12 g g = lωf ωf = (√2 1) . √2 − 2! 24 ⇒ s 5 − l We could also have calculated the kinetic energy as CM CM T = Ttrans + Trot 1 2 1 CM 2 1 1 1 2 2 = mvCM + I22 ωf = + l ωf 2 2 2 2 6 1 = l2ω2 = T origin. 3 f rot 2 ˙ l CM ml Now use v = Rθ to find vCM = √2 ωf , and on p. 422 of Goldstein we can find I22 = 6 . We thus obtain the same answer as above. Problem 3 (Goldstein 5.17) Take M as the mass of the cone, (x1′ , x2′ , x3′ ) as the principal axes going through the CM, which 3 is at (0, 0, a = 4 h) in the body system. x3′ is the cone axis, x2′ is perpendicular to the plane going through x3′ and the segment OA, and x1′ is orthogonal to the plane of x2′ and x3′ . As stated in the problem, R is the radius, h is the height and α is the half-angle of the cone. Due to the condition of no-slipping, the angular velocity of the cone, ~ω, is directed along the instantaneous axis of rotation which is the line of contact between the surface of the cone and the surface on which the cone is rolling (i.e. along the segment OA, see figure). x3 x′1 O a x2 α θ CM x R ′3 x1 x′2 A ~ω (a) The kinetic energy (T ) can be calculated either using the center of mass as origin and taking as reference frame the system of principal axes described above, or using a system of axes parallel to that one but with origin at the apex of the cone O, which is a fixed point for the body. In the first case, the kinetic energy is given by the sum of two terms: the kinetic energy of the center of mass (i.e. the kinetic energy of a pointlike object of mass M located at the CM of the cone and moving as CM moves) and the kinetic energy of the body as rotating about the center of mass (expressed in terms of the principal moments of inertia and the component of the angular velocity in the principal axes frame or body frame), CM CM T = Ttrans + Trot kinetic energy 1 1 T CM = = Mv2 = Ma2 cos2 αθ˙2. trans of the CM 2 CM 2 ˙ In the above equation we have used that vCM = a cos αθ as it rotates about ˆx3. This implies that ω (angular velocity about OA) and θ˙ are related. Thus we also have vCM = a sin αω. Combining these two relations we find v ω = CM = θ˙ cot α. (1) a sin α The components of ~ω in the body frame are cos2 α ~ω =(ω sin α, 0,ω cos α)=(θ˙ cos α, 0, θ˙ ). sin α This follows from our choice of x2′ and because of the relation (1). Therefore: 1 1 T CM = I ω2 + I ω2 rot 2 1 1 2 3 3 2 I = I = 3 M R2 + h = 1 2 20 4 I = 3MR2. n 3 10 2 4 1 3 2 h 2 ˙2 1 3 2 cos α ˙2 = M R + cos αθ + MR 2 θ 2 20 4 ! 2 10 sin α 2 4 CM CM 1 2 2 ˙2 1 3 2 h 2 ˙2 1 3 2 cos α ˙2 T = Ttrans + Trot = Ma cos αθ + M R + cos αθ + MR 2 θ 2 2 20 4 ! 2 10 sin α 9 3 1 3 = Mh2 cos2 αθ˙2 + Mh2 tan2 α + cos2 αθ˙2 + Mh2 cos2 αθ˙2. 32 40 4 20 3 To get this last line we used a = 4 h, R = h tan α, and the next line uses the trigonometric identity, 2 1 tan α = 2 1. cos α − 9 3 3 3 3 T = Mh2 cos2 αθ˙2 Mh2θ˙2 Mh2 cos2 αθ˙2 + Mh2 cos2 αθ˙2 32 40 − 40 4 20 3 = Mh2θ˙2(1+5cos2 α).
Load more

Recommended publications
  • The Experimental Determination of the Moment of Inertia of a Model Airplane Michael Koken [email protected]
    The University of Akron IdeaExchange@UAkron The Dr. Gary B. and Pamela S. Williams Honors Honors Research Projects College Fall 2017 The Experimental Determination of the Moment of Inertia of a Model Airplane Michael Koken [email protected] Please take a moment to share how this work helps you through this survey. Your feedback will be important as we plan further development of our repository. Follow this and additional works at: http://ideaexchange.uakron.edu/honors_research_projects Part of the Aerospace Engineering Commons, Aviation Commons, Civil and Environmental Engineering Commons, Mechanical Engineering Commons, and the Physics Commons Recommended Citation Koken, Michael, "The Experimental Determination of the Moment of Inertia of a Model Airplane" (2017). Honors Research Projects. 585. http://ideaexchange.uakron.edu/honors_research_projects/585 This Honors Research Project is brought to you for free and open access by The Dr. Gary B. and Pamela S. Williams Honors College at IdeaExchange@UAkron, the institutional repository of The nivU ersity of Akron in Akron, Ohio, USA. It has been accepted for inclusion in Honors Research Projects by an authorized administrator of IdeaExchange@UAkron. For more information, please contact [email protected], [email protected]. 2017 THE EXPERIMENTAL DETERMINATION OF A MODEL AIRPLANE KOKEN, MICHAEL THE UNIVERSITY OF AKRON Honors Project TABLE OF CONTENTS List of Tables ................................................................................................................................................
    [Show full text]
  • Explain Inertial and Noninertial Frame of Reference
    Explain Inertial And Noninertial Frame Of Reference Nathanial crows unsmilingly. Grooved Sibyl harlequin, his meadow-brown add-on deletes mutely. Nacred or deputy, Sterne never soot any degeneration! In inertial frames of the air, hastening their fundamental forces on two forces must be frame and share information section i am throwing the car, there is not a severe bottleneck in What city the greatest value in flesh-seconds for this deviation. The definition of key facet having a small, polished surface have a force gem about a pretend or aspect of something. Fictitious Forces and Non-inertial Frames The Coriolis Force. Indeed, for death two particles moving anyhow, a coordinate system may be found among which saturated their trajectories are rectilinear. Inertial reference frame of inertial frames of angular momentum and explain why? This is illustrated below. Use tow of reference in as sentence Sentences YourDictionary. What working the difference between inertial frame and non inertial fr. Frames of Reference Isaac Physics. In forward though some time and explain inertial and noninertial of frame to prove your measurement problem you. This circumstance undermines a defining characteristic of inertial frames: that with respect to shame given inertial frame, the other inertial frame home in uniform rectilinear motion. The redirect does not rub at any valid page. That according to whether the thaw is inertial or non-inertial In the. This follows from what Einstein formulated as his equivalence principlewhich, in an, is inspired by the consequences of fire fall. Frame of reference synonyms Best 16 synonyms for was of. How read you govern a bleed of reference? Name we will balance in noninertial frame at its axis from another hamiltonian with each printed as explained at all.
    [Show full text]
  • Sliding and Rolling: the Physics of a Rolling Ball J Hierrezuelo Secondary School I B Reyes Catdicos (Vdez- Mdaga),Spain and C Carnero University of Malaga, Spain
    Sliding and rolling: the physics of a rolling ball J Hierrezuelo Secondary School I B Reyes Catdicos (Vdez- Mdaga),Spain and C Carnero University of Malaga, Spain We present an approach that provides a simple and there is an extra difficulty: most students think that it adequate procedure for introducing the concept of is not possible for a body to roll without slipping rolling friction. In addition, we discuss some unless there is a frictional force involved. In fact, aspects related to rolling motion that are the when we ask students, 'why do rolling bodies come to source of students' misconceptions. Several rest?, in most cases the answer is, 'because the didactic suggestions are given. frictional force acting on the body provides a negative acceleration decreasing the speed of the Rolling motion plays an important role in many body'. In order to gain a good understanding of familiar situations and in a number of technical rolling motion, which is bound to be useful in further applications, so this kind of motion is the subject of advanced courses. these aspects should be properly considerable attention in most introductory darified. mechanics courses in science and engineering. The outline of this article is as follows. Firstly, we However, we often find that students make errors describe the motion of a rigid sphere on a rigid when they try to interpret certain situations related horizontal plane. In this section, we compare two to this motion. situations: (1) rolling and slipping, and (2) rolling It must be recognized that a correct analysis of rolling without slipping.
    [Show full text]
  • Work, Energy, Rolling, SJ 7Th Ed.: Chap 10.8 to 9, Chap 11.1
    A cylinder whose radius is 0.50 m and whose rotational inertia is 10.0 kgm^2 is rotating around a fixed horizontal axis through its center. A constant force of 10.0 N is applied at the rim and is always tangent to it. The work done by on the disk as it accelerates and rotates turns through four complete revolutions is ____ J. Physics 106 Week 4 Work, Energy, Rolling, SJ 7th Ed.: Chap 10.8 to 9, Chap 11.1 • Work and rotational kinetic energy • Rolling • Kinetic energy of rolling Today • Examples of Second Law applied to rolling 2 1 Goal for today Understanding rolling motion For motion with translation and rotation about center of mass ω Example: Rolling vcm KKKtotal=+ rot cm 1 2 1 2 KI= ω KMvcm= cm rot2 cm 2 UM= gh Emech=+KU tot gravity cm 2 A wheel rolling without slipping on a table • The green line above is the path of the mass center of a wheel. ω • The red curve shows the path (called a cycloid) swept out by a point on the rim of the wheel . • When there is no slipping, there are simple vcm relationships between the translational (mass center) and rotational motion. s = Rθ vcm = ωR acm = αR First point of view about rolling motion Rolling = pure rotation around CM + pure translation of CM a) Pure rotation b) Pure translation c) Rolling motion 3 Second point of view about rolling motion Rolling = pure rotation about contact point P • Complementary views – a snapshot in time • Contact point “P” is constantly changing vtang = 2ωPR vA = ωP 2Rcos(ϕ) A v = ω R φ cm P R ωP φ P vtang = 0 vRRcm==ω Pω cm ∴ωP = ωcm ∴α pcm= α Angular velocity and acceleration are the same about contact point “P” or about CM.
    [Show full text]
  • Mechanik Solution 9
    Mechanik HS 2018 Solution 9. Prof. Gino Isidori Exercise 1. Rolling cones In the following exercise we consider a circular cone of height h, density ρ, mass m and opening angle α. (a) Determine the inertia tensor in the representation of the axes of Fig.1. (b) Compute the kinetic energy of the cone rolling on a level plane. (c) Repeat the exercise, only this time the cone's tip is fixed to a point on the z axis such that its longitudinal axis is parallel with the plane. Figure 1: The cone with the reference frame for Exercise 1(a). Solution. (a) We start with the inertia around the z-axis. To this end, we have to compute: Z ZZZ 2 2 3 Iz = ρ (x + y )dV = ρ r dz dr dφ Z 2π Z R Z h π = ρ dφ r3dr dz = ρhR4 (S.1) 0 0 hr=R 10 3 = mh2 tan2 α : 10 where we used R = h tan α and m = πhR2ρ/3 in the last step. By the symmetry of the problem, the other two moments of inerta are identical to each other. So it suffices to compute: Z ZZZ 2 2 2 2 2 I1 = ρ (y + z )dV = ρ (r sin φ + z )r dz dr dφ Z 2π Z R Z h = ρ dφ rdr (r2 sin2 φ + z2)dz 0 0 hr=R (S.2) π = ρ hR2(R2 + 4h2) 20 3 = mh2 tan2 α + 4 : 20 1 Figure 2: The rolling cone in part a) of the problem.
    [Show full text]
  • Dynamics of a Rolling and Sliding Disk in a Plane. Asymptotic Solutions, Stability and Numerical Simulations
    ISSN 1560-3547, Regular and Chaotic Dynamics, 2016, Vol. 21, No. 2, pp. 204–231. c Pleiades Publishing, Ltd., 2016. Dynamics of a Rolling and Sliding Disk in a Plane. Asymptotic Solutions, Stability and Numerical Simulations Maria Przybylska1* and Stefan Rauch-Wojciechowski2** 1Institute of Physics, University of Zielona G´ora, Licealna 9, PL-65–417 Zielona G´ora, Poland 2Department of Mathematics, Link¨oping University, 581 83 Link¨oping, Sweden Received October 27, 2015; accepted February 15, 2016 Abstract—We present a qualitative analysis of the dynamics of a rolling and sliding disk in a horizontal plane. It is based on using three classes of asymptotic solutions: straight-line rolling, spinning about a vertical diameter and tumbling solutions. Their linear stability analysis is given and it is complemented with computer simulations of solutions starting in the vicinity of the asymptotic solutions. The results on asymptotic solutions and their linear stability apply also to an annulus and to a hoop. MSC2010 numbers: 37J60, 37J25, 70G45 DOI: 10.1134/S1560354716020052 Keywords: rigid body, nonholonomic mechanics, rolling disk, sliding disk 1. INTRODUCTION The problem of a disk rolling (without sliding) in a plane is integrable [1, 4, 9, 15]. It is described in the Euler angles by four dynamical equations and this system admits three additional integrals of motion. It can be in principle reduced to a single quadrature. The difficulty is, however, that only the energy integral is given explicitly, while two other integrals, of angular momentum type, are defined implicitly as constants of integrations for the Legendre equation for the dependence of ω3 on cos θ [22] and [9, §8].
    [Show full text]
  • Rolling Motion: Experiments and Simulations Focusing on Sliding Friction Forces
    Rolling motion: experiments and simulations focusing on sliding friction forces Pasquale Onorato, Massimiliano Malgieri and Anna De Ambrosis Dipartimento di Fisica Università di Pavia via Bassi, 6 27100 Pavia, Italy Abstract The paper presents an activity sequence aimed at elucidating the role of sliding friction forces in determining/shaping the rolling motion. The sequence is based on experiments and computer simulations and it is devoted both to high school and undergraduate students. Measurements are carried out by using the open source Tracker Video Analysis software, while interactive simulations are realized by means of Algodoo, a freeware 2D-simulation software. Data collected from questionnaires before and after the activities, and from final reports, show the effectiveness of combining simulations and Video Based Analysis experiments in improving students’ understanding of rolling motion. Keywords Rolling motion, friction force, collisions, video analysis, simulations, and billiard. Introduction As it is well known, rolling motion is a complex phenomenon whose full comprehension involves the combination of several fundamental physics topics, such as rigid body dynamics, friction forces, and conservation of energy. For example, to deal with collisions between two rolling spheres in an appropriate way requires that the role of friction in converting linear to rotational motion and vice-versa be taken into account (Domenech & Casasús, 1991; Mathavan et al. 2009; Wallace &Schroeder, 1988). In this paper we present a sequence of activities aimed at spotlighting the role of friction in rolling motion in different situations. The sequence design results from a careful analysis of textbooks and of research findings on students’ difficulties, as reported in the literature.
    [Show full text]
  • Rolling Rolling Condition for Rolling Without Slipping
    Rolling Rolling simulation We can view rolling motion as a superposition of pure rotation and pure translation. Pure rotation Pure translation Rolling +rω +rω v Rolling ω ω v -rω +=v -rω v For rolling without slipping, the net instantaneous velocity at the bottom of the wheel is zero. To achieve this condition, 0 = vnet = translational velocity + tangential velocity due to rotation. In other wards, v – rω = 0. When v = rω (i.e., rolling without slipping applies), the tangential velocity at the top of the wheel is twice the 1 translational velocity of the wheel (= v + rω = 2v). 2 Condition for Rolling Without Big yo-yo Slipping A large yo-yo stands on a table. A rope wrapped around the yo-yo's axle, which has a radius that’s half that of When a disc is rolling without slipping, the bottom the yo-yo, is pulled horizontally to the right, with the of the wheel is always at rest instantaneously. rope coming off the yo-yo above the axle. In which This leads to ω = v/r and α = a/r direction does the yo-yo move? There is friction between the table and the yo-yo. Suppose the yo-yo where v is the translational velocity and a is is pulled slowly enough that the yo-yo does not slip acceleration of the center of mass of the disc. on the table as it rolls. rω 1. to the right ω 2. to the left v 3. it won't move -rω X 3 4 Vnet at this point = v – rω Big yo-yo Big yo-yo, again Since the yo-yo rolls without slipping, the center of mass The situation is repeated but with the rope coming off the velocity of the yo-yo must satisfy, vcm = rω.
    [Show full text]
  • Implementation of Rolling Wheel Deflectometer (RWD) in PMS and Pavement Preservation
    TECHNICAL REPORT STANDARD PAGE 1. Report No. 2. Government Accession No. 3. Recipient's Catalog No. FHWA/11.492 4. Title and Subtitle 5. Report Date Implementation of Rolling Wheel Deflectometer August 2012 (RWD) in PMS and Pavement Preservation 6. Performing Organization Code LTRC Project Number: 09-2P SIO Number: 30000143 7. Author(s) 8. Performing Organization Report No. Mostafa Elseifi, Ahmed M. Abdel-Khalek , and Karthik Dasari 9. Performing Organization Name and Address 10. Work Unit No. Department of Civil and Environmental Engineering Louisiana State University 11. Contract or Grant No. Baton Rouge, LA 70803 12. Sponsoring Agency Name and Address 13. Type of Report and Period Covered Louisiana Department of Transportation and Final Report Development 2009-2011 P.O. Box 94245 14. Sponsoring Agency Code Baton Rouge, LA 70804-9245 LTRC 15. Supplementary Notes Conducted in Cooperation with the U.S. Department of Transportation, Federal Highway Administration 16. Abstract The rolling wheel deflectometer (RWD) offers the benefit to measure pavement deflection without causing any traffic interruption or compromising safety along tested road segments. This study describes a detailed field evaluation of the RWD system in Louisiana in which 16 different test sites representing a wide array of pavement conditions were tested. Measurements were used to assess the repeatability of RWD measurements, the effect of truck speeds, and to study the relationship between RWD and falling weight deflectometer (FWD) deflection measurements and pavement conditions. Based on the results of the experimental program, it was determined that the repeatability of RWD measurements was acceptable with an average coefficient of variation at all test speeds of 15 percent.
    [Show full text]
  • Life on a Merry-Go-Round
    Life on a Merry-Go-Round An Examination of Relativistic Rotating Reference Frames Nathaniel I. Reden Submitted to the Department of Physics of Amherst College in partial fulfillment of the requirements for the degree of Bachelor of Arts with Distinction. Faculty advisor Kannan Jagannathan May 6, 2005 Acknowledgements I would like to thank Jagu for his help in probing the subtleties of space and time, and for always sending me in a useful direction when I hit a block. His guidance was integral to the integrals (and other equations) contained herein. The Physics Department of Amherst College taught me everything I know, and I am grateful to them. I would also like to thank my family, especially my parents, for their love and support, and my grandfather, Herman Stone, for inspiring me to study the sciences. They have always encouraged me to push myself and are willing to lend a helping hand whenever I need it. Wing, Devindra, Lucia, Maryna, and Ina, my suitemates, are owed thanks as well for helping me relax and feel at home, and for not playing video games all the time. The slack was picked up by Matt, whose lightheartedness has kept me optimistic even in the darkest times. To all who have made me laugh this past year, thank you. Lastly, I would like to thank Vanessa Eve Hettinger for her warmth and love. She fended off distractions and took good care of me, and she is always there when I need a hug. I don’t know what this year would have been like without her.
    [Show full text]
  • In Use Determination of Aerodynamic and Rolling Resistances of Heavy-Duty Vehicles
    sustainability Article In Use Determination of Aerodynamic and Rolling Resistances of Heavy-Duty Vehicles Dimitrios Komnos 1,2, Stijn Broekaert 3 , Theodoros Grigoratos 3 , Leonidas Ntziachristos 2 and Georgios Fontaras 3,* 1 FINCONS Group, 20871 Vimercate, Italy; [email protected] 2 Mechanical Engineering, Aristotle University of Thessaloniki, 54124 Thessaloniki, Greece; [email protected] 3 European Commission Joint Research, 21027 Ispra, Italy; [email protected] (S.B.); [email protected] (T.G.) * Correspondence: [email protected] Abstract: A vehicle’s air drag coefficient (Cd) and rolling resistance coefficient (RRC) have a sig- nificant impact on its fuel consumption. Consequently, these properties are required as input for the certification of the vehicle’s fuel consumption and Carbon Dioxide emissions, regardless of whether the certification is done via simulation or chassis dyno testing. They can be determined through dedicated measurements, such as a drum test for the tire’s rolling resistance coefficient and constant speed test (EU) or coast down test (US) for the body’s air Cd. In this paper, a methodology that allows determining the vehicle’s Cd·A (the product of Cd and frontal area of the vehicle) from on-road tests is presented. The possibility to measure these properties during an on-road test, without the need for a test track, enables third parties to verify the certified vehicle properties in order to preselect vehicle for further regulatory testing. On-road tests were performed with three heavy-duty vehicles, two lorries, and a coach, over different routes. Vehicles were instrumented with wheel torque sensors, wheel speed sensors, a GPS device, and a fuel flow sensor.
    [Show full text]
  • Rolling Down a Ramp with No FricOn
    → → → → → → → Fnet = ∑Fi = ma τ net = τ i = r× F = r× m a = m r× (r×α )= Iα { } ∑ € € Work and Rotational Kinetic Energy W = ΔKE rot W = ΔKE trans NET Work done ON system 1 2 2 1 2 2 = 2 I(ω f −ω i ) = 2 m(v f − vi ) θ 2 Rotational work, x2 W = ∫ τ net dθ fixed axis rotation€ θ1 € W = ∫ Fdx 1− D motion x1 (if torque is const) = τ (θ f −θi ) dW d € dW d Power, P = = (τθ) = τω P = = (F • x ) = F •v fixed axis rotation€ dt dt dt dt € € We can compare linear variables with rotational variables x θ v ω s = rθ a α v = rω Δt Δt T a = rα F τ T m I The same can be done for work and energy: € € For translational€ systems For rotational systems W = F ⋅ x W = τ ⋅θ 1 1 KE = mv 2 KE = Iω 2 2 2 € € All points on wheel move All points on wheel Combination of “pure with same ω. All points on move to the right with same rotation” and “pure outer rim move with same linear velocity vcom as center translation” linear speed v = v . of wheel com v = ω ×r Note at point P: vector sum of velocity = 0 (point of stationary contact) at point T: vector sum of velocity = 2v (top moves twice as fast as com) € com 1 2 1 2 2 Icomω + 2 Mvcom = KEtot Note: rotation about COM and translation of COM combine for total KE Remember: v =ωr € € € com Example: Rolling down a ramp with no fricon Object, with mass m and radius r, roles from top of incline plane to bottom.
    [Show full text]