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MAE_242_Lec4.ppt-Engineering Mechanic - Dynamic

MAE_242_Lec4.ppt-Engineering Mechanic - Dynamic

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MAE_242_Lec4.ppt-Engineering Mechanic - Dynamic

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  1. MAE 242 Dynamics – Section I Dr. Kostas Sierros

  2. Announcement • A.PowerPoint lecture notes are now posted in: • http://www.mae.wvu.edu/~cairns/teaching.html • Username: cairns • Password: materials • B.Help session for next Tuesday(4th Sep.) (5:15-6:00 pm) in room 401 ESB • C.Homework to hand in next Tuesday • (The 4th of September) • Quiz for next Thursday • (The 6th of September)

  3. Problem 1

  4. Problem 2

  5. Problem 3

  6. Kinematics of a particle: Objectives • Concepts such as position, displacement, velocity and acceleration are introduced • Study the motion of particles along a straight line. Graphical representation • Investigation of a particle motion along a curved path. Use of different coordinate systems • Analysis of dependent motion of two particles • Principles of relative motion of two particles. Use of translating axis

  7. Lecture 4 • Kinematics of a particle (Chapter 12) • -12.7-12.8

  8. Material covered • Kinematics of a particle • Curvilinear motion: Normal & tangential components and cylindrical components • Next lecture; Absolute dependent motion. Analysis of two particles • …and…Relative motion. Analysis of two particles using translating axis

  9. Today’s Objectives • Students should be able to: • Determine the normal and tangential components of velocity and acceleration of a particle traveling along a curved path. • Determine velocity and acceleration components using cylindrical coordinates

  10. Normal and tangential components I When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known,normal (n)and tangential (t) coordinatesare often used In the n-t coordinate system, theorigin is located on the particle(the originmoves with the particle) Thet-axisistangentto thepath (curve)at the instant considered, positive in the direction of the particle’s motion Then-axisisperpendicularto thet-axiswith the positive direction toward the center of curvature of the curve

  11. Normal and tangential components II The positive n and t directions are defined by theunit vectorsunandut, respectively Thecenter of curvature, O’, always lies on theconcave side of the curve. Theradius of curvature, r, is defined as the perpendicular distance from the curve to the center of curvature at that point Theposition of the particleat any instant is defined by the distance, s, along the curve from a fixed reference point.

  12. Velocity in the n-t coordinate system Thevelocity vectoris alwaystangentto the path of motion (t-direction) The magnitudeis determined by taking thetime derivativeof thepath function, s(t) v=vutwhere v = ds/dt Here v defines themagnitudeof the velocity (speed) and utdefines thedirection of the velocity vector.

  13. . Here v represents the change in the magnitude of velocity andutrepresents the rate of change in the direction ofut. . After mathematical manipulation, the acceleration vector can be expressed as: . a=vut+ (v2/r)un=atut+anun Acceleration in the n-t coordinate system I Acceleration is the time rate of change of velocity: a = dv/dt = d(vut)/dt = vut + v ut . ·

  14. Acceleration in the n-t coordinate system II There are two components to the acceleration vector: a= atut+ anun The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity. at = v or at ds = v dv . Thenormalorcentripetal componentis always directed toward the center of curvature of the curve. an = v2/r Themagnitudeof the acceleration vector is a = [(at)2 + (an)2]0.5

  15. Special cases of motion I There are some special cases of motion to consider 1)The particle moves along astraight line. r => an = v2/r = 0 => a = at = v . Thetangential componentrepresents thetime rate of changein themagnitudeof thevelocity. 2) The particle moves along a curve atconstant speed. at = v = 0 => a = an = v2/r . The normal componentrepresents thetime rate of changein thedirectionof the velocity.

  16. 4)The particle moves along a path expressed as y = f(x). The radius of curvature, r, at any point on the path can be calculated from r = ________________ [ 1 + (dy/dx)2 ]3/2 d2y/dx 2 Special cases of motion II 3)The tangential component of acceleration isconstant, at = (at)c. In this case, s = so + vot + (1/2)(at)ct2 v = vo + (at)ct v2 = (vo)2 + 2(at)c(s – so) As before, so and vo are the initial position and velocity of the particle at t = 0

  17. Three dimensional motion If a particle moves along aspace curve, the n and t axes are defined as before. At any point, thet-axisistangentto thepathand then-axispointstowardthecenter of curvature.The plane containing the n and t axes is called the osculating plane. A third axis can be defined, called the binomial axis, b. The binomial unit vector,ub,is directedperpendicular to the osculating plane, and its sense is defined by thecross productub=utxun. There is no motion, thus no velocity or acceleration, in the binomial direction.

  18. Curvilinear motion: Normal & tangential components …Example 12.16 !!!! http://c-more.automationdirect.com/images/sample_box_conveyor.gif

  19. Curvilinear motion: Cylindrical components (12.8) Applications The cylindrical coordinate system is used in cases where the particle moves along a 3-D curve. www.cim.mcgill.ca (spiral motion) www.cim.mcgill.ca

  20. Cylindrical components We can express the location of P in polar coordinates asr=rur. Note that the radial direction, r, extends outward from the fixed origin, O, and the transverse coordinate, q, is measured counter-clockwise (CCW) from the horizontal.

  21. The instantaneous velocity is defined as: v = dr/dt = d(rur)/dt dur . v = rur + r dt Velocity (Polar coordinates) Using the chain rule: dur/dt = (dur/dq)(dq/dt) We can prove that dur/dq = uθ so dur/dt = quθ Therefore: v = rur + rquθ . . . . Thus, the velocity vector has two components: r, called the radial component, and rq, called the transverse component. The speed of the particle at any given instant is the sum of the squares of both components or v = (r q )2 + ( r )2 . . .

  22. Acceleration (Polar coordinates) The instantaneous acceleration is defined as: a = dv/dt = (d/dt)(rur + rquθ) . . After manipulation, the acceleration can be expressed as a = (r – rq2)ur + (rq + 2rq)uθ . .. . .. . . .. The term (r – rq2)is the radial acceleration or ar. The term (rq + 2rq) is the transverse acceleration or aq .. . . . .. . .. . The magnitude of acceleration is a = (r – rq2)2 + (rq + 2rq)2

  23. . . . Velocity: vP = rur + rquθ + zuz Acceleration: aP = (r – rq2)ur + (rq + 2rq)uθ + zuz . . .. . .. .. Cylindrical coordinates If the particle P moves along a space curve, its position can be written as rP = rur + zuz Taking time derivatives and using the chain rule:

  24. HOMEWORK HIBBELER 12-6 page 15 12-53 page 27 12-60 page 29 12-85 page 45 12-87 page 46 12-96 page 48 Reminder: The homework will be collected next Tuesday (9/4) in class. Homework preparation rules: http://www.mae.wvu.edu/~cairns/Syllabus_MAE242.doc

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