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MAE 242 Dynamics – Section I Dr. Kostas Sierros

MAE 242 Dynamics – Section I Dr. Kostas Sierros. Make – Up Midterm Exam. 12.9, 12.22, 12.26, 12.42, 12.53, 12.65, 12.71, 12.83, 12.84, 12.100, 12.111, 12.153 13.5, 13.15, 13.34, 13.45, 13.51 14.10, 14.15, 14.17, 14.65. 355 ESB FRIDAY EVENING 8-9:15. No calculators with storing capacity

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MAE 242 Dynamics – Section I Dr. Kostas Sierros

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

  2. Make – Up Midterm Exam 12.9, 12.22, 12.26, 12.42, 12.53, 12.65, 12.71, 12.83, 12.84, 12.100, 12.111, 12.153 13.5, 13.15, 13.34, 13.45, 13.51 14.10, 14.15, 14.17, 14.65 355 ESB FRIDAY EVENING 8-9:15 • No calculators with storing capacity • Be careful with UNITS – Practice • Free body diagrams – Resolve forces • Explain why are you doing things • Full marks will be awarded for FULLY explained solutions • Do not use random formulae but ONLY the relevant ones • READ THE QUESTIONS CAREFULLY

  3. Problem 1

  4. Problem 2

  5. Problem 3

  6. Example Given:A rod assembly rotates around its z-axis. The mass C is 10 kg and its initial velocity is 2 m/s. A moment and force both act as shown (M = 8t2 + 5 N·m and F = 60 N) Find:The velocity of the mass C after 2 seconds Plan:Apply the principle of angular impulse and momentum about the axis of rotation (z-axis)

  7. t t 2 2 ò ò Angular impulse: + = (8t2 + 5) dt + (0.75)(3/5)(60) dt = (8/3)t3 + 5t + 27t = 85.33 N·m·s M dt (r x F) dt t t 1 1 2 2 ò ò 2 0 0 0 Example (continues) Solution: Angular momentum:HZ = r x mv reduces to a scalar equation. (HZ)1 = 0.75(10)(2) = 7.5(2) and (HZ)2 = 0.75(10)(v2) = 7.5v2 Apply the principle of angular impulse and momentum. 7.5(2) + 85.33 = 7.5v v = 13.38 m/s

  8. Planar kinematics of a rigid body Chapter 16 Chapter objectives • To classify the various types of rigid-body planar motion • To investigate rigid body translation and analyze it • Study planar motion • Relative motion analysis using translating frame of reference • Find instantaneous center of zero velocity • Relative motion analysis using rotating frame of reference

  9. Lecture 13 • Planar kinematics of a rigid body: • Rigid body motion, Translation, Rotation about a fixed axis • -16.1-16.3

  10. Material covered • Planar kinematics of a rigid body : • Rigid body motion • Translation • Rotation about a fixed axis • …Next lecture…continue with Chapter 16

  11. Today’s Objectives • Students should be able to: • 1. Analyze the kinematics of a rigid body undergoing planar translation or rotation about a fixed axis

  12. Applications Passengers on this amusement ride are subjected to curvilinear translation since the vehicle moves in a circular path but always remains upright. If the angular motion of the rotating arms is known, how can we determine the velocity and acceleration experienced by the passengers? Does each passenger feel the same acceleration?

  13. Applications (continued) Gears, pulleys and cams, which rotate about fixed axes, are often used in machinery to generate motion and transmit forces. The angular motion of these components must be understood to properly design the system. How can we relate the angular motions of contacting bodies that rotate about different fixed axes?

  14. Rigid body motion (section 16.1) There are cases where an objectcannotbe treated as a particle. In these cases thesizeorshapeof the body must be considered. Also,rotation of the body about its center of mass requires a different approach. For example, in the design of gears, cams, and links in machinery or mechanisms, rotation of the body is an important aspect in the analysis of motion. We will now start to studyrigid body motion. The analysis will be limited toplanar motion. A body is said to undergo planar motion when all parts of the body move along paths equidistant from a fixed plane.

  15. Four stroke engine Otto Cycle

  16. Planar rigid body motion There arethreetypes of planar rigid body motion.

  17. Planar rigid body motion (continues) Translation:Translation occurs if every line segment on the body remains parallel to its original direction during the motion. When all points move along straight lines, the motion is calledrectilineartranslation. When the paths of motion are curved lines, the motion is calledcurvilineartranslation.

  18. Planar rigid body motion (continues) Rotation about a fixed axis:In this case, all the particles of the body, except those on the axis of rotation, move alongcircular pathsin planes perpendicular to the axis of rotation. General plane motion:In this case, the body undergoesboth translation and rotation. Translation occurs within a plane and rotation occurs about an axis perpendicular to this plane.

  19. Planar rigid body motion (continues) An example of bodies undergoing the three types of motion is shown in this mechanism. The wheel and crank undergorotation about a fixed axis. In this case, both axes of rotation are at the location of the pins and perpendicular to the plane of the figure. The piston undergoesrectilinear translationsince it is constrained to slide in a straight line. The connecting rod undergoescurvilinear translation, since it will remain horizontal as it moves along a circular path. The connecting rod undergoesgeneral plane motion, as it will both translate and rotate.

  20. Rigid body motion – Translation (16.2) The positions of two points A and B on a translating body can be related by rB = rA + rB/A where rA& rB are the absolute position vectors defined from the fixed x-y coordinate system, and rB/A is the relative-position vector between B and A. Thevelocity at B is vB = vA+ drB/A/dt . Now drB/A/dt= 0 since rB/A is constant. So, vB = vA, and by following similar logic, aB = aA. Note, all points in a rigid body subjected to translation move with thesame velocity and acceleration.

  21. Rigid body motion – Rotation about a fixed axis (16.3) Angular velocity, , is obtained by taking the time derivative of angular displacement:  = d/dt (rad/s) + Similarly,angular accelerationis  = d2/dt2 = d/dt or  = (d/d)+rad/s2 When a body rotates about a fixed axis, any point P in the body travels along acircular path.The angular position of P is defined by q. The change in angular position, d, is called the angular displacement, with units of either radians or revolutions. They are related by 1 revolution = 2 radians

  22. Rigid body motion – Rotation about a fixed axis (16.3 continued) If the angular acceleration of the body isconstant,a =aC, the equations for angular velocity and acceleration can be integrated to yield the set ofalgebraicequations below. w = wO + aCt q = qO + wOt + 0.5aCt2 w2 = (wO)2 + 2aC (q – qO) qO and wO are the initial values of the body’s angular position and angular velocity. Note these equations are very similar to the constant acceleration relations developed for therectilinearmotion of a particle.

  23. The magnitude of the velocity of P is equal to wr (the text provides the derivation). The velocity’s direction is tangent to the circular path of P. Rigid body rotation – Velocity of point P In thevector formulation, the magnitude and direction ofvcan be determined from thecross productofwandrp.Hererpis a vector from any point on the axis of rotation to P. v= wxrp=wxr The direction ofvis determined by the right-hand rule.

  24. Rigid body rotation – Acceleration of point P The acceleration of P is expressed in terms of itsnormal(an) andtangential(at)components. In scalar form, these are at = a r and an = w2 r. Thetangential component, at, represents the time rate of change in the velocity'smagnitude. It is directedtangent to the path of motion. The normal component, an, represents the time rate of change in the velocity’sdirection. It is directedtowardthecenterof the circular path.

  25. Rigid body rotation – Acceleration of point P (continued) The magnitude of the acceleration vector is a = (at)2 + (an)2 Using thevector formulation, the acceleration of P can also be defined by differentiating the velocity. a= dv/dt=dw/dt xrP+wxdrP/dt =axrP+wx (wxrP) It can be shown that this equation reduces to a= axr–w2r= at+an

  26. •To determine themotion of a point, the scalar equations v = w r, at = a r, an = w2r , and a = (at)2 + (an)2 can be used. Rotation about a fixed axis - Procedure •Establish asign conventionalong the axis of rotation. •If a relationship is known between anytwoof the variables (a, w, q, or t), the other variables can be determined from the equations: w = dq/dt a = dw/dt a dq = w dw •If a isconstant, use the equations for constant angular acceleration. •Alternatively, thevector form of the equations can be used(withi,j, kcomponents). v= wx rP =w xr a= at +an =axrP +wx (wx rP) =axr–w2r

  27. Example Given:The motor M begins rotating at w = 4(1 – e-t) rad/s, where t is in seconds. The radii of the motor, fan pulleys, and fan blades are 1 in, 4 in, and 16 in, respectively. Find: The magnitudes of the velocity and acceleration at point P on the fan blade when t = 0.5 s. Plan: 1) Determine the angular velocity and acceleration of the motor using kinematics of angular motion. 2) Assuming the belt does not slip, the angular velocity and acceleration of the fan are related to the motor's values by the belt. 3) The magnitudes of the velocity and acceleration of point P can be determined from the scalar equations of motion for a point on a rotating body.

  28. 1) Since the angular velocity is given as a function of time, wm = 4(1 – e-t), the angular acceleration can be found by differentiation. am = dwm/dt = 4e-t rad/s2 Example (continues) Solution: When t = 0.5 s, wm = 4(1 – e-0.5) = 1.5739 rad/s, am = 4e-0.5 = 2.4261 rad/s2 2) Since the belt does not slip (and is assumed inextensible), it must have the same speed and tangential component of acceleration at all points. Thus the pulleys must have the same speed and tangential acceleration at their contact points with the belt. Therefore, the angular velocities of the motor (wm) and fan (wf) are related as v = wm rm = wf rf => (1.5739)(1) = wf(4) => wf = 0.3935 rad/s

  29. Themagnitudeof the acceleration of P can be determined by aP = (an)2 + (at)2 = (2.477)2 + (9.704)2 = 10.0 in/s2 Example (continues) 3) Similarly, the tangential accelerationsare related as at = am rm = af rf => (2.4261)(1) = af(4) => af = 0.6065 rad/s2 4) The speed of point Pon the fan, at a radius of 16 in, is now determined as vP = wfrP = (0.3935)(16) = 6.30 in/s Thenormal and tangential components of accelerationof point P are calculated as an = (wf)2 rP = (0.3935)2 (16) = 2.477 in/s2 at = af rP = (0.6065) (16) = 9.704 in/s2

  30. For those doing the make-up exam tomorrow… 355 ESB FRIDAY EVENING 8-9:15

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