Chapter 16

1. Chapter 16

2. RIGID BODY MOTION: TRANSLATION & ROTATION (Sections 16.1-16.3) Objectives : To analyze the kinematics of a rigid body undergoing planar translation or rotation about a fixed axis.

3. Methodology • Understand 1D translational motion • Use analogy: • To relate rotation with translation • To relate general motion of rigid body with relative motion between two particles using vector (tensor) notations.

4. APPLICATIONS Passengers ride are subjected to curvilinear translation. Given the angular motion of the rotating arms, determine v and a experienced by the passengers? Does each passenger feel the same acceleration? How can we relate the angular motions of contacting bodies that rotate about different fixed axes?

5. PLANAR KINEMATICS OF A RIGID BODY There are cases where an object cannot be treated as a particle. In these cases the size or shape of 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 and links in machinery or mechanisms, rotation of the body is an important aspect in the analysis of motion.

6. PLANAR RIGID BODY MOTION There are three types of planar rigid body motion. Translation: when every line segment on the body remains parallel to its original direction during motion. When all points move along straight lines: motion is rectilinear, when the paths are curved: motion is curvilinear translation.

7. PLANAR RIGID BODY MOTION(continued) Rotation about a fixed axis: all particles of the body, except those on the axis of rotation, move along circular paths in planes perpendicular to the axis of rotation. General plane motion: the body undergoes bothtranslation and rotation. Translation occurs within a plane and rotation occurs about an axis perpendicular to this plane.

8. PLANAR RIGID BODY MOTION(continued) Example of body undergoing the three types of motion: The wheel and crank (A and B) undergo rotation about a fixed axis. Both axes of rotation are at the location of the pins and perpendicular to the plane of the figure. The piston (C) undergoes rectilinear translation. The connecting rod (D) undergoes curvilinear translation, since it will remain horizontal as it moves along a circular path. The connecting rod (E) undergoes general plane motion, as it will both translate and rotate.

9. RIGID-BODY MOTION: TRANSLATION The positions of two points A and B on a translating body can be related by rB = rA + rB/A The velocity at B is vB = vA+ drB/A/dt . Now drB/A/dt= 0 since rB/A is constant. So, vB = vA, also: aB = aA. Note, all points in a rigid body subjected to translation move with the same velocity and acceleration.

10. Angular velocity, , is the time derivative of angular displacement:  = d/dt (rad/s) + Similarly, angular acceleration is  = d2/dt2 = d/dt or  = (d/d) rad/s2 RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS When a body rotates, point P travels along a circular path. The angular position of P is defined by q. d is the angular displacement, with units of either radians or revolutions (1 revol. = 2 radians)

11. Time issues: • Clockwise versus counterclockwise • Design of stairs • Time of days, months • Hour time?!

12. RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS (continued) If a is constant (aC) by analogy: w = wO + aCt q = qO + wOt + 0.5aCt2 w2 = (wO)2 + 2aC (q – qO) qO and wO are initial angular position and angular velocity.

13. The velocity of P: v= wr, its direction is tangent to the circular path of P. RIGID-BODY ROTATION: VELOCITY OF POINT P In vector formulation, the magnitude and direction of v can be determined from the cross product of w and rp.rp is a vector from any point on the axis of rotation to P. v = w x rp = w x r The direction of v is determined by the right-hand rule.

14. RIGID-BODY ROTATION: ACCELERATION OF POINT P The acceleration of P is expressed in terms of its normal (an) and tangential (at) components. In scalar form, these are at = a r and an = w2 r. The tangential component, at, represents the time rate of change in the velocity's magnitude. It is directed tangent to the path of motion. The normal component, an, represents the time rate of change in the velocity’s direction. It is directed toward the center of the circular path.

15. The magnitude of the acceleration vector is a = (at)2 + (an)2 RIGID-BODY ROTATION: ACCELERATION OF POINT P (continued) Using the vector formulation, the acceleration of P is a = dv/dt = dw/dt x rP + w x drP/dt = a x rP + w x (w x rP) It can be shown that this equation reduces to a = a x r – w2r = at + an

16. 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 2.5cm, 10cm, and 40cm, respectively. Find: The magnitudes of the velocity and acceleration at point P on the fan blade when t = 0.5 s.

17. 1) The angular velocity is given as a function of time, wm = 4(1 – e-t), thus: am = dwm/dt = 4e-t rad/s2 EXAMPLE (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. 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)(2.5) = wf(10) => wf = 0.3935 rad/s

18. EXAMPLE (continued) 3) Similarly, the tangential accelerations are related as at = am rm = af rf => (2.4261)(2.5) = af(10) => af = 0.606rad/s2 4) The speed of point P on the fan, at a radius of 40cm, is now determined as vP = wfrP = (0.3935)(40) = 15.8 cm/s The normal and tangential components of acceleration of P are an = (wf)2 rP = (0.3935)2 (40) = 6.2 cm/s2 at = af rP = (0.6065) (40) = 24.3 cm/s2

19. GROUP PROBLEM SOLVING Given: Starting from rest when s = 0, pulley A (rA = 50 mm) is given a constant angular acceleration, aA = 6 rad/s2. Pulley C (rC = 150 mm) has an inner hub D (rD = 75 mm) which is fixed to C and turns with it. Find: The speed of block B when it has risen s = 6 m.

20. aB = 0.15 m/s2 GROUP PROBLEM SOLVING(Solution) 1) Assuming the belt is inextensible and does not slip, aC = 2 rad/s2 2) Assuming the cord attached to block B is inextensible and does not slip, the speed and acceleration of B is

21. vB = 1.34 m/s GROUP PROBLEM SOLVING (continued) 3) Since aA is constant,

22. End of 16.1-3 Let Learning Continue

23. ABSOLUTE MOTION ANALYSIS (Section 16.4) Objective: To determine the velocity and acceleration of a rigid body undergoing general plane motion using an absolute motion analysis. Note: to be read by student if interested in geometry. In class we only use vector (tensor) methodology.

24. EXAMPLE 1 Given: Two slider blocks are connected by a rod of length 2 m. Also, vA = 8 m/s and aA = 0. Find: Angular velocity, w, and angular acceleration, a, of the rod when q = 60°.

25. reference q A sA EXAMPLE 1 (Solution by geometry)) In class solution by vectors By geometry, sA = 2 cos q By differentiating with respect to time, vA = -2 w sin q Using q = 60° and vA = 8 m/s and solving for w:w = 8/(-2 sin 60°) = - 4.62 rad/s Differentiating vA and solving for a, aA = -2a sin q – 2w2 cos q = 0 a = - w2/tan q = -12.32 rad/s2

26. EXAMPLE 2 (Solution by geometry)) In class solution by vectors Solution: xP = 60 cos q + (225)2 – (60 sin q)2 vP = -60w sin q + (0.5)[(225)2 – (60sin q)2]-0.5(-2)(60sin q)(60cos q)w vP = -60w sin q – [0.5(60)2sin2qw] / (225)2 – (60 sin q)2 At q = 30°, w = 150 rad/s and vP = -5550mm/s Given: Crank AB rotates at a constant w = 150 rad/s Find: Velocity of P when q = 30°

27. GROUP PROBLEM SOLVING How to deal with flexible bodies! Given: The w and a of the disk of the dimensions shown. Find: The velocity and acceleration of cylinder B in terms of q.

28. GROUP PROBLEM SOLVING (Solution) Law of cosines: = aB

29. End of 16.4 Let Learning Continue

30. RELATIVE MOTION ANALYSIS: VELOCITY (Section 16.5) • Objectives: • Describe the velocity of a rigid body in terms of translation and rotation components. • Perform a relative-motion velocity analysis of a point on the body.

31. APPLICATIONS As block A moves to the left with vA, it causes the link CB to rotate counterclockwise, thus vB is directed tangent to its circular path. Which link undergoes general plane motion? How to find its angular velocity? Gear systems are used in many automobile automatic transmissions. By locking or releasing different gears, this system can operate the car at different speeds.

32. = Disp. due to translation drB = drA + drB/A Disp. due to rotation Disp. due to translation and rotation RELATIVE MOTION ANALYSIS: DISPLACEMENT A body undergoes a combination of translation and rotation (general motion) Point A is called the base point, has a known motion. The x’-y’ frame translates with the body, but does not rotate. The displacement of B:

33. = + RELATIVE MOTION ANALYSIS: VELOCITY The velocity at B is :(drB/dt) = (drA/dt) + (drB/A/dt)or vB = vA + vB/A Since the body is taken as rotating about A, vB/A = drB/A/dt = w x rB/A Here w will only have a k component since the axis of rotation is perpendicular to the plane of translation.

34. RELATIVE MOTION ANALYSIS: VELOCITY (continued) vB = vA + w x rB/A When using the relative velocity equation, points A and B should generally be points on the body with a known motion. Often these points are pin connections in linkages. Both points A and B have circular motion since the disk and link BC move in circular paths. The directions of vA and vB are tangent to the circular path of motion.

35. RELATIVE MOTION ANALYSIS: VELOCITY (continued) vB = vA + w x rB/A When a wheel rolls without slipping, point A has zero velocity. Furthermore, point B at the center of the wheel moves along a horizontal path. Thus, vB has a known direction, e.g., parallel to the surface.

36. EXAMPLE 1 Given: Block A is moving down at 2 m/s. Find: The velocity of B at the instant  = 45.

37. EXAMPLE 1 (Solution) vB = vA + wAB x rB/A vB i= -2 j + (wkx (0.2 sin 45 i - 0.2 cos 45 j )) vB i= -2 j + 0.2 w sin 45 j + 0.2 w cos 45 i Equating the i and j components gives: vB = 0.2 w cos 45 0 = -2 + 0.2 w sin 45 Solving: w = 14.1 rad/s or wAB= 14.1 rad/s k vB = 2 m/s or vB = 2 m/s i

38. EXAMPLE 2 Given:Collar C is moving downward with a velocity of 2 m/s. Find: The angular velocities of CB and AB at this instant.

39. EXAMPLE 2 (solution) Link CB. Write the relative-velocity equation: vB = vC + wCB x rB/C vB i= -2 j + wCB k x (0.2 i - 0.2 j ) vB i = -2 j + 0.2wCB j + 0.2 wCB i By comparing the i, j components: i: vB = 0.2 wCB => vB = 2 m/s i j: 0 = -2 + 0.2 wCB => wCB = 10 rad/s k

40. EXAMPLE 2 (continued) Link AB experiences only rotation about A. Since vBis known, there is only one equation with one unknown to be found. vB = wAB x rB/A 2 i = wAB k x (-0.2 j ) 2 i = 0.2 wAB i By comparing the i-components: 2 = 0.2 wAB So, wAB = 10 rad/s k

41. GROUP PROBLEM SOLVING Given: The crankshaft AB is rotating at 500 rad/s about a fixed axis passing through A. Find: The speed of the piston P at the instant it is in the position shown.

42. GROUP PROBLEM SOLVING (solution) 1) First draw the kinematic diagram of link AB. Applying the relative velocity equation vB = -50 j m/s

43. GROUP PROBLEM SOLVING (continued) 2) Now consider link BC. Applying the relative velocity equation: vC = -50 j m/s

44. End of 16.5 Let Learning Continue

45. INSTANTANEOUS CENTER (IC) OF ZERO VELOCITY (Section 16.6) To be read by student if interested in geometry. Note: every body has an instantaneous center of zero velocity. Think of motion of planets and an instantaneous center for the universe

46. End of 16.6 Let Learning Continue

47. RELATIVE MOTION ANALYSIS: ACCELERATION (Section 16.7) Objectives: a) Resolve the acceleration of a point on a body into components of translation and rotation. b) Determine the acceleration of a point on a body by using a relative acceleration analysis.

48. APPLICATIONS The forces delivered to the crankshaft, and the angular acceleration of the crankshaft, depend on the speed and acceleration of the piston in an automotive engine. How can we relate the accelerations of the piston, connection rod, and crankshaft in this engine?

49. = + dv dv dv / B A B A dt dt dt These are absolute accelerations of points A and B. They are measured from a set of fixed x,y axes. This term is the acceleration of B wrt A. It will develop tangential and normal components. RELATIVE MOTION ANALYSIS: ACCELERATION The equation relating the accelerations of two points on the body is determined by: The result is aB= aA+ (aB/A)t+ (aB/A)n

50. Graphically: aB = aA+ (aB/A)t + (aB/A)n = + RELATIVE MOTION ANALYSIS: ACCELERATION The relative tangential acceleration component (aB/A)t is ( x rB/A) and perpendicular to rB/A. The relative normal acceleration component (aB/A)n is (-2rB/A) and the direction is always from B towards A.