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Today’s Objectives : Students will be able to:

RELATIVE MOTION ANALYSIS: VELOCITY. Today’s Objectives : Students will be able to: 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. In-Class Activities :

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Today’s Objectives : Students will be able to:

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  1. RELATIVE MOTION ANALYSIS: VELOCITY • Today’s Objectives: • Students will be able to: • 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. In-Class Activities: • Translation and Rotation Components of Velocity • Relative Velocity Analysis

  2. APPLICATIONS As the slider block A moves horizontally to the left with vA, it causes the link CB to rotate counterclockwise. Thus vB is directed tangent to its circular path. Which link is undergoing general plane motion? How can its angular velocity, , be found?

  3. APPLICATIONS (continued) Planetary gear systems are used in many automobile automatic transmissions. By locking or releasing different gears, this system can operate the car at different speeds. How can we relate the angular velocities of the various gears in the system?

  4. = Disp. due to translation drB=drA+drB/A Disp. due to rotation Disp. due to translation and rotation RELATIVE MOTION ANALYSIS (Section 16.5) When a body is subjected to general plane motion, it undergoes a combination of translation and rotation. Point A is called the base point in this analysis. It generally has a known motion. The x’-y’ frame translates with the body, but does not rotate. The displacement of point B can be written:

  5. = + RELATIVE MOTION ANALYSIS: VELOCITY The velocity at B is given as :(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.

  6. RELATIVE MOTION ANALYSIS: VELOCITY 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. Here 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 known since they are always tangent to the circular path of motion.

  7. RELATIVE MOTION ANALYSIS: VELOCITY(continued) vB = vA + w x rB/A When a wheel rolls without slipping, point A is often selected to be at the point of contact with the ground. Since there is no 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.

  8. PROCEDURE FOR ANALYSIS The relative velocity equation can be applied using either a Cartesian vector analysis or by writing scalar x and y component equations directly. Scalar Analysis: 1. Establish the fixed x-y coordinate directions and draw a kinematic diagram for the body. Then establish the magnitude and direction of the relative velocity vector vB/A. 2. Write the equation vB = vA + vB/A and by using the kinematic diagram, underneath each term represent the vectors graphically by showing their magnitudes and directions. 3. Write the scalar equations from the x and y components of these graphical representations of the vectors. Solve for the unknowns.

  9. PROCEDURE FOR ANALYSIS (continued) Vector Analysis: 1. Establish the fixed x-y coordinate directions and draw the kinematic diagram of the body, showing the vectors vA, vB, rB/A and w. If the magnitudes are unknown, the sense of direction may be assumed. 2. Express the vectors in Cartesian vector form and substitute into vB = vA + w x rB/A. Evaluate the cross product and equate respective i and j components to obtain two scalar equations. 3. If the solution yields a negative answer, the sense of direction of the vector is opposite to that assumed.

  10. READING QUIZ 2. In the relative velocity equation, vB/A is A) the relative velocity of B with respect to A. B) due to the rotational motion. C) xrB/A . D) All of the above.

  11. EXAMPLE Given: Block A is moving down at 2 m/s. Find: The velocity of B at the instant  = 45. Plan: 1. Establish the fixed x-y directions and draw a kinematic diagram. 2. Express each of the velocity vectors in terms of their i, j, k components and solve vB = vA + w x rB/A.

  12. EXAMPLE (continued) Solution: vB = vA + wAB x rB/A vBi= -2 j + (wkx (0.2 sin 45 i - 0.2 cos 45 j )) vBi= -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

  13. EXAMPLE II Given:Collar C is moving downward with a velocity of 2 m/s. Find: The angular velocities of CB and AB at this instant. Plan: Notice that the downward motion of C causes B to move to the right. Also, CB and AB both rotate counterclockwise. First, draw a kinematic diagram of link CB and use vB = vC + wCB x rB/C. (Why do CB first?) Then do a similar process for link AB.

  14. EXAMPLE II (continued) Solution: Link CB. Write the relative-velocity equation: vB = vC + wCB x rB/C vBi= -2 j + wCB k x (0.2 i - 0.2 j ) vBi = -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

  15. EXAMPLE II (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

  16. 2 ft V=15 ft/s O A CONCEPT QUIZ 1. If the disk is moving with a velocity at point O of 15 ft/s and  = 2 rad/s, determine the velocity at A. A) 0 ft/s B) 4 ft/s C) 15 ft/s D) 11 ft/s 2. If the velocity at A is zero, then determine the angular velocity, . A) 30 rad/s B) 0 rad/s C) 7.5 rad/s D) 15 rad/s

  17. 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. Plan: 1) Draw the kinematic diagram of each link showing all pertinent velocity and position vectors. 2) Since the motion of link AB is known, apply the relative velocity equation first to this link, then link BC.

  18. wAB rB/A B A • • 100 mm y vB x GROUP PROBLEM SOLVING (continued) Solution: 1) First draw the kinematic diagram of link AB. Link AB rotates about a fixed axis at A. Since wAB is ccw, vB will be directed down, so vB = -vBj. Applying the relative velocity equation with vA = 0: vB = vA + wAB x rB/A -vBj = (500 k) x (-0.1 i + 0 j) -vBj = -50 j + 0 i Equating j components: vB = 50 vB = -50 j m/s

  19. C vC 500 mm rC/B wBC y B vB x GROUP PROBLEM SOLVING (continued) 2) Now consider link BC. Since point C is attached to the piston, vC must be directed up or down. It is assumed here to act down, so vC = -vCj. The unknown sense of wBC is assumed here to be ccw: wBC = wBCk. Applying the relative velocity equation: vC = vB + wBC x rC/B -vCj = -50 j + (wBCk) x (0.5 cos60 i + 0.5 sin60 j) -vCj = -50 j + 0.25wBCj – 0.433wBCi i: 0 = -0.433wBC => wBC = 0 j: -vC = -50 + 0.25wBC => vC = 50 vC = -50 j m/s

  20. vA 1. Which equation could be used to find the velocity of the center of the gear, C, if the velocity vA is known? A) vB = vA + wgear x rB/A B) vA = vC + wgear x rA/C C) vB = vC + wgear x rC/B D) vA = vC + wgear x rC/A 2. If the bar’s velocity at A is 3 m/s, what “base” point (first term on the RHS of the velocity equation) would be best used to simplify finding the bar’s angular velocity when  = 60º? A) A B) B C) C D) No difference. B 4 m  A C ATTENTION QUIZ

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