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INSTANTANEOUS CENTER OF ZERO VELOCITYToday’s Objectives:Students will be able to:1. Locate the instantaneous center ofzero velocity.2. Use the instantaneous center todetermine the velocity of any pointon a rigid body in general planemotion.In-Class Activities: Check Homework Reading Quiz Applications Location of theInstantaneous Center Velocity Analysis Concept Quiz Group Problem Solving Attention QuizDynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

READING QUIZ1. If applicable, the method of instantaneous center can be usedto determine the of any point on a rigid body.A) velocityB) accelerationC) velocity and accelerationD) force2. The velocity of any point on a rigid body is tothe relative position vector extending from the IC to thepoint.A) always parallelB) always perpendicularC) in the opposite directionD) in the same directionDynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

APPLICATIONSThe instantaneous center (IC)of zero velocity for thisbicycle wheel is at the pointin contact with ground. Thevelocity direction at any pointon the rim is perpendicular tothe line connecting the pointto the IC.Which point on the wheel has the maximum velocity?Does a larger wheel mean the bike will go faster for thesame rider effort in pedaling than a smaller wheel?Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

APPLICATIONS (continued)As the board slides down the wall (tothe left), it is subjected to generalplane motion (both translation androtation).Since the directions of the velocitiesof ends A and B are known, the IC islocated as shown.How can this result help you analyzeother situations?What is the direction of the velocity of the center of gravity ofthe board?Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

INSTANTANEOUS CENTER OF ZERO VELOCITY(Section 16-6)For any body undergoing planar motion, there always exists apoint in the plane of motion at which the velocity isinstantaneously zero (if it is rigidly connected to the body).This point is called the instantaneous center (IC) of zerovelocity. It may or may not lie on the body!If the location of this point can be determined, the velocityanalysis can be simplified because the body appears to rotateabout this point at that instant.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

LOCATION OF THE INSTANTANEOUS CENTERTo locate the IC, we use the fact that the velocity of a point on abody is always perpendicular to the relative position vector fromthe IC to the point. Several possibilities exist.First, consider the case when velocity vAof a point A on the body and the angularvelocity w of the body are known.In this case, the IC is located along theline drawn perpendicular to vA at A, adistance rA/IC vA/w from A.Note that the IC lies up and to the rightof A since vA must cause a clockwiseangular velocity w about the IC.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

LOCATION OF THE INSTANTANEOUS CENTER(continued)A second case occurs when thelines of action of two nonparallel velocities, vA and vB,are known.First, construct line segmentsfrom A and B perpendicular tovA and vB. The point ofintersection of these two linesegments locates the IC of thebody.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

LOCATION OF THE INSTANTANEOUS CENTER(continued)A third case is when the magnitude and direction of two parallelvelocities at A and B are known. Here the location of the IC isdetermined by proportional triangles.As a special case, note that if the body is translating only(vA vB), then the IC would be located at infinity. Then wequals zero, as expected.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

VELOCITY ANALYSISThe velocity of any point on a body undergoing general planemotion can be determined easily, often with a scalar approach,once the instantaneous center of zero velocity of the body islocated.Since the body seems to rotate aboutthe IC at any instant, as shown in thiskinematic diagram, the magnitude ofvelocity of any arbitrary point isv w r, where r is the radial distancefrom the IC to the point.The velocity’s line of action isperpendicular to its associated radialline.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

EXAMPLE IGiven: A linkage undergoingmotion as shown. Thevelocity of the block, vD,is 3 m/s.Find: The angular velocitiesof links AB and BD.Plan: Locate the instantaneous center of zero velocity of linkBD and then solve for the angular velocities.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

EXAMPLE I (continued)Solution: Since D moves to the right, it causes link AB torotate clockwise about point A. The instantaneous center ofvelocity for BD is located at the intersection of the linesegments drawn perpendicular to vB and vD. Note that vB isperpendicular to link AB. Therefore we can see that the IC islocated along the extension of link AB.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

EXAMPLE I (continued)Using these facts,rB/IC 0.4 tan 45 0.4 mrD/IC 0.4/cos 45 0.566 mSince the magnitude of vD is known,the angular velocity of link BD can befound from vD wBD rD/IC .wBD vD/rD/IC 3/0.566 5.3 rad/sLink AB is subjected to rotation about A.wAB vB/rB/A (rB/IC)wBD/rB/A 0.4(5.3)/0.4 5.3 rad/sDynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

EXAMPLE IIGiven: The center O of the gear set rolls with vO 6 m/s.The gear rack B is fixed.Find: The velocity of point A on the outer gear.Plan: Locate the IC of the smaller gear. Then calculate thevelocities at A.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

EXAMPLE II (continued)Solution:Note that the gear rollswithout slipping. Thus, theIC is at the contact pointwith the gear rack B.The angular velocity of the wheel can be found fromw vO/rO/IC 6/0.3 20 rad/s ( or CW)The velocity at A will bevA w rA/IC (-20 ) k ( 0.6 i 0.3j) (6 i 12 j) m/svA 62 122 13.4 m/sq tan-1(12/6) 63.4 Dynamics, Fourteenth EditionR.C. HibbelerqCopyright 2016 by Pearson Education, Inc.All rights reserved.

CONCEPT QUIZ1. When the velocities of two points on a body are equal inmagnitude and parallel but in opposite directions, the IC islocated atA)B)C)D)infinity.one of the two points.the midpoint of the line connecting the two points.None of the above.2. When the direction of velocities of two points on a body areperpendicular to each other, the IC is located atA)B)C)D)infinity.one of the two points.the midpoint of the line connecting the two points.None of the above.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

GROUP PROBLEM SOLVINGGiven: Rod CD is rotatingwith an angularvelocity wCD 4 rad/sCCW.Find: The angular velocitiesof rods AB and BC.Plan: This is an example of the second case in the lecture notes.Since the direction of Point B’s velocity must beperpendicular to AB, and Point C’s velocity must beperpendicular to CD, the location of the instantaneouscenter, I, for link BC can be found.Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

GROUP PROBLEM SOLVING (continued)Solution:Draw kinematic diagrams for Link CD and Link AB:Link CD:vCLink AB:wABrCDwCD 4 rad/svC wCD (rCD) 4 (0.5) 2 m/sDynamics, Fourteenth EditionR.C. HibbelerrABvBvB wAB (rAB) m/sCopyright 2016 by Pearson Education, Inc.All rights reserved.

GROUP PROBLEM SOLVING (continued)With the results of vB and vC, the IC for link BC can be located.Kinematic diagram for Link BC:vCvBwBC30 rB/ICrC/IC ICrC/IC (0.4) tan30 rC/IC 0.2309 mrB/IC 0.4 / cos30 rB/IC 0.4619 mThen, the kinematics gives:vC wBC (rC/IC) 2.0 wBC (0.2309 )wBC 8.66 rad/svB wBC (rB/IC) wAB (rAB) 8.66 (0.4619 ) wAB (1)wAB 4.0 rad/sDynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

ATTENTION QUIZ1. The wheel shown has a radius of 15 in and rotates clockwiseat a rate of w 3 rad/s. What is vB?A) 5 in/sB) 15 in/sC) 0 in/sD) 45 in/s2. Point A on the rod has a velocity of 8 m/s to the right.Where is the IC for the rod?A) Point A. CB) Point B.C) Point C.D) Point D.Dynamics, Fourteenth EditionR.C. HibbelerD Copyright 2016 by Pearson Education, Inc.All rights reserved.

Dynamics, Fourteenth EditionR.C. HibbelerCopyright 2016 by Pearson Education, Inc.All rights reserved.

with an angular velocity w CD 4 rad/s CCW. Find: The angular velocities of rods AB and BC. This is an example of the second case in the lecture notes. Since the direction of Point B’s velocity must be perpendicular to AB, and Point C’s velocity must be perpendicular to CD, the locati