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these components are andTheir sense of directionis established from the diagram such thatacts perpendicular to in accordance with the rotationalmotion of the body, and is directed from BtowardsA.*Acceleration Equation.•Represent the vectors in graphicallyby showing their magnitudes and directions underneath eachterm. The scalar equations are determined from the xandycomponents of these vectors.aB=aA+1aB>A2t+1aB>A2n1aB>A2nArB>A,1aB>A2t1aB>A2n=v2rB>A.1aB>A2t=arB>A1aB>A2n1aB>A2taB=aA+A*rB>A-v2rB>AaB=1aB2t+1aB2n.aA=1aA2t+1aA2nrB>A.A,V,aB,aA,vBvAV*The notation may be helpful in recallingthatAis assumed to be pinned.aB=aA+1aB>A1pin22t+1aB>A1pin22nThe mechanism for awindow is shown.Here CArotates about a fixed axisthroughC,and ABundergoes general planemotion.Since point Amoves along a curved pathit has two components of acceleration, whereas pointBmoves along astraight track and thedirection of its accelerationis specified.ACB(aA)n(aA)taBV,A
366CHAPTER16PLANARKINEMATICS OF ARIGIDBODY16The rod ABshown in Fig. 16–27ais confined to move along theinclined planes at AandB. If point Ahas an acceleration of and a velocity of , both directed down the plane at the instantthe rod is horizontal, determine the angular acceleration of the rod atthis instant.SOLUTION I (VECTOR ANALYSIS)We will apply the acceleration equation to points AandBon the rod.To do so it is first necessary to determine the angular velocity of therod. Show that it is dusing either the velocityequation or the method of instantaneous centers.Kinematic Diagram.Since points AandBboth move alongstraight-line paths, they have nocomponents of acceleration normalto the paths.There are two unknowns in Fig. 16–27b, namely,and Acceleration Equation.aB=aA+A*rB>A-v2rB>Aa.aBv=0.283 rad>s2 m>s3 m>s2EXAMPLE16.14Carrying out the cross product and equating the iandjcomponentsyields(1)(2)Solving, we havedAns.SOLUTION II (SCALAR ANALYSIS)From the kinematic diagram,showing the relative-accelerationcomponents and Fig. 16–27c, we haveEquating the xandycomponents yields Eqs. 1 and 2, and the solutionproceeds as before.caBa45°d=c3 m>s2c45°d+ca110 m2cd+c10.283 rad>s22110 m2;daB=aA+1aB>A2t+1aB>A2n1aB>A2n,1aB>A2ta=0.344 rad>s2aB=1.87 m>s2a45°aBsin 45°= -3 sin 45°+a1102aBcos 45°=3 cos 45°-10.283221102aBcos 45°i+aBsin 45°j=3 cos 45°i-3 sin 45°j+1ak2*110i2-10.28322110i210 mBA(a)vA2 m/saA3 m/s24545xyAB4545aA3 m/s2rB/Av0.283 rad/saB(b)AAB10 mrB/A(aB/A)tarB/A(aB/A)nv2rB/Av0.283 rad/s(c)AFig. 16–27
16.7RELATIVE-MOTIONANALYSIS: ACCELERATION36716EXAMPLE16.15At a given instant, the cylinder of radius r, shown in Fig. 16–28a, has anangular velocity and angular acceleration Determine thevelocity and acceleration of its center Gand the acceleration of thecontact point at Aif it rolls without slipping.