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Dynamic Analysis of Mechanical Systems Kinematics 1 3.3 3D Kinematics of Rigid Bodies 3.3.1 Fixed Axis Rotation A shown in the figure, let’s consider a rigid body that rotates with angular velocity about a fixed axis n-n. In the Cartesian frame O XYZ , the position of a point P on the body can be located by a position vector r . As you can see, the path of point P is an arc in the plane normal to In a small time interval t , the body swings an small angle t = . This leads to the change of r ( ) magnitude direction sin sin sin t t t = = = r r r r r r r r According to the definition of derivative, the velocity of point P can be expressed as ( ) 0 lim t t t t → = = = rr v r (1) ⊥ ⊥v v r， Taking time derivative of Eq.(1) by the chain rule leads to the acceleration of P ( ) t n = = + = + a a a r r v r r (2) where = —- the angular acceleration of the body. t = a r —-the tangential component ( )n = a r —-the normal component Dynamic Analysis of Mechanical Systems Kinematics 2 3.3.2 Fixed Point Rotation As shown in the Figure, rotation of a body about a fixed point can be regarded as the rotation about an axis called the instantaneous axis. Thus, α can be decomposed into the following form ⊥= +α α α (3) where α is aligned with the rotation axis, representing the change rate of in magnitude, while ⊥α is normal to the rotation axis, reflecting the change rate of in direction. 3.3.3 General Motion The general motion of a body can be visualized as a combination of the translation along with a point (the reference point) on the body and the rotation about an axis passing through the point. Let A and B are two points on a body. If we take B as the reference point, the position vector of A can be represented by A B AB= +r r r , =AB BAr (4) Taking time derivative of Eq.(4) gives A B AB= +v v v ABv — the velocity of A relative to B. For pure translation A B AB= =v v v 0 For pure rotation with angular velocity about an axis passing through B B A AB AB= = = v v v r0 For general motion A B AB= + v v r (5) Taking time derivative of Eq.(5) leads to the acceleration of A. ( )A B AB AB= + + a a r r (6) Dynamic Analysis of Mechanical Systems Kinematics 3 For 2D motion in the x y plane, because kω = and kα = , Eqs.(5) and (6) can be simplified as ( )( ) 2 A B AB A B AB AB B AB AB = + = + + = + v v k r a a k r k k r a k r r (7) Eq.(7) can also be expressed by using 2D coordinate vector notation. Let Ar and Br are the 2D coordinate vectors of points A and B. Then， AB A B= r r r can be expressed as cos sin AB AB AB AB l l = = = = r r u r u r ， ， (8) where = ABl r is the length of ABr , u is the unit vector of ABr , is the position angle of ABr measured from the x axis. Thus, Eq.(7) can be rewritten as 2 2 A B AB B A B AB AB B l l l = + = + = + = + v v Qr v Qu a a Qr r a Qu u (9) 0 1 1 0 = = = Q， ， whereQ is a 2 2 rotation matrix that rotates a 2D coordinate vector an angle /2 anticlockwise about the z axis such that ⊥r Qr or T 0=r Qr . In general, let 1r and 2r be two vectors lying in the x y plane, it is easy to see that the hybrid product ( )1 2 k r r can be represented by ( ) ( ) T1 2 2 1 2 1 = k r r r k r r Qr (10) The 2D coordinate vector notation will be used in kinematic and dynamic modeling and analysis of planar systems Dynamic Analysis of Mechanical Systems Kinematics 4 Sample Problem 1 (2D Problem) For a 4R four bar linkage shown in the figure, assume link 1 rotates with a constant angular velocity ( )1 rad s . Determine: the angular velocity ( 2 , 3 ) and angular acceleration ( 2 , 3 ) of link 2 and link 3 in terms of 1 ; velocity and acceleration of point 2P on link 2 and point 3P on link 3 in terms of 1 . Solution: In the Cartesian frame A xy , the position loop closure constraint equation can be represented by B CB D CD+ =r r r r 0 (1) We express these vectors in the form of 2D coordinate vectors 1 2 1 1 1 2 2 2 1 2 3 0 0 0 3 3 3 3 cos cos , sin sin cos1 , sin0 B CB D CD l l l l l l l l = = = = = r u = r u = r u r u = Taking time derivative of Eq.(1) leads to the velocity constraint equation B CB C+ =v v v 0 or 1 1 1 2 2 2 3 3 3+l l l =Qu Qu Qu 0 (2) It can be seen that there are two unknowns, i.e. 2 and 3 in Eq.(2) which also contains two scalar equations. In order to determine 2 explicitly, we take dot product on both sides of Eq.(2) with 3u and note that 3 3⊥u Qu . Thus, we have T T 1 1 3 1 2 2 3 1+ 0l l =u Qu u Qu (3) Rearranging Eq.(3) yields the explicit expression of 2 in terms of 1 ( ) ( ) T 1 3 11 3 1 2 1 1T 2 3 2 2 3 2 sin sin ll l l = = u Qu u Qu (4) Similarly, we take dot product on both sides of Eq.(2) with 2u , and note that 2 2⊥u Qu . This leads to the explicit expression of 3 in terms of 1 ( ) ( ) T 1 2 11 2 1 3 1 1T 3 2 3 3 3 2 sin sin ll l l = = u Qu u Qu (5) Dynamic Analysis of Mechanical Systems Kinematics 5 Taking point B as a reference point of link 2, the velocity of point 2P can be determined by 2 2 21 1 1 2 P P B P Bl l = +v Qu Qu (6a) ( ) ( ) 2 2 2 2 2 2 2 2 2 cos = , = sin P B P B P B P B P B l + = + r r u r Substituting Eq.(4) into Eq.(6a) leads to 2 2 2 T 1 3 1 1 1 1T 2 3 2 P P B P B l l l l = u Qu v Qu Qu u Qu (6b) In the same manner, taking point D as a reference point of link 3, the velocity of point 3P can be expressed in term of 1 . ( ) ( ) 3 3 3 3 3 3 3 3 3 3 T 1 2 1 3 1T 3 2 3 3 3 3 3 cos = , = sin P P D P D P D P D P D P D P D P D P D l l l l l = = + = + u Qu v Qu Qu u Qu r r u r (6c) Acceleration Analysis: Differentiating the velocity constraint equation with respect time gives the acceleration constraint equation 2 2 2 2 2 2 3 3 3 1 1 1 2 2 2 3 3 3l l l l l = + Qu Qu u u u (7) Taking dot product on both sides of Eq.(7) with 3u and 2u respectively, leads to 2 T 2 T 2 1 1 1 3 2 2 2 3 3 3 2 T 2 3 2 2 T 2 2 T 1 1 1 2 2 2 3 3 2 3 3 T 3 2 3 l l l l l l l l + = + = u u u u u Qu u u u u u Qu (8) To this end, the acceleration of point 2P and 3P can be readily obtained by differentiating Eq.(6) with respect time. 2 2 22 2 3 3 33 3 2 2 1 1 1 2 2 2 3 3 P B P B P D P D P P B P B P P D P D l l l l l = + = a u Qu u a Qu u (9) Important Remarks: In the velocity and acceleration analysis of four bar planar linkages, there are always two unknowns involved. It is advantageous to solve one of the unknowns explicitly by using the dot product technique. This technique is particularly efficient in the case when only the output motion is of interest. Dynamic Analysis of Mechanical Systems Kinematics 6 Sample Problem 2(3D problem) The bar AB of a spatial mechanism as shown is connected by ball socket joints to collars sliding on the fixed bars CD and EF. The point B moves with a constant velocity (m/s)20iv =B . Determine the velocity and acceleration of the collar A and the angular velocity and angular acceleration of bar AB. Solution: Setup the coordinate system as shown. Velocity Analysis Velocity of point A can be expressed by either of two ways DAAA v ev = where AD AD DA rr rr e = —- the unit vector of bar CD. A B AB= + v v r where B Bv=v i x y z = + +i j k —-the angular velocity of bar AB. The velocity constraint equation can be formulated as ABBDAA vv rie += (1) Now, we have four unknowns but only three equations are available. Note that the ball socket joint restricts the rotation about the axis of bar AB. So, an additional constraint equation can be generated 0AB =r (2) In order to solve Av , we take dot product on both sides of Eq.(1) with ABr and noting that ( ) 0AB AB =r r . This leads to A AB DA B ABv v=r e r i Hence, Av can be obtained by AB A B AB DA v v= r i r e (3) In order to solve ω , we rewrite Eq.(1) as ier BDAAAB vv = (4) and we take cross product on both sides of Eq.(4) with AB r . This leads to ( ) ( )ierrr BDAAABABAB vv = (5) The use of the Lagrangian identity (see Eq.(10) in Chapter 2) leads to ( ) ( ) ( )AB AB AB AB AB AB = r r r r r r (6) y z x 400 mm 300 mm A B C D E F ( )200, 200, 400 mm Bv Av Dynamic Analysis of Mechanical Systems Kinematics 7 Substituting Eq.(2) and (6), gives ( ) 2 AB AB AB =r r r (7) Substituting Eq.(7) into Eq.(5), finally results in ( )ier r BDAAAB AB vv = 2 1 (8) Numerical Calculation of Velocity Analysis ( ) kji kji rr rr e 3 1 3 2 3 2 1.02.02.0 )3.04.0(2.02.0 222 1.0 + = + + + = = AD AD DA Given iv 20=B , kir 3.04.0 + =AB the velocity of A can be evaluated by Eq.(3) 0.4 20 48 m/s 2 1 0.4 0.3 3 3 AB A B AB DA v v = = = + r i r e From Eq.(8) we have ( )ier r BDAAAB AB vv = 2 1 3 1 48 3 2 4820 3 2 48 3.004.0 3.04.0 1 22 + = kji ( )rad/s2.5140.4.38 163212 3.004.04 kji kji ++= = Acceleration Analysis The acceleration of point A can be expressed in two ways A A DAa=a e ( ) ABABBA rraa + += where 0B =a because Bv is a constant vector. x y z = + +i j k —- the angular acceleration of AB Hence, the acceleration constraint equation can be formulated as ( )A DA AB ABa = + e r r (9) Since ( ) ( ) ( ) 2 AB AB AB AB = = r r r r Then 2 A DA AB ABa = e r r (10) Taking dot product on both sides of Eq.(10) with ABr and noticing that ( ) 0AB AB =r r , leads to 2 2 A AB DA ABa = r e r Hence, Aa can be obtained by Dynamic Analysis of Mechanical Systems Kinematics 8 2 2 AB A AB DA a = r r e (11) In order to solve , we rewrite Eq.(10) as 2 AB A DA ABa = +r e r (12) We take cross product on both sides of Eq.(12) with AB r and note that AB AB =r r 0 . This gives ( )AB AB A AB DAa = r r r e Expanding ( ) ( ) ( )AB AB AB AB AB AB = r r r r r r and noting that 0AB =r , finally results in ( )2 1 A AB DA AB a= r e r (13) Do the numerical calculation of acceleration by yourself. Important Remarks: In velocity and acceleration analysis of four bar spatial mechanisms, there are normally four unknowns involved. One is the output/input velocity or acceleration along/about an axis. The others are three components of angular velocity or angular acceleration of the connecting rod. The procedure to solve the problem of this kind can be divided into four steps: (1) Formulate the loop closure constraint equation. (2) Formulate the additional constraint equation according to the joint type at each end of the connecting rod. This is the key point in dealing with the problem of this kind (3) Solve the output/input velocity or acceleration by using the projection technique (dot product). (4) Solve the angular velocity and acceleration of the connecting rod by using Lagrangian identity. Please solve the relevant the problem in the sample sheet. Dynamic Analysis of Mechanical Systems Kinematics 9 3.4 Moving Coordinate System According to Eq.(5) and Eq.(6) in Section 3.3, the velocity and acceleration of point A on a body are given by ( ) A B AB A B AB AB = + = + + v v r a a r r And we assume that the coordinates of these vectors are evaluated in an inertial reference frame O XYZ . However, it would be more convenient if we express the coordinates of these vectors in a moving frame as the inertial matrix of a body in the moving frame becomes a constant matrix (see 3D Dynamics in Chapter 4). As shown in the figure, by taking point B on the body as the origin, let’s setup a moving frame B xyz that rotates with angular velocity . Note that the B xyz maybe or maybe not attached to the body. If the B xyz is attached to the body, = , otherwise . In the B xyz , can be represented by kji zyx ++= (1) where i , j , k are the unit vectors of three axes of the B xyz , , x y z , are the coordinates of expressed in the B xyz . It is important to note that i , j , k rotate with angular velocity , therefore, the time derivatives of i , j , k can be represented as kkjjii = = = , , Taking time derivatives of Eq. (1) by the chain rule, leads to angular acceleration of the body expressed in the moving frame B xyz ( ) d d x y z x y z x y z x y z x y z xyzt = + + + + + = + + + + + = + + + = + i j k i j k i j k i j k ω i j k (2) Remarks: The first term in Eq.(2) represents the change rate of in amplitude, while the second term represents the change rate of in direction with respect to . If the moving frame is body-fixed, then = 0 . When the moving frame is not attached to the body, it is convenient to represent by rel += (3) rel —the angular velocity of the body relative to the moving frame. Eq,(3) is referred to as Addition Theorem of angular velocity of a rigid body. Dynamic Analy

sis of Mechanical Systems Kinematics 10 According to the addition theorem, angular velocity of the end link (link n) of an n-DOF serial kinematic chain can be expressed as the sum of the angular velocity of link i relative to link 1i , 1 1 n n i i i = = ω (4) where , 1 1i i i i = ω ω ω . IF the chain is composed of n revolute joints as shown in the figure, we can also obtain the general expression of the velocity of point 1nA + on link n by using the addition theorem (see the complementary reading below). 1 , -1 1 1 n n n A i i i i i i i i + = = = = v q ω q s (5) where i —the magnitude of , 1i i ω is —the unit vector of , 1i i ω whose direction is coincident with thi joint axis 1i n iA A+=q — the position vector from 1nA + to an arbitrary point iA on the thi joint axis. ATTENTION: The coordinates of all vectors should be expressed in the same frame when the vector addition is made. Dynamic Analysis of Mechanical Systems Kinematics 11 Complementary readings Consider an DOFn ( )6n serial kinematic chain, which is composed of n revolute joints and n movable links. Let i and is be the magnitude and unit vector of the thi revolute joint connecting link i with link 1i , and +1i i iA A=r be the position vector from iA to +1iA where iA is an arbitrary point on the thi joint axis. Determine the velocity of point 1nA + on the end link n . Consider first the case where 2n = 2 1 11 1 1 1 1 as A A A = + = =v v ω r s r v 0 (1) 3 2 2 2 2 2 2 1 A A A i i i= = + = + v v ω r v s r (2) Substituting Eq.(1) into Eq.(2), results in ( ) ( ) ( ) 3 2 1 2 1 2 1 1 2 1 1 1 1 1 1 2 1 2 2 2 1 2 2 2 2 2 1 A i i i i i i i i i i i i i i i i i i = = = = + = + + = + + + = + + = v s r s r s r s r r r s r s r r r s r s r r s r q s (3) where ( )1 1 2 3 1A A= + =q r r , 2 2 3 2A A= =q r Combining Eq.(3) with the addition theorem 2 2 1 i i i = = ω s ,leads to 3 2 12 A i i i i i = = v q s sω (4) In general 1 1 1 n n n A i i i i i i iin + = = = = = v q s ξ ξ sω (5) ξ ——The twist of link n about point 1nA + . iξ ——The unit twist of joint i about point 1nA + Conclusion: The twist of link n about point 1nA + can be expressed as the linear combination of unit twists of all joints about point 1nA + . Think: The case where a prismatic joint is used. Implementation of acceleration analysis. Dynamic Analysis of Mechanical Systems Kinematics 12 Sample Problem 1 The bar OA with the length of l rotates about the x-axis with constant rate . The entire assembly rotates about the z-axis with constant rate . For the given angle , determine the angular velocity and acceleration of bar OA, and the velocity and the acceleration of point A. Solution Establish the moving frame xyzO as shown. The angular velocity and acceleration analysis of bar OA, The angular velocity of the moving frame k = The angular velocity of the bar OA relative to the moving frame i =rel Thus, the angular velocity of the bar can be expressed as ki += Because and are constants, taking time derivative of leads to the angular acceleration of the bar OA ( ) jkikω =+ = = The velocity and acceleration analysis of point A The position vector of point A can be represented by kjr sincos ll += For a fixed point rotation, the velocity of point A can be expressed as kji kji rv cossincos sincos0 0 lll ll + == = The acceleration of A can be represented by = + a r v where ( ) ikjjr sinsincos lll =+ = ( ) kji kji v sincossin cossincos 0 222 lll lll + = = The acceleration of point A can finally be obtained by ( ) kjia sincossin2 22 lll + = z O l x y Dynamic Analysis of Mechanical Systems Kinematics 13 Sample Problem 2 A shown in the figure, the bent bar is rigidly attached to the vertical shaft which rotates with a constant rate 0 .The circular disk with radius R is pinned to the bent bar and undergoes a pure roll on the horizontal surface. Determine the disk’s angular velocity disk and angular acceleration disk Solution Setup a moving frame B-xyz fixed to the bent bar where the x axis is coincident with horizontal part of the bar and the y axis is coincident with rotation axis of the vertical shaft. Since the B xyz is attached to the bar, then 0bar = = j While, the angular velocity of the disk relative to the B xyz can be formulated by ( )cos sinrel rel = +i j rel — the magnitude rel is the unknown. According to addition theorem, the angular velocity of the disk can be expressed as ( )0+ cos sindisk bar rel rel rel = = + +i j Now, we need to establish a constraint equation for solving rel . Because of the rolling without slipping, point C is the instantaneous center of zero velocity of the disk. This means C A disk CA= + =v v r 0 where ( )0 0 ( cos ) sin ( cos ) A bar AB h b b h b = = + = + v r j i j k and the position vector from A to C can be formulated by sin cosCA R R = r i j Thus 0 0( cos ) cos sin 0 sin cos 0 C A disk CA rel relh b R R = + = + + + = v v r i j k k 0 0 Expanding this equation and equating its k component, leads to 0 cos sinrel h b R R = + Since the coordinates of disk in the B xyz are constants, ( )0+ cos sindisk bar rel rel rel = = + +i j the angular acceleration of the disk can finally be obtained by ( )( )0 0 0 cos sin cos disk disk rel rel rel = = + + = j i j k Dynamic Analysis of Mechanical Systems Kinematics 14 Sample Problem 3 As shown in the figure, the collar and clevis A are given a constant upward velocity 0.2m/s. That causes the ball end of the bar AB to slide in the radial slot in a rotating disk. Suppose the disk rotates at constant rate 2 rad/s = , determine the angular acceleration α of the bar AB when the z-coordinate of A reaches 75 mmz = . Solution: Establish the moving frame xyzO attached to the disk as shown in the figure. The angular velocity of the xyzO is given by =Ω k (1) The angular velocity of bar AB relative to the xyzO is given by rel rel = ω i (2) The angular velocity of bar AB is given by rel rel = + = +ω Ω ω i k (3) In order to determine rel , the velocity constraint equation due to the loop closure of the mechanism can be formulated by B A BA= + v v ω r (4) where BA B A y z= = r r r j k (5) A z=v k (6) B y y= +v j j (7) Substituting = j Ω j and =Ω k into Eq.(7), gives B y y = v j i (8) Substituting Eqs.(5), (6) and (8) into Eq.(4) and equating the j and k components allows rel to be expressed as rel rel z y y z = = ， (9) Then, substituting the first expression into Eq.(3) leads to the expression of the angular velocity of the bar AB in terms of z , y and rel z y = + = +ω Ω ω i k , 2 2y l z= (10) Hence, taking time derivative of ω leads to the angular acceleration of the bae AB ( ) ( ) 2 2 2 2 zy z zy z y y y y zy z zy z y y y y = = = = i i i Ω i i k i i j (11) Finally, substituting ( )y z y z= , 22 zly = , and the assumed date 0.25 m/sz = , 75 mmz = , 2 rad/s = , 125 mml = into Eq.(11), results in 23 4 (rad/s )= i j (12) Dynamic Analysis of Mechanical Systems Kinematics 15 Sample Problem 4 As shown in the figure, the motor turns the disk at the constant rate 30 rad/sp = . The motor is also rotating about the horizontal axis BO at a constant rate 2 rad/s = . Simultaneously, the entire assembly is rotating about the vertical axis C-C at a constant rate 8 rad/sq = . For the instant 30 = , determine the angular velocity and angular acceleration of the disk. Solution According to the addition theorem, the angular velocity of the disk diskω can be expressed as / /disk shaft motor shaft disk motor= + +ω ω ω ω (1) shaftω — the angular velocity of the horizontal shaft BO /motor shaftω —-the angular velocity of the motor relative to the horizontal shaft BO /disk motorω ——the angular velocity the disk relative to the motor Setup the moving frame xyzO attached to the motor as shown such that motor=Ω ω and /rel motor shaft=ω ω . In the xyzO kiω cossin qqshaft += /motor shaft = ω j , /disk motor p=ω i Thus sin cosmotor q q = = +ω Ω i j k and ( )/ sin cosdisk disk motor p q q = + = + +ω Ω ω i j k (2) It can be seen from Eq. (2) that p , q and are constants but varies with time. So, taking time derivative of diskω gives the angular acceleration of the disk diskα ( ) ( ) ( ) d d d sin cos d d d cos sin sin cos sin cos cos cos sin disk motor diskp q q t t t q q q q p q q q pq p q = + + + = + + = + + α i j k ω ω i j k i k i j k (3) Substituting the assumed data into this equation, finally results in ( ) 2 cos cos sin (8 2 .866) 30 8 .866 2 (30 8 .5) 13.4 208 52 (rad/s ) disk q pq p q = + + = + + = + + α i j k i j k i j k (4) Dynamic Analysis of Mechanical Systems Kinematics 16 Sample Problem 5 The figure shows a 5-DOF serial robot. The arm 1 rotates about the z-axis at the constant rate 1 , the arm 2 rotates about x-axis at constant rate 2 = ，the arm 3 rotates about the axis 21 OO at constant rate 3 , and the arm 4 rotates about the axis which is through 2O and momentarily parallel to the x-axis at the constant rate =4 . Finally, the arm 5 rotates about the axis AO 2 at the constant rate 5 . For the given configuration as shown, evaluate the angular velocity and angular acceleration of the arm 5 in the moving frame 1O xyz attached to the arm 1. Solution: According to the addition theorem, the angular velocity of the arm 5 can be expressed by 5 5 , 1 1 i i i = = ω ω (1) where , 1i i ω denotes the angular velocity of arm i relative to arm 1i ( 1~5i = ) At the instant shown, , 1i i ω evaluated in the 1O xyz can be formulated by 1,0 1 =ω k 2,1 2 = =ω i i 3 2 3 3cos sin = +ω j k， (2) 4 3 4 = = ω i i， ( ) ( )5,4 5 5cos sin = + ω j k Substituting these expressions into Eq.(1) leads to ( ) ( ) ( ) ( )( ) ( )( ) 5 1 3 3 5 5 3 5 1 5 cos sin cos sin cos cos sin = + + + + + = + + + + ω k i j k i j k i j k (3) Do the angular acceleration analysis of the robot by yourself Dynamic Analysis of Mechanical Systems Kinematics 17 Complementary readings Based upon the velocity and acceleration expressions of point A on a rigid body provided that point A and point B are on the same body, AB ABBA v rvv += (1) ( ) AB A B AB AB= + + a a r r a (2) we now consider more general case where point A and point B are not on the same body. As seen from the figure, setup a moving frame B xyz and take point B as the reference point. The point A can be expressed as A B AB= +r r r , AB x y z= + +r i j k (3) where , ,x y z are the coordinates of vector ABr expressed in the B xyz , , ,i j k are the unit vectors of three axes of the B xyz Taking time derivative of Ar , leads to AB x y z x y z= + + + + +v i j k i j k , , = = = i Ω i j Ω j k Ω k Thus AB AB rel= +v Ω r v (4) where rel x y z= + +v i j k ——velocity of point A relative to the B xyz as observed in the B xyz . Thus, the velocity of point A can be expressed as A B AB rel= + +v v Ω r v (5) Taking time derivative of Av by the chain rule leads to the acceleration of point A dd d d d d relAB A B AB t t t = + + + vΩ r a a r Ω where ( ) d d AB AB AB rel t = = + r Ω Ω v Ω Ω r v d d rel rel rel t = + v a Ω v , rel x y z= + +a i j k Thus ( ) 2A B AB AB rel rel= + + + +a a Ω r Ω Ω r Ω v a (6) If =Ω ω , and rel relv a are zero vectors. Thus, Eq.(5) is identical to Eq.(1), and Eq.(6) is identical to Eq.(2) Dynamic Analysis of Mechanical Systems Kinematics 18 Example 6 The motor housing and its bracket rotate about the Z-axis at the constant rate 3 rad/s= . The motor shaft and disk have a constant rate 8 rad/sp = with respect to the motor housing in the direction shown. Given 30= , determine the velocity and acceleration of point A at the top of the disk at the instant shown. Solution Setup the moving frame B xyz attached to the motor cos sin = K i j k , cos = K j i , sin =K k i Note that there is no relative rotation between the motor housing and bar OB, so Ω=Ω K Velocity of A is given by A B AB rel= + +v v Ω r v where , ,B OB Y OB YΩ r Ωr= = v K J i ( ) ( ) , , , ,cos sin AB AB y AB z AB y AB z Ω r r Ω r r = + = + Ω r K j k i ( ), , ,rel AB AB y AB z AB zp r r pr= = + =v p r j j k i Thus ( )( ), , , ,cos sin A B AB rel AB y AB z OB Y AB zΩ r r r pr = + + = + + v v Ω r v i Do the acceleration analysis by yourself using Eq.(6). O C