56 Kinematics of a rigid body in plane motion In Fig. 5.10 the instantaneous centre for member BC is found to be the intersection of AB and DC, since the velocity of B is perpendicular to AB and the velocity of C is perpendicular to CD . If the velocity of B is known then (5 * 8) VB OC VE we=-=-=- I2B 12C IZE Each point on link CB is, instantaneously, rotating about 12. Figure 5.9 Returning to Fig. 5.6 and assuming that OAB is anticlockwise and of given magnitude, we can 5.6 Velocityimage place the points a, b and d on the diagram (Fig. 5.9). Note that U and dare the same point as there is no relative velocity between A and D. TO construct the Point C we must view the motion of c from two vantage Points, namely D and B. Since DC is of fixed length, the only motion of C relative to D is perpendicular to DC; hence we draw dc perpendicular to DC. Similarly the velocity of c relative to B is perpendicular to CB; hence we draw bc perpendicular to BC. The intersection of these two lines locates c. The angular velocity of CB is obtained from v~/B/CB (clockwise). The direction of rotation is determined by observing the sense of the velocity of C relative to B and remembering that the relative velocity is due only to the rotation of CB. Note again that angular velocity is measured with respect to a plane and not to any particular point on the plane. 5.5 Instantaneous centre of rotation Another graphical technique is the use of instantaneous centres of rotation. The axes of rotation of DC and AB are easily seen, but BC is in general plane motion and has no fixed centre of rotation. However, at any instant a point of zero velocity may be found by noting that the line joining the centre to a given point is perpendicu- lar to the velocity of that point. If the velocity diagram has been constructed for two points on a rigid body in plane motion, then the point on the velocity diagram for a third point on the link is found by constructing a triangle on the vector diagram similar to that on the space diagram. Hence in our previous example a point E situated at, say, one third of the length of BC from c will be represented on the velocity diagram by a point e such that ce/cb = 5, as shown in Fig. 5.9. More generally, see Fig. 5.11, since ab is perpendicular to AB, ac is perpendicular to AC and bc is perpendicular to BC, triangle abc is similar to triangle ABC. Figure 5.10 Figure 5.1 1 Problems with sliding joints In the mechanism shown in Fig. 5.12, the block or slider B is free to move in a slot in member AO. In order to construct a velocity diagram as shown in Fig. 5.13, we designate a point B’ fixed on the link A0 coincident in space with B. The velocity of B relative to C is perpendicular to CB, the velocity of B’ relative to 0 is perpendicular to OB ’ and the velocity of B relative to B ’ is parallel to the tangent of the slot at B. Figure 5.12 5.9 Simple spur gears 57 -,- " Figure 5.13 The two mechanisms used as examples, namely the four-bar chain and the slidercrank chain, employ just two methods of connection which are known as turning pairs and sliding pairs. It is remarkable how many mechanisms are con- structed using just these simple arrangements. Figure 5.17 5.7 Acceleration diagrams The complete acceleration diagram for the Having constructed the velocity diagram, it is now mechanism can now be constructed as shown in possible to draw the relevant acceleration Fig. 5.17 (see also example 5.1). The acceleration diagram. The relative acceleration between two of C is given by the line ac and the angular points is shown in polar co-ordinates in Fig. 5.14. acceleration of CB is given by cc'/CB (clockwise), since cc ' = hcB CB. 5.8 Acceleration image In the same way that the velocity of a point on a rigid body may be constructed once the velocities of any two other points are known, the I acceleration can be found from the known accelerations of two other points. uUA = o rl erl + hrl eel Figure 5.14 2 If AB is of fixed length, then only two depends on the angular velocity, which is known from the velocity diagram, and the other term depends on the angular acceleration, which is unknown in magnitude but is in a direction perpendicular to AB. components remain (see Fig. 5.15). One term am = o 2 r2er2 + hrzee2 ._ Figure 5.18 From Fig. 5.18, the angle between (IUA and Figure 5.15 rNc is Referring to the four-bar chain shown in Fig. 5.6 and given the angular acceleration of link AB, the acceleration vector of B relative to A may be drawn (Fig-5.16)- Note carefully the directions of the accelerations: B is accelerating centripetally towards A. arctan (2) = arctan (3) which is independent of rl. The angle between QUA and aB,C is therefore the Same as the angle between rl and r2; hence the triangle abc in the acceleration diagram is similar to triangle ABC. 5.9 Simple spur gears When two spur gears, shown in Fig. 5.19, mesh together, the velocity ratio between the gears will be a ratio of integers if the axes of rotation are Figure 5.16 58 Kinematics of a rigid body in plane motion (5.11) @A- wC rB %-@C rA - or Figure 5.19 fixed. If the two wheels are to mesh then they must have the same circular pitch, that is the distance between successive teeth measured along the pitch circle must be the same for both wheels. If Tis the number of teeth on a wheel then the circular pitchpc is rDlT, where D is the diameter of the pitch circle. The term ‘diametral pitch’ is still used and this is defined as P = TID. Another quantity used is the module, m = DIT. The number of teeth passing the pitch point in unit time is 27rwT, so for two wheels A and B in mesh l@ATAl= Iw~TBI (5.9) @A DB TB % DA TA = - or the minus sign indicating that the direction of rotation is reversed. Figure 5.21 that is the motion relative to the arm or carrier is independent of the speed of the arm. For example, if oc = 0 we have the case of a simple gear train where (5.12) OA rB % rA _- - Figure 5.22 U W Figure 5.22 shows a typical arrangement for an epicyclic gear in which the planet is free to rotate on a bearing on the carrier, which is itself free to rotate about the central axis of the gear. If the carrier is fixed, the gear is a simple gear train so that the velocity ratio Figure 5.20 Figure 5.20 shows a compound gear train in which wheel B is rigidly connected to wheel C; thus % = wc . The velocity ratio for the gear is OD %% _- - - TS @s % 0s TA @A @C @A - @A =($)(-zi)= - wA% = (3)( -2) = - Tc TA TD TB (5.10) Note that the direction of rotation of the annulus is the same as that of the planet, since the annulus is an internal gear. Also, we see that the number of teeth on the planet wheel does not affect the velocity ratio - in this case the planet is said to act an an idler. If the carrier is not fixed, then the above velocity ratio is still valid provided the angular speeds are relative to the carrier; thus 5.10 Epicyclic motion If the axle of a wheel is itself moving on a circular path, then the motion is said to be epicyclic. Figure 5.21 shows the simplest type of epicyclic motion. If no slip occurs at P, the contact point, then the velocity of P is given as VPlOl = VO2/01+ VPl02 (5.13) @A-@C -3 - hence WArA=@C(rA+rB)-(L)SrB @S @C TA 5.1 1 Compound epicyclic gears 59 If two of the speeds are known then the third may be calculated. In practice it is common to fix one of the elements (i.e. sun, carrier or annulus) and use the other two elements as input and output. Thus we see that it is possible to obtain three different gear ratios from the same mechanism. 5,ll Compound epicyclic gears In order to obtain a compact arrangement, and also to enable a gearbox to have a wider choice of selectable gear ratios, two epicyclic gears are often coupled together. The ways in which this coupling can occur are numerous so only two arrangements will be discussed. The two chosen are common in the automotive industry and between them form the basis of the majority of automatic gearboxes. Simpson gear train In the arrangement shown in Fig. 5.23(a), the two sun wheels are on a common shaft and the carrier of the first epicyclic drives the annulus of the second. This second annulus is the output whilst the input is either the sun wheel or the annulus of the first epicyclic. This design, used in a General Motors 3-speed automatic transmission, provides three forward gears and a reverse gear. These are achieved as follows. First gear employs the first annulus as input and locks the carrier of the second. Second gear again uses the first annulus as input but fixes the sun wheel shaft. Third is obtained by locking the first annulus and the sun wheel together so that the whole assembly rotates as a solid unit. Reverse gear again locks the second carrier, as for the first gear, but in this case the drive is via the sun wheel. Figure 5.23(b) shows a practical layout with three clutches and one band brake which carry out the tasks of switching the drive shafts and locking the second carrier or the sun wheel shaft. To engage first gear drive is applied to the forward clutch and the second carrier is fixed. In normal drive mode this is achieved by means of the one-way Sprag clutch. This prevents the carrier from rotating in the negative sense, relative to the drive shaft, but allows it to free-wheel in the positive sense. This means that no engine braking is provided during over-run. To provide engine braking the reverse/low clutch is engaged in the lock-down mode. For second gear the reverseAow clutch (if applied) is released and the intermediate band brake is applied, thus locking the sun wheel. For third gear the intermediate band is released and the direct clutch activated hence locking the whole gear to rotate in unison. For reverse gear the forward clutch is released, then the direct clutch and the reverse/low clutch are both engaged thus only the second epicyclic gear is in use. The operation of the various clutches and band brakes is conventionally achieved by a hydraulic circuit which senses throttle position and road speed. The system is designed to change down at a lower speed than it changes up at a given throttle position to prevent hunting. Electronic control is now used to give more flexibility in changing parameters to optimise for economy or for performance. To determine the gear ratios two equations of the same type as equation 5.13 are required and they are solved by applying the constraints dictated by the gear selected. A more convenient set of symbols will be used to represent rotational speed. We shall use the letter A to refer to the annulus, C for the carrier and S for the sun, also we shall use 1 to refer to the first simple epicyclic gear and 2 for the second. In this notation, for example, the speed of the second carrier will be referred to as C2. For the first epicyclic gear and for the second epicyclic gear (5.14) (5.15) Where R is the ratio of teeth on the annulus to teeth on the sun. In all cases S2 = S1 and C1 = A2 = wo , the output. With the first gear selected C2 = 0 and Al = oi, the input. From equation 5.14 S1 = -wi X R1 + wo(l+ R1 ) and from equation 5.15 S1 = -wo x R2 wo (1 + R1+ R2 1 R1 Eliminating S1 wi = thus the first gear ratio = wi/wo = (1 + R1 + Rt)/R1 With second gear selected S1 = 0 and wi is still AI . 60 Kinematics of a rigid body in plane motion Figure 5.23(a) Figure 5.23(b) From equation 5.14 0 = wo(l + R1 ) - wi x R1 thus the second gear ratio qlwg = (1 + R1)/R1 Summarising we have GEAR 15t (1 + R1+ R2)lRI 2nd GEAR RATIO The third gear is, of course, unity. For the reverse gear C2 = 0 and wi = S1 so from (1 + R1 )lRl equation 5.17 3rd 1 milog = - R2 Reverse -R2 5.1 1 Compound epicyclic gears 61 Figure 5.23(c) Figure 5.23(d) Ravigneaux gearbox The general arrangement of the Ravigneaux gear is shown in Fig. 5.23(c). This gear is used in the Borg Warner automatic transmission which is to be found in many Ford vehicles. In this design there is a common planet carrier 62 Kinematics of a rigid body in plane motion Discussion examples Example 5.1 The four-bar chain mechanism will now be analysed in greater detail. We shall consider the mechanism in the configuration shown in Fig. 5.24 and determine vc, z+, oz, w3, aB, ac, aE , ;2 and h, , and the suffices 1, 2, 3 and 4 will refer throughout to links AB, BC, CD and DA respectively. and the annulus is rigidly connected to the output shaft. The second epicyclic has two planets to effect a change in the direction of rotation compared with a normal set. In the actual design, shown in Fig. 5.23(d), the first planet wheel doubles as the idler for the second epicyclic gear. When first gear is selected, the front clutch provides the drive to the forward sun wheel and the common carrier is locked, either by the rear band brake in lock-down mode or by the free- wheel in normal drive. For second gear the drive is still to the forward sun wheel but the reverse sun wheel is fixed by means of the front band brake. For top gear drive both suns are driven by the drive shaft thereby causing the whole gear train to rotate as a unit. For the reverse gear the rear clutch applies the drive to the reverse sun wheel and the carrier is locked by the rear band brake. For the first gear the input wj= S2 and C1 = C2 = 0, the output wo = Al = AZ. So, from equation 5.15, Figure 5.24 Velocities In general, for any link PQ of length R and rotating with angular velocity w (see Fig. 5.25(a)) we have, from equation 2.17, SZ=R2XA therefore wi/wo = S21A = R2 For second gear S2 is the input but SI = 0 From equation 5.14 0 = -AX R1 + (1 + RI)C and from equation 5.15 S2 = R2 X A + C(l- R2) Elimination of C gives S2 = R2 + A X RI X (1 - R2)/(1+ R1) R1 +R2 w~IwO = S2IA = ___ 1+Rl thus The top gear ratio is again unity. Reverse has C= 0 with input S1 so from equation 5.14 Sl=-RlXA giving the gear ratio witwo = S11A = -R1. Summarising we have GEAR GEAR RATIO 1st R2 2nd (Rl + R2)/(1+ R1) 3rd 1 Reverse -R1 Figure 5.25 VQfp = Rer + Roee If PQ is of fixed length then R = 0 and VQ/P has a magnitude Rw and a direction perpendicular to the link and in a sense according the the direction of 0. Velocity diagram (section 5.4). Since II is constant, the magnitude of vBIA is wllI and its direction is perpendicular to AB in the sense indicated in Fig. 5.25(b), so we can draw to a suitable scale the vector ab- which represents Z)B/~. The velocity of C is determined by considering the known directions of vUB and VUD Discussionexamples 63 Link Velocity Direction Sense Magnitude (ds) Line AB %/A LAB \ (AB)wl = (0.15)12 = 1.8 ab BC Z)C/B LBC ? (BC)o, = ? bc CD ~crr) ICD ? (CD) w3 = ? cd From the concept of the velocity image we can find the position of e on bc from be BE bc E _- - Thus be = 1.28 - =0.337ds t:0) and by noting that (see equation 2.24) The magnitude of + is ae and this is found (i) ve1ocity. There are sufficient data to draw the Znstantaneous centre (section 5.5). In Fig. 5.28, ve1ocity triang1e representing equation (i) I, the instantaneous centre of rotation of BC, is at (Fig. 5.26). the intersection of AB and CD. The triangle IBC From this figure it can be Seen that the location rotates instantaneously about I. From the known of point C on the velocity diagram is the direction of vB, the angular velocity of the intersection of a line drawn through b perpen- triangle is clearly Seen to be clockwise. dicular to BC and a line drawn through a, d perpendicular to DC. By scaling we find that the magnitude of dc is 1.50 ds and thus vUA = 'uC = 1.50 ds 14" from the diagram to be 1.63 ds. Thus VUA = %/A -k VUB and vUA = vUD since A and D each have zero VE = 1.63 d~ 20" The magnitude of 02 is VB wl(AB) 12(0.15) - = 6.7 rads Cr);!=-= IB IB and q = -6.7 k rads 0.27 The magnitude of w;? is bc 1.28 BC 0.19 w;?=- =- =6.7rad/s To determine the direction, we note that vuB, the velocity of C relative to B is the sense from b to c (and that %IC is in the opposite sense) so that BC is rotating clockwise (see Fig. 5.27). Thus % = -6.7 k rads The magnitude of q is where k is the unit vector coming out of the page. cd 1.5 CD 0.15 03=-= - 10 rads and the direction is clearly anticlockwise, so that o3 = 10 k rads e- - The magnitude of vc is VC = %(IC) = 6.7(0.225) = 1.50 ds and the sense is in the direction shown. The magnitude of q is VC 1.47 = 9.8 rads w3=-=- CD 0.15 and the sense is clearly anticlockwise so that w3 = 9.8k rads 64 Kinematics of a rigid body in plane motion Point E lies on link BC so that the instant- aneous centre for E is also I. The magnitude of % is and the sense is in the direction shown. are obviously due to inaccuracies in drawing. Accelerations For any link PQ of length R, angular velocity w and angular acceleration h (see Fig. 5.29) we have, from equation 2.18, + = %(IE) = 6.7(0.245) = 1.64 m/s The discrepancies between the two methods The magnitude of 4 is C’C 4.7 BC 0.19 4 = - = - = 24.7 rads2 To determine the sense of 4 we note that the normal component of urn is c’c in the sense of c’ to c; thus BC has a clockwise angular accelera- tion. uQIP = (R - Ro2) e, + (Rh + 2Ro) ee & = -24.7k rads2 If PQ is of fixed length then R = R = 0 and uQIP has one component of magnitude Rw2 always in the sense of Q to P and another of magnitude Rh, perpendicular to PQ and directed according to CD 0.15 the sense of h. Acceleration diagram (section 5.7). See Fig. 5.30. The radial and normal components of UB/A are both known, and summing these gives the total acceleration uB since A is a fixed point can find the position of e on bc from (ab’ + b’b = ab in the diagram). The radial directed from C to B. The normal component of uUB is perpendicular to BC but is as yet unknown in magnitude or sense. Similar reasoning applies to uUD. However we have enough data to locate point c on the acceleration diagram shown in Fig. 5.30. The magnitudes and directions of UB and uc are Similarly we find that the magnitude of ;3 is 28 187rads2 o3=-=-= C”C and the sense is anticlockwise, ;3 = 187krad/s2 From the concept of the acceleration image we -+-+ 9 be BE bc BC _- component of uuB has a magnitude of l2 %2 and is Thus be = 0.99 - =0.260m/s2 (;io) The magnitude and direction of uE are taken from the diagram and we find taken directly from the diagram. - UE = auA = ae = 24.2 m/s2 45” 46“ 43” 2 + UB = uB/A = ab = 22.0 m/~ uc = ac/D = dc = 31.6 m/~ + Link Acceleration Direction Sense Magnitude Line aB/A (radial) [(AB A/ l1oI2 = 0.15(12)2 = 21.6 ab‘ aB/A (normal) LAB 7 11 hi = 0.15(35) = 5.25 b‘b uuB (radial) IIBC A/ 12~2~ = 0.19(6.7)2 = 8.53 be’ AB { BC { CD { UC/D (normal) ICD ? 13h3 = ? c”c UC/B (normal) IBC ? 124 = ? c’c UQD (radial) llCD L 13w32 = 0.15(10)2 = 15.0 de’’ Vector-algebra methods Vector algebra can be used in the solution of mechanism problems. Such methods are a powerful tool in the solution of three-dimensional mechanism problems but usually take much longer than graphical methods for problems of plane mechanisms. They do, however, give a systematic approach which is amenable to computer programming. An outline of a vector-algebra solution to the present problem is given below. Students who are following a course leading to the analysis of three-dimensional mechanisms should find this a useful introduction and are encouraged to try these techniques on a few simple plane mechan- isms. values of d is consistent with the links BC and CD joining at C, and one of the values of c corresponds with the mechanism being in the alternative position shown dotted in Fig. 5.31. The vector Z2 can then be found from equation (ii). The results are Z1 = (0.075Oi+O.l299j) m Z2 = (0.1893i+O.l58Oj) m l3 = (0.0350i - 0.14571’) m Now, vC = %+vUB and, from equation 5.3, v,=0,xl,+O,xZ, Vc = 03 x (-13) 01 x z1 +O, x z, + 03 x z3 = 0 also (iv) (VI Equating the two expressions for vc 7 Writing w1 = 12k, 02 = *k and o3 = u3k7 and carrying out the vector products in equation (v), gives From Fig. 5.31 we note that (-1.559-0.01580, + O.1457~3)i + (0.9 + 0.18930, + 0.035 65w3)j = 0 Z, + Z2 + l3 + 1, = 0 (ii) and the vectors 1, and l3 can be determined by first evaluating angles 13, and O3 by the methods of The vector I1 = I1 (cos 61 i+ sin Od) is known Equating the coeffjcients of i and j to zer~ and solving for O, and w3, we find normal trigonometry and then writing O, = -6.634 Z2 = 12 (cos 62i + sin 6d) Z3 = l3 (cos 63 i - sin 63j) Alternatively we can write Z2 = 12e2 = l2 (ai + bj) Z3 = 13e3 = 13(ci+dj) and w3 = 9.980 Using % = w1 X Zl and equation (iv) leads to l%l = d[(1.559)2+(0.9)2] = 1.800ds % = -1.5593+0.9jm/s and and determine the values of a, b, c and d. Noting that (iii) vc = -(1.4533+0.3558j) m/s lvcl = d[(1.453)2+(0.3558)2] = 1.497 m/s A quicker way of finding vc, if 02 is not d = kd(1- c2) and substituting in equation (ii) with z4 = -14i and insertion of numerical values gives required, iS to note that Since DC/B is perpendicu- lar to BC, we can write 0. 190e2 = (0.225 - 0.180~) i vC/B-z2 =o -[0.1299fO.l80d(l -c2)U or (vc-%).Z2 = 0 Taking the modulus of this equation eliminates e2 and rearranging and squaring we find two values for c, each with two corresponding values of d from equation (iii). Only one of each pair of and carrying out the dot product we find % is known and writing from equation (iv) vc = o3kx (-0.035 Oli+O.l457j) [...]... resultant of the external forces on a body is equal to the product of the total mass and the acceleration of the centre of mass We must now consider the effect of the positions of the lines of action of the applied forces, remembering that the acceleration of the centre of mass is the same whether or not the line of action of the resultant passes through the centre of mass Consider initially a group of particles... module of 4 mm 5.15 74 Kinematics of a rigid body in plane motion Figure 5. 54 a) For the numbers of teeth given in Fig 5. 54, show that the number of teeth on wheel G is 52 b) Determine the overall speed ratio of the gearbox 6 Kinetics of a rigid body in plane motion 6.1 General plane motion In this chapter we consider the motion of a rigid body in general plane motion, by which we mean that the centre of. .. =- (6.3) or moment of external forces = C moment of (mass x acceleration) = moment of the rate of change of momentum = rate of change of the moment of momentum We may n ~ ~ k e of the definition of the use centre of mass and, by writing ri = rc + P i (see Fig 6.21, equation 6.3 becomes c r i x Fi = crG x + CrG x mipi + C p i x mifG+ C p i x mipi = rG x mifG ~f~ + C p i x mipi (6 .4) (6-1) 1 Figure 6.2... rotates about the fixed axis at 0 a~ = - [rcosowo2 + a (cos4%c2 - sin4(jBc )I i -b [sin4%c2 + C O S ~ ( ; ) B ~ ~ ~ VG = [ - r ~ i n 8 w ~ + a s i n 4 o ~ ~b c- s 4 % ~ 1 j []i0 [ i: where sin4 = (r/l)sin 0 and o~~ = I) -4 57 Figure 5 .46 shows one of the cylinders C of a petrol engine The crankshaft AB is rotating anticlockwise at a constant speed of 3000 rev/min about A The piston E which slides in cylinder... = 2. 84 gives R2 = Ri(2. 84) - 1- R1 = 1.67 X 2. 84 - 1 - 1.67 = 2.07 The reverse gear ratio is numerically equal to R2 = 2.07 The diameter of the sun wheel plus twice that of the planets must equal the diameter of the annulus For a meshing gear train all gears will have the same diametral pitch, that is the ratio of the number of teeth to the diameter is constant It now follows that the number of teeth... centre of the connecting rod Figure 5 .48 For angle DAB = 30°, determine (a) the velocities of E and G and the angular velocity of BD; (b) the accelerations of E and G and the angular acceleration of BD Solve this Problem graphically and check Your results from the formulae of the previous question At the instant when the door is in the position shown, the trunnion block has an upward velocity of 0.75... impeller blade of a ‘centrifugal’ pump The radius of C ~ ~ a t u P e the r of blades at the tip is 150 mm The impeller has an angular velocity of 30 rads clockwise and an angular aCCeleration of 0.01 rads’ in the same sense At the blade tip the particle has, relative to the impeller blade, a f tangential velocity o 15d s and a tangential acceleration of 10 m/s2 Figure 5 .47 Figure 5 .49 5.8 Problems... 5.1 In the mechanism shown in Fig 5 .40 , AB is rotating anticlockwise at 10 rads When 8 = 45 ”, determine the angular velocity of link BDC and the velocities of C and D 5 .4 Part of the control system for an engine is illustrated in Fig 5 .43 At the instant when the beam OA passes through the horizontal position, its angular speed w is found to be 1.1rads The motion of the point A is transmitted through... same cylinder When angle BOA = 60°, find the velocities and accelerations of each piston k vc = [ - r s i n O w o + I ~ i n 4 % ~ ] i 5 9 An ‘up-and-over’ mechanism for a garage door comprises two identical units of the type shown in Fig 5 .48 , mounted one on each side of the door Each (jBc = sec4 - sinOwoZ- sin40Bc2 k unit consists of a trunnion block T which runs on two 0.1 m diameter rollers in a vertical... epicyclic by taking the diameter of the annulus to be the same as the first gear so that, assuming the same diametral pitch, both annuli will have the same number of teeth, that is 85 This means that Ts2 = 8512.07 = 41 , to the nearest whole number The actual ratio 85 /41 = 2.07 to two places of decimals The number of teeth on the planet = (85 - 41 )/2 = 22 In this gear the number of teeth on the sun is a prime . %(IE) = 6.7(0. 245 ) = 1. 64 m/s The discrepancies between the two methods The magnitude of 4 is C’C 4. 7 BC 0.19 4 = - = - = 24. 7 rads2 To determine the sense of 4 we note that. acceleration) = moment of the rate of change of = rate of change of the moment of or momentum momentum We may n~~ke use of the definition of the centre of mass and, by writing. [-rsinOwo+I~in4%~]i VG = [-r~in8w~+asin4o~~]i- [bc0s4%~1j (jBc = sec4 - sinOwoZ - sin40Bc2 k ~c = -[~~~s~w~~+I(cos~~~~-sin~~~~]i a~ = - [rcosowo2 + a (cos4%c2 - sin4(jBc )I