96 Energy 7.8 The power equation If the work-energy relationship is written for a small time interval At, then we have AW = A(k.e. + V) + A(1osses) Dividing by At and going to the limt At-0 leads to dW d d - = - (k.e. + V) +-(losses) dt dt dt (7.30) or, power input equals the time rate of change of the internal energy plus power ‘lost’. Let us consider a simple case of a single particle acted upon by an external force P and also under the influence of gravity, then P-v = -[4rnv-v+rngz] = mv-a+rngi d dt If the motion is planar, v=Xi+ik a=xi+zk and P= Pxi+P,k so that PxX + P,i = rn (Xx + iz) + mgi Px+Pztana = rntan2cr+1)z+rngtancr Hence z may be found without considering the workless constraints. or P2cosa,- Plcoscrl = 0 (7.31) (7.32) Figure 7.15 condition for equilibrium is If z = xtancr (Fig. 7.14) then dividing by x gives From the free-body diagram (Fig. 7.16) the CP=P,+P,= w=o and P1 sin cyl + P2sin cr2 - rng = 0 are not apparent, but the reader is asked to be patient as later examples will show some of its rewards. A virtual displacement is defined as any small displacement which is possible subject to the constraints. The word virtual is used because the displacement can be any displacement and not necessarily an actual displacement which may occur during some specific time interval. The notation used for a virtual displacement of some co-ordinate, say u, is Fu. This form of delta is the same as is used in mathematics to signify a variation of u; indeed the concepts are closely related. The work performed by the forces in the system over this displacement is the virtual work and is given the notation 621r. Conditions for equilibrium Let us first consider a single particle which is free to move in a vertical plane subject to the action of two springs as shown in Fig. 7.15. Figure 7.14 Figure 7.16 7.9 Virtual work and Fz respectively, then The concept of virtual work is one which saves a considerable amount of labour when dealing with complex structures, since there is no need to dismember the structure and draw free-body diagrams. Basically we shall be using the method as an alternative way of presenting the conditions for equilibrium and also to form a basis for the discussion of stability. In the early stages of understanding the principle the main advantages If equations 7.31 and 7.32 are multiplied by 6x i P,cosa2Fx- P~COScr,FX = 0 P,sincrlFz+P2sina2Fz-rngFz= 0 or CP-FS = 0 (7.33) These are the equations for the virtual work for the arbitrary displacements Fx and 6z - note, arbitrary displacements. In both cases we may state that the virtual work done by the forces over an arbitrary small displacement from the equilib- rium position is zero. If a system comprises many particles then the total virtual work done on all particles over any virtual displacement (or combination of displace- ments) is zero when the system is in equilibrium. Principle of virtual work We may now state the principle of virtual work as follows. If a system of particles is in equilibrium then the virtual work done over any arbitrary displacement, consistent with the con- straints, is zero: W=O (7.34) Application to a system with a single degree of freedom Consider a rigid body freely pinned at A and held in equilibrium by a spring attached at B (Fig. 7.17). This body has one degree of freedom, that is the displacement of all points may be expressed in terms of one displacement such as 8, the angular rotation. If the spring is unstrained when AB is horizontal, then in a general position the active forces are the weight and the spring force; the forces at the pin do no work if friction is negligible. For a small displacement 68, the displacement of G is a 68 (Fig. 7.18) and the virtual work is W = Mg (a 68COS 8) - kR8 (R de) For equilibrium, w = o = (Mgacos8-kR2e)68 Mgacos 8 - kR28 = 0 and, as 68 is arbitrary, (7.35) 7.9 Virtual work 97 If, as in the previous example, the forces are conservative then equation 7.34 may be inter- preted as W=W,+W,+W' (7.36) gravitational potential energy) elastic strain energy) W' = virtual work done by external forces where WG = - SVG = -(variation of WE = WE = -variation of and Therefore W=W'-6vE-6VG=o or W'=6(VE+VG) (7.37) Reworking the last problem, 0 = 6[-Mg~sin8+4k(R8)~] 0 = [ - M~UCOS 8 + kR28] 68 Stability Consideration of some simple situations shown in Fig. 7.19 will show that not all equilibrium configurations are stable. However, we cannot always rely on common sense to tell us which cases are stable. We have demonstrated that for equilibrium W = 0, but further consideration of 98 Energy the value of W as the virtual displacement becomes large will lead to the conclusion that if W becomes negative then the force will be in the opposite direction to the displacement , showing that the forces are tending to return the system to the equilibrium configuration, which is therefore one of stable equilibrium. In mathematical notation, for stability S(W)<O or 62W<~ (7.38) Looking at our previous case once again, therefore 6(W) = (-Mgasine- kR2)(Se)2 state is stable. configuration is defined by w = (Mgacose- kR2e)6e For 0<8<7r it is seen that any equilibrium For a conservative system, the equilibrium 6V= 0 and stability is given by S(W)<O 6(6V)>O But, since W = -SV, If V can be expressed as a continuous function of 8, then av ae av=-ae and a2v ae2 a2v=- e se=-(6e)2 ae a iaV ae ) Hence for equilibrium av ae -=o a2v ae2 and - >O for stability (7.39) Systems having two degrees of freedom The configuration of a system having two degrees of freedom can be defined by any two indepen- dent co-ordinates q1 and q2. The virtual work for arbitrary virtual displacements 6ql and 6q2 may be written in the form (7.41) W = QI 6qi + Q2 642 Since the virtual displacements are arbitrary, we may hold all at zero except for one and, as W = 0 for equilibrium, we have Qi =O and Q2 = 0 The stability of a system having two degrees of freedom will be discussed for a conservative system. It will be remembered that constant forces are conservative, therefore the majority of cases may be considered to be of this type. If the independent co-ordinates - referred to as generalised co-ordinates - are q1 and q2, then the total potential energy (gravitational plus strain) is V = V(ql ,q2); hence av av a41 a42 6V = - 6ql +- 6q and, since SV = 0 for equilibrium, we have av av _- - -=o a41 a42 (7.42) For stable equilibrium we must have @V>O for all possible values of 6ql and Sq2. The second variation may be written or, since a2V/aql aq2 = a2V/aq2aq1, then It is clear that, if 6q2 = 0, then a2v ->0 %I2 and, if 6ql = 0, then a2v ->O a4z2 These are necessary conditions for stability, but not sufficient. To fully define stability, a2V must be >O for any linear combination of 6ql and 6q2. a2v ->0 ad a2v ->0 a422 a2v a2v > 2 and ~ ad ad (aqd:dVq2) >O , - Consider the conservative system shown in Fig. 7.20; the active forces, real and fictitious, are shown in Fig. 7.21. W = SV- m,xl ax1 -m2x2Sx2 - ZeSe = 6[thl2 + mlgxl - m2gxz + const.] + (7.44) 100 Energy I [ ti) r2 [ (3 r: I (3 r1 = k - x2 +mlg- -m2g Sx2 or m2+ml - +> i2+k - x2 = m2g - ml (:)g (7.46) This approach does not involve the internal forces, such as the tension in the ropes or the workless constraints, but these may be brought in by dividing the system so that these forces appear as external forces. Equation 7.46 could have been derived by the application of the power equation with a similar amount of labour, but for systems having more than one degree of freedom the power equation is not so useful. Discussion examples Example 7.1 A block of mass m can slide down an inclined plane, the coefficient of friction between block and plane being p. The block is released from rest with the spring of stiffness k initially compressed an amount x, (see Fig. 7.22). Find the speed when the block has travelled a distance equal to 1 .&, . Figure 7.22 Solution If a free-body-diagram approach is used to solve this problem, the equation of motion will be in terms of an arbitrary displacement x (measured from, say, the initial position) and the acceleration R. Integration of this equation will be necessary to find the speed. If an energy method is used, consideration of the initial and final energies will give the required speed. The two methods are compared below. a) Integration of equations of motion. The free-body diagram (Fig. 7.23) enables us to write the following equations: [CF, = mYG1 WCOS~-N=O .‘.N= WCOS~ (i) [CFx = m.fG] Wsina-pN- T, = mx (ii) If we measure x from the initial position, the spring tension T, is given by T, = k(x - x,) and we shall be integrating between the limits 0 and 1.2~~. We could, on the other hand, choose to measure x from the position at which there is no force in the spring, giving T, = kx, and the limits of integration would be from -x, to +O.&,. Using the former, (i) and (ii) combine to give Wsina-pWcosa-k(x-x,) = m.f (iii) Since we are involved with displacements, velocities and accelerations, the appropriate form for R is vdvldx: the direct form f = dv/dt is clearly of no help here. Hence equation (iii) becomes 1.kc 0 I { W(sin a - pcos a) - k (x - x,)} dx = [‘mvdv 0 I:&= {w(sin a - pcos a) + kx, } x - tkx2 = m[+o2]. [ 0 { mg (sin a - p cos a) + kx, } 1 .kc - ik (1. h,)’ = imv2 and thus v can be found. The reader should check this result by measuring x from some other position, for instance the position at which the spring is unstrained, as suggested previously. b) Energy method. Since energy is lost due to the friction, we use equation 7.29 (see Fig. 7.24): [work d~ne],,,,,,~ = [k.e. + VG + VE]~ - [k.e. + VG+ VGll + ‘losses’ (iv) where the ‘losses’ will be pN( 1.2~~) as explained in section 7.6. For the general case of both p and N varying, this loss will be Jb2”cpNdx. None of the external forces does any work, according to our definitions, and thus the left-hand side of If, however, the block had been following a known curved path, the spring tension T, could have been a complicated function of position giving rise to difficult integrals, possibly with no analytical solution. The energy method requires only the initial and final values of the spring energy and so the above complication would not arise. Variation in N could cause complications in both methods. In some cases the path between the initial and final positions may not be defined Figure 7.24 at all; here it would not be possible to define T, as a general function of position. An energy method would give a solution directly for cases equation (iv) is zero. It should be pointed out that the correct result where friction is negligib]e (see, for example, can be obtained by treating the friction force as problem7.2). external to the system and saying that this force does negative work since it opposes the motion. Example 7m2 The left-hand side of equation (iv) would then be See Fig. 7.25(a). The slider B of maSS rn is wou1d be Omitted' This is a common way Of fixed to the slider engages with the slot in link dealing with the friction force but is not OA. The moment of inertia of the link about o is Io and its mass is M, the mass centre being a considered to be a true energy method. Kinetic energy. In the initial position (x = 0) the distance a from 0. The spring of stiffness k is speed and thus the k.e. are zero. In the final attached to B and is unstrained when 8 = 0. position (x = 1.2~~) the k.e. is frnv2, from equation 7.26. Gravitational energy, V,. The datum for measuring gravitational energy is arbitrary and we may take as a convenient level that through the initial position; thus the initial g.e. is zero. Since the block then falls through a vertical distance of 1.2xCsina, the final gravitational energy is, from equation 7.27, -rng(l.2xcsina). Strain energy, VE . In the initial position, the Spring is compressed an amount x, and thus, from equation 7.28, the strain energy is fkx:. In the final position the spring is extended b an amount O.&, and so the final s.e. is fk(0.2~~) . Note that only the gravitational energy can have a negative value. Equation (iv) becomes -pN(1*2rC) and the 'losses' tem On the right constrained to move in vertical guides. A pin P Figure 7-25 The system is released from rest at 8 = 0 under the action of the torque Q which is applied to link OA. ne variation of Q with e is shown in Fig. 7.25(b). Determine the angular speed of OA when 8 = 45", neglecting friction. 9 O = [frnv2 - rng ( 1 .&,sin a) + fk (0.2x:)I - [0 + 0 + dhC2 J + pN(1.2~~) We still need a free-body diagram to determine that N = mgcosa, as in equation (i), and then v can be found directly. For this particular problem there is little to choose between the free-body-diagram approach and the energy method. In the energy method we avoided the integration of the first method, which however presented no difficulty. Solution This problem has been approached in example 6.5 by drawing two free-body diagrams and writing two equations of motion involving the 102 Energy a - I@, = ;IO I~IoA~ + fmVg2 + Mg- 32 d2 +mgl+Ikl']-[(I] (ii) Before we can evaluate wOA we need to express l1 [ contact force at the pin P. Since we are here concerned only with the angular velocity at a given position, and details of internal forces are not required, an energy method is indicated and will be seen to be easier than the method of Chapter 6. Equation 7.29 becomes VB in terms of 00~. Since y = ltan 8, (work done),,,,,,, = [k.e. + vG + vE]2 - [k.e. + VG + VEII (i) since there are no losses. The left-hand side of this equation is the work done by the external and is thus Jf4 Qd8. This is simply the area under the curve of Fig. 7.25(b), which is found to be (11/32) re,. The normal reaction N between the slider and the guides is perpendicular to its motion and the force R in the pin at 0 does not move its point of application: thus neither of these forces does work (see Fig. 7.26). VB = dyldt = 1 sec2 8d8ldt = lsec2 8wOA and at 8 = d2, VB = 21IiIOA. Substitution in (ii) gives forces or couples on the system during the motion 11 - TQ, = &(Io + 4ml2) wOA2 32 +g M-+ml ++k12 K2 1 from which IiIO~ can be found. Comparison of this method with the free-body- diagram approach and the difficulty associated with integrating the equation of motion shows the superiority of the energy method for this problem. What if the force S on the pin P has been required? This force does not appear in the energy method, but this does not mean that the energy method is of no help. Often an energy method can be used to assist in determining an unknown acceleration and then a free-body- diagram approach may be employed to complete the solution. Kinetic energy. As the mechanism is initially at Example 7.3 rest, the initial k.e. is zero. Since the motion of OA is rotation about a fixed axis, the final k.e. of OA, from equation 7.8, is 41000A2. The slider B has no rotation and its final kinetic energy, from equation 7.7, is simply fmvB2. Gravitational energy, VG. We will take as datum levels the separate horizontal lines through the mass centres of link and slider when 8 = 0 and thus make the initial value of VG zero. When 8 = 45" the mass centre of the link has risen through a height a/d2 and that of the slider Figure 7.27 through a height I, and so the final value of VG is CD has a moment of inertia about D of 6 kg m2 Mgald2 + mgl. and its mass is 4.5 kg. BC has a moment of inertia Strain energy, VE . Initially the strain energy is about its mass centre E of 1.5 kg m2 and its mass is zero and in the final position the spring has been 4 kg. At the instant when both AB and CD are compressed an amount I; the final value of VE is vertical, the angular velocity of AB is 10 rads and thus Ik1'. its angular acceleration is 50 radls2, both measured in an anticlockwise sense. A four-bar chain ABCD with frictionless joints is shown in Fig. 7.27. Substituting in (i) gives Neglecting the inertia of AB, determine the From the velocity diagram we see that vE = -lOi ds; the component of UE in the same direction is aE-i = -(50- 25/d3) ds2. Substitut- ing into the power equation (1) gives torque T which must be applied at A to produce the above motion. Solution The velocity of B is ox3 = 10k x lj = - 1Oi ds. The velocity diagram is shown in Fig. 7.28 and it can be seen that link BC is not rotating (I$ = 0). T10 = 4(10)(50 - 25/d3) + 6(5)(25) and Example 7.4 A slider-crank chain PQR is shown in Fig. 7.30 in its equilibrium position, equilibrium being main- tained by a spring (not shown) at P of torsional stiffness k. Links PQ and QR are of mass m and 2m and their mass centres are at GI and G2 respectively- The slider R has a mass M. The moment of inertia of PQ about P is Zp and that of QR about G2 is ZG2; also, PGI = GIQ = QG2 = G2R = a. T = 217.2 N m . 1om/s b, c, e a, d Figure 7.28 The kinetic energy of link BG is thus $MvE2 and that of CD, for fixed-axis rotation about D, is $ZD~2. 'It would clearly not be correct to write the power equation (section 7.8) as d dt T-w = To = - (4MvE2 + $Zk2) since 4MvZ is not a general expression for the k.e. of link BC (it is the particular value when AB is vertical). As ZI, and a, do not have the same direction, the correct power equation is d dt Figure 7.30 Find the frequency of small oscillations of the system about the equilibrium position, 0 = eo, since += 0. Solution Equations of motion for the links can of course be obtained from a free-body-diagram approach, but this would involve the forces in the pins and would be extremely cumbersome. Use of the power equation leads directly to = 5k rads and thus the required result. In this case we have power = d(energy)/dt = 0, since the energy is constant for the conservative system and clearly no power is fed into or taken out of the system. Let the link PQ rotate clockwise from the equilibrium position through a small angle /3 as +=-=-=2 cc' 50 5 rads2 shown in Fig.7.31. The new positions of the Tow= To = -((~M+-Z)E+~I~~+$ZD$~) = M+ 'aE + ID44 (i) neglecting friction. The acceleration of B is QB = [-l(5O)i- 1(10)j2] m/s2 From the velocity diagram we find $ = 10W2 ac = [ -242- 2(5)2j] ds2 The acceleration diagram is shown in Fig. 7.29 and 6; is given by CD 2 various points are shown by a prime. Figure 7.29 Kinetic energy. Link PQ has fixed-axis rotation about P and its k.e. is thus 4Zp@'. By symmetry, the angular speed of QR is also /3. The k.e. of QR is given by 4ZG2fi2 + 4(2rn)~~~~, where 104 Energy d dt = - [3acos (eo + p)i - asin(Oo + pb] = -3asin(eo + p)Bi - acos(eo + B)j The k.e. of slider R is MvR2 where d+ d vR=-(PR’) =-[4acos(Oo+p)i] dt dt = -hsin(e,+p)/% Gravitational energy. A convenient datum level is the horizontal through PR. The mass centre of each of the links PQ and QR is at a height asin(eo + p) above the datum and their gravita- tional energy is thus, from equation 7.27, VG = mgasin(80++) + (2m)ga~in(8~+p) The slider R moves the datum level and thus has no gravitational energy with respect to this level. Strain energy. The couple applied by the spring to the link PQ in the equilibrium position is clockwise and equal to kyo , where yo is the angle of twist. As link PQ rotates clockwise through an angle p, the angle of twist is reduced to (yo-p) and thus the strain energy, from equation 7.28, is Ik(y0-p)’. The total energy E is thus E = {k.e.} + {VG} + {VE} = { +zPB + $z,, B + f(2m) x [9a’sin’(~~ + p) + a’cos’( eo + p)] 8’ + IM. 16u2sin2(8, + p) B’ > + {3mgasin(~o++)} + {bk(yO ~)’} = constant (since the system is conservative) Since the above is a general expression for the energy, it can be differentiated to give the power equation. The term fi’ arises which is negligible for small oscillations. We note that, since p is small, sin( 0, + p) = sin 0, and cos(8, + p) =cos eo, but these approximations must not be made before differentiating. After dividing throughout by B, we find, since b2 is small, ZB + kp = k yo - 3mga cos 80 (9 where Z = Z, + ZG2 + 2m (9a’ sin’ eo + a2cos2 0,) + 16Ma2sin2 0,. It can be shown by the method of virtual work, or otherwise that kyo = 3mgacos eo so that equation (i) reduces to ZB+ kp = 0. Thus, for small p, the motion about the equilibrium position is simple harmonic with a frequency of (1/27r)d( k/Z) . Example 7.5 The mechanism shown in Fig. 7.32 is in equilibrium. Link AB is light and the heavy link BC weighs 480 N, its mass centre G being midway between B and C. Friction at the pins A and C is negligible. The limiting friction couple Qf in the hinge at B is 10 N m. Figure 7.32 Pin C can slide horizontally, and the horizontal force P is just sufficient to prevent the collapse of the linkage. Find the value of P. Solution This problem has been solved earlier in Chapter 4 (example 4.3). There a free-body diagram was drawn for each of links AB and BC and the unknown forces were eliminated from the moment equations. It will now be solved by the method of virtual work and the two methods will be compared. If in the virtual-work method we treat forces due to gravity and springs and friction as being externally applied, the total virtual work done may then be equated to zero. In order to obtain the correct sign for the virtual work done by the internal friction couple Qf in the present problem, we may use the following rule: the virtual displacements must be chosen to be in the same direction as the actual or impending displacements and the virtual work done by friction is given a negative sign. Applying this rule, we let the virtual displace- ment of C be 6x to the right, since this is the direction in which it would move if the mechanism were to collapse. If a mechanism has a very small movement, the displacement vector of any point on the mechanism will be proportional to the velocity vector. Thus we can draw a small-displacement diagram which is identical in form with the corresponding velocity diagram. This results in very lengthy means of solution, whereas the virtual-work method disposes of the problem relatively quickly (see, for example, problem). Example 7.6 A roller of weight W is constrained to roll on a circular path of radius R as shown in Fig. 7.34. The centre c of the roller is connected by a Spring of stiffness k to a pivot at 0. The position of the roller is defined by the angle 8 and the Spring is UnStretched when 8 = 90"- Fig. 7.33, where ab is drawn perpendicular to AB, bc is drawn perpendicular to BC and oc is of length Sx. Since the weight W acts vertically downwards and the vertical component (hg) of the displace- ment of G is also downwards, the virtual work done by W is positive and given by + W(hg). The virtual work done by P is -P(oc), since the force P is opposite in sense to the assumed virtual displacement. The virtual work done by the friction couple Qf is -Qf IS+ I, where IS+ I is the magnitude of the change in the angle ABC. AB rotates clockwise through an ang1e ab/AB and BC rotates a) Show that the position 8 = 0 is one of stable anticlockwise through an angle bc/BC. [If equilibium only if W/(Rk) > 0.293. course the angular speed of AB would have been given by ab/AB, and so on.] The change in the angle ABC is thus Figure 7-34 Fig. 7-33 had been a ve1ocity diagram then Of b) If W/(Rk) = 0.1, determine the positions of stable equilibium. Solution The strain energy V, in the spring is zero when centre C is at B. We can also make the gravitational energy VG zero for this position by ab bc a+=-+- AB BC taking AB as the datum level. Summing the virtual work to zero gives W(hg) - P(oc) - Qf - +- = 0 (:: 3 The virtual displacements ab, bc and hg are Figure 7.35 From Fig. 7.35, the stretch in the spring is OC - OB = 2R cos (8/2) - Rd2 and C is a vertical distance Rcos8 below AB. Thus, using equations 7.27 and 7.28, the total potential energy Vis given scaled directly from the diagram to give 480(0.1875 SX) - P(SX) 0.8386~ 0.451 Sx which, on dividing throughout by SX, gives by P = 40.0 N. Comparing the virtual-work solution of this problem with that of the normal staticdfree-body- diagram approach of Chapter 4, it can be Seen that here we are not concerned with the forces at A and B and the vertical component of the force at C. However, for this simple problem the more straightforward approach of Chapter 4 is to be preferred. = R2k{sin8[W/(RR)-1] The virtual-work method comes into its own when many links are connected together. In such cases, drawing separate free-body diagrams for each link and writing the relevant equations is a -10 ~ ( 0.2235 +E) = o v= v,+v, = - WR cos 8 + +k [2R cos( 8/2) - Rd2I2 The equilibrium positions are given, from dV/d8 = WRsinO+ k[2Rcos(8/2)- Rd2] equation 7-39, by x [ - R sin( 0/2)] + d2sin(8/2)) = o (i) a) NOW, One solution to equation (i) is clearly 8 = 0. [...]... shaft The numbers of teeth on the annulus, each planet and the sun wheel are 200, 50 and 100 respectively The axial moment of inertia of the sun wheel and the associated shafting is 0. 06 kgm2 and that of the spider is 0.09 kg m2 Each planet has a mass of 2.0 kg and an axial moment of inertia of 0.0025 kg m2 The centres of the Planet wheels are at a radius of 120mm from the central axis of the gear 7.10... the speed of the module just prior to impact with the lunar surface (a) neglecting the variation of g with height and (b) taking into account the variation of g Take the value of g at the surface of the Moon to be 1 .62 N/kg and the effective radius of the Moon to be 1.74 x lo3 km 7 .6 A four-bar linkage consists of three similar uniform rods AB, BC and CD as shown in Fig.7.42 Each has a length of 0.5m... v1) (8.21) 8 .6 The rocket in free space The rocket is a device which depends for its operation on the ejection of mass, and again the mechanics is best understood by considering the rate of change of momentum Figure 8.4 Referring to Fig 8.4 and using the following notation: mo mf me v vj rit = mass of rocket structure = mass of fuel = mass of exhaust = velocity of the rocket = velocity of the jet relative... the the the the Solve part (a) of problem 6. 25 by an energy method 7. 16 Figure 7.50 In the system shown in Fig 7.50, the spring has a stiffness of 60 0N/m and is unstrained when its length is 0.15 m If the roller R has a mass of 3 kg, determine the value or values of x for an equilibrium configuration State whether the equilibrium is stable or unstable 7.21 A uniform rod of mass m and length 1 can pivot... ZG (8 .6) (8-7) which is an expression of the conservation of moment of momentum The term ‘angular momentum’ is often used in this context but is not used in this book since the term suggests that only the moments of momentum due to rotation are being considered whereas, for example, a particle moving along a straight line will have a moment of momentum about a point not on its path 8.4 Impact of rigid... given by COS(W)= d2/1.8 = 0.7 86 hence, 8 = k 76. 4" The type of stability at these two positions can be confirmed by substituting for e in equation (ii): d2V/d82 = R2k[0.235(0.1 - 1) 0.707(0.7 86) ] = R2k[-0.211 + 0.5 561 > O + Thus, at 8 = f 76. 4", the system is in stable equilibrium Example 7.7 ~~~i~~ the erection of a structure, three beams are connected as shown in Fig 7. 36 Beam ABC may be considered... rear wheel are equal and independent of the separate speeds of the wheels b) See Fig.7. 46 The car is being driven along a plane curve at a constant speed such that the path of each wheel centre is a circular arc L and R are the left- and right-hand rear wheels and RL and RR are the corresponding radii of the paths Denoting the angular speeds of L and R and the cage of the differential gear by y , and... travels a distance of 1cm Figure 7.42 Initially the mechanism is held with AB vertical, and in this position the spring exerts a clockwise couple of 80 N m on AB If the mechanism is then released, what is the angular speed of AB when 6 = 30”? A roller of radius R has an axial moment of inertia I and a mass m Initially the roller is at rest and then it is pulled along the ground by means of a horizontal... 7.44 Determine the natural frequency of each system for small oscillations An electric locomotive develops a constant power output of 4MW while hauling a train up a gradient of slope a,.cSin (1170) The maSS of train and locomotive is 1x 1 06 kg The rotational kinetic energy is 10 per cent of the translational kinetic energy n e resistance to motion per unit maSS of train is given by 7.13 R = (12.8+0.138~)10-'N/kg... exists between the velocity of approach and the velocity of recession of the points of contact Velocity of approach = vl Velocity of recession = (u2+ y a ) - v2 (8.11) Rewriting equations 8.8 to 8.10, VI- 0 2 M m (8.12) = -u2 (8.13) (8.14) Substituting for u2 and ( y a ) in equation 8.14 gives (01- V 2 ) ( V 1 + 02) = m - (81- V 2 l 2 M +-maL(vl - v# ZG Figure 8.2 Conservation of linear momentum gives . of WE = WE = -variation of and Therefore W=W'-6vE-6VG=o or W' =6( VE+VG) (7.37) Reworking the last problem, 0 = 6[ -Mg~sin8+4k(R8)~] 0 = [ - M~UCOS 8 + kR28] 68 . g. Take the value of g at the surface of the Moon to be 1 .62 N/kg and the effective radius of the Moon to be 1.74 x lo3 km. 7 .6 A four-bar linkage consists of three similar uniform. and that of the spider is 0.09 kg m2. Each planet has a mass of 2.0 kg and an axial moment of inertia of 0.0025 kg m2. The centres of the Planet wheels are at a radius of 120mm