PowercanPlrol3ccstimatcddirectlyholn~rgy~lorcxamplc when pumping known quaatities ofwater. This is ptiadiuly d where detailed bbody diagrams end fora ady& would be very di@dt, and thea an avoided. Thc basic input powcr rcquirtd mwt be at bast qual to the output power, so a low Wton input power epa udy be cakdared quite essily. If information is availabk on the eWcmy of the systdm components, then an estimate of the actual input powar can be made. Transport vehicks use power @nut clcrodynsmic drag and roWq resis- Iance, end intsrmittently against the brakbmlrss and to podm pormtial energy when climbing hills. Powa is Plso quidfor Ornoreiclation to provide kinetic energy. Thc instantmco~~power requirement at the wheel8 can be estimated by a dgnamic analysis of the vehicle, and the CII~~power can then bc athared with idonnation on component ct8- &my. Hsna tbrate of fuel consumption may be catimnted,

6 lmpulse and momentum

6.1 lntrod~~lngImpulse and momentum There is one more fundamental tool of Dynamics that I want to show you in this course. This is the use of the concepts of impulse and mo- mentum. The linear momentum G of an object in translation is defined as the product of its mass and velocity:

Momentum is a vector - it is in the same direction as the velocity. The impulse of a force is I-IF dt which is simply Ft for a constant force over a given time t. The change of the momentum inside a chosen system is equal to the impulse on that system, so G,=C,+J Impulse is therefore a transfer of momentum, and is also a vector. For motion in a straight line

As a simple example, consider dropping a 2 kg brick, releasing it with an initial velocity of 0.5 m S-' downwards. Neglecting the aerodynamic force there is only one significant force, the weight (Figure 53). which I shaU call F. The fact that it is a weight force is of no special significance. What is the velocity of the brick after a time t = 2 S? The initial momentum Figure 53 along y is

Note that the units of momentum arc kg m S-!. Over a period of 2 S, the impulse of the (constant) force F = mg along y is J=Ft=19.6Nx2sJ=39.2NsJ The units of impulse are Ns, but IN is lkgxlms-', so INS= 1 kg m S-'. So the units of impulse and momentum ap,and J=39.2Ns=39.2kgms-' The magnitude of the final momentum is G2=Gl+J=1+39.2=40.2kgms-' so mu2 = 40.2 kg m S-' v, = G2/m so 8,=20.lms-'l This simple example could easily have bm solved by the methods of Block 4, and because it is so simple it does not dyshow the benefits of the impulse-momentum method. However, there are certain sorts of problems which are very much more easily solved by using impulse-momentum, and some of these comprise the remaining parts of Section 6. What the example does show is that the general method is to keep account of the momentum in a chosen system, and that this involves calculating momentum and impulse (i.e. momentum transfer). You should recognize this as very similar to keeping track of the kinetic energy and work in the work-energy method. However, momentum is a vector, unlike energy, so it is necessary to take careful account of direction, using co- ordinate axes. Impulse and momentum can be resolved into components and dealt with in each direction separately, just like force and acceleration, and can have negative components. As you have just seen, at first appearance impulse and momentum have different units, N s and kg m S-' nspcctively, but these can be reconciled through the dehition of the newton, 1 N = 1 kg X l m S-'. Actually this is similar to energy and work. Kinetic energy (frnv') units are kg m' S-' and work (Fs)units are N m. Again these are equivalent to each other, and we had a common name for both, the joule, J. There is no special name for momentum-impulse units (perhaps there should be) so we must use kg m S-' or N S. I shall freely interchange these without further ex- planation. You should do the same.

Calculate the magnitude of the momentum of each of the following, ex- pressing them in both forms of units. (a) A 0.1 kg apple at 4 m S-' (b) An 80 kg man at 2 m S-' (C) A 900 kg car at 30 m S-'

Figure 54 For the 0.1 S kg machine component in Figure 54, calculate the magnitudes of the total momentum, and the components of momentum in the x and y directions.

SA086 An aircraft jet engine has a thrust of IS0 kN. What impulse is produced in 30 S? SA066 A very small rocket motor used for controlling a satellite is fired giving the thrust variation shown in Figure 55. (a) Estimate the impulse. (b) The satellite being controUed has a mass of 800 kg. Estimate the speed change.

To choose the be~tsystem for impulse-momentum analysis requires care. The best system is usually diKerent from the one that would be chosen for a work-energy approach. The reason for this difference lies mainly in the way in which force transfers work and impulse. A force can only do work by moving, but it can 'do impulse' all the time. Therefore, when the system is separated from its environment, every external force imparts impulse (although some of these may be negligible because of size, as on a normal free-body diagram). On the other hand, momentum cannot appear or disappear because of internal forces, unlike energy. This is because all the internal forces are equal and opposite Wewton 3 pairs' that must operate for equal times. Even if they have unequal movements and do unequal work, they cannot produce momentum. In this respect momentum is simpler than energy, it can only appear in the system by transfer across the boundary, and cannot be produced internally. On the other hand, in the selection of an energy system our boundary could cut non-moving forces as these do no work. However, the non-moving forces do have impulses, so you must be careful. When I was advising you on the selection of free bodies for static analysis in Unit 4,I said that you must cut the unknown force that you want to find, but should cut a minimum of other unknowns This is also true for momentum system choice. If it is final velocities that are bciig estimated rather than forces, it is often posibk to avoid forces altogether. I shall show you how to approach two particular types of problem for which the impulse-momentum method is especially useful. In the first type, the variation of force with time is known (as in the satellite example of SAQ 86). The second type of problem involves impacts or explosions. Perhaps you are wondering whether in impulse-momentum analysis there is an equivalent to power in work-energy analysis. Power is the flow of energy per second, with units of newton metre per second, N m S-'. The 'flow' of momentum or impulse has units of newton second per second, i.e. N S S-', which is simply newtons. The unit of momentum flow or impulse flow is the same as the unit of force. To look at it another way, force is the flow of impulse or momentum. It is sometimes useful to think about momentum flow as being analogous to power and energy flow in the last section. Wherever there is a force, there is a momentum flow.

SA0 87 My weight is 883 N. (a) Draw a free-body diagram of me, standing still (air forces neglected). (b) Is there any energy flow? (c) Are there any impulses acting on me over a time interval7 (d) Use impulse-momentum considerations to calculate the floor's force on me, instead of ordinary static equilibrium. 6.2 The baslc Impulse-momentum problem In the basic impulse-momentum problem there is a clearly de6ned object. The momentum of the object and the total impulse acting on the object can be directly related by C1=C1+1 which for the object of mass m is

If F is known as a function of time then the momentum, and hence velocity change, can easily be found. This is not a very common problem in prac- tical engineering because the forces are not normally known beforehand as functions of time. One case in which it can be useful is when the force is constant. For example, dry sliding friction is often treated as a constant force. In this case the equation is mul = mul - Ft, SO any one unknown can be found.

The wheels of a car of mass 880 kg lock on tarmac at a speed of 30 m S-'. Estimate the time required to stop.

8duHon Choose the car as the system (Figure 56). Weight = mg = 8633 N. Estimating p = 0.9 (tyre on tarmac) gives F,. 7770 N. Initial momentum = mul = + 26.40 kN s Final momentum = mul = 0 Impulse - Ft = - 7770 X t

One practical case in which a non-constant force is likely to be known as a funotion of time is in rocketry. The thrust of a rocket depends mainly on the flow of fuel constituents which are controlled by pumps and valves whilst speed and position effects an minor and often negligible. SAQ 86 represents an example of this type. If velocities are already known then the method may be used to find the time average of a force. Example 10 A 30 tonne spacecraft undergoes a test after repairs on a control rocket (l tonne is 1000 kg). The rocket is fired for 2.0 s and the resulting speed change is measured to be 0.02 m S-'. Find the average thrust.

Solution Choose the spacecraft as the system. Initial momentum mul = 0 (speed measured relative to this condition)

Final momentum mu, = 30 X 10' kg X 0.02 m S-' = 600 N s

Impulse Ft = F X 2.0 s for average force F mu, = mv, + Ft 600=0+2.0x F F = 300 N average.

In simple questions of the types just shown, the solution is just as easily found by the methods of Block 4. Really the difference in the methods is just a matter of whether the force is integrated before dividing by the mass, or afterwards, in which case it is the acceleration that is integrated in the kinematics. However, these problems do illustrate the use of impulse-momentum in a simple context.

SA0 88 A 120 kg motorbike plus 80 kg rider is to accelerate from zero to 5 m S-' in 0.8 S. Estimate the average force required.

SA0 U# In an orbit-adjusting manoeuvre a 4 tonne satellite requiring a velocity change of 2.5 m S-' is to fire a rocket engine of thrust 230 N. Estimate the required duration of firing.

6.3 Impact problems In some cases, for example during collisions, the variation of force with time may be almost impossible to measure with any useful degree of accuracy. In this case the F = mii method of Block 4 is clearly not appli- cable. Useful calculations are nevertheless possible by working with the total impulse of the force over the collision period. However, such prob- lems can be more directly tackled by a choice of system which includes both objects. The momentum loss of one object must he momentum gain of the other because they must exert equal and opposite impulses on each other. Alternatively, it can be said that the total momentum of the pair of objects remains constant. Of course for this to be 'true' we need to know that the external forces on the system must not contribute significant momentum. As a specific example consider one car crashing into another (Figure 57). before u~ UB .-..-..-..--...... -..- + As the system I shall choose everything inside the dashed box. Now it is quite likely that one or both cars would have its brakes firmly on in these circumstances. If we were calculatmg the deceleration of car A just before after VA Vs impact we would certainly not ignore the friction forces - they would actually he the cause of the deceleration. However, the collision occurs 4 over a very short period of time - the momentum is transferred quickly. X In a collision it is generally true that there is a very large force acting for Figure 57 a short time - giving a large impulse during a short time. (The con- 50 sequential damage that often occurs is clear evidence of this.) During the short impact period the usual operating forces are overshadowed by the impact forces, so it is reasonable to neglect all but the impact forces during the short contact period of the collision, which is likely to be only a fraction of asond. For my chosen system then, during the short period of the impact the external forces have negligible impulse, so the momentum content of the system is constant in any given direction. Along X: m,u, + m.u. = m~v~+ m.vB Unfortunately this is not yet sufficient to give a prediction of the velocities just after collision. There are two unknowns: the velocities v, and 4.We have only one equation. The failure of the momentum method alone to give us the solution arises from the good physical reason that the outcome of the collision is not pre-determined solely by the initial masses and velocities. When two tennis balls collide they rebound, but two pieces of putty of the same initial mass and velocity as the tennis balls will stick together. Both collisions comply with the momentum laws, but in the case of the putty the energy has been mostly lost, used up in deforming and heating the puny. The tennis balls retain most of the energy, only losing a little to heat. In order to reach a solution to the problem we need an estimate of the fractional energy loss, or equivalent information. This is usually expressed in terms of the c@cient of restitution, e, which is de fined as the ratio of separation speed to approach speed, so separation speed e = approach speed

or separation speed = e X approach speed In my car example: (v. - v,) = e(u, - U,)

(a) What is the coefficient of restitution in a collision il very little energy is lost? (b) What is the coefficient of restitution of two pieces of putty?

Once we know e then we have two equations, and v, and v, can be determined. Unfortunately, there lies the true difficulty. In practical prob- lems we can usually only make a very rough estimate of e because it depends upon the material, the shape and the speed of the objects. All that the idea of dcient of restitution does for us is to conveniently condense our ignorance into one number. Howevm, practising engineers do not have the luxury of choosing the problems they have to solve, and making an estimate of e is often the best we can do. In many respects coefficient of restitution can be likened to coefficient of friction - it is an attempt to represent complex behaviour by a single number. It is difficult to predict and prone to inaccuracy, but it is better than nothing. What values of e can be expected in practice? It must lie in the range b to 1. A hard impact results in large stresses and strains so a lot of the energy may go into plastic strain (i.e. non-rewverable deformations) of the object, or actually breaking it, giving low e. Typical car collisions cause extensive permament deformation of bodywork and hence a low e Table 2 for which zero can be useful approximation. In a light impact the yield Typical coefficients of restitution stress may not be reached so e may be quite high, approaching I for very gentle contact. Materials like putty have such a low yield stress that e can Glass on glass 0.95 be taken as zero regardless of the other object involved. For interest rather Steel on steel 0.7 than practical use, Table 2 shows some approximate values of e for solid Wood on wood 0.5 spheres in direct central impact. Even in these relatively well defined Lead on lead 0.15 conditions, values can vary 20% or more. Although this section is headed 'Impact problems' the method can also be used for objects separating, for example by an explosion. There is no convenient name for this so I shall coin the term 'expact' for the opposite of impact. In this case the objects have the same initial velocity, so it is like an impact problem with a zero coefficient of restitution, but running backwards in time. A typical example is a gun firing a bullet. The mo- mentum method gives one relationship between the final velocities, but again more information is needed. Normally this will be observation of one of the final velocities, or else data on the energy available.

UA U8 before "LFlm-C - 4 X -VA v, after Figure S8

Notes on the procedure Steps I and 2: The choice of co-ordinate axes is crucial. For impact prob- lems, the initial velocities must be given relative to some known reference (e.g. a road) and it is usually best to consider the axes fixed to that. For expact problems it is usually most convenient to measure velocities rela- tive to the initial common velocity of the parts. Step 3: Always mark the velocities in the positive co-ordinate direction. If a velocity is physically in the other direction then it is negative. Step 4: J = 0. We are neglecting external impulse for a short period. Step 5: For a zero coefficient of restitution it may be more convenient to specify the single final velocity on the original diagram. Step 5 is then unneassary. A l520 kg vehicle travelling at 30 m S-' crashes into the rear of a 1780 kg one travelling at 15 m S-'. Estimate the velocities immediately after the collision using a coefficient of restitution estimate of 02.

Solulfon Steps 1,2,3 Figure 59 shows the system with velocities immediately won UA us 4 4 before and alter. r------v Step 4 G, =G, ;-l L - - a m,u, + m~u,= m,v, +m+, anm vs Step 5 Restitution e = 0.2 'L

Step 6 so u.=u,+3 Initial momentum mAuA+ mBuB= 72 300 N s Substituting in equation from step 4: 72300= 15200A+1780(vA+3) 72MO=3N)OoA+5340 v, = 20.3 m S-' q = 23.3 m S-' Figure 60 shows the result.

Example 12 2nd sage The rocket of Figure 61, having a total mass 20 tonnes, has just completed liring of its first stage. An explosive charge separates the first stage (now 12 tonnes) from the second, so that the specd of separation of the two parts is 4.2 m S-'. What extra speed is imparted to the second stage? aoluifm Steps 1,2 and 3 Sce Figure 62. step 4 0 = m,v, + m.4 (v, and q unknown) Step 5 - v, = 4.2 m s- ' (given) v,=%-4.2 Step 6 O= I2000(~~-4.2)+8000v. 0=20OOOy-m400 uB = 50400/20000 = 2.52 m S-' is the extra speed of the second stage. 4 (If required, v, = I!,, -4.2 = - 1.68 m S-') X

A 40 tome spscecraft docks with a 600 tonne satellite, arriving at a spd of 0.30 m S-' relative to the satellite (Figure 63). The two craft lock to- gether on docking. (a) What is the w&cimt of restitution? (b) Estimate . the resulting change in specd of the satellite.

SA0 02

An 80 kg man standing in a stationary 20 kg boat jumps for the jetty, C giving himself a horizontal aped relative to the boat of2 m S-'. Estimate 0.3 S-' his horizontal speed relative to the jetty. -m 6.4 Summary Impulse-momentum methods are appropriate to the solution of specific problems. Two main types are covered here: 1 Force is known as a function of time; find the velocity change of an object. Or if velocities are known, find the average force or time required. 2 Impact problems. The brief interaction of two objects coming together or flying apart, using a system including both objects. In each type, the method used is to keep account of the momentum in a chosen system. Momentum may enter a system by mass flow or by the application of a force. In either case the units of momentum flow rate are newtons. In the solution of impact problems, additional information such as the coefficient of restitution or the energy loss is normally required. Coefficient of restitution figures are of limited accuracy.

Closure

In this Block you have been introduced to two principal new ideas - energyjwork and momentumjimpulse. Keeping account of the quantity of these in a chosen system provides a powerful method of solving problems. Energy transfer by mechanical means occurs when a force acts on a mov- ing object. Hence the work-energy method is particularly useful where relationships between force and position are known (for example, springs). Momentum transfer occurs whenever a force acts (a force existing through time). Hence the impulse-momentum method is particularly useful where relationships between force and time are known. This method is particu- larly useful in the analysis of impactlexpact problems. There has not been time to cover angular momentum here (gyroscopes, etc.). An important idea evolving from energy is that of power - the rate at which energy or work is done. This is of great practical importance be- cause the size of an engine or motor depends on the power it is required to produce. The historical development of the ideas of momentum and energy is very interesting. Newton worked exclusively with the idea of momentum, but in 1686 the German philosopher Leibniz published a paper in the journal Acta Erudirorium criticizing the idea of momentum. The concept of poten- tial energy was already appreciated although not with that name. Leibniz argued that a given amount of potential energy put into two bodies of different mass did not produce equal momentum. He pointed out that the product of mass and velocity squared was, however, equal. Hence Leibniz introduced the basic idea that we now know as kinetic energy, although the factor of one half in )muZ was yet to follow. The main result of these suggestions was bitter controversy. It was fifty years before the differences were resolved by the French academic d'Alembert who pointed out that the enerev-, was related to the distance moved bv the force whilst the momentum depended upon the time that the force acted. If you are finding it a struggle to acquire a clear understanding of some of the ideas in this course, then it may be some consolation to you to know that some of the finest intellects ever known had to work very hard indeed to produce these ideas in the first place. Tat am refamnce Utim Active force 4.1 Fora that does work Active fora diagram 4.1 Diagram with active forcm only Brake power 5.3 Mechanical power dissipated at a brake Brake thmal efeuency 5.3 For an engine, the engine test brake power divided by fuel po- ckdcicnt of nstitution 6.3 On impact, relative separation speed divided by relative approach spad ERcctive inertia (or mm) 5.4 Fora divided by ~cccleration,which may be pater than real mans, cg because of wheel rotation mency 5.3 Usually, usdul output power divided by input power Like impact, but with qtionfollowing initial steady state Fuel power Fuel mean flow rate times M heat of combustion Gravitational potential Energy by virtue of position in gravitational field cnergy Integral of fora component over time Indicated power Engine power calculated from measured cylinder pressure readings (on an 'indicator') Indicated thermal cffictnoy 5.3 Indicated power divided by fuel power Kinctic mcrgy 21 Mschanical c~lergyd body motion Linear momentum 6.1 Momentum (m)of ution Potential energy 3 Energy because of shape or position Power 5.1 Rate of work or energy Strain energy 3 Potential energy assooiatcd with shape Work 1.1 Force component times distance moved Working fora 4.1 Same as 'active fora' Answers to Self-Assessment Questions

SA0 l SA0 7 W=Fs, which is in Nm (a) Only rolling friction does work: ~10.02 F-180N W=-9W (h) p=0.9 F-8100N W=-405kJ (c) p=0.1 F-900N W=-45kJ (d) Only rolling friction does work: p=0.02 F-180N W=-9W (However the brakes do reduce the car's kinetic energy, discwad later.) Figvn 64 Velocities refafire to initial 140~ity (e) Only rolling friction docs wok W = - 9 W (I) Only rolling friction docs wok W = - 9 W (The engine adds to the dskinetic energy, but The weight docs zero work as there is no movement in this is not work done by the tyre friction fora, the direction of the weight force. discussed later.) SAP 8 (a) At 0.1 m extension, the applied force is 20 N. The average fora during the extension is 10 N.

@) At 0.2 m extension, the applied fora is 40 N. Extending from 0.1 m to 0.2 m the average fora is 340 + 20) = U) N. The extra extension is 0.2 -0.1 -0.1 m.

Alternatively, you could have used the integral expression W - IF ds. SA0 9 From the areas under the curves, my estimates are: Work done on the band in extending it: W. = 0.74 J Figure 66 Work doac on the band during wntdon: W.= -0.661 Work 'lost' - 0.08 J (Actually this heats up the band slightly.)

Work done by the weight force and reaction force is SAP 10 wo(no vertical motion). (a) W=MB=+60x2n=377J (h) -377 J

SA0 11 The component of fora in the direction of motion of the application point is constant at F ms 30" = 43.3 N. SA0 I The disun~moved is s = rf) = 0.76 X n12 = 1.1 W m, so The wheel is not rolling so them is no work doac the work done is distorting new &OM of tbe tym. There i8 just a W~43.3~1.194=51.7J sliding friction fom. Alternatively, M = 43.3 X 0.76 = 32.9 N m 8 = n/2 = 1.57 :ad W=MB-51.71 SA0 1s (a) I = mk' = 844 kg m2 s=lOm K = fIw' = 6.08 M1 @)M8=6.08x106

8 = 2500 rcv = 15.708 X 103 rad S= lOm total M=KI8=387Nm "70 Figure 67 (a) From Table l, I = fmr2 = 3.375 kg m' Along y: K = fIw2= 16.9 kJ T-mg=rna (b) From Table 1, I = &m' + &m12 = 0.232 kg m' T=973 N K = fIw2= 1.16 kJ (a) The work done on the load by the rope is W- + 973 X 10 = 9.73 kl SA0 20 (b) The work done by the rope on the dnun is (a) W, =240 X 2n/60=25.13 rad S-' W- -973 X 10- -9.73k.J Initial energy: K, = flw: = 6.3 kl Final energy: K, -6.3 - 2.4= 3.9 kJ SA0 19 K2=fIw: so w$=2K,/I (a) W=+MN W, = 19.75 rad S-' (= 189 rcv min-') @) W = + 60 N (Not zero!) @) 2.4 kJ of energy must be added, which must equal SA0 l4 the work done by the torque, MO. Approximating each case as a rigid body with the 8 = 2400148 = 50 rad = 7.96 rev same swd at each point: (a) K=fxO.W3~2~=0.006J=6mJ SA0 21 (a) (v& =or = 60 X 0.05 = 3 m S-' (b) K=fx0.1 x4'=0.8J @) It is shown as a slender rod, i.e. of negligible thickness

(neglecting extra energy of swinging arms and legs) I = hml' = 26.7 X 10-6 kg m' (d) K-f X 1400~20'-28OW (C) K-fmu2+fIw2 =0.144+0.046 = 0.192 1 (d) I. =bl' + m(f1)' = fm12

l,, = 0.1067 X 10-' kg m' (C) K fI,,w2=0.192 1 SA0 l6 - (a) kg X m' SA0 91 (h) fIm2 =kg m' rad2 S" (a) Midway between the sphercs =kg m' s-~ (b) The spheres are small, so their contribution to I for the complete asaembly can be approximated =kgms-'xm by md2 where d is the distance to the centre of the =N m complete assembly. -1 I=2~8M)x2.5~=lOx103kgm' which arc the units of energy. Note that the red' can The tube is light, so it contributes negligible I. be deleted, unlike other units. (I must ask you to take this on trust.) (d) U, = or= 5.5 m S-' (e) 2xfrnu'-2xfx800~5.5'=24.2W No, a 'rigid body' has a single value of m and a single value of a. The change fiom I, to I. compensates for the change from MO to MO. SA0 23 SAoa

(a) 1=2~~'=2~14xO.Z'=I.355kgm~ (a) So00 J kg-' X 20 X 10-S kg= 100 J o = v/r = 5/0.28 = 17.86 rad a-' @) If you could convert all this energy into kinetic energy of the stone (you cannot do it completely) +mu1 = l00 (b) The fora will do negative work of 566 J on the axle unit v=63ms-' Fs = 566 This is about 140 miles per hour.

1001 SA0 24 -I kg The total kinetic energy comprises translation energy 100 J kg-' - of the body plus translation and rotation energy of (fifty times as much steel as rubber) the wheels. The translation energy of the body plus wheels is $mua where m is the total mass. In addition SA0 51 there will be the rotational energy of the wheels (a) The initial kinetic energy is K =4 x flo2where I is the value for one wheel. K,=fmv1=50kJ I = mk' = 12 X (0.22)' = 0.5808 kg m' This must be taken up as strain energy of the o S-' = v/r = 3610.28 = 128.6 rad springs: 2xfke'=fmo2

The wheel rotation contributes about 4% of the total, so the simple estimate tmv' wiU usually be about 4% =0.111 lower than the more accurate estimate. e = 0.333 m

SA0 2s (b) At a compression of 0.5 m, the strain energy IN From the areas under the curve: S=2xfke1=112.5kJ (a) Loading work done on the rubber W = + 0.74 J +mu1 =S (b) Unloading work done by the rubber W = + 0.66 J v' = 2S/m = (c) Lost energy = 0.74 - 0.66 0.08 J v=3 m S-' (d) Returned €raftion = 0.6610.74 = 0.89 SA0 S2 (i.e. storage efficiency89%) In the first position the spring length is 8Olcos 45" = 113.1 mm. 8 N After the 30" movement, the spring length is 80 mm (a) k=-=80Nm-' 0.1 m (equilateral triangle). (b) At e = 0.04 m, F = 3.2 N Average force = 1.6 N W = 1.6 N x 0.04 m = 0.064 J (c) At e = 0.08 m, F = 6.4 N Average force = H3.2 + 6.4) =4.8 N W = 4.8 N x 0.04 m = 0.192 J Figure 68 The extension of the spring changes fiom 63.1 mm to 30 mm. The work done by the spring is the decrease of strain energy. SAoa W=jke:-+keg= f6.2 J (a) Work done on spring equals increase in strain (The work done by the spring is positive - it has lost energy energy.) W = tkef = f X 2 X (10)' X (0.02)' = + 0.4 J (b) ~=fke:-jke:=1.6-0.4=+1.2J SA033 SA0 38 (a) This is a compression spring - it is used in a (a) mgh=1.%2J condition where its stressed length is less than its free length. The extension is, therefore, negative, @) If this became kinetic energy but this does not change the sign of the strain +mua - mgh energy S = fkea. In the closed position e = - 25 mm: v= 198 m S-' S, = 1.25 J At this speed aerodynamic drag would be considerable, In the open position e = - 40 mm: and the hailstone would reach a much lower terminal S' = 3.2 J speed. This is fortunate, for otherwise this type of weather could be very hazardous. To estimw the terminal sped The work input required is S, -SI = 1.95 J you would need a value for the drag wcffiient. @) Neglecting the spring mass and also friction, the energy will Convert to kinetic energy of the valve: SA0 39 +mu' = 1.95 J (a) The potential energy released at bottom dead centre is v=Il.4ms-' 0.8 X 9.81 X (0.4 - 0.4 X cos 30")

SA034 + 2 X 9.81 X (0.85 - 0.85 X cos 30') = 2.655 J (a) Work done by the weight force is mgh (= 200 J) Thii bmcskinetic energy (neglecting drag) +mu2 = mgh v' = 2gh v= 15.3 m s-l @) Fs=fmu2=ZOOJ Figure 69 F=4kN Thin will be the kinetic energy +Iflow'.so

8AQ S5 W' = 2 X 2.655/1.6 (a) Work done on the ball is W = 36 J. This will be the W= 1.822 rad a-' initial kinetic energy. (b) The energy lost is the di&rence between the potential @) Neglecting drag, the 36 J will be removed by negative energies at the two extreme positions where K = 0. work done on the ball by gravity. Relative to the bottom position, at 15' the potential mgh = 36 energy is h-21.6111

+ 2 X 9.81 X (0.85 - 0.85 X cos IS') = 0.675 J SA0 38 Energy lost is 2.655 - 0.675 = 1.980 1 (a) K=fmvZ=3.125 X 109 J=3.125 GJ (b) The potential energy V = mgh = 11.77 GJ (C) The engine burning fuel. (d) Drag. (relative to a position at sea level)

SA0 37 SAQ 41 (a) Height change h = 40 sin 8" = 5.567 m Water has fallen to fill the spaa previously mupied by the wood. This water has given up gravitational potential Work done by weight W = mgh (~49.15kJ) energy, transferred to the wood through the work of the This will become kinetic energy buoyancy force. fmv' =mgh

v= 10.5 m S-'

(b) Some is lost 88 metal distor&ion, heat and noise, some remains as kinetic energy in both cm,etc. (4) K, = flo: = 480 1 (5) K2=+lo:=12000J (6) W = MO (M to be found) (7) Kz=K,+W 12W0=480+Mx50 (8) M = (12000 - 480)/50 = 230.4 N m

Figure 70 (2) The car (a) The weight mg did work mgh = + 5.5 kJ. (3) (b) The buoyancy B did work - Bh = - 14 kl. -motion (c) Neglecting drag the other vertical force needed is (B -mg) acting downwards. This will do work (B-mg)x h=Bh-mgh = 14-5.5 kl Figure 73 = 8.5 kJ

SA0 19 Each wheel rotates O = sir = 10/0.28 = 35.7 rad - Work is done against friction on each wheel W=-MO=-500~35.7=-17.85W For both wheels the total energy change on the car is

W = - 35.7 kJ SAQ 48 (1) yes (2) The car (3) SA045 motion Workiig through the steps of the procedure: c3F (I) Yes. the work-cnergy procedure is appropriate. (2) The swimmer is my system. Figure 74 (3) See Figure 71. n (4) K, = +mu2 (4) K,=0 5) K,= 0 (5) KZ= fmv2 (6) W = - Fs = - pmgs (p = 0.1 estimated on ice) (6) W = + mgh (= 3.43 k0 (7) K2=KI+W (7) K2=Kl+W 0 = f mu' - pmgs + m0 (8) S = vz/2pg Figure 71 =115m (alongway!)

SAQ 48 (I) Yes SAQ 48 (1) (2) Both pulleys plus belt (2) The flywheel

motion 20 rad S" 100 rads-'

Figure 75 Figure 72 (5) K1 -+l*@:++l& + +m2 where m and v ~IC the mass and sped of thc Mt. The bclt energy is simply +mva kopw all puts of thc bclt have cped v. cu,=60nds-'

v=(u,rA-60~0.05-3m S-'

m, -v/r. = 310.1 = 30 rad S-' (3) Ki=18+22.5+4.5-451 -motion (6) W = MO (M to be found, 8 ;- 30 rad) engine resistance (7) Neglect stmh energy of Mt. S2=S1=0 K,=K,+W c257 45-0+Mx30 Prevn 7.3 (4) Kl = fmv' 10320 J (8) M - 45/30 - 1.5 N m - (5) K,=O SA050 (6) For a trawl of r the engine rotatw 33.1 X s radians (1) ye-5 WE=-MO (--36~33.1xs--l191.6~~ (2) The mate W, - -p,mgs (- - 126.5s) (3) motion W=W8+Ws (=-1318.1s) ('I) KI-KI+W P 0-10320-1318.1s FUIun 76 (8) a= 7.83 m (4) K1=5J (5) K, = +mu: (v2 to he determined) (6) W, = 196.2 J (2) The complete system W,=-~~xcos30" (jl-0.05) --l71 W = 196.2 - 17 - 179.2 ('I) KI-KI+W +mu: = 5 + 179.2 = 184.2 (8) v2 - 6.07 m S-' (acaptable) MOH (1) y- (2) The car (3) -motion

F- -77 = 47 (v2 to he found) (4) K110 (6) W-mgh-30~9.81 X 14=+4120J (5) K8 - +mua= 10320 J (6) Thc total hohntal push is H - 400 N. F1 = kmg- 126.5 N. m annot be neglected w"=Ha (-400 xs) W,- -Fla= - (- - 126.5 X a) W = (H - = 273.5s m0 W MOdl (1) Ye& (2) and (3)an for SAQ 53 Part (a) (4) K, - 67.7 J (1) y= (5) K1 -0 (2) flynm (6) WW - mgh - M x 9.81 X 0.5 - 147.2 J The Wintorque M wiU do work W.--MO (Mtobedanmincd) 8 = S/? = 0.slo.u - 2 rad W.--2M W - 147.2 - 2M (7) KI-KI+W 0 = 67.7 + 147.2 - 2M (8) M-107.4Nm%llONm

MOB Part (a) (1) y= (2) The burll* (3) motion Part (b)

(4) K,=O (5) K,-+m'-811 J (6) W-Fs (F W avwagc lorw) = 0.75F (7) Ka=K,+W fma=0+Fs (S) F - $r#d/a- 1082 N Thsare~oftbermrIscsofthcbullctirA=rr'-3&5mma. U~F-pA,p=F/A=28MNm-'in thcg~pmure. Part (b)

pm (c) We naod to bow il the work .g.iart Woni. *gni8cant duriDg startiry ihc lpriq alnuioo ia 60 mm, which on thc 6 mm ndiw shaft giva 10 dam

NOW This U only an estimate. I as sec one pssible rmon for tha Mien M bs uuh(Pntia4ly worm.The spriw tension will change the q5pmting fom required of tk bosrlags, which durins ka running support only tk weight ofthe fiywheol. The wcigbt is about 2 N whercurtbe~umlpriqtso*onU60N.The.ygugc @qtsndonis30N.u,ilIgtinutcthatthcfrictioa louVininerrPrinpraportiontoiho~~lord(.nd thatlallyhm~'~m?,thenW- (3012) X (-0.72 ml) = - 10.8 ml. This is now about a half SAOW of one pmt of the spring energy supplied, which in (a) Zero. ordinary engineering terms is probably still small enough to neglect. In practice I could tentatively call this loss (h) (-40 cos 4S0)(0.3)= - 8.49 W negligible, but in the back of my mind I would be ready (0) (- 2 cos 600)(7) = - 7 W to reassess it in the light of further information because I have made a very crude estimate of the effect of bearing (d) (10-')(0.8) = 0.8 mW load. [I did not expect you to give this complete an (e) (820 cos 60')(1.2) = 492 W answer to the SAQ, but simply to calculate -0.72 mJ.] (I) (-72 cos 60")(40) = - 1.44 kW

SA0 El (a) Along y: T-mg= mu (2) The car

v=0.4ms-' a=6m8-'

Figure 84 (4) KI=fmv'+4xflw1 W = v/r = 3010.28 - 107.1 rad S-l

(6) Each wheel rotates a distana B = s/r = 214.3 rad. For a braking torque M on each wheel, the work done on the system is W=-4MB (C) It incnaaes the kinetic energy. (d) No. The tension is constant but the speed is varyin& so Tu varies. Noto: The average fora on the bottom of a wheel is SA0 62 F = 48010.28 = 1714 N (a) The average wheel normal reaction is motion

N=880 X 9.8114-2158N The friction dcient necessary is

This is plausible for a tyre on a good road. However, the braking torque would have to be distributed amongst the wheels according to their normal reactions. Figure 85

Power lost to drag is DV = 60 X 20 = l200 W Power lost to rolling rcsistana is SA0 59 Flu=20x20=400W (a) P=600 W,an increaseof300 W For these foras the power Bow is -1600 W total. (b) 300Wx60~=18000J=I8kJ At steady speed, power input required is l600 W. Total (b) From the car engine, via the tow bar. Time Extra power power (c) 10min=600s 18000/600= 30W 330 W (d) 3 min= 180s 18000/180= 100 W 400W (c) 30s 18000/30= 600W 900W (f) 10s 18000/10 = 1800 W 2100W (S) Is 18000/1 =18000W 18300W

(h) The agreement is good down to 30 S. maa ~yBjwtarPtioofpowerqnoithasao tSnwd01u.Its unit is tbc numba I. -087 (a) PO- qP, = 0.92 X 6.2 kW = 5.70 kW (a) P, = Pdq - W92 = 43s kW (e) Pl - Pa/q = 1.2p.84 = 1.43 kW a411 (B) 43 X 025 = 10.8 kW @) 10.8 X 0.9 - 9.7 kW (b) nte man pub in 14.8 W. wbicb pullr with @gible loll~thcroptotbcfmtpuney.Themti'm (c) 9.7 X 0.9 = 8.7 kW hero is tan&&d m tbrop* W tbpower maim (d) 4 = 8.7143 = 0.20 in the rope and &a on down to the sceond pulley. Hem the pulley is moving, and the power is moved orq-~~~~~~~~~~~~030 wmpbly fmm tbc row 14.8 W pullr into the load (no powm Bon in tBe 0thpk of rnh9.8 W goa SAOlj toloadpotentialcnergy,thead~flawof5W [a) P, - Po/q - 6p.74 = 8.l kW be@ acmnulated aa kinetic energy. @) P0=qP,-0.74x8-S.9kW

Becaurc the bdt is in $Lon it tramnits power -to to the direction of motion.

atan (a) At a stcsdy spscd the cable tendon equals the weight: T=1200 N POW iuput to load kmcabb -1200x0.5=6wW (b) Gearbox input powa = 600I0.8 = 750 W (c) Motor input power= 750/0.75 - loo0 W

=on (a) 30rnSx1000kgm-a=30Mg

W66 (6) Some power now ruturn8 in the bottom of the pan (c) Eseh kiloysmiee of water acquires potential belt, W uh poww must be transmitfed in the top Y=m#h-I x9.81 X 18=176.61 m. D. i.& UY potmtial m hawse is 176.6 J Q-'. Tbo gobin* W-l d1101a W

P-0.3472 &S-' X 176.4 1 kg-' -61.3 W (d) Thc pump input is P = 61318.7 - 876 W (e) Ths mooor input power is P- 87.6/0.8 = 110 W (a) Kinetic enagy = +m' = 18 J for one kilogramme, i.e. 18 J kg-'. At a flow rate of 0.3472 kg S-' the power is D, P, and mg sin B are all forces and hence in N 0.3472 X 18 = 6.25 W. I kgm' (b) Increase in motor input power is so (m + I/r2) adds comatly to m. (kg) m.a-kgm sC1=N

SAOM (a) The potential energy per kilogramme is (a) mgh/m=gh=2943 1 kg-' -motion This provides the kinetic energy (+muz)for l kg of water, +U' +u2=gh U = (2gh)'I2 Figure 89 (b) Generator input power - 5010.91 = 54.95 MW q=+p$=240Nm-' Turbine input power= S4.95/0.88= 62.44 MW D=CqA=0.37~240X 1.8-160 N The energy content of the water was 2943 J kg-', P.=p,,N=275N (unchanged) so the flow rate (using the dot notation) must be For zero acceleration 62.44 X 10' J S-' T-D-FR=O m= 2943Jkg-' T=D+FB=435N = 2.12 X lWkg S-' (21 tonne S-') (b) Whml diameter is 0.56 m. so r = 0.28 m which is a great deal of water.

SAO m (a) 5 gallons =2.27 X 10-a m3 M = Tr =435 X 0.28 = 121.8 N m is the axle torque (60.9 N m on each of two wheels) (b) 5 gallons = 15.9 kg h = 15.9130 = 0.53 kg S-' (C) Neglecting any changes in the eficiencies (of course they will be diflennt, but without further data this is the best we can do) the fuel-to-wheel power etliciency is This is a very high power. A petrol station with twenty pumps could dispcnss up to about 500 MW q = 0.3 X 0.95 X 0.9 X 0.9 = 0.23 (unchanged) instantaneously - the output of a power station. This high power for dueUing is a substantial advantage of chemical fuel ovm, say, electricity for vehicle propulsion. The complete electrical system of my house is fused for 70 amps, which corresponds to about 17 kW. Correspondingly it would take about I2 hours of the fuU (C) Range = - = 20 =23,8 bkg-1 70 amps to provide the mfuellig energy of 30 s of petrol. h 0.84 X 10- Of course then arc many factors to consider, hut the (l) The range bap improved by 45% (from 16.4 km kg-'). rapid chemical fuelling rate possible i6 very convenient. The new value is about 47 miles per gallon. 20 m S-' is about the optimum sped for a typical car. Those with very small engines can benefit by reducing speed to 15 m S-'. Thersucmforthamli@~Ltha~blc SAOl* rcmdyluDdc A given d reduction in drab .mount (a) P = 120 kW X 0.95 X 0.92 X 0.9 = 94.4 kW Mwilldoromwhwo&roifyo~~~vo~ raLullcsyoudotlMcthcdbt.llec (b) P=Mm+lom Accordin#toourmoddatvmyIowrpkdthc M@ (U -0) ndt~md~toFI=27SNwithn@@Isdng, = Mu/r wbichyou~eapcetto~~vmy~~ Howtwr, the cegb GIllaot ogaote &tly at vcq =n low power outputs (i.e. its brake thermal sfecicy T-P/# drtniorataL to bdoa, sbopt I5 m E-' thsn h little tobcIpiacd.h~,tbc~cangctwoneifyou = 4.7 kN go too slowly. (wb*hwollldodybcporibhaiththaideal~ ntio).

SA088 (a) O.lx4-OAkgm8-'-0.4Ns (bl 80~2-160kgms-'-l60Ns (C) 90ilx30-27000kgm~-~=27kN1

MO# Tadm~umG-0.15~3-0.45N~ G,=0.4SasW-O.390Ns G,=-0.4SlinW--0225Ns

SA086 Ft=lSOx lO'xM=4.5~ IO6N8-4.5MNa

MOH (a) FromtheimlgldgI)M.cu~e, J%tf Ns @) J/m- 111800 = 0.014 m S-' Slow

nYn 92 (b) No. (C) YI)*OM due to my weight, one to the hrrPrction. (d) Inthaet,thctooal@hnpul&mmebNt-Wt. I#y~dmomwtudl~b~,ro Nf-Wf=O N=W-883 N srhkhiajustthc~i.e.thefnr+uwlyr*isalroa mamentm Bow adyds. uollll nol-4 mup-lOOONs, 8-0.80 This L about 10 mike pcr gallon, It is to low becaw I)M. ~arLclim~ahin~~fl&atthcbctivelnUa=lnU%+Fl lLer d the force term). IQM = 0 +Q.&F F-=1000~.0=12MN SA0w Purt (a) Zero Part (b) (1A (2A (3)

Measure all speeds relative to the initial speed of the aatcllite. Hence U. - 0 (4) GI G1 mAuA= (mA+ na)v (v to be found) Only v M unknown. (5) Not narssary.

Since U. =0, ths speed change of the satellite M v - 0 = v

mlniw to

X - after Figure 94 Index active force pule active force diagram aemdynamic drag Linetic energy of combined motion bearing friction of m&n brake power of manslation brake thermal efficiency buoyancy force linear momentum load-deflection curve coefficient of restitution coefficient of rolling friction momentum collision Newton-3 pairs effective inertia effective mass Parallel Axes Theorem eff~ciency potential energy expact power force-displacement curve radius of gyration fuel consumption rolling resiswee fuel power scmod moment of maps gravitational potential energy S1 units horsepower impact problems impact-expact procedure watt impulse work indicated power working force indicated thermal efficiency work-energy procedure