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.