nm0796: okay shall we start sf0797: yes nm0796: yep okay well you'll see that today we're we have a silent visitor in the front row what's actually happening today is that the lecture is being recorded and i think i mentioned to you last week the purposes of this so you know what's happening er by and large i think we'll carry on as normal and ignore what's going on over here and er so that's where we are if you can carry on like that we'll work in a normal way er what i wanted to say to you this morning is that i think w-, today's session is going to be the last of the lecture sessions for the course or if it isn't there'll only be about ten or fifteen minutes next week and we will then come back and look at some example sheets now so far i believe you've had three example sheets am i right sm0798: yep nm0796: have you had example sheet four sm0799: yep nm0796: right well then i've got today for you example sheets five and six which makes the complete set so i'll give those out at the end of the lecture so apart from that we're working on i think page five of your printed notes set three and just to bring you back to where we were last week we had looked at the example which was a quite a complicated one at the top of that page er to do with the motion of the rocker arm and that was the last thing we went through so that's that's where we are at and now we're going to proceed a little bit further er through these notes so the first thing we're going to do is take a look at a problem which involves rotational momentum and that's the next problem that's on the list which is example three-E and what we're doing here is looking at the rotational equivalence of a collision so where previously we've looked at collisions of particles one particle striking another one body striking another er here we're looking at collisions that occur in rotation which is not nearly such a common idea and it's a concept that perhaps is a little bit difficult to perceive what we are going to be doing here is looking at a problem where a clutch of an engine is put in so that what we're ha-, what is happening is there's an a an engine with its flywheel has a certain amount of rotational momentum and then this is suddenly connected to the drive shaft and the wheels er of a vehicle so in fact you have a rotational impact between the engine system on the one hand and the driven system on the other hand and what happens is that the total rotational momentum of the system is conserved although as we shall see the energy is some of the energy is lost in the process so it works in exactly the same way as linear momentum problems work so let's just take a quick look at what the problem says then er the engine of a car is revved up to five-thousand R-P-M which is slightly unconventional units but i'll say a bit more about that in a minute the clutch is then engaged over a short time interval to get the vehicle moving from rest we'll presume that the vehicle's in first gear the effective moment of inertia in a flywheel engine is given the driven system is rather less five times less in first gear er do note that when you've got a system like that when you're in second gear or third gear or fourth gear the effective moment of inertia of that system becomes greater as you go up through the gears that in fact is quite a a complicated thing to work out and you have to know something about gear boxes and gear ratios but we're not concerned with that here er determine the speed of the engine after clutch sm0800: do we do that nm0796: yes next year determine the speed of the engine after clutch engagement and the amount of energy lost okay so there's our problem let's take a look at how we solve this one then now i'm going to work in some slightly unorthodox units here just for convenience er if you don't like what i'm doing er by all means convert this problem for yourself into S-I units but i think when you see me go through this you'll realize that er the mixed units that i'm using are actually very convenient for the first part of this problem so h-, here we have then first of all a determination of the angular momentum the I-theta- dot of the engine before you engage the clutch the angular velocity the theta-dot is five- thousand remember that's in R-P-M er the moment of inertia is point-five which is in standard S-I units multiply the two together and you get a figure of two- thousand-five-hundred in some hybrid units and provided i stick with these hybrid units throughout everything will be all right as i say you can convert to S-I if you want to er after the so-called impact in other words once the clutch has been engaged then you've got the two systems rotating together at some speed which we're about to determine what has happened is that the moment of inertia of the total system has gone up to be the sum of the two so the total moment of inertia is point-five plus point-one for the driven system so we've got moment of inertia point-six this then is an expression for the angular momentum after engagement and if you equate this expression to two- thousand-five-hundred presuming that we have conservation of angular momentum then the theta-dot works out to be four-one-six-seven and because we're using these hybrid units we know that that must be also in R-P-M so that is a very straightforward calculation i hope you will agree er very similar to linear momentum conservation now let's have a look at the energy the second part of the problem tell me if i move this up too far now at this point i've chickened out and i've gone into S-I units 'cause i want to work in straightforward joules er in other words S-I energy units er and here what i'm doing is working out the energy before kinetic energy before half-I-theta-dot-squared and the energy after so this is an expression then for half- I of the engine moment of inertia of the engine times the theta-dot-squared and you'll see that i've made a conversion from five-thousand R-P-M into radians per second and five-thousand divide by sixty tells me how many radians per sorry how many revs per second i've got and a rev is two-pi-radians so the conversion factor here is two-pi-over-sixty then square it if you put your calculators around that one you get this figure for the kinetic energy sixty- eight- odd-thousand joules er after the clutch has been engaged we've now got an increased moment of inertia to point-six but a decreased speed to four-one- six-seven according to the calculation we've just done and that will tell us the energy kinetic energy of the system after the clutch engagement so you see that we've got a loss of something like twenty per cent of the energy in that process which is probably a realistic sort of loss where does where does the energy go to sf0801: heat sm0802: nm0796: heat yeah and where's the heat going to appear sm0803: on the clutch plates nm0796: on the clutch plates right okay so that then is a fairly straightforward sort of problem isn't it but the the key thing here is to see the parallel here between linear particle motion and rotational motion and we did a very similar thing for that trivial problem to do with railway trucks colliding with each other if you remember it all right moving on then just in passing there is a chart at the back of your notes showing er the comparison between translational motion and rotational motion i don't want to dwell on that or to describe it in any detail but i would like you er some time in your own time just to take a look at that and see what is said there there are one or two comments about the vector er the vector properties of some of these quantities and as we said at the beginning of the course you can ignore these for the moment but there are one or two little oddities there they won't concern you just look at the fact er the key fact that there is a one to one correspondence between what happens in linear particle motion and what happens in rotational motion i know i've said that about ten times but people do sometimes forget it even though i say it ten times well now finally what we're going to come on and look at is er the general type of motion in which we have got a mixture of linear motion according to your second set of notes and rotational motion which is according to these notes and sometimes this can become a little bit complex er the key features of this and the key things that you need to remember are first of all that you need to work as at the centre of gravity of the system that is the safest possible thing to do just occasionally though secondly there is a fixed centre of rotation for a a set of bodies or a body and we'll see one of those in a minute in that case it's sometimes more convenient to work at the centre of rotation but you have to be a little bit careful how you handle that it's always safe to work at the centre of gravity all right so the first problem we're d-, we're going to look at this morning is example three-F at the bottom of that page er this one is a relatively straightforward problem of the what i call the direct kind the direct kind if you remember is where you know forces applied and you can work out accelerations or vice versa you know the motion and you have to work out the forces this particular one is in the first category what we've got here is a thing like a bobbin or a cotton reel resting on the surface a piece of cord ar-, is wound around it and we pull on the cord and the bobbin is going to roll along the surface that's the nature of the problem and what we have to do is work out what that motion is going to be under the conditions that are applied now thi-, this problem does get a little bit complicated for two reasons can you anybody think what those two reasons might be no sm0804: the mass has been reduced has it nm0796: we're not going we're not going to take that into account it could be a problem yes if the cord were heavy we would have to take that into account but we'll assume it's a light cord sm0805: as it unwinds so the radius reduces nm0796: no we'll assume the cord's all on one radius okay well i-, just i'll tell you what they are two things first of all you don't know whether this thing is going to slip on the surface or roll so somehow or other we've got to cope with that and the way in which we cope with it is we assume one thing work out what the answers are going to be and if they turn out to be daft we've made the wrong assumption and then we go back and try the other assumption er usually if you get it wrong something quite inconsistent will appear and you know that you've got it wrong so that's the first problem slipping or rolling er the second problem is which way is it going to roll if it's rolling let's assume it's going to roll is it is this bobbin going to roll to the right or is it going to roll to the left sm0806: to the left nm0796: well that's you see the interesting thing 'cause seeing that you think it's going to unwind so the bobbin's going to have to go to the left in fact that's impossible it always goes to the right if you pull on anything in one direction it will go in that direction unless there's some other reaction that stops it doing so this thing will roll to the right and you'll notice that i put that at the bottom there well you may not believe that but just think about it okay so let's then see how we are going to tackle the problem as with any of these sort of problems we'll work through this in a fairly well defined order perhaps before i do that i should just recite the main features of the problem to you which are written into it er first of all it's two-hundred millimetre radius o-, or point-two of a metre mass twenty-five kilograms it tells you the radius of gyration at one-seventy-five millimetres poi-, point-one-seven-five metres tells you the radius of the groove which appears also on the picture underneath the force applied is twenty newtons and what you're being asked is to calculate the angular acceleration and it tells you the coefficient of friction if you need to know it between this bobbin and the surface right so at the top here of the solution first two things i write down is first of all the mass of the bobbin and secondly i need to calculate the moment of inertia because we're actually told the radius of gyration and remember that the moment of inertia about the centre is going to be mass times radius of gyration squared M-K-squared so that moment of inertia is going to be twenty-five times point-one-seven-five-squared and that result is given here now i'm going to make my big assumption i'm going to assume that this thing rolls and doesn't slip and i've put a note there that i must check at the end that that is a reasonable assumption to make or at least check for consistency we'll see how we do that when we get there and then the next and a final preliminary to looking at the dynamics is to actually look at the kinematics i haven't actually written that word on this overhead but usually i do you might like to just write kinematics here remember that kinematics is looking just at the motion nothing to do with forces or anything of that kind so i'm assuming it's rolling here is a picture of what is happening if it's rolling there is an instantaneous centre of rotation at the point of contact between the bobbin and the surface so in other words looking at that bobbin it's just rolling about this point of contact at that instant course as soon as it rolls a little distance the point of instantaneous centre is going to move round the rim but it's always at the point of contact so if we assume that in this rolling the linear motion of the centre of gravity is X to the right and the angular motion angular displacement is theta er angular motion is not confined to a single point on the body if you've got what's called the rigid body in other words it doesn't change its shape an angular motion is the same for any point on that body so in fact the angular displacement about the instantaneous centre is theta as well which means that the centre is going to move by the corresponding value of R- theta R is point-two of a metre so there is a relationship here between the X at the centre and the angular movement and X is going to be point-two-theta that's the R and that's the theta the other part of the kinematics that's important is if we differentiate this twice so X- double-dot is then going to be point-two-theta-double-dot so that's the kinematics now we move on to the dynamics and we need the free-body diagram for this bobbin remember what this is it must have all forces and moments shown including the inertia force and the inertia moment and according to me there are six such forces only one of which is explicitly applied the only applied force to this thing is the twenty newtons down here all of the others come about either by virtue of contact with other poin-, other bodies hence the reaction and the frictional traction force and i've put a little note here that this frictional traction force ought to be less than mu-R if it isn't less than mu-R then this thing is going to slip so our assumption is wrong so one of the checks that we have to do at the end is to check whether F actually is less than mu-R then the other forces there are two inertia forces the M-X-double-dot which must be in the opposite sense to positive-X or positive-X-double-dot and the I- theta-double-dot in the opposite sense to positive-theta and last but not least the weight of the object the M-G downwards which acts at the centre of gravity well as always it's important to get that picture right once you've got that picture right and complete then everything else should be easy but it's always that first step where things go wrong you either forget one of the forces you ought to put in or you get one of them wrong there are there are too many of them so i hope that you agree that i have got the correct number of forces here remember the things you have to check inertia forces wherever the body contacts something else and any applied forces three types and they're all there so that is your picture and i'll try and keep this on the screen at the same time as showing you the equations so you can see where they've come from can you can you see that all right er so the first and simple one is the horizontal equilibrium from the picture that's just disappearing at the top so the t-, the twenty is balanced by the er M-X-double-dot and the F so i've written it this way round er F equals F to the left is twenty to the right minus M-X-double-dot to the right the M is twenty-five kilograms the X-double-dot we saw from kinematics is point- two- theta-double-dot so a fairly simple expression for F comes out of there and then the other equation that we need is the moment equation i'll take that off i think well you've got the picture in front of you haven't you i think i'll take the picture away and then this can go higher up you can see what's going on right so i've i've taken moments about the centre of the bobbin now you'll notice that several of the forces actually pass through that centre so they don't have moments which is handy er the ones that do are the inertia moment I-theta-double-dot which is anti anticlockwise is it yep and the twenty times the radius of the groove which is anticlockwise balanced by the friction or tractional force of F times that radius point-two none of the other forces have a moment about them in the middle and then i substitute directly for F from what we've found up here and put some numbers in so there's the value of I that i calculated right at the beginning that comes to one-point- five F here goes into there and you end up there then with an equation which is solely in terms of the angular acceleration so we can work out what it is i haven't sho-, shown the detailed algebra but that tells you what the result comes out to be so that's the solution to the problem except that we must now check our assumption w-, and if you remember what we need to check is that the traction force is less than mu-R well the traction force from the equation for the horizontal equilibrium is twenty minus five- theta-double-dot and that comes to about thirteen newtons so a traction force if it's rolling is thirteen newtons the value of the reaction is the same as the rate which is twenty-five kilograms times nine-point-eight-one in newtons which is ab-, more or less two-hundred-and-forty-five so mu-R is point- one times that which is twenty-four-and-a-half newtons so it is indeed the case that the traction force is less than mu-R it's about a half of it so in fact the assumption that it's rolling is okay right so what we've seen there then is a mixture of linear and rotational motion a-, and it's had a slight complication because of this business of the rolling or slipping which may not appear in many problems but you often get a little complication of that sort to deal with right so now if we turn over then to the final page of the notes page six and here is a problem in the next category that i usually look at you remember we have three k-, types of problems direct problems of the sort we've just solv-, solved the impact momentum type problems which i'm about to do and thirdly the energy conservation of energy type problems which is the final one so there are two problems on this page one is to do with impact momentum on a mixed body in which we've got rotational and linear momentum going on together and the second problem is of energy conservation problem the the thing that i've chosen for this next problem is what is sometimes known as the cricket bat problem those of you who play cricket will be familiar with the fact that when you are batting and you strike the ball if the ball doesn't hit the right place on the bat it stings your hands and the reason for that is that you have an impulsive reaction at your hands the ball striking the bat i-, causes an impulse a linear impulse on the bat and because of the the way in which equilibrium of imp-, of impulses occurs there will be in general er an impulsive reactions where you're hanging on to it the hinge if you like of the bat except in certain very special circumstances if you happen to get the ball hitting the bat in the right place it doesn't sting and such a a place is called the centre of percussion and you can easily work out from the dynamics where that position happens to be in the case of a cricket bat it's probably about that much from the bottom of the bat which if you look at the way in which cricket bats are shaped is where the bat is actually thickest and chunkiest and that's really where you should hit it if the ball comes up and hits it er higher up the bat then you're going to suffer the ball goes even higher you'll need your guard on but there you are okay so let's have a look see what the problem says i've done this in very general terms that picture in the notes is not meant to be a cricket bat i think i can draw a bit better than that [laugh] but essentially it's the same problem what we've got here is a body that is hinged at the top at O centre of gravity is somewhere further down distance L beneath and we're actually going to strike the thing with an impulse at a point H below the hinge er and there are one or two bits missing on the diagram for which my apologies so i'd like you to just add the first thing that's missing is that there's an arrow round there with nothing on it that should have theta-dot put against it that is the angular motion that's going to occur after the impact now the impact is sh-, shown at the point A here by a line it's a very defective drawing i'm sorry about this th-, there should be an arrowhead there on there and also that should be called P P is going to be what i call the impact the impulse it's not a force it's an impulse and then over at the right of the diagram you see hanging in mid-air er V and theta- dot or actually i think it says theta but there should be a dot over it er there should be again an arrow like that to show that theta-dot is clockwise as it is up here and V is going to be to the left so you want an arrow on there otherwise the picture is wonderful sm0806: what's an impulse nm0796: pardon sm0806: what's an impulse nm0796: an impulse if you remember i hope you will do when i tell you is a sho-, a a force of short d-, duration applied to a body and the impulse is the integral of that force over time and that impulse applied to a body causes a change in linear momentum remember that that was in our our second set of notes okay right just getting the bones out of the example i'm going halfway down it er let's the body be struck by an impulse P at A which we've shown on the diagram find find the value of H for A to be the centre of percussion now what that actually means is that we've got to find H such that the reaction there which also you could add to your diagram if you want to what you want is to have H such that R is zero so let's i'll just write under here so you don't have a misapprehension P and R are impulses not forces all right can we just then set off into the problem can i take that one away have you got what you need from that right i don't think you need to draw the picture again you've got it in your notes so don't waste time doing that er key key thing here again is that we've got mixed motion the centre of gravity is moving linearly to the left er but er we've also got this rotation theta-dot and the point O is fixed at the top or assume to be fixed once again let's look at the kinematics first si-, the simple motion the the value of velocity V to the left at the centre of gravity is going to be the corresponding value of R- theta-dot like it was for the previous problem only here the R is the distance L so L-theta-dot taken from the centre rotation at the top tells you the value of V and now we do a rather similar thing to the previous problem we look at the linear motion and then the angular motion in the previous problem it was linear equilibrium of forces and moment equilibrium but you need to do the two because we've got mixed motion so the first first thing is looking at the linear impulse the total linear impulse on this object is primarily the P to the left but we must also see that there is the reaction R at the top which is to the right so the total linear impulse on the body is P minus R and that's going to result in the linear momentum of this thing at the sin-, centre of gravity so it's going to be equal to M-V i've used the kinematic straightaway to find out an expression for P so that's a fairly simple one slightly more complicated the angular impulse now we haven't actually used this before er the angular impulse let me just recap on that for you the moment of an impulse gives you a change in angular momentum so what i'm doing here is taking a moment of the impulses P and R about the centre of gravity i'm working at the centre of gravity so P gives you er a cl-, a clockwise moment of impulse of P times the distance H-minus-L and the R gives you a also a clockwise moment of impulse of R times L so you've got these two terms in the expression for the moment of impulse and that will result in a change in angular momentum I-theta-dot so what we're er assuming here is that the object starts stationary but is struck and then ends up with these velocities so we've got a change of linear momentum and a change of angular momentum both determined by these er impulse equations so that's what happens in general and now the problem becomes easy because what i'm going to do is to see what the value of H has got to be for R-equals-zero so i simply put R- equals-zero in those two equations that we've got these two here that one put R- equals-zero this one put R-equals-zero and what will then happen is that a value of H the height will be defined well how have i do-, done this let's just have a quick look this first equation here comes from this one er P P is going to be M-L-theta-dot from there so putting that into there M-L-theta-dot times H-minus-L plus nought is I-theta- dot and you'll see that the theta-dots cancel out and if we substitute I-equals-M-K- squared then the Ms cancel out as well and so you get a purely geometric equation for H the value of H is in terms of the radius of gyration and the L distance from the hinge to the centre of gravity so this is a rather general abstract sort of analysis er probably more usually this would be couched in terms of a very specific problem er an object of a specific size and mass and so on but what you've got here is a general formula that will enable you to work out centre of percussion for any object you care to choose and what the problem f-, says in its final sentence is apply the theory to find the centre of percussion for a uniform bar hinged at one end so a uniform bar at hinged at one end we have to decide what we mean by K and L so this takes us back to something that i asked you to remember couple of weeks ago i can get this lot to stay on board i'll move it up in a minute if you can't see the bottom right so if you've got a uniform bar length big L the the little L in our picture is the distance from the hinge to the centre of gravity in other words from the top of the bar to a point halfway down it so the little L that we had in the general formula is going to be the big L divided by two it's half the length distance from the end to the middle and then the formula that i asked you to remember was that the moment of inertia of a bar hinged about one end was M-L-squared-over-twelve so K-squared sorry about the centre i mean what am i talking about er moment of inertia about the centre for a bar is M-L- squared-over-twelve so the K-squared is big-L-squared-over-twelve and all that remains to work out the corresponding value of H is to substitute these two into that formula and what you f-, find is that the centre of percussion is about two-thirds of the length and the that corresponds fairly well with a cricket bet doesn't it although a cricket bat isn't actually a uniform bar all right now we're going to cheat example three-H is one which i'm not going to re-, do on the overhead at all what i'm going to do here is give you a handout for your notes and er then superficially i will work through this over the next two minutes er this problem is one in the final category of making use of the principle of conservation of energy and it's also closely related to er a piston mechanism used in an internal combustion engine so looking first of all at the picture in the notes if i can find them sorry er what you see here is a link A-B which corresponds with a connecting rod in an engine the force P applied would correspond to the force applied by exploding gases on a piston er B-C is the crank on the crankshaft so what is happening is that B-C is going round and round er connecting rod is following it at B but point A moves in a straight line in other words it moves up and down the cylinder so you can see the mechanism can you right so the thing is actually starting with a rotational velocity of ten radians per second for B-C in other words the rotation rate of the crankshaft and a constant force is applied through half a cycle of the engine hundred newtons er that means that that force moves through a distance of how much i-, if B-C goes all the way th-, round through a hundred-and-eighty degrees how far has point A moved sm0807: nought-point-four metres nm0796: yes point-four metres so the work done by P is a hundred newtons times point-four which is forty newtons forty joules so the work you're putting in is forty joules and that work is going to result in an increase in the kinetic energy of the mechanism of the same amount forty joules and the interesting feature of this type of calculation is that you are not actually working out anything to do with accelerations and what have you during the course of the process if you did that it gets very complicated 'cause you've got to take all the angles into account and this is one of the great virtues of using the conservation of energy as a principle you simply look at the initial state and the final state you don't worry at all about what happens in between right so let's have a quick look then at the script which i've just given out principle at the top now there's a useful formula given here which you can work out for yourself if you have a bar hinged at one end fixed at one end with a linear velocity V at the other end the kinetic energy of that bar is M-V-squared-over- six that's a little magic formula er it actually is very useful in this problem because you can use it at every stage in what happens er but you can easily prove it for yourself the third stage the initial kinetic energy point B is moving down at two metres per second that arises from the fact that B-C is going round at ten radians per second it's the R-theta-dot so that velocity V applies to both rods at that position the left hand rod A-B fits the category of paragraph two with a V of whatever it is two metres per second and it also applies to the other one the crank so we can very easily use the M-V-s squared-over-six for both to tell us what the total kinetic energy is and that's at the end of section three you'll see that the kinetic energy initially is quite small about two joules er this works because the centre of gravity does not have any horizontal component of velocity for this particular problem in either the initial or the final state if it did working out the kinetic energy gets a bit more complicated and then the final kinetic energy point B has gone all the way round through a hundred-and-eighty degrees and now it's moving upwards with a different velocity which is unknown V but we can use the same formula same formula is going to apply to work out kinetic energy except we simply put V instead of two in the expression above five is work done which we've just talked about six is making use of the principle at the top which tells us that V is nine-point-four-seven so it's gone up from two metres per second downwards at the beginning to nine-point-four-seven upwards at the end and then theta-dot comes directly out of that by the same means so actually when you get into it it's a very simple problem isn't it looks complicated to start with but once you've got a few tricks in your armoury it's very straightforward thank you that will do us for today now ne-, next week what we're going to do is take a look at some of the example sheets er i've got two more of these here for you today can you take one each of those and next week i want you to come prepared and ready to ask some questions about things that you've had a try at so if you could put a little bit of effort into looking at perhaps the earlier example sheets er particularly example sheet four and we'll have a look at that next week and in the final week of term we'll look at these two all right any anything else