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<?xml version="1.0"?>

<!DOCTYPE TEI.2 SYSTEM "base.dtd">





<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>


<availability><p>The British Academic Spoken English (BASE) corpus was developed at the

Universities of Warwick and Reading, under the directorship of Hilary Nesi

(Centre for English Language Teacher Education, Warwick) and Paul Thompson

(Department of Applied Linguistics, Reading), with funding from BALEAP,

EURALEX, the British Academy and the Arts and Humanities Research Board. The

original recordings are held at the Universities of Warwick and Reading, and

at the Oxford Text Archive and may be consulted by bona fide researchers

upon written application to any of the holding bodies.

The BASE corpus is freely available to researchers who agree to the

following conditions:</p>

<p>1. The recordings and transcriptions should not be modified in any


<p>2. The recordings and transcriptions should be used for research purposes

only; they should not be reproduced in teaching materials</p>

<p>3. The recordings and transcriptions should not be reproduced in full for

a wider audience/readership, although researchers are free to quote short

passages of text (up to 200 running words from any given speech event)</p>

<p>4. The corpus developers should be informed of all presentations or

publications arising from analysis of the corpus</p><p>

Researchers should acknowledge their use of the corpus using the following

form of words:

The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

British Academy and the Arts and Humanities Research Board. </p></availability>




<recording dur="00:49:06" n="6410">


<respStmt><name>BASE team</name>



<langUsage><language id="en">English</language>



<person id="nm0796" role="main speaker" n="n" sex="m"><p>nm0796, main speaker, non-student, male</p></person>

<person id="sf0797" role="participant" n="s" sex="f"><p>sf0797, participant, student, female</p></person>

<person id="sm0798" role="participant" n="s" sex="m"><p>sm0798, participant, student, male</p></person>

<person id="sm0799" role="participant" n="s" sex="m"><p>sm0799, participant, student, male</p></person>

<person id="sm0800" role="participant" n="s" sex="m"><p>sm0800, participant, student, male</p></person>

<person id="sf0801" role="participant" n="s" sex="f"><p>sf0801, participant, student, female</p></person>

<person id="sm0802" role="participant" n="s" sex="m"><p>sm0802, participant, student, male</p></person>

<person id="sm0803" role="participant" n="s" sex="m"><p>sm0803, participant, student, male</p></person>

<person id="sm0804" role="participant" n="s" sex="m"><p>sm0804, participant, student, male</p></person>

<person id="sm0805" role="participant" n="s" sex="m"><p>sm0805, participant, student, male</p></person>

<person id="sm0806" role="participant" n="s" sex="m"><p>sm0806, participant, student, male</p></person>

<person id="sm0807" role="participant" n="s" sex="m"><p>sm0807, participant, student, male</p></person>

<personGrp id="ss" role="audience" size="s"><p>ss, audience, small group </p></personGrp>

<personGrp id="sl" role="all" size="s"><p>sl, all, small group</p></personGrp>

<personGrp role="speakers" size="14"><p>number of speakers: 14</p></personGrp>





<item n="speechevent">Lecture</item>

<item n="acaddept">Engineering</item>

<item n="acaddiv">ps</item>

<item n="partlevel">UG1</item>

<item n="module">Rigid body dynamics</item>





<u who="nm0796"> okay shall we start </u><u who="sf0797" trans="latching"> yes </u><pause dur="0.7"/> <u who="nm0796" trans="pause"> yep <pause dur="3.4"/> okay well you'll see that today we're <pause dur="0.3"/> we have <pause dur="1.2"/> a silent visitor in the front row <pause dur="0.9"/> what's actually happening today <pause dur="0.4"/> is that the lecture is being <pause dur="0.5"/> recorded <pause dur="0.4"/> and i think i mentioned to you last week <pause dur="0.2"/> the purposes <pause dur="0.3"/> of this so you know what's happening <pause dur="0.6"/> # by and large i think we'll carry on as normal <pause dur="0.3"/> and ignore what's going on over here <pause dur="1.0"/> and # <pause dur="1.1"/> so <pause dur="0.7"/> that's where we are if you can <pause dur="0.9"/> carry on like that <pause dur="0.3"/> we'll work in a normal way <pause dur="0.6"/> # what i wanted to say to you this morning <pause dur="0.3"/> is that i think <trunc>w</trunc> <pause dur="0.5"/> today's session is going to be <pause dur="0.3"/> the last <pause dur="0.3"/> of the lecture sessions <pause dur="0.4"/> for the course <pause dur="0.6"/> or if it isn't there'll only be about ten or fifteen minutes next week <pause dur="0.4"/> and we will then <pause dur="0.6"/> come back and look at some example sheets <pause dur="0.2"/> now so far <pause dur="0.5"/> i believe you've had <pause dur="0.8"/> three example sheets am i right </u><pause dur="1.9"/> <u who="sm0798" trans="pause"> yep </u><u who="nm0796" trans="latching"> have you had example sheet four </u><u who="sm0799" trans="latching"> <gap reason="inaudible" extent="1 sec"/> yep </u><pause dur="0.5"/> <u who="nm0796" trans="pause"> right <pause dur="0.6"/> well then i've got <pause dur="1.2"/> today for you example sheets five <pause dur="0.3"/> and six which makes the complete set <pause dur="0.6"/> so i'll give those out <pause dur="0.5"/> at the end of <pause dur="1.0"/> the lecture <pause dur="1.3"/>

so apart from that we're <pause dur="0.4"/> working on <pause dur="0.4"/> i think page five <pause dur="1.1"/> of your printed notes <pause dur="1.1"/> set three <pause dur="2.9"/> and just to bring you back to where we were last week we had looked at the <pause dur="0.5"/> example <pause dur="1.1"/> which was a quite a complicated <pause dur="0.2"/> one <pause dur="0.7"/> at the top of that page <pause dur="0.7"/> # to do with the motion of the rocker arm and that was the last thing <pause dur="0.6"/> we went through <pause dur="1.0"/> so that's that's where we <pause dur="0.4"/> are at <pause dur="0.3"/> and now we're going to proceed a little bit further <pause dur="0.9"/> # <pause dur="0.3"/> through these notes <pause dur="0.3"/><vocal desc="clears throat" iterated="n"/><pause dur="0.4"/> so the first <pause dur="0.3"/> thing we're going to do is take a look <pause dur="1.0"/> at <pause dur="0.5"/> a problem which involves rotational momentum <pause dur="1.5"/> and that's the next <pause dur="1.0"/> problem that's on the list <pause dur="1.6"/> which is example three-E <pause dur="1.6"/> and what we're doing here <pause dur="0.4"/> is looking at the rotational <pause dur="0.3"/> equivalence of a collision <pause dur="0.5"/> so where <pause dur="0.2"/> previously we've looked at collisions of particles one particle <pause dur="0.4"/> striking another <pause dur="0.7"/> one body striking another <pause dur="0.9"/> # here we're looking at <pause dur="0.2"/> collisions that occur in rotation <pause dur="0.6"/> which is not nearly such a common <pause dur="0.3"/> idea <pause dur="0.6"/> and it's a concept that perhaps is a little bit difficult to perceive <pause dur="0.7"/> what we are going to be

doing <pause dur="0.5"/> here is looking at a problem <pause dur="0.4"/> where a clutch of an engine <pause dur="0.5"/> is put in <pause dur="0.2"/> so that <pause dur="0.6"/> what we're <trunc>ha</trunc> what is happening is there's an <pause dur="0.2"/> a an engine with its flywheel <pause dur="0.2"/> has a certain amount of rotational momentum <pause dur="0.6"/> and then this is suddenly connected <pause dur="0.5"/> to the drive shaft <pause dur="0.3"/> and the wheels <pause dur="0.5"/> # <pause dur="0.2"/> of a vehicle <pause dur="0.5"/> so in fact you have a rotational impact between the engine system on the one hand <pause dur="0.6"/> and the driven system on the other hand <pause dur="0.6"/> and what happens is that the total <pause dur="0.4"/> rotational momentum <pause dur="0.6"/> of the system is conserved <pause dur="0.4"/> although as we shall see <pause dur="0.4"/> the energy <pause dur="0.7"/> is some of the energy is lost <pause dur="0.2"/> in the process <pause dur="1.4"/> so it works <pause dur="0.5"/> in exactly the same way <pause dur="0.4"/> as linear <pause dur="0.3"/> momentum <pause dur="0.4"/> problems work <pause dur="1.3"/> so let's just take a quick look at what the <pause dur="0.2"/> problem says then <pause dur="1.0"/> # <reading>the engine of a car is revved up to five-thousand R-P-M</reading> <pause dur="0.6"/> which is slightly unconventional units <pause dur="0.4"/> but i'll say a bit more about that in a minute <pause dur="1.0"/> <reading>the clutch is then engaged over a short <pause dur="0.3"/> time interval <pause dur="0.4"/> to get <pause dur="0.2"/> the vehicle moving from rest</reading> <pause dur="1.8"/> we'll presume that the vehicle's in

first gear <pause dur="2.2"/> the effective <pause dur="0.8"/> moment of inertia <pause dur="0.4"/> in a flywheel engine is given <pause dur="1.1"/> the driven system is rather less <pause dur="1.7"/> five times less <pause dur="1.1"/> in first gear <pause dur="0.6"/> # do note that when you've got a system like that when you're in second gear or third gear or fourth gear <pause dur="0.7"/> the effective <pause dur="0.8"/> moment of inertia of that system <pause dur="0.5"/> becomes greater <pause dur="0.6"/> as you go up through the gears <pause dur="2.4"/> that in fact is quite <pause dur="0.2"/> a a complicated thing <pause dur="0.3"/> to work out and you have to know something about gear boxes and gear ratios <pause dur="0.2"/> but we're not concerned with that here <pause dur="2.2"/> # determine the speed of the engine after clutch </u><u who="sm0800" trans="overlap"> do we do that <gap reason="inaudible" extent="1 sec"/> </u><u who="nm0796" trans="latching"> yes next year <pause dur="1.6"/> determine the speed of the engine after clutch engagement and the amount of energy lost <pause dur="0.3"/> okay so there's our <pause dur="0.4"/> problem <pause dur="5.1"/><kinesic desc="puts on transparency" iterated="n"/><vocal desc="clears throat" iterated="n"/> let's take a look at how we solve this one then <pause dur="8.1"/> now i'm going to work in some <pause dur="0.4"/> slightly unorthodox units here <pause dur="0.5"/> just for convenience <pause dur="0.7"/> # if you don't like what i'm doing <pause dur="1.6"/> # <pause dur="0.9"/> by all means <pause dur="1.0"/> convert this problem for yourself <pause dur="0.2"/> into S-I <pause dur="0.2"/> units <pause dur="1.4"/> but i think when you see me go through this <pause dur="0.2"/>

you'll realize that # <pause dur="0.2"/> the mixed units that i'm using <pause dur="0.3"/> are actually very convenient for the first <pause dur="0.2"/> part of this problem <pause dur="5.8"/> so <trunc>h</trunc> here we have then <pause dur="3.2"/> first of all <pause dur="0.3"/> a determination <pause dur="0.3"/> of the <pause dur="0.4"/> angular momentum the I-theta-<pause dur="0.2"/>dot <pause dur="0.7"/> of the engine before <pause dur="0.3"/> you engage the clutch <pause dur="3.2"/> the <pause dur="0.4"/> angular velocity the theta-dot is five-thousand remember that's in R-P-M <pause dur="1.0"/> # the moment of inertia is point-five <pause dur="0.3"/> which is in standard S-I units <pause dur="0.5"/> multiply the <pause dur="0.9"/> two together <pause dur="0.4"/> and you get a figure of two-thousand-five-hundred <pause dur="0.4"/> in some <pause dur="0.3"/> hybrid <pause dur="0.2"/> units <pause dur="0.7"/> and provided i <pause dur="0.2"/> stick with these hybrid units throughout <pause dur="0.6"/> everything will be all right <pause dur="1.6"/> as i say you can convert to S-I if you want to <pause dur="1.6"/> # after <pause dur="0.3"/> the so-called impact in other words <pause dur="0.7"/> once the clutch has been engaged <pause dur="0.3"/> then you've got the two systems rotating <pause dur="0.4"/> together <pause dur="0.9"/> at some speed <pause dur="0.6"/> which we're about to determine <pause dur="1.4"/> what has happened is that the <pause dur="0.2"/> moment of inertia of the total system has gone up <pause dur="0.7"/> to be the sum of the two <pause dur="0.3"/> so <pause dur="0.2"/> the total moment of inertia is point-five <pause dur="0.5"/> plus <pause dur="0.2"/> point-one <pause dur="0.9"/> for the driven system <pause dur="3.2"/> so

we've got <pause dur="0.2"/> moment of inertia point-six <pause dur="2.3"/> this then is an expression for the angular momentum after <pause dur="0.4"/> engagement <pause dur="0.7"/> and if you <pause dur="0.3"/> equate <pause dur="0.9"/> this expression to two-thousand-five-hundred <pause dur="1.3"/> presuming that we have <pause dur="0.4"/> conservation of angular momentum <pause dur="0.4"/> then <pause dur="0.4"/> the theta-dot works out <pause dur="1.1"/> to be four-one-six-seven <pause dur="0.3"/> and because we're using these hybrid units we know that that must be <pause dur="0.4"/> also <pause dur="0.5"/> in R-P-M <pause dur="5.5"/> so that is a very straightforward <pause dur="0.4"/> calculation i hope you will agree <pause dur="1.3"/> # <pause dur="0.2"/> very similar to <pause dur="0.6"/> linear momentum conservation <pause dur="3.8"/> now let's have a look <pause dur="0.9"/> at the energy <pause dur="0.6"/> the second part of the problem <pause dur="0.8"/><kinesic desc="adjusts transparency" iterated="y" dur="9"/> tell me if i move this up too far <pause dur="6.6"/> now at this point i've chickened out and i've gone into S-I units 'cause i want to work in <pause dur="0.3"/> straightforward joules <pause dur="0.6"/> # in other words S-I energy units <pause dur="0.7"/> # <pause dur="0.2"/> and here what i'm doing <pause dur="0.2"/> is working out the energy before <pause dur="0.5"/> kinetic energy before <pause dur="0.4"/> half-I-theta-dot-squared <pause dur="0.3"/> and the energy after <pause dur="1.1"/> so <pause dur="1.0"/> this is an expression then for <pause dur="0.2"/> half-<pause dur="1.1"/>I of the engine moment of inertia of the engine <pause dur="0.3"/> times the theta-dot-squared <pause dur="0.2"/>

and you'll see <pause dur="0.6"/> that i've made a conversion <pause dur="0.4"/> from five-thousand R-P-M <pause dur="0.4"/> into radians per second <pause dur="1.4"/> and five-thousand divide by <pause dur="0.6"/> sixty <pause dur="0.4"/> tells me how many radians per <pause dur="0.5"/> sorry how many revs per second i've got <pause dur="0.8"/> and a rev is two-pi-radians <pause dur="0.3"/> so <pause dur="0.4"/> the conversion factor here is <pause dur="0.3"/> two-pi-over-sixty <pause dur="1.8"/> then square it <pause dur="1.9"/> if you put your calculators around that one <pause dur="0.4"/> you get this figure for the kinetic energy sixty-eight-<pause dur="0.8"/>odd-thousand <pause dur="0.2"/> joules <pause dur="2.1"/> # after the clutch has been engaged we've now got <pause dur="0.7"/> an increased <pause dur="0.5"/> moment of inertia to point-six <pause dur="0.6"/> but a decreased <pause dur="0.2"/> speed to four-one-six-seven <pause dur="0.7"/> according to the calculation we've just done <pause dur="1.5"/> and that will tell us <pause dur="0.8"/> the energy <pause dur="0.8"/> kinetic energy of the system <pause dur="0.9"/> after <pause dur="1.0"/> the clutch engagement <pause dur="1.8"/> so you see that we've got <pause dur="0.8"/> a loss of something like twenty per cent <pause dur="0.6"/> of the energy in that process <pause dur="4.2"/> which is probably a realistic <pause dur="0.6"/> sort of loss <pause dur="0.4"/> where does where does the energy go to </u><pause dur="0.7"/> <u who="sf0801" trans="pause"> heat </u><pause dur="0.7"/> <u who="sm0802" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u><u who="nm0796" trans="overlap"> heat yeah <pause dur="0.6"/> and where's the heat going to <pause dur="0.2"/> appear </u><pause dur="1.0"/> <u who="sm0803" trans="pause"> on the clutch plates </u><u who="nm0796" trans="latching"> on the clutch plates right <pause dur="0.5"/> okay <pause dur="5.0"/> so that

then is a fairly straightforward sort of problem isn't it but <pause dur="0.2"/> the the key thing here <pause dur="0.3"/> is to see the parallel <pause dur="0.4"/> here between linear <pause dur="0.3"/> particle motion <pause dur="0.3"/> and rotational motion <pause dur="0.7"/> and we did a very similar thing for that trivial problem to do with <pause dur="0.3"/> railway trucks colliding with each other if you remember it <pause dur="2.3"/> all right moving <pause dur="0.5"/> on then <pause dur="1.2"/> just in passing <pause dur="0.3"/> there is a chart at the back of your notes showing # <pause dur="0.4"/> the comparison between <pause dur="0.6"/> translational motion <pause dur="0.3"/> and <pause dur="1.1"/> rotational motion <pause dur="0.4"/> i don't want to dwell on that <pause dur="0.6"/> or to describe it in any detail <pause dur="0.4"/> but i would like you <pause dur="1.0"/> # some time in your own time <pause dur="0.3"/> just to take a look at that <pause dur="1.3"/> and see what is said there there are one or two comments <pause dur="0.3"/> about the vector <pause dur="1.0"/> # <pause dur="0.4"/> the vector properties of some of these quantities <pause dur="0.3"/> and as we said at the beginning of the course <pause dur="0.2"/> you can ignore these for the moment but there are one or two little oddities there <pause dur="0.3"/> they won't concern you <pause dur="1.0"/> just look at the <pause dur="0.5"/> fact <pause dur="0.4"/> # <pause dur="0.3"/> the key fact <pause dur="0.3"/> that <pause dur="0.4"/> there is a one to one <pause dur="0.4"/> correspondence between what happens in linear particle motion <pause dur="0.2"/> and what happens in rotational motion <pause dur="0.5"/>

i know i've said that about ten times but <pause dur="0.6"/> people do sometimes forget it <pause dur="0.2"/> even though i say it ten times <pause dur="2.3"/> well now finally what we're going to come on and look at <pause dur="0.3"/> is # the general <pause dur="0.2"/> type of motion <pause dur="0.3"/> in which we have got <pause dur="0.3"/> a mixture <pause dur="0.6"/> of linear motion <pause dur="0.5"/> according to your second set of notes and rotational motion <pause dur="0.5"/> which is according to these notes <pause dur="0.4"/> and sometimes this can become <pause dur="0.4"/> a little bit complex <pause dur="1.0"/> # the key features of this <pause dur="1.4"/> and the key things that you need to remember <pause dur="0.3"/> are first of all <pause dur="0.5"/> that you need to work <pause dur="0.7"/> as at the centre of gravity of the system <pause dur="0.4"/> that is the safest possible thing to do <pause dur="1.5"/> just occasionally though secondly <pause dur="1.0"/> there is a fixed centre of rotation for a <pause dur="0.4"/> a set of <pause dur="0.9"/> bodies or a body <pause dur="0.5"/> and we'll see one of those in a minute <pause dur="0.5"/> in that case it's sometimes more convenient <pause dur="0.3"/> to work <pause dur="0.2"/> at the centre of rotation <pause dur="0.4"/> but you have to be a little bit careful <pause dur="0.3"/> how you handle that <pause dur="0.6"/> it's always safe to

work at the centre of gravity <pause dur="4.6"/> all right so the first <pause dur="0.6"/> problem we're <trunc>d</trunc> we're going to look at this morning is example three-F at the bottom of that page <pause dur="0.7"/> # this one <pause dur="0.6"/> is a <pause dur="0.6"/> relatively straightforward <pause dur="0.7"/> problem of the <pause dur="0.2"/> what i call the direct kind <pause dur="0.3"/> the direct kind if you remember <pause dur="0.5"/> is where you know <pause dur="0.4"/> forces applied and you can work out accelerations <pause dur="1.4"/> or vice versa <pause dur="0.4"/> you know the motion and you have to work out the forces <pause dur="0.8"/> this particular one <pause dur="0.4"/> is in the first category <pause dur="0.6"/> what we've got here is a thing like a bobbin or a cotton reel <pause dur="0.2"/> resting on the surface <pause dur="0.5"/> a piece of <pause dur="0.2"/> cord <trunc>ar</trunc> <pause dur="0.3"/> is wound around it and we pull on the cord <pause dur="0.5"/> and the bobbin is going to roll <pause dur="0.5"/> along the surface <pause dur="1.8"/> that's the nature of the problem <pause dur="1.5"/> and what we have to do <pause dur="0.9"/> is work out what that motion is going to be <pause dur="0.4"/> under the <pause dur="0.3"/> conditions that are applied <pause dur="3.0"/> now <trunc>thi</trunc> this problem <pause dur="0.4"/> does get a little bit <pause dur="0.3"/> complicated for two reasons can you anybody think what those two reasons might be <pause dur="7.5"/> no </u><pause dur="0.7"/> <u who="sm0804" trans="pause"> the mass has been reduced <pause dur="0.7"/> has it </u><pause dur="1.4"/> <u who="nm0796" trans="pause"> we're not going we're

not going to take that into account it could <pause dur="0.4"/> be a problem yes if the cord were heavy <pause dur="0.6"/> we would have to take that into account but we'll assume it's a light cord </u><pause dur="3.1"/> <u who="sm0805" trans="pause"> as it unwinds <unclear>so the</unclear> radius reduces </u><pause dur="1.2"/> <u who="nm0796" trans="pause"> no <pause dur="1.1"/> we'll assume the cord's all on one radius <pause dur="2.5"/> okay <pause dur="0.2"/> well <trunc>i</trunc> <pause dur="0.4"/> just i'll tell you what they are <pause dur="0.2"/> two things <pause dur="0.3"/> first of all <pause dur="1.1"/> you don't know whether this thing <pause dur="0.5"/> is going to slip on the surface or roll <pause dur="1.2"/> so somehow or other we've got to <pause dur="0.4"/> cope with that <pause dur="0.4"/> and the way in which we cope with it <pause dur="0.4"/> is we assume one thing <pause dur="0.7"/> work out what the answers are going to be <pause dur="0.6"/> and if they turn out to be daft <pause dur="0.2"/> we've made the wrong assumption <pause dur="0.4"/> and then we go back <pause dur="0.6"/> and try the other assumption <pause dur="1.1"/> # usually if you get it wrong <pause dur="0.2"/> something quite inconsistent will appear <pause dur="0.5"/> and you know that you've <pause dur="0.2"/> got it wrong <pause dur="1.3"/> so that's the first problem <pause dur="0.3"/> slipping or rolling <pause dur="2.0"/> # the second problem is <pause dur="0.2"/> which way is it going to roll if it's rolling let's assume it's going to roll <pause dur="1.2"/> is it is this bobbin going to roll <pause dur="0.3"/> to the right <pause dur="0.2"/> or is it going to roll to the left </u><pause dur="2.1"/> <u who="sm0806" trans="pause"> to the left </u><pause dur="1.1"/> <u who="nm0796" trans="pause"> well

that's <pause dur="0.2"/> you see the interesting thing 'cause <pause dur="0.2"/> seeing that you think <pause dur="1.1"/> it's going to unwind so the bobbin's going to have to go to the left <pause dur="0.6"/> in fact that's impossible <pause dur="0.6"/> it always goes to the right <pause dur="0.3"/> if you pull on anything in one direction <pause dur="0.7"/> it will go <pause dur="0.6"/> in that direction unless there's some other reaction <pause dur="0.3"/> that stops it doing so <pause dur="1.1"/> this thing will roll to the right <pause dur="3.7"/> and you'll notice that i put that <pause dur="1.9"/> at the bottom there <pause dur="2.7"/> well you may not believe that <pause dur="1.8"/> but just <shift feature="voice" new="laugh"/> think about it <shift feature="voice" new="normal"/><pause dur="1.5"/><kinesic desc="changes transparency" iterated="y" dur="10"/> okay <pause dur="0.5"/> so <pause dur="1.6"/> let's then see <pause dur="0.3"/> how <pause dur="0.6"/> we are going to <pause dur="1.0"/> tackle the problem <pause dur="3.6"/> as with any of these sort of problems we'll work through this in a <pause dur="0.7"/> fairly well <pause dur="0.4"/> defined order <pause dur="5.5"/> perhaps before i <pause dur="0.2"/> do that i should just recite <pause dur="0.5"/> the main features of the problem to you which are <pause dur="0.5"/> written into it <pause dur="0.6"/> # first of all <pause dur="1.4"/> it's two-hundred millimetre radius <pause dur="1.3"/> <trunc>o</trunc> or point-two of a metre <pause dur="0.9"/> mass twenty-five kilograms <pause dur="1.1"/> it tells you the radius of gyration <pause dur="0.2"/> at one-seventy-five millimetres <pause dur="0.3"/> <trunc>poi</trunc> point-one-seven-five metres <pause dur="1.6"/> tells you the <pause dur="1.0"/> radius of the groove <pause dur="0.3"/> which

appears also on the <pause dur="0.5"/> picture underneath <pause dur="1.0"/> the force applied is twenty newtons <pause dur="1.5"/> and <pause dur="1.0"/> what you're being asked is to calculate the angular <pause dur="0.4"/> acceleration <pause dur="1.6"/> and it tells you <pause dur="0.2"/> the coefficient of friction if you need <pause dur="0.3"/> to know it <pause dur="0.3"/> between <pause dur="0.6"/> this bobbin <pause dur="0.3"/> and the surface <pause dur="5.3"/> right so <pause dur="0.3"/> at the top here of the solution <pause dur="0.2"/> first two things i write down is first of all the mass <pause dur="0.7"/> of <pause dur="0.6"/> the bobbin <pause dur="0.7"/> and secondly <pause dur="0.3"/> i need to calculate <pause dur="0.2"/> the moment of inertia because we're actually told <pause dur="0.4"/> the radius of gyration <pause dur="0.9"/> and remember that the <pause dur="0.2"/> moment of inertia about the centre is going to be <pause dur="0.3"/> mass <pause dur="0.2"/> times radius of gyration squared M-K-squared <pause dur="0.9"/> so <pause dur="0.6"/> that <pause dur="0.4"/> moment of inertia <pause dur="0.4"/> is going to be twenty-five times point-one-seven-five-squared <pause dur="0.4"/> and that <pause dur="0.3"/> result is given here <pause dur="3.8"/> now i'm going to make my big assumption <pause dur="1.2"/> i'm going to assume <pause dur="0.9"/> that <pause dur="1.1"/> this thing rolls <pause dur="1.7"/> and doesn't slip <pause dur="1.7"/> and i've <pause dur="0.2"/> put a note there that i must check <pause dur="0.5"/> at the end <pause dur="0.5"/> that that is a reasonable assumption to make <pause dur="2.3"/> or at least check for consistency <pause dur="1.4"/> we'll see how <pause dur="0.3"/> we do that when

we get there <pause dur="4.0"/> and then the next <pause dur="0.6"/> and a final <pause dur="0.3"/> preliminary to looking at the dynamics <pause dur="0.3"/> is to actually look at the kinematics i haven't actually written that word on this overhead but <pause dur="0.4"/> usually i do <pause dur="0.4"/> you might like to just <pause dur="0.6"/> write <pause dur="0.2"/> kinematics here <pause dur="0.3"/> remember that kinematics is looking <pause dur="0.3"/> just at the motion <pause dur="0.5"/> nothing to do with forces or <pause dur="1.1"/> anything <pause dur="0.3"/> of that kind <pause dur="2.3"/> so i'm assuming it's rolling <pause dur="0.8"/> here is a picture of what <pause dur="0.3"/> is happening <pause dur="0.5"/> if it's rolling there is an instantaneous <pause dur="0.4"/> centre <pause dur="0.2"/> of rotation <pause dur="0.7"/> at the point of contact between <pause dur="0.4"/> the bobbin and the surface <pause dur="1.6"/> so in other words <pause dur="0.5"/> looking at that <pause dur="0.4"/> bobbin it's just rolling about this point of contact <pause dur="1.0"/> at that instant <pause dur="0.3"/> course as soon as it rolls a little distance <pause dur="0.8"/> the point of instantaneous centre is going to move round the rim but it's always at the point of contact <pause dur="1.8"/> so <pause dur="2.1"/> if we assume that in this rolling the <pause dur="0.2"/> linear motion of the centre of gravity is X <pause dur="0.8"/> to the right <pause dur="1.1"/> and the <pause dur="0.6"/> angular motion <pause dur="0.4"/> angular displacement is theta <pause dur="1.8"/> # angular motion is not <pause dur="0.4"/> confined to a single point on

the body <pause dur="0.4"/> if you've got <pause dur="0.3"/> what's called the rigid body in other words it doesn't <pause dur="0.6"/> change its shape <pause dur="0.5"/> an angular motion <pause dur="0.6"/> is the same for any point on that body <pause dur="1.4"/> so in fact the angular displacement about the instantaneous centre is theta as well <pause dur="0.6"/> which means that <pause dur="0.2"/> the centre is going to move by <pause dur="0.3"/> the corresponding <pause dur="0.2"/> value of R-theta <pause dur="1.4"/> R is point-two of a metre <pause dur="1.6"/> so <pause dur="0.4"/> there is a relationship here <pause dur="0.3"/> between the X <pause dur="0.4"/> at the centre <pause dur="0.4"/> and the angular movement <pause dur="0.6"/> and X is going to be point-two-theta that's the R <pause dur="0.9"/> and that's the theta <pause dur="3.8"/> the other part of the kinematics that's important <pause dur="0.7"/> is if we differentiate this twice <pause dur="0.5"/> so X-<pause dur="0.2"/>double-dot <pause dur="0.8"/> is then going to be <pause dur="0.4"/> point-two-theta-double-dot <pause dur="2.5"/> so that's the kinematics <pause dur="4.8"/> now we move on to the dynamics and we need the free-body diagram <pause dur="0.6"/> for this <pause dur="0.8"/> bobbin <pause dur="0.5"/> remember what <pause dur="1.0"/> this is it must have all <pause dur="0.5"/> forces <pause dur="0.3"/> and moments shown <pause dur="0.5"/> including the inertia <pause dur="0.3"/> force and the inertia moment <pause dur="2.0"/> and according to me there are six <pause dur="0.5"/> such forces <pause dur="3.8"/> only one of which is <pause dur="0.4"/> explicitly applied the only applied

force to this thing is the twenty newtons down here <pause dur="0.9"/> all of the others come about either <pause dur="0.4"/> by virtue of contact <pause dur="0.4"/> with other <pause dur="0.2"/> <trunc>poin</trunc> other <pause dur="0.2"/> bodies <pause dur="1.0"/> hence <pause dur="0.9"/> the <pause dur="0.3"/> reaction <pause dur="0.7"/> and <pause dur="0.3"/> the frictional <pause dur="0.2"/> traction force <pause dur="3.0"/> and i've put a little note here that this frictional traction force <pause dur="0.2"/> ought to be less than mu-R <pause dur="2.2"/> if it isn't less than mu-R <pause dur="0.5"/> then this thing is going to slip <pause dur="0.6"/> so our assumption is wrong <pause dur="1.0"/> so <pause dur="0.5"/> one of the checks that we have to do at the end <pause dur="1.3"/> is to check <pause dur="0.6"/> whether <pause dur="0.5"/> F actually is less than <pause dur="0.2"/> mu-R <pause dur="0.7"/>

<vocal desc="clears throat" iterated="n"/><pause dur="0.5"/><vocal desc="clears throat" iterated="n"/><pause dur="6.4"/> then <pause dur="0.9"/> the other <pause dur="0.3"/> forces there are two inertia forces the <pause dur="0.9"/> M-X-double-dot which must be in the opposite sense to positive-X <pause dur="0.2"/> or positive-X-double-dot <pause dur="0.5"/> and the I-theta-double-dot <pause dur="0.7"/> in the opposite sense <pause dur="0.3"/> to positive-theta <pause dur="2.8"/> and <pause dur="0.3"/> last <pause dur="0.5"/> but not least <pause dur="0.9"/> the weight of the object the M-G <pause dur="0.6"/> downwards which acts at the centre of gravity <pause dur="3.4"/> well as always <pause dur="0.3"/> it's important to get that <pause dur="0.4"/> picture right <pause dur="0.3"/> once you've got that picture <pause dur="0.2"/> right and complete <pause dur="0.8"/> then everything else should be easy <pause dur="0.6"/> but it's always that first step <pause dur="0.2"/> where <pause dur="0.4"/> things go wrong <pause dur="1.1"/> you either forget one of the forces you ought to <pause dur="0.2"/> put in <pause dur="1.4"/> or you get one of them wrong <pause dur="0.4"/> there are there are too many of them <pause dur="1.5"/> so i hope that you agree <pause dur="1.0"/> that i have got <pause dur="0.3"/> the correct number of forces <pause dur="0.2"/> here <pause dur="0.4"/> remember the things you have to check inertia forces <pause dur="0.4"/> wherever the body contacts something else <pause dur="0.5"/> and any applied forces <pause dur="0.3"/> three <pause dur="0.4"/> types and they're all there <pause dur="6.6"/> so that is your picture <pause dur="1.9"/><kinesic desc="puts on transparency" iterated="n"/> and

i'll try and keep this on the screen at the same time as showing you <pause dur="0.9"/> the equations <pause dur="3.2"/> so you can see where they've come from <pause dur="16.7"/> can you can you see that all right <pause dur="1.3"/> # so the first and simple one is the horizontal <pause dur="0.3"/> equilibrium from the picture that's just disappearing <pause dur="0.4"/> at the top <pause dur="1.3"/> so the <trunc>t</trunc> the twenty <pause dur="0.7"/> is balanced by <pause dur="2.4"/> the <pause dur="0.2"/> # <pause dur="0.4"/> M-X-double-dot and the F so i've written it this way round <pause dur="0.6"/> # F <pause dur="0.6"/> equals <pause dur="0.6"/> F to the left is twenty to the right <pause dur="0.4"/> minus <pause dur="0.2"/> M-X-double-dot to the right <pause dur="2.8"/> the <pause dur="2.5"/> M is twenty-five kilograms <pause dur="0.8"/> the X-double-dot <pause dur="0.5"/> we saw from kinematics is point-two-<pause dur="0.2"/>theta-double-dot <pause dur="0.3"/> so <pause dur="1.2"/> a fairly simple expression for F <pause dur="0.4"/> comes out of there <pause dur="3.0"/> and then the other equation that we need <pause dur="0.4"/> is the moment <pause dur="1.1"/> equation <pause dur="7.6"/> i'll take that off <pause dur="1.5"/> i think <pause dur="8.5"/> well you've got the picture in front of you haven't you <pause dur="0.8"/><event desc="takes off transparency" iterated="n"/><kinesic desc="adjusts transparency" iterated="y" dur="12"/> i think i'll take the picture away and then this can go higher up <pause dur="0.4"/> you can see what's going on <pause dur="8.1"/> right so i've i've taken moments about the centre <pause dur="1.9"/> of <pause dur="0.2"/> the bobbin <pause dur="0.3"/> now you'll notice that several of the forces actually pass through that

centre <pause dur="0.3"/> so they don't have moments which is handy <pause dur="1.0"/> # the ones that do <pause dur="0.9"/> are the inertia moment I-theta-double-dot <pause dur="0.2"/> which is <pause dur="0.6"/> anti <pause dur="1.4"/> anticlockwise is it <pause dur="4.2"/> yep <pause dur="0.8"/> and the twenty <pause dur="0.4"/> times the <pause dur="0.6"/> radius <pause dur="0.3"/> of the groove <pause dur="1.3"/> which is anticlockwise <pause dur="0.3"/> balanced by <pause dur="1.5"/> the friction or tractional force of F times that radius point-two <pause dur="0.4"/> none of the other forces <pause dur="0.3"/> have a moment about them <pause dur="0.2"/> in the middle <pause dur="2.6"/> and then i substitute directly for F from what we've found up here <pause dur="2.8"/> and put some numbers in <pause dur="0.3"/> so there's the value of I <pause dur="0.3"/> that i calculated <pause dur="0.4"/> right at the beginning <pause dur="1.2"/> that comes to one-point-five <pause dur="1.5"/> F <pause dur="0.4"/> here goes into there <pause dur="1.7"/> and you end up there then <pause dur="0.3"/> with <pause dur="0.2"/> an equation which is <pause dur="0.4"/> solely in terms of the angular acceleration so we can work out what it is <pause dur="0.8"/> i haven't <trunc>sho</trunc> shown the detailed algebra <pause dur="0.3"/> but that tells you <pause dur="0.7"/> what the result comes out to be <pause dur="1.9"/> so that's the <pause dur="1.1"/> solution to the problem <pause dur="0.3"/> except that we must now check <pause dur="0.3"/> our assumption <pause dur="1.3"/> <trunc>w</trunc> and if you remember <pause dur="0.8"/> what we need to check <pause dur="1.4"/> is that the traction force is less than mu-R <pause dur="2.6"/> well the

traction force from <pause dur="0.6"/> the equation for the horizontal equilibrium is twenty minus <pause dur="1.1"/> five-<pause dur="0.4"/>theta-double-dot <pause dur="2.0"/> and that comes to about thirteen <pause dur="0.3"/> newtons <pause dur="0.4"/> so a traction force <pause dur="0.4"/> if it's rolling is thirteen newtons <pause dur="2.2"/> the value of the reaction is the same as the rate which is twenty-five kilograms times nine-point-eight-one <pause dur="0.9"/> in newtons which is <trunc>ab</trunc> <pause dur="0.6"/> more or less two-hundred-and-forty-five <pause dur="1.6"/> so mu-R is point-one times that <pause dur="0.2"/> which is <pause dur="1.0"/> twenty-four-and-a-half newtons <pause dur="2.5"/> so <pause dur="0.3"/> it is indeed the case that the traction force <pause dur="0.5"/> is <pause dur="0.7"/> less than mu-R <pause dur="0.4"/> it's about a half of it <pause dur="1.0"/> so in fact the assumption <pause dur="0.5"/> that it's rolling <pause dur="0.8"/> is okay <pause dur="9.2"/> right so what we've seen there then is a <pause dur="0.4"/> mixture of linear <pause dur="0.3"/> and rotational motion <pause dur="0.9"/> <trunc>a</trunc> and it's <pause dur="0.2"/> had a slight complication because of this business of the rolling or slipping <pause dur="0.2"/> which may not appear in <pause dur="0.8"/> many problems <pause dur="0.3"/> but you often get a little <pause dur="0.3"/> complication of that sort to deal with <pause dur="6.7"/> right so now if we turn over then <pause dur="0.6"/> to the final page of the notes <pause dur="0.7"/> page six <pause dur="12.8"/> and <pause dur="0.7"/> here is a problem in

the <pause dur="0.4"/> next <pause dur="0.2"/> category that i <pause dur="0.3"/> usually <pause dur="0.8"/> look at <pause dur="0.3"/> you remember we have three <trunc>k</trunc> types of problems <pause dur="0.5"/> direct problems of the sort we've just <trunc>solv</trunc> solved <pause dur="1.3"/> the <pause dur="0.2"/> impact momentum type problems which i'm about to do <pause dur="0.6"/> and thirdly <pause dur="0.4"/> the <pause dur="0.2"/> energy <pause dur="0.6"/> conservation of energy type problems <pause dur="0.4"/> which <pause dur="0.4"/> is the final one so there are two problems on this page <pause dur="0.2"/> one is to do <pause dur="0.3"/> with <pause dur="0.4"/> impact <pause dur="0.3"/> momentum <pause dur="0.8"/> on a mixed body in which we've got rotational and linear momentum going on together <pause dur="1.0"/> and the second problem <pause dur="0.5"/> is of energy <pause dur="0.4"/> conservation problem <pause dur="3.6"/> the <pause dur="0.3"/> the thing that i've chosen for <pause dur="0.2"/> this next problem is what is sometimes known as the cricket bat <pause dur="0.2"/> problem those of you who play <pause dur="0.3"/> cricket <pause dur="0.7"/> will be familiar <pause dur="0.6"/> with the fact that when you <pause dur="0.7"/> are batting and you strike the ball <pause dur="0.3"/> if the ball doesn't hit the right place <pause dur="0.4"/> on the bat <pause dur="0.3"/> it stings your hands <pause dur="0.6"/> and the reason for that <pause dur="0.3"/> is that you have an impulsive <pause dur="0.5"/> reaction at your hands the ball striking <pause dur="0.3"/> the bat <pause dur="0.4"/> <trunc>i</trunc> <pause dur="0.6"/> causes an impulse <pause dur="0.3"/> a linear impulse on the bat <pause dur="0.7"/> and <pause dur="0.5"/> because of the <pause dur="0.2"/> the way in

which equilibrium of <trunc>imp</trunc> of <pause dur="0.4"/> impulses occurs <pause dur="0.3"/> there will be <pause dur="0.2"/> in general <pause dur="0.7"/> # an impulsive reactions where you're hanging on to it the hinge if you like <pause dur="0.3"/> of the bat <pause dur="1.0"/> except in <pause dur="0.3"/> certain very special circumstances if you happen <pause dur="0.5"/> to get the ball hitting the bat in the right place <pause dur="0.5"/> it doesn't sting <pause dur="0.7"/> and such a <pause dur="0.3"/> a place is called the centre of percussion <pause dur="0.5"/> and you can easily work out <pause dur="0.3"/> from the dynamics <pause dur="0.2"/> where that position <pause dur="0.8"/> happens to be <pause dur="0.4"/> in the case of a cricket bat <pause dur="0.4"/> it's probably about that much from the bottom of the bat <pause dur="0.5"/> which if you look at the way in which <pause dur="0.2"/> cricket bats are shaped <pause dur="0.3"/> is where the bat <pause dur="0.5"/> is actually thickest and chunkiest and that's really where you should hit it <pause dur="0.6"/> if the ball comes up and hits it <pause dur="0.6"/> # <pause dur="0.3"/> higher up the bat then you're <pause dur="0.6"/> going to suffer <pause dur="2.0"/> the ball goes even higher you'll need your <pause dur="0.4"/> guard on <pause dur="0.2"/> but there you are <pause dur="3.2"/> okay so let's have a look see what the problem says <pause dur="1.1"/> i've done this in very general terms that <pause dur="0.3"/> picture in the notes <pause dur="0.4"/> is not meant to be a cricket bat <pause dur="2.8"/> i

think i can draw a bit better than that <pause dur="0.3"/><vocal desc="laugh" iterated="n"/><pause dur="1.0"/> but essentially it's the same <pause dur="0.4"/> problem what we've got here <pause dur="0.3"/> is a body that is <pause dur="0.2"/> hinged <pause dur="0.7"/> at the top at O <pause dur="1.6"/> centre of gravity <pause dur="0.4"/> is somewhere <pause dur="0.3"/> further down distance L beneath <pause dur="0.6"/> and we're actually going to strike <pause dur="0.5"/> the thing <pause dur="0.3"/> with an impulse <pause dur="0.4"/> at <pause dur="0.6"/> a point H below the hinge <pause dur="1.0"/> # and there are one or two bits missing on the diagram <pause dur="1.8"/> for which my apologies <pause dur="3.2"/> so i'd <pause dur="0.2"/> like you to just add <pause dur="8.1"/><kinesic desc="writes on transparency" iterated="y" dur="8"/> the first thing that's missing <pause dur="0.6"/> is that there's an arrow round there <pause dur="0.4"/> with nothing on it <pause dur="0.2"/> that should have theta-dot <pause dur="3.7"/><kinesic desc="writes on transparency" iterated="y" dur="3"/> put against it that is the angular motion that's going to occur <pause dur="0.5"/> after <pause dur="0.3"/> the impact now the impact is <trunc>sh</trunc> <pause dur="0.3"/> shown at the point <pause dur="0.2"/> A here by a line <pause dur="1.3"/> it's a very defective drawing i'm sorry about this <trunc>th</trunc> there should be an arrowhead there <pause dur="1.1"/><kinesic desc="writes on transparency" iterated="y" dur="1"/> on there and also that should be called P <pause dur="0.6"/><kinesic desc="writes on transparency" iterated="y" dur="1"/> P <pause dur="0.5"/> is going to be <pause dur="0.5"/> what i call the impact the impulse <pause dur="0.2"/> it's not a force it's an impulse <pause dur="3.1"/> and then over at the right of the diagram you see <pause dur="0.7"/> hanging in mid-air <pause dur="0.7"/><kinesic desc="writes on transparency" iterated="y" dur="3"/> # <pause dur="0.2"/> V and theta-<pause dur="0.2"/>dot <pause dur="1.1"/> or actually i think it says

theta but there should be a dot over it <pause dur="1.1"/> # <pause dur="0.3"/> there should be <pause dur="0.5"/> again an arrow like that <kinesic desc="writes on transparency" iterated="y" dur="1"/> to show that theta-dot is clockwise as it is up here <pause dur="2.0"/> and V <pause dur="0.8"/> is going to be to the left <pause dur="3.3"/> so you want an arrow on there <pause dur="1.9"/><kinesic desc="writes on transparency" iterated="y" dur="1"/> otherwise the picture is wonderful </u><pause dur="0.5"/> <u who="sm0806" trans="pause"> what's an impulse </u><pause dur="0.5"/> <u who="nm0796" trans="pause"> pardon </u><pause dur="0.2"/> <u who="sm0806" trans="pause"> what's an impulse </u><pause dur="1.1"/> <u who="nm0796" trans="pause"> an impulse if you remember <pause dur="0.6"/> i hope you will do <pause dur="0.2"/> when i tell you <pause dur="0.6"/> is <pause dur="1.5"/> a <trunc>sho</trunc> a a force of short <trunc>d</trunc> duration <pause dur="0.2"/> applied to a body <pause dur="0.4"/> and the impulse is the integral of that force over time <pause dur="1.1"/> and that <pause dur="0.4"/> impulse applied to a body <pause dur="0.3"/> causes a change in <pause dur="0.2"/> linear momentum <pause dur="1.3"/> remember that <pause dur="1.0"/> that was in our our second <pause dur="0.6"/> set of notes <pause dur="2.3"/> okay <pause dur="6.6"/> right just getting the bones out of the <pause dur="0.2"/> example i'm going halfway down it <pause dur="0.6"/> # let's the body be struck by an impulse P at A which we've shown on the diagram <pause dur="1.8"/> find <pause dur="0.7"/> <reading>find the value of H for <pause dur="0.3"/> A to be the centre of percussion</reading> <pause dur="0.4"/> now what that actually means <pause dur="0.5"/> is that we've got to find <pause dur="0.8"/> H <pause dur="0.3"/> such that <pause dur="0.6"/> the reaction <pause dur="0.7"/> there <pause dur="0.7"/> which <pause dur="0.2"/> also you could add to your diagram if you want to <pause dur="3.2"/>

what you want is to have H such that R is zero <pause dur="3.4"/> so let's i'll just write under here so you don't <pause dur="1.0"/> have a misapprehension <pause dur="1.5"/><kinesic desc="writes on transparency" iterated="y" dur="9"/> P and R <pause dur="4.6"/> are impulses not forces <pause dur="17.2"/><vocal desc="clears throat" iterated="y" dur="1"/><pause dur="2.9"/> all right can we just then <pause dur="1.0"/> set off into the problem <pause dur="1.3"/> can i take that one away have you <pause dur="0.8"/> got what you need from that <pause dur="2.6"/><kinesic desc="changes transparency" iterated="y" dur="12"/> right <pause dur="1.9"/><vocal desc="clears throat" iterated="n"/><pause dur="19.7"/> i don't think you need to draw the picture again <pause dur="1.8"/> you've got it in your notes <pause dur="0.8"/> so don't waste time doing that <pause dur="2.9"/> # key key thing here again is that we've got mixed motion the centre of gravity <pause dur="0.8"/> is moving linearly to the left <pause dur="0.7"/> # but # we've also got this rotation <pause dur="0.7"/> theta-dot <pause dur="2.0"/> and the point O is fixed <pause dur="0.2"/> at the top <pause dur="0.7"/> or assume to be fixed <pause dur="3.9"/> once again let's look at the kinematics first <pause dur="0.2"/> <trunc>si</trunc> the simple motion <pause dur="0.6"/> the <pause dur="0.6"/> the value of velocity V to the left <pause dur="0.5"/> at the centre of gravity <pause dur="1.7"/> is going to be the corresponding value of R-theta-dot like it was for the previous problem <pause dur="0.5"/> only here <pause dur="0.4"/> the R is the distance L <pause dur="2.3"/> so L-theta-dot <pause dur="0.2"/> taken <pause dur="0.3"/> from the centre rotation at the top <pause dur="0.4"/> tells you the value of V <pause dur="7.6"/>

and now we do <pause dur="0.5"/> a rather similar thing <pause dur="0.4"/> to the previous problem we look at the linear <pause dur="0.2"/> motion <pause dur="0.3"/> and then the <pause dur="0.4"/> angular motion <pause dur="0.5"/> in the previous problem it was <pause dur="0.7"/> linear equilibrium of forces and <pause dur="1.4"/> moment equilibrium <pause dur="1.2"/> but you need to do the two because we've got mixed motion <pause dur="0.7"/> so the first <pause dur="2.6"/> first thing <pause dur="0.5"/> is looking at the linear impulse the total linear impulse on this object <pause dur="0.3"/> is <pause dur="0.5"/> primarily the P to the left <pause dur="0.4"/> but we must <pause dur="0.5"/> also <pause dur="0.4"/> see that there is the reaction R at the top <pause dur="0.4"/> which is to the right so <pause dur="0.3"/> the total linear impulse on the body is P minus R <pause dur="1.1"/> and that's going to result <pause dur="1.1"/> in the linear momentum <pause dur="0.6"/> of this thing at the <trunc>sin</trunc> centre of gravity <pause dur="1.0"/> so it's going to be equal to M-V <pause dur="2.6"/> i've used the kinematic straightaway <pause dur="0.3"/> to find out an expression for P <pause dur="1.9"/> so that's a fairly simple one <pause dur="1.6"/> slightly more complicated the <pause dur="0.4"/> angular <pause dur="0.3"/> impulse now we haven't actually used this before <pause dur="2.2"/> # the angular <pause dur="0.4"/> impulse <pause dur="0.7"/> let me just recap on that for you <pause dur="0.8"/> the moment <pause dur="0.2"/> of an impulse <pause dur="0.2"/> gives you <pause dur="0.4"/> a change <pause dur="0.4"/> in angular momentum <pause dur="2.2"/>

so <pause dur="0.2"/> what i'm doing here is taking <pause dur="0.8"/> a moment <pause dur="1.8"/> of the impulses P and R about the centre of gravity i'm working at the centre of gravity <pause dur="0.6"/> so P <pause dur="1.8"/> gives you <pause dur="0.4"/> # a <trunc>cl</trunc> <pause dur="0.3"/> a clockwise moment of impulse of P times the distance H-minus-L <pause dur="0.8"/> and the R gives you <pause dur="0.5"/> a <pause dur="0.2"/> also a clockwise <pause dur="0.4"/> moment of impulse <pause dur="0.5"/> of R times L <pause dur="1.1"/> so you've got these two terms <pause dur="4.2"/> in the expression for the moment of impulse <pause dur="1.0"/> and that <pause dur="1.0"/> will result in a change in angular momentum <pause dur="1.0"/> I-theta-dot <pause dur="0.6"/> so what we're # assuming here is that the <pause dur="0.4"/> object starts stationary <pause dur="0.3"/> but is struck <pause dur="0.4"/> and then ends up with these velocities <pause dur="1.4"/><vocal desc="clears throat" iterated="n"/><pause dur="0.2"/> so we've got a change of linear momentum and a change of angular momentum <pause dur="0.3"/> both determined <pause dur="0.4"/> by these <pause dur="0.4"/> # impulse equations <pause dur="4.7"/> so that's what happens in general <pause dur="7.0"/> and now the problem becomes easy because what i'm going to do <pause dur="0.3"/> is to see <pause dur="0.5"/> what the value of H has got to be <pause dur="0.2"/> for R-equals-zero <pause dur="1.4"/> so i simply put R-equals-zero in those two equations that we've got <pause dur="1.8"/> these two here <pause dur="0.2"/> that one <pause dur="0.2"/> put R-equals-zero <pause dur="0.4"/> this one put R-equals-zero <pause dur="2.0"/> and

what will then happen is <pause dur="0.4"/> that <pause dur="1.4"/> a value of H <pause dur="0.7"/> the height will be defined <pause dur="4.2"/> well how have i <trunc>do</trunc> <pause dur="0.2"/> done this let's just have a <pause dur="0.2"/> quick <pause dur="0.2"/> look <pause dur="0.8"/> this <pause dur="0.3"/> first equation here comes from this one <pause dur="2.1"/> # P <pause dur="0.3"/> P is going to be M-L-theta-dot from there so <pause dur="0.4"/> putting that into there <pause dur="0.5"/> M-L-theta-dot times H-minus-L <pause dur="0.3"/> plus nought <pause dur="0.6"/> is I-theta-<pause dur="0.7"/>dot <pause dur="1.4"/> and you'll see that the theta-dots cancel out <pause dur="2.1"/> and <pause dur="1.9"/> if we substitute I-equals-M-K-squared then the Ms cancel out as well <pause dur="3.3"/> and so you get a purely <pause dur="0.5"/> geometric <pause dur="0.2"/> equation <pause dur="0.7"/> for H <pause dur="0.6"/> the value of H is in terms <pause dur="0.3"/> of the radius of gyration <pause dur="1.1"/> and <pause dur="0.6"/> the L <pause dur="1.2"/> distance from the hinge <pause dur="0.4"/> to the <pause dur="0.2"/> centre of gravity <pause dur="6.1"/> so this is a rather general <pause dur="0.4"/> abstract <pause dur="0.5"/> sort of analysis <pause dur="1.9"/> # probably more usually <pause dur="0.3"/> this would be couched in terms of a <pause dur="0.3"/> very specific problem <pause dur="0.4"/> # an object of a specific size <pause dur="0.4"/> and mass <pause dur="0.7"/> and so on <pause dur="1.0"/> but what you've got here <pause dur="0.3"/> is a general formula <pause dur="0.5"/> that will enable you to work out centre of percussion for any object you care to choose <pause dur="1.1"/> and what the problem <trunc>f</trunc> says in its final sentence is <pause dur="0.3"/> apply the theory <pause dur="0.3"/> to find

the centre of percussion for a uniform bar <pause dur="0.6"/> hinged at one end <pause dur="0.6"/> so a uniform bar at <pause dur="0.4"/> hinged at one end we have to <pause dur="0.6"/> decide what we mean by K <pause dur="0.4"/> and L <pause dur="10.3"/> <kinesic desc="puts on transparency" iterated="n"/> so this takes us back to something that i asked you to remember <pause dur="0.7"/> couple of weeks ago <pause dur="9.1"/> i can get this lot to stay on board <pause dur="9.0"/> i'll move it up in a minute if you can't see the bottom <pause dur="3.5"/> right so if you've got a uniform bar <pause dur="0.2"/> length <pause dur="0.5"/> big L <pause dur="2.1"/> 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 <pause dur="0.2"/> to a point halfway down it <pause dur="0.6"/> so the little L <pause dur="0.8"/> that we had in the general formula <pause dur="0.3"/> is going to be the big L divided by two it's half the length <pause dur="0.3"/> distance from the end to the middle <pause dur="3.7"/> and then the formula that i asked you to remember <pause dur="1.1"/> was that <pause dur="0.2"/> the moment of inertia of a bar hinged about one end was M-L-squared-over-twelve <pause dur="0.7"/> so <pause dur="0.5"/> K-squared <pause dur="4.9"/> sorry about the centre i mean what am i talking about <pause dur="0.7"/> # moment of inertia about the centre for a bar is M-L-squared-over-twelve so the K-squared <pause dur="0.3"/> is big-L-squared-over-twelve <pause dur="1.6"/> and

all that remains to <pause dur="0.2"/> work out the <pause dur="0.2"/> corresponding value of H <pause dur="0.3"/> is to substitute these two <pause dur="0.4"/> into that formula <pause dur="1.5"/> and what you <trunc>f</trunc> <pause dur="0.3"/> find <pause dur="0.4"/> is that the centre of percussion <pause dur="0.4"/> is about two-thirds <pause dur="0.2"/> of the length <pause dur="0.4"/> and the that <pause dur="0.3"/> corresponds fairly well with a cricket bet doesn't it <pause dur="0.8"/> although a cricket bat isn't actually a uniform bar <pause dur="8.6"/> all right <pause dur="1.8"/> now we're going to cheat <vocal desc="clears throat" iterated="y" dur="1"/><pause dur="0.9"/> example three-H <pause dur="0.6"/> is one which i'm not going to <trunc>re</trunc> <pause dur="0.4"/> do on the overhead at all <pause dur="1.4"/> what i'm going to do here <pause dur="0.2"/> is give you a handout for your notes <pause dur="3.1"/> and # <pause dur="2.0"/> then superficially i will work <pause dur="0.4"/> through this <pause dur="1.5"/> over the next two minutes <pause dur="14.3"/><event desc="passes out handouts" iterated="n"/> # this problem is one in the final category of making use <pause dur="0.4"/> of the principle of conservation of energy <pause dur="3.1"/> and it's also <pause dur="0.8"/> closely related to # a piston <pause dur="1.9"/> mechanism <pause dur="0.3"/> used in an internal combustion engine <pause dur="0.4"/> so looking first of all at the picture in the notes <pause dur="1.4"/> if i can find them <pause dur="4.2"/> sorry <pause dur="1.2"/> # what you see here <pause dur="0.6"/> is <pause dur="1.1"/> a link A-B which corresponds with a <pause dur="0.2"/> connecting rod <pause dur="0.5"/> in an engine <pause dur="0.9"/> the force P applied <pause dur="0.2"/> would <pause dur="0.5"/> correspond <pause dur="0.4"/> to the <pause dur="0.7"/> force

applied by exploding gases on a piston <pause dur="0.9"/> # B-C is the crank <pause dur="0.3"/> on the crankshaft <pause dur="0.3"/> so what is happening is that B-C is going round and round <pause dur="0.6"/> # connecting <pause dur="0.2"/> rod <pause dur="0.4"/> is following it at B <pause dur="0.3"/> but point A <pause dur="0.2"/> moves in a straight line in other words it moves up and down the cylinder <pause dur="0.3"/> so you can see the mechanism can you <pause dur="1.5"/> right <pause dur="0.4"/> so <pause dur="0.2"/> the thing is actually starting <pause dur="0.5"/> with <pause dur="1.0"/> a rotational velocity of ten radians per second for B-C in other words the rotation <pause dur="0.3"/> rate of the crankshaft <pause dur="1.1"/>

and <pause dur="0.3"/> a constant force is applied <pause dur="0.2"/> through <pause dur="0.3"/> half a cycle of the engine <pause dur="1.3"/> hundred newtons <pause dur="1.1"/> # that means that that force moves <pause dur="0.3"/> through a distance of how much <pause dur="2.5"/> <trunc>i</trunc> <pause dur="0.4"/> if B-C goes all the way <trunc>th</trunc> <pause dur="0.2"/> round through a hundred-and-eighty degrees <pause dur="0.4"/> how far <pause dur="0.3"/> has point <pause dur="0.2"/> A moved </u><pause dur="0.2"/> <u who="sm0807" trans="pause"> nought-point-four metres </u><pause dur="0.2"/> <u who="nm0796" trans="pause"> yes <pause dur="0.2"/> point-four metres <pause dur="1.4"/> so <pause dur="0.6"/> the work done by P is a hundred newtons times point-four which is forty newtons <pause dur="0.2"/> forty <pause dur="0.5"/> joules <pause dur="1.2"/> so the work you're putting in is forty joules <pause dur="0.2"/> and that work is going to result <pause dur="0.4"/> in an increase <pause dur="0.5"/> in the kinetic energy of the mechanism of the same amount <pause dur="0.2"/> forty joules <pause dur="2.4"/> and <pause dur="0.3"/> the interesting <pause dur="0.4"/> feature of this type of calculation <pause dur="0.4"/> is that you are not actually working out anything to do with accelerations and what have you <pause dur="0.3"/> during the course of the process <pause dur="0.3"/> if you did that it gets very complicated 'cause you've got to take all the angles into account <pause dur="0.8"/> and this is one of the great virtues of using <pause dur="0.5"/> the

conservation of energy <pause dur="0.4"/> as a principle you simply look at the initial state <pause dur="0.2"/> and the final state <pause dur="1.0"/> you don't worry at all <pause dur="0.2"/> about what happens in between <pause dur="3.1"/> right so let's have a quick look then <pause dur="0.6"/> at the script <pause dur="0.3"/> which i've just given out <pause dur="1.8"/> principle at the top <pause dur="0.4"/> now there's a useful formula given here which you can work out for yourself <pause dur="0.3"/> if you have a bar hinged at one end <pause dur="1.6"/> fixed at one end with a linear velocity V at the other end <pause dur="0.3"/> the kinetic energy of that bar is M-V-squared-over-six <pause dur="1.8"/> that's a little magic formula <pause dur="0.9"/> # <pause dur="0.2"/> it actually <shift feature="voice" new="laugh"/> is very useful <shift feature="voice" new="normal"/> in this problem <pause dur="0.3"/> because you can use it at every stage in what happens <pause dur="1.6"/> # <pause dur="0.6"/> but you can easily prove it for yourself <pause dur="2.7"/> the third stage the initial kinetic energy <pause dur="0.8"/> point B is moving down <pause dur="0.4"/> at two metres per second <pause dur="0.4"/> that <pause dur="0.3"/> arises from the fact that B-C is going round at ten radians per second it's the <pause dur="0.2"/> R-theta-dot <pause dur="1.3"/> so <pause dur="1.1"/> that velocity V applies to both rods at that position <pause dur="0.6"/> the left hand rod A-B <pause dur="0.8"/> fits the category of <pause dur="0.9"/> paragraph two <pause dur="0.2"/> with a V of <pause dur="1.0"/> whatever it is two

metres per second <pause dur="0.3"/> and it also applies to the other one the crank <pause dur="1.1"/> so we can very easily use the <trunc>M-V-s</trunc> squared-over-six <pause dur="0.2"/> for both to tell us what the total kinetic energy is <pause dur="0.6"/> and that's at the end of section <pause dur="0.2"/> three <pause dur="0.6"/> you'll see that the kinetic energy initially is quite small about two joules <pause dur="1.9"/> # this <pause dur="0.2"/> works because the centre of gravity does not have <pause dur="0.3"/> any <pause dur="0.3"/> horizontal <pause dur="0.2"/> component of velocity <pause dur="0.5"/> for this particular problem <pause dur="0.4"/> in either the initial or the final state <pause dur="0.5"/> if it did <pause dur="0.3"/> working out the kinetic energy gets a bit more complicated <pause dur="2.0"/> and then the final kinetic energy <pause dur="0.3"/> point B has gone all the way round through a hundred-and-eighty degrees <pause dur="0.2"/> and now it's moving upwards with a different velocity which is unknown V <pause dur="0.4"/> but we can use the same formula <pause dur="0.6"/> same formula is going to apply <pause dur="0.4"/> to work out <pause dur="0.2"/> kinetic energy <pause dur="0.6"/> except we simply put V <pause dur="0.5"/> instead of two <pause dur="0.7"/> in the expression above <pause dur="1.6"/> five is work done which we've just talked about <pause dur="1.5"/> six is making use of the principle at the top <pause dur="1.4"/> which tells us that V

is nine-point-four-seven <pause dur="0.3"/> so it's gone up from two metres per second <pause dur="0.2"/> downwards at the beginning <pause dur="0.4"/> to nine-point-four-seven upwards at the end <pause dur="2.5"/> and <pause dur="0.6"/> then theta-dot comes directly out of that <pause dur="0.5"/> by the same <pause dur="0.2"/> means <pause dur="0.6"/> so actually when you get into it it's a very simple problem isn't it <pause dur="0.3"/> looks complicated to start with <pause dur="0.3"/> but once you've got a few tricks <pause dur="0.4"/> in your armoury <pause dur="0.8"/> it's very straightforward <pause dur="4.7"/> thank you that will <pause dur="0.2"/> do us for today<pause dur="0.4"/> now <trunc>ne</trunc> next week what we're going to do <pause dur="0.3"/> is take a look at some of the example sheets <pause dur="1.1"/> # i've got <pause dur="3.8"/> two <pause dur="0.2"/> more of these here for you today <pause dur="0.3"/> can you <pause dur="0.5"/> take one each of those <pause dur="1.0"/> and next week i want you to come <pause dur="0.4"/> prepared and ready to ask some questions <pause dur="0.5"/> about things that you've had a try at <pause dur="0.4"/> so <pause dur="0.4"/> if you could put a little bit of effort <pause dur="0.5"/> into looking at perhaps the earlier <pause dur="0.2"/> example sheets <pause dur="0.8"/> # particularly example sheet four <pause dur="0.9"/> and we'll have a look at that next week <pause dur="0.8"/> and in the final week of term <pause dur="0.2"/> we'll look at these two <pause dur="5.3"/> all right any anything else