 All right, well we got two more things to cover and and there there there's hardly anything to do with these because what we did with the With the particle motion, we only have to add to a little tiny bit and then we can keep going with it So today we wrap up our work on the work energy equation We only have a little bit to add to it little tiny thing here Bob. You're looking forward to that You don't like to work in the equation I like it. It's so easy to use. It's just an accounting problem If you remember what it was We had four terms in it The work remember is that done by outside forces? We can add a little bit to that just with the The stuff we've done now that we've been working on some rigid bodies that will take Two things now the first thing we've always had the fact that work can be done by a force Moving through a distance in the same direction as that force remember that's what the dot product does for us It throws out the part that's perpendicular to the direction of motion Keeps the part that's parallel to the direction of motion even accounts for the fact that Those could be in opposite directions as would be for breaking forces By the way, this is of course the net force on the object But you can do it you can do that calculation for the individual forces and add them together Or you can add the forces together and then do this for the resultant and you'll get exactly the same answer so So those for those so inclined. It's a very flexible Method as I think is so much of the rest of the work energy method because you can do it You can put all the disparate parts together however you wish The only part we can add to it now is the fact that we can have objects that are having work done to them During rotation and so that might be something like well if we we had some object and we applied Say a Couple to it That would cause it to rotate Then that couple would do work. Oh, I don't want to change colors not yet That couple would do work by moving an object to a rotational distance which would then be d theta so it's still very much like Like the linear Work energy term work terms right in that it's still the The force doing the turning which would be a couple or even a moment if we have one Let's put that in just in case. Oh, and by the way, this is also the net couple or moment Times the distance angular distance and this too is actually a dot product But for our two-dimensional problems, it's just simply the couple times the direction in which it moves Including the possibility here that those two might be in opposite directions of each other the couple could be acting one way And the displacement angular displacement of any acting the other So that's that's just a possible small addition that we might have a problem. No great worries I hope we'll we'll treat it when we get to them the only other term that has any real change to it is the kinetic energy term Because this will now involve a possibility that there's a change in translational kinetic energy and Or a change in rotational Kinetic energy might have both might have one or the other Might have neither in some problems. So it just depends on all the problems But again, it's it's very flexible if you have a couple things Translating you can calculate the kinetic energy for them individually and just add them all up together You can do the same thing for rotational things The only difference is We do have the possibility That an object is in pure rotation So we might know where a center of gravity is but it happens to be rotating about some other point a clear that this is a Condition of pure rotation when we have this kind of thing it Is a little bit different in that not only Do we have the translational motion because the center of? mass is moving But we have the fact that we have mass rotating and its distribution with respect to that point a Also plays a factor in this But it's not a terribly big deal to handle leader. We look at it like this. We know that the change in translational Well, there might be a little bit simpler way to do this. Let me do it this way. I'll do just the Kinetic energy itself and then you can do it for two different points and get the change in kinetic energy It'll just make the algebra out on the board a little bit simpler Because it's the same whether it's the t2 or the t1 we're talking about So there's no sense putting both of them up there when we've got a couple things going So the translational kinetic energy Would be our typical one-half and v square type terms because the center of gravity itself has some Linear velocity, but then there's also a rotational component to this this solid That we'd also have to take into account So we could use our straightforward a rotational kinetic energy term and Would do it exactly like that, but if we work with that a little bit remember our kinetic our sorry kinematics relation that The velocity of the center of gravity has to do with how far away it is from that center and how fast It's rotating so we can take out vg and put in omega d In its place That gives us both terms with just the angular velocity in it So we don't have to actually figure out the velocity of the center Massage we don't want to we're just looking for the angular velocity or if we just have that that's all we need to use But we can rework this a little bit let's see one half is a common term to both then Omega squared is also a common term to both so I'll take the one half out the front and the Omega squared out the back that looks right. What's that term in parentheses should look familiar That's the parallaxes theorem giving us Moment of inertia with respect to that point of rotation point a Which then gives us a second option? If we have that moment of inertia We can calculate the kinetic energy by just using that if the object is improved pure rotation. This is only if that point a Is stationary at least for that instant if you think back though, we had the method Determining what them what the instantaneous center of zero velocity was and then using that for Calculations if we happen to have or can't find the moment of inertia with respect to the instantaneous center We could do the calculation from there also, though remember that When an object was rotating without slipping the contact point at the bottom was a point that was not moving at that instant That's an also a point of instantaneous zero velocity So if you have an object that's rolling you have to know it's over the inertia with respect to that point You can then Use just that much simpler term but either one of these Will work. It's not that it's They're exclusive of each other. They're just the different forms of the very same thing And it just kind of depends on what you need and or what you have in a problem as to which is the easier to use Do is the easier that's what I do because I'm professionally lazy Then the other two terms don't change they're still just as as they were Just have to be careful if an object is rotating like this then its center of gravity is Changing height and there will be a gravitational potential energy term with just that rotation If it doesn't go an entire complete rotation giving it no change So the other two terms exactly the same note change No sweat We're ready to go Okay, so a little problem here kind of like kind of like a garage door problem. We've got a Slider constrained to sliding in the horizontal direction and Some kind of link Connecting that to Another slider Constrained to slide in yet another direction perpendicular to the first things up there So it looks beautiful now it does Where we'll take this arm this a bit of link arm here to be uniform and slender Just like me point four meters long ten kilogram mask and then you know if you're if you You take a look at your garage door you it's it's got usually a spring on it to assist with the closing So we'll go on there such that the Spring constant is 800 newtons per meter and It's unstretched when the door is Open when this angle here Is zero so the doors all the way up That way it's easiest to reach up pull the thing down that stretches the spring then It'll kind of hold in place and then when you want to open it It's a little easier to get it going goes up to the top so we want to find the angular velocity of the Door itself and of course then if we wanted to we can find the speed of the one end coming up We needed right after that, but find the angular velocity When at rest so partly open picture something Just how fast is it moving then when it gets to be fully open position? Of course, you want to make sure that it doesn't slam up there break the window the well you guys Since you're trying to sneak sneak in from being out too late at night You don't want the door to make a big clanging noise when you open and wake up your parents They come running upstairs, and you're grounded for two weeks You're not worried about being grounded for two weeks Bobby you probably get in the trouble Oh, hang on with Bobby for a while What's your question If it's if the scene is closing so theta's becoming smaller, I guess what's becoming smaller We want to find out what the what the oh sorry find omega when When theta equals zero When it rested 33 so Was that your question? Or was there still another one? So you're saying it begins at 30. Yeah starts and starts as about where it is here at rest Then is it allows it start to close is fine. I think we have lots of the at 30 degrees Yeah, you should have said you should have said zero and then left early for the day But I would call your grants and ground to do for two weeks All right remember work energy problems work real well when springs are involved when Gravitational position changes evolve when general position changes are involved. So Let's start with that Again, I recommend though are anybody ever listens keep it in its little compartment So it's a bunch of small problems. You don't screw up the minus signs. Don't screw up the units Things just seem a lot simpler If you do so Any of those zero so the problem just starts getting smaller right from the beginning or the work term Remember what goes in there is any outside forces any non-conservative Forces remember what the conservative forces are Yeah, these are two conservative forces the spring and the gravity because if we return to the original position There's no difference in and the action of those things everything returns to exactly like it was before That's not true with any other terms if we if we're going to look at friction in these channels Friction is nothing you can get back. No matter how many times you repeat it. Sorry professor I'll run you but that's the truth of it. So yeah that first term zero no no No forces in there. There are motorized garage doors We have to take that into account if we had one, but we're keeping the problem. This is an acoustic garage door So there's no No motor attendant to it I Know 10 kilograms wait, we'll take the sliders You better it's already on the board game we're committed We're committed to the work energy probe. We're halfway through it. We're on the second term already all right T2 minus T1 any of those zero T1 of course starts at rest down to the third degree so that one's zero but The T2 then remember is very much like the thing we just looked at where the center of gravity Will have some contribution to the angular Some contribution to the kinetic energy Changed but so will the fact that this isn't a massless arm and it itself is going through a change So you might want to do however, you prefer to do the split that T2 into the translational part and the rotational part one half that'd be squared plus one half I G Omega squared where that omega there is the thing we're looking for because that's omega 2 since that's t2 So I'm put a little t on a little 2 on both of those you want however you might recognize the that When theta does equal zero when we're looking for this last little moment As it as it finally reaches the horizontal position What's the velocity of this point? Let's call it point a that's slider a Slider b. What's the velocity of slider a at the instant when the piece reaches the top? It'll be zero so we can use well and I'll have a picture up there any more at the instant here at at Theta equals zero our point of interest then VA equals zero and We can best Say we have a possibility of Using that as our kinetic energy term if We know the moment of inertia or can find it about point a with respect to point a You're grounded. That's three weeks now buster Do we know the moment of inertia of the object? We're talking about in motion about point a With respect to point a and since we're considering the sliders is massless. We're talking about the little arm here Do we know it's moment of inertia about point a? We do it's one of the things in the in the book or some of you might even remember it off-hand It's one-third ML squared where L is the length of the slender rod M's it's massive course If we do it about the center, it's one-twelfth And if you take one-twelfth ML squared plus md squared Or d is half of L you end up with with exactly that anyway Well, we've got M. We've got L that simplifies this term this whole remember This is the change in kinetic energy term because t1 for all objects was zero So this all is just based on whatever happens at point. Oh, what about the fact that? Before it reaches theta equals zero Point a is moving Does that negate all of this stuff? Where I said well since point a is not moving at that instant We can do this as a simple rotation about that point It's before that is rotating before that point a is moving Remember making the jump from here to here dependent on point a not moving So can we do this the second the last week of school you have to stand up you have to take a stand here Bobby will stand there Alex This VA only holds when theta equals zero before that that's not true Can we make the jump from here to here which dependent upon a not moving? Jake says, yeah, yeah Duby says, yeah, Frank Sure Just cuz it's already up there you kind of feel like but I can erase it Even though I hardly ever make mistakes at the board That yeah, of course we can't remember this is only dependent upon what's going on at point two Not between what's going on between one and two all we care about is the instant of point two of our concern What's happening and at that instant blossom is not moving so we can make this step right here and We've got all those pieces. So that's a really easy number to calculate Because yeah, we have we have the mass the length everything else comes up to be point two six seven Omega two squared and the units will be Newton meters If Omega two is in the usual ratings per second, so we know whatever number we come up with it's got to be that You could do the spring term because that has nothing to do with this rotational stuff You can also do the kinetic energy turn So let me let me collect while you're doing that only collect our our on our evolving work energy term you know the change in kinetic energy will be point two six seven omega two squared so you do the change in kinetic energy terms Sorry to change a potential energy terms and We'll see if we agree and then put the whole thing together for the quick solution See how we do want to split them up among yourselves Because one is so much easier than the other you want to jump in there take a claim here You're not looking at me. There must be Alex or ready to kill. How does that make you feel? Sox doesn't come. Yeah, anybody in here can kill you it could do Yeah, but it'd take me longer You know that you guys are gonna have to go a couple rounds No, I'm working out in the semester So why? I Work on gold where they're staring dispensers of the locker room I use the building itself You know Bob Think I'm gonna start awarding you some extra credit points for those times when I do hear what you said Spec showing this class. How do you mean disrespect golds champ like that? Of course, they have enough weight You haven't seen those guys undressed in the locker room just that's one of the effects of steroids How we doing Jake? Brain on a weekend already. This is all they gotta do. This is all that's on your mind right now Just this little tiny bit here Pat You don't pay with it. Look at that blank page Oh Don't forget watch your minus signs watch your units here one of those get screwed up Come on somebody come check with poor Frank. He can't get up I There's there's no sense going beyond either one of these two things and before you check with anybody because if you get if you're Those you go beyond it. You're just doing the wrong problem Whoa good question for the height for the potential energy term We only have well, we have two things changing height the slider is too, but remember we're treating that as massless Masses inconsequential compared to everything else. So what do you use for? Delta H For the height here It is indeed the change in height of the Center of gravity is just like you said half the length times times the 30 degrees sine 30 which is which is what point one it's up and it's pasta and We'll give you units of Newton meters right there as it should So I do that's 90 degrees in a slip out of your chair, which was great last week I really appreciate that I'm still thinking about that was such an awesome move Right. Is that what you got? Bobby that makes sense then Can anybody do this in their head? I'm going to the calculator for that count for that ten times point one And the units are newton meters. So that's our delta VG term to get it Very agree with it. Is it? Let's see Vee two minus Vee one Which are either of those zero is Is the spring ever at a zero energy condition? Which means a completely unstressed or uncompressed at? Unstretched when theta equals zero which is our final position. So Vee two is zero You didn't actually do that. You might have missed the minus sign Did anybody? One half k Dell squared up. Sorry. Bell one squared minus one half k del one squared 800 and Dell one the amount of stretch in the spring at the initial point Not as complete Total length, which we don't know anyway. We only know where it's unstressed. We don't know its total length And so that would be 0.4 times sine 30. Oh squared. Yeah, you're only going to forget there miss you might miss the minus sign here you're not careful with this and It is quite common in this square So that's minus 16 newton meters. Same thing you got everybody You have minus 16 and so now you've got a very simple way We can just call some dope down there and help come to solve this now Almost anybody walks by I could do it and you got Huh, why don't you get no extra credit for comments? It's nine comments that I don't actually hear Likewise or counterclockwise So how do I want to designate that? All right, hopefully fairly straightforward Again remembered that easiest two easiest things to do is miss this minus sign and just then forget to square Which is an operational thing if you miss this minus sign you're doing a completely different problem I don't think your bosses are happy when you do a completely different problems All right, any question before we clean up Do another one. This is a side view not top view. So get ready. Are you serious this early? You get out of class anytime you want get up and go because I can catch him We've got a hinge bar here Not quite to scale a little too difficult to draw that Pinned at the bottom free hinge there It's two meters long And at the midpoint Attach a spring here at the top. We have a mass of Radius does not significantly big compared to the length of the Entire rod itself. So we'll just take that as a concentrated mass So some of the other details then rods two meters long the mass up there's four kilograms and The rod is nine Distance from the wall the tenth of a meter Again, the uniform slender rod Two meters long. There's an initial angular velocity of three gradings per second bring the length is A quarter of a meter. That's the initial situation at that instant this has an angular velocity In that direction It does that's exactly what it is. Oh, so it's initial It's initially compressed a little bit. It's rest length is out here somewhere We attached it push the bar to the wall to compress the spring initially That's the compressed length that's that's how long it is right now So L1 is this 0.1 meters And remember That Dell is the difference in those which of course we're going to need for that the potential energy term over there So the deal is you need to figure out What spring constant Will allow this to just come to a stop at horizontal Find K at horizontal do this on your car roof so if you don't guess right and it hits your car and kind of bashing your car roof you've got a moonroof a Shatter right through What a drive Bob is a red with a blue moonroof But that's not it I'm not putting my car under something you guys design okay So if the spring is too strong It's not going to make it to the bottom if it's too light It's going to hit the bottom with some residual velocity So find K such as just Not moving at the bottom there Bob you can make this it'll get out of class question if you just go slow enough You know if it takes you a full half hour to do this problem Yeah, why'd you go like your tuition dollars weren't used appropriately just say why didn't I get everything from manning? I possibly could have when I had the chance About that hammock the hammock reading your dynamics book And it's like getting better. What your rod is that? Looks like a battle ship That's like a bad one First I thought we were drawing a rocket Rocket even Frank's not confused by that drawing are you Frank? No, it's a good one You don't have to treat them separately remember that's the beauty of this you can calculate each of the parts separately if you want or you can count calculate each of the parts Collectively however you want to do this the work energy equation Accommodates your individual preferences what the registration of a register for fall t2 is zero Yeah, we wanted to just come to a rest there. No angular velocity. No linear velocity either So yeah, that one's here. Oh by the way, I assume you already hit that the word term is zero I hope we're doing no work today They think this is the linear velocity. Well, it depends that's on how you have concentrated mass there that is the Term we need to worry about here You might recognize that this is a situation one of those situations like we just had where it's in pure rotation About we can label that as point a So we've got the minus sign can we do Omega one squared, which is just a given But can we do that Since point a itself is not moving Let's see. Oh got the minus got the one half then I Hey remember when you have a compound body and need to find the moment of inertia But you can do the moment of inertia of the two separately So I guess we do what we call that the rod events I are I am I'll make it one straight Moment of inertia of the rod about with respect to point a of course we have that That's the one third type of thing. We just used and that has in it. So the moment of inertia about a Different point other than its center of mass But what about the about the Like out the mass itself We don't have the moment of inertia of a small mass With respect to just some point off in space anywhere in the book. I think they don't think we do So we'll have to use the parallel axis theorem on it where M is of course the mass itself And what is D? The distance the mass is from the point of rotation or the axis we're interested in so that we've got that's the two meters What's the moment of inertia of? The mass what kind of looks like a sphere is that in the book there? The moment of inertia of a sphere Well, what did I say about the size of it didn't I say something I? I Said it's radius with small compared to everything else in the problem So that means its own moment of inertia is Essentially zero Doesn't matter no different particle has no moment of inertia a Concentrated mass or a sphere of negligible radius has no moment of inertia No appreciable moment of inertia about its own center of gravity So we've got all those terms then and so As long as you remember your square and your minus sign you should have all the pieces there For that one well, we'll need to dealt to be G term we can put in very much space for it on the board So finding the center of mass is still work, right? Yeah, yeah, yeah, but Well, you do it remember the the location of the center of mass is equal to the position of the individual centers of mass times their mass Where x is the? Distance of the individual center of mass from whatever point you're interested in which would be problem a would be the easiest over Tomas So yeah, you could place it. Maybe you wouldn't be surprised. I'm sure to find it Probably somewhere in here And when you put that in to here Then you pretty much get the same thing you're still going to have to do the individual moments of inertia I Guess yeah, you still have to know wait around that and g delta h Again, that's two things that are doing this two things going through a change in height So just do them separately rod and The mass are each Undergoing a change in height and not the same one the only thing they share is actually g But what numbers you get for those pieces and then we can look at this spring Takes a little bit more work for not much You got all three terms Frank one check on Your mom so Scary there and that looks right. No, that's not right That was right. Well, no you're doing okay, right? Oh That's what I missed Anybody have Delta T Well The thing is I don't know how necessarily you do the most the only way to do the moment of inertia Of the compound bodies to do that anyway, so it's not going to really help that sound that Yeah, then you can't really do anything with it once you found it maybe submit it for extra credit We've got some of these terms to share Frank you okay? Yeah, well, we have a mega squared that shouldn't Omega one square Yeah, that's the distance that mass is From the point of interest so that's that is yet the full two years for your elastic energy terms still not correct Got Delta T Delta BG everybody's getting 167 new meters Yes our negative 167 because it drops we've got all all the pieces in there so anybody have that Delta T term yet This is the negative 167 26 so we can make our equation Zero well you had what? For Delta T Yes, that's what I got to We're here and you got to catch that negative. We got to make the squares. There's a couple of several of them in there Got the Delta BG term in agreement Delta BG We don't need to Delta Oh Comes from here, which is why I strongly suggest you make this what seems like a trivial step so Bobby you're asking for Delta VE Which remember we'll have the K term that we're looking for that's the only unknown the whole problem Which is good. We don't have one equation work energy equation. You have Delta VE Check Okay, it's Yeah, well once you get Delta VE if you're right Okay, just follow us Know a little bit different What the double check? Delta VE you have Delta VE Okay, let's let's put that one up looking for Two squared minus Bell one squared neither of which are zero is that right? because initially it's compressed and Finally, it's at length so Initial Dell We have That's going to be what minus point one five, right? That's the Mount Squoosh Put into the spring just get things started then Dell to Let's see that's going to be the spring from here down to The midpoint here So that that is going to be L2 the full length of the spring at the finish of the problems that itself is A triangle one meter By 1.1 meters No, it's only to the midpoint or one meter by 1.1 meters. So the hypotenuse is L2 Freeman on that picture that triangular picture Point three nine. No, no, it's not the point one minus. It's the point two five minus so one point two Four squared. Is that right? original one five completely changed on The spring the the direction in which it's pulling is completely different now doesn't matter This is like the kinetic energy term where the direction of the velocity doesn't matter the direction of the spring doesn't matter colon Something wrong guys something wrong up there. I was right with the one point four nine And then we take for the original length point two five off of that And so we've got all those terms Because if it's if we do do it that way even though it's not important here But if we do it this way then compression will give us a negative number Extension will give us a positive number This is the only reason for doing it that way, but we square it anyway, so the minus sign doesn't matter So how do you do one point two four? That's that's the final length Minus the minus the original length point two five the final length is one point four nine minus the point two five Okay fingers all right, and so that comes a point five seven five four Is it positive? And then there's our unknown in there too as long as K has units of Newton's per meter last little bit so Bobby you were a little different on that term, but you're okay now. We're good. Okay, but what went wrong make yourself a good drawing Full foot boy is anything I guess. Okay, let me get out of class question Do you want to get out of class? Wrong answer. We'll stay do another problem That was your chance to answer the question wrong. I have another one if you want it If not You can start your weekend Yeah Sure, you stay and do a little extra work for everybody else. You didn't get bonus points This isn't this isn't like a salary job where you get a point pay the net amount Set amount no matter how much work you do This is like a more like hourly where this is peace work. I'll pay you for what you do Yeah, for it You want to scoot with Bob Bob's head now? We might arrive We think I'll be here you guys all want one Sure Well, I'll give you Bob's points You're what oh, okay, I thought you said I'm leaving All right This one the trouble is it's gonna be in terms of the variables You you guys you guys all did this you guys all did this problem. Anyway, it was one of our Our physics labs where we're at a pulley then with a weight coming from it And we figured out we the way we did it we figure out both the moment of inertia of the disc and The bearing torque that was slowing it down to the sum of the torques, but we did that as a Straight kinetics problem some of the torques equals I alpha so we'll do this as a Work energy problem, right? So that is of radius r mass mp This will put a B on this math or you can use lowercase and uppercase if you can make them distinct enough That'll be a mass and B It falls Through Delta H if you remember when we did this in physics one We had to figure out the tension from the acceleration of the block that you time Then you use the tension to figure out the torque and lights do this as a work energy problem You can treat it all as one system, and you don't care you don't care what the tension in the line is because it's interior to the system and So starting from rest find Velocity of the block as it hits the bottom And it'll be in terms of our mp and B Delta H Whatever else you want to do it is a work energy problem Yeah, well remember when you have an extra unknown like that you can use the kinetics equations and The kinetic equation is that the equals are omega. So if you have omega you can get rid of it in terms of the B Nope just starts from rest just like we did in lab Last year and some of you may want to come back to it again with all Just something to do separate items which we could do then we have to have attention to figure out how much work it does But then when we add it all together the two work terms on that would cancel anyway because the tension doesn't have Problem with this crutches, didn't we? But it's so nice you it's supplied that that in class Real-life experience for about center. That's the only place and that's the easiest one that's right out of the book Just use IG Sorry Oh in the picture let me see Which a disc is just very skinny one And notice that the length of it doesn't matter in IZZ And so that's what you want to have the mass moments of worship. I love what you use Substitute for omega r squared b squared for omega squared and that's not That's all you need Fun times equals RV the American guard beam That's what they're thinking about your summer trip with your family Okay So close it's like some negative science. Exactly because Delta H is negative. So that's the square root of a negative number I brought it over and divide it out The other side be able to sign Missing another negative sign somewhere else. All right, there's lots of them in here Lots of happy little negative sense Frank. Thank you. Got it Cylinder what's what's IG it's one half m r squared All right Well, let's see. Let's see if I can figure out where I know that yeah, but you know, there's got to be one because Delta H is negative Yeah, however, I'm saying from the physics viewpoint You know, there's got to be one there because you got to have two negative signs under the radical And you only have one So no negative sign here. So it must be in here The same No, because you say that I Could have just drawn it on the other side Still the same problem. So I have the Delta T term as all positive. So do you where that one is not come on? There's a negative Delta H. No, no negative for the Delta H goes in there with it So so if you're writing it that way what you can then Delta H is always positive So use the minus to indicate whether So now you've got your minus So it's like saying g equals negative 920 at one That's why I was pretty much doing with Delta H All right On this points a 100,000 easy plus the five five Can I spend these like that staples to get more of these? Yeah What There's no books yeah with the stripes on the front and they all Yeah, there's so many lines Because the lines are there You bought wide rule so that's what you got that's about the line space Yeah, she's still shopping for your life in the sixth grade you gotta go to your own shop She works at Yeah, okay, so you're dealt a positive if you're just putting in the magnet to the Delta H Not its direction as well It's fairly simple In the end the whole bunch of stuff cancels it's only got the two masses and Delta H and G the T Is a good idea to make that a lot simpler Because you can eliminate BV or Omega Yeah, so don't eliminate it, so eliminate Omega this becomes a lot simpler No, oh wait No Yeah, kind of unit's work Yeah, well Sort We'll all extend the bonus point deadline