 All right, we're looking at Kinetics we started with the rectilinear, so we'll do a real quick one on that Just to remind us a bit of what the deal was plus bring something else into it. We Had it one time before so imagine a Couple sets of pulleys here This one's gonna get good. So do your very best on this one It's easier. I think I've just found out over the years to draw on the pulleys first and Then put in all the ropes and the light So there's the pulleys Frictionless cart, that's what the wheels stand for on a 20 degree 20 degrees slope Take a second to get the pulleys in there Actually, I like this problem. It's like this Other stuff 30 kilogram cart on perfectly frictionless wheels This down here Is 10 kilograms and that slope is 20 degrees Core the cable Start from this hub Goes up over that one and then back to An anchor at that hub not too bad Okay All right starting from rest neglecting all friction and The mass of the pulleys now in a couple weeks We might not necessarily neglect the mass of the pulleys because not only are these masses accelerating Linearly Translationally, but you're gonna have to spin the pulleys up to some kind of speed and that's going to take some of the For some of the energy that's in the problem, but for now we'll take the pulleys to be massless and There to be no friction in the system so starting from rest want to find then the acceleration of the two Blocks in the problem. So What do we got any ideas what to do? Well, I don't know if we so formally call it that but most certainly the the length of the cable joining these assuming It's Stretchless, which we wouldn't assume in strength of materials, but we can still assume here Assume that the length of the cables is constant So from some reference point that doesn't move We want something about how the cables are how the how each of them relates to the length of the cable so maybe call that x a and Maybe this xb will work remember we don't have to go all the way down to the piece Because none of the constants that are left over matter So we don't care about this little bit of cable over there or any of the cables around the loops or The fact that our xb is coming up a little bit short of the blocks doesn't matter because however fast the Pulleys move the blocks themselves move as well If you remember we set up the length of the cable in Terms of those and you might pick other reference points, but I think that's probably as easy as anything So it looks like we've got two x a plus three xb Plus a bunch of constants That are of no concern Because we don't actually want the length of that cable But the time rate of change of the length of the cable which of course is zero because it's constant length so just from that then you get two x double dot a equals minus three x Double dot b for the accelerations Is that about where everybody got to kind of quick? Trouble is And by the way, of course those are the accelerations we're looking for just in a slightly different Notation What's the trouble here? No, that's not so much the trouble The the minus sign just means if one of them is getting Shorter on that side. The other one's going to be getting longer on that side. So that's no Concern, but what's the trouble was solving the problem? We Well, yeah, it's the rate of rich the rate at which these are changing The trouble is of course, that's two unknowns and one equation. We only have one equation so we need a second equation that doesn't necessarily bring in any more unknowns and At least refers to one of those if not both Suggestions Well, the only other place we've got any help this is this is all kinematics The only other place we can look is at the kinetics in other words bringing in the forces in the problem. So the forces are going to come in and some free-bodied diagram So each one of these has some weight to it We're just drawing in the weight vectors What else All right I'm not here Well, you guys put that in you put the tension in on those two free-bodied diagrams remember that with massless Pulleys with no friction in the system the tension is the same everywhere in the cables When we take into account the mass of the cable the pulleys themselves in a little bit the tension will not be the same Throughout the cable be different on one side of the pulley than the other for each and every pulley in the system, but we'll take the Tension to be the same in all of the cables at this time. Well, it's only one cable, but all sections of the cables So there's clearly some effect of the tension here If the tension in the cable is say T How much is that force on block A? And how much is this force on block B if the tension in the cable we call T which is unknown Is that T? It's two T and this is Three T In fact, that's the point of a blocking tackle system is to get with one rope more force applied When we're needed Is that it? Block A has a normal force on it because it's in contact with the slope So perpendicular to that Something like that. So we're looking for the acceleration of A and B We're not sure which direction and B, but that's the type of thing we're looking for so we're not sure Which direction but it might be Something like that it might be that they're in the opposite direction, but we do know they're related by equation one there So we have another unknown T now in the equation in the problem so We need not one more equation now. We need two more Those two unknowns but now we've got T as an unknown. So we're gonna need at least three equations Where else do we go for extra equations? Who's got a smartphone has the eBay app? Pick one up there. Somebody has extra equations and put them on eBay Nobody's pulling out. Where are we gonna get two more equations from me? We don't need to go there. We're gonna get two more equations But three unknowns So far T is unknown If we had mass In the pulleys then those two T do we wouldn't have this Simplicity of T being on both sides of there So where else are we gonna get an equation? Well, that's one equation so the mass some of the forces Time equals the mass times acceleration. So there's one equation Well ends unknown so do we need where are we gonna get a third? And do we need a fourth equation now that n comes into it with some of the forces on that? Ends one of the forces work. Well, maybe let's work on this one because It doesn't happen to have a normal force in it. So That can help us a bit 3t minus mbd equals mb x double dot b That's pretty useful that didn't introduce any more unknowns Okay, I don't With respect to this Yep However These two do agree but that would be negative so we can put a negative in to Go ahead and go with this and that will keep this native the integrity of this negative as well All right, so there's is that a Legitimate equation to use No extra unknowns but We're still short one unknown at least or one equation We can do the same for this But we're gonna have the normal force So what's the deal? What do we do? Travis you have another idea you help with with that one The normal force is an unknown Indeed, but it's not in the direction is perpendicular direction of motion So it's not really of concern. It's not one of the equations of motion Because it's not the direction in which we're moving so if we sum the forces in the a direction we get minus 2t plus Let's see. Where's the 20 degrees? That's the 20 degrees. So this is plus ma g sine 20 equals ma x double dot a doesn't involve Any new unknowns doesn't involve the normal force Does relate the the last of the accelerations so now you've got three equations three unknowns Since it's not an algebra class. I'll help you with it. You can double check That you get these plus I'll put them up and you interpret them for me X double dot a is a little over one meter per second squared X double dot b minus 0.6 8 to per second squared and it wasn't asked for but there it is T comes out to be 35 newtons Let go from rest. We'd see those accelerations in this system idealized as frictionless and with massless pulleys Which way does the system actually move? Either x goes down and b comes up or the other way around Which is it gonna vote for other way around Which means x goes a goes up b comes down well x double dot a is positive that means x a is Increasing so From rest then a is going to go down and B is going to go up so as I Hypothesized with little sketches there That's indeed what it does All right questions before I clear that one Joey. You okay with that one? Are you? Best question is now for a raise it Okay, that's a straight rectilinear coordinates now if we look at kinematics in Curvilinear coordinates again planar curvilinear coordinates We've got some other Things we can do planar curvilinear Portnates the kinetics of planar planar curvilinear coordinates kinetics the Equations of motion well if we use the cartesian coordinates system We again have two equations Of motion two equations of motion Plus any of the kinematics we might want to bring in in fact that's true on any of these problems To get enough equations. We often need to use both the kinetics and the kinematics in the normal tangential coordinate system we have Again just Two equations, but there's a little bit more opportunity for For Bringing in the kinematics if we need it because we know that the normal acceleration is The centripetal acceleration and that the tangential Is the Timerated changes of velocity the velocity itself is always tangential so we can add those little bits to it Could be that we know more parts to that Since we can also bring in If remember this is best for known Path with some kind of curvature. That's where the road comes from we can also add in other parts to it such as Omega if if we know something about how the Radius line to the center of curvature is moving or Are omega or a? Tangential equals our Alpha or any combinations thereof Which there could be several of them so no sense writing them all down, but that's a That's in normal tangential coordinates Plus any of their kinematics which for the most part would be something like that or any of their of the kinematics along the path itself as well Then we had that one other coordinate system. We looked at the polar coordinate system where we have r and Theta Coordinates that one is a little bit more involved in that. We also had a couple other directions A couple other components we went through these so not to belabor it are double dot minus r Theta dot squared if you remember this is very good for a for motion about a fixed Point as as would be with a radar gun or something like that And we set these up. I guess about a week ago Those Polar coordinates with any of the other kinematic equations that might be needed Which could be these but might have some other parts to it and remember some of these were also Easy to tie with sometimes the X and Y notation Remember this one works good for type of a satellite tracking problem and the plane itself Its motion might or easily be described in an XY notation So you can combine the two if needed if and when needed all right, so that's that's a Little bit of review just taking what we've had already and tacking it on to the ethical domain Describes all the kinetics of what we're doing here so far. All right, so we'll look at a couple problems now Imagine a plane Flying on some curved path of some kind Radius two miles at least local radius two miles remember sometimes these are taken as the Curvature the local curvature even though the path does not have to be circular everywhere But it's someone certain particular point it can be approximated by a circle 400 miles per hour and we want to find the angle at which The plane must Fly around that corner. You know if you go around a corner So there's the Row direction I guess and then this direction would be Z and if we're looking either in front of or behind the plane We see it banking as it went around the corner. So this is this this is the same view Only from the back of the plane now as it goes around banks around the corner That makes that kind of noise so what we want to come up with here is given corner Path of that radius at that speed What is the angle at which the plane must bank any? Any suggestions of Well, we have three coordinate systems we can use Cartesian Normal tangential or polar the Z Axis straight up and down isn't going to help much because all three of them could have that as it's third dimension Plus we're not that's just part of the viewpoint of it. No, we need that generally for curved paths either the normal tangential or the Polar coordinate systems are going to work a little bit better The normal tangential tends to be a little bit simpler. We don't have all those extra components in there So if we use the normal tangential That'll be the tangential direction. Remember the normal direction is towards the center of curvature So that would put then in this picture That normal direction and that as well and there's a Z there What's next? We're looking at kinetics So It's gonna involve some forces What forces are on the plane jump in and take the easy one first dude The way they'll presumably weigh you something Whatever that might be Don't have anything for that. What other forces might there be? Well there if it's going in a circle there must be some Centripetal force directed towards the center of that circle or it wouldn't go in a circle Where does that centripetal force come from? Remember the deal with forces They all have to come from something real. We can't say there must be a centripetal force because it's going in the circle Because there is a centripetal force. It's going in a circle So we've got to find what causes that force There's got to be some force Directed towards the center in the end direction or it's not going to go in the circular path Do you already said are you gonna say the same thing David? We'll ask him How does the lift act The lift remember is a pressure differential between the air above the wings being a less pressure than the angle below the wings So the lift tends to act perpendicular to the wings There's also of course some drag force, but we're not going to worry about that We're assuming it's got its jets on is going to constant tangential constant tangential velocity And in fact, there's the theta that we need and it's that component of the lift towards the center in the normal direction this applies the Centripetal force needed to fly in a circular plan. In fact, that's why they do have to bank Because they need to turn that lift over and use some of it for the centripetal force Now what? Any other forces other than a drag and a thrust in the tangential we Tangential direction, which we'll assume just cancel each other. Cancel each other if it's flying at constant velocity David We'll assume the plane is maintaining a constant value Sure Level Flight corner of flight through there Well, therefore in other words, these are all the forces we have and then you can now solve in the Two coordinate directions the tangential direction Which will assume to be zero and isn't going to tell us anything more than the drag equals the thrust anyway So that's not going to help us much So we'll go to the normal direction Any forces in the normal direction? Of course, that's not mg But it is L Sign theta Remember we're looking for theta. So that's the centripetal Force there. What's that equal to? There's there's nothing that balances that You know in the end direction. So what do we do? We just put in a centripetal force there On a z direction if we break L into its two components What must this be equal to? This is the centripetal force Which equals m? v squared Over row. We have v. We have row. We don't have m and We don't have L. So then what what we're going to have to do the z direction The sum of the forces in the z direction If it's in level flight, we know those to be equal to zero And so we can then say that L Cosine theta equals mg so You could solve for L Put it in here The ends will then cancel Which means for any plane at that speed in that kind of corner. It's going to bank At whatever theta we come up with L is mg Over cosine theta Time sine theta v squared over row So good they must cancel we get What tan theta? equals V squared over row G You just work out is that okay? The unit don't work out means we screw something up Something else might have messed up, but at least that'll tell us if we messed up meters squared per second squared meters times meters per seconds rest of the units cancel, so we're okay and All of those pieces now We've got and so we can get the bank angle for the plane Comes out to be when you watch all your units little over 45 degrees for That speed that corner Chris That's how I came up That's the spirit because we can go back and watch all the tapes and see that I don't ever do anything wrong up here skip things I don't know that never happens should be Should be pretty straightforward Don't forget to be this piece in miles per hour, so you're gonna need it in something else 587 feet per second Would work because then all the units if you have G and 32.2 feet per second squared Should work out exactly don't use 9.81 Yeah, we have English units Yeah, sometimes you're looking at the harder stuff and you don't pay attention enough to the easy stuff All right, we'll do a related problem only with a little more That we can throw into it the first part is very very similar to this the second part Isn't quite throwing a little bit extra, so oh And this is perfect because the Daytona 500 is coming up, you know They've got those great banked tracks on which they run or will never see another day on our heart so They're there we're looking at a car from the back as it goes around the great banked turn of Daytona No, I just I'm just always amazed following a car Nascar stickers and they're still breathing over the loss of daylight. I guess because I still see those I Don't know. That's all I know about it. Nothing more. All right, so at a track speed of 100 feet per second Huh and a radius of curvature of 600 feet What is the Bank angle such that Friction between the tires and the road surface are not of concern in other words if the if the the banking was made of greasy ice The car could still go this speed around that corner and not Slide down or slide up the banking So to maintain a 600 foot radius what? Banking what bank angle is required? Very similar to the last problem we did So you you kick that out, and then we're going to add to it a little bit just to make life Danica Patrick hit the wall yesterday Yeah That go get one better looking than Dale Earnhardt was It is an RPI Graduate female who's also on the NASCAR circuit. It's not quite a little bit. But then that is all I know about NASCAR That's the limit right there grew up within smelling distance of the Coors brewery. So I know of Coors lights All about to very very similar to the problem. We just did So you guys check it Assuming no lift on these cars like we had lift on the plane. So what's the difference? Where's the centripetal force come from now? Where's the centripetal force come from now? There's no left even though some of these well the NASCAR and that wings on Indy cars have wings That's those are for negative lift anyway Yeah Well, I think so unless that'd be a great mystery novel sneak in flip their wing over because the Wings is put upside down to increases the normal force which increases the friction Which increases the speed with which they can go around the track So what supplies the normal force in this problem? You drew it there. I mean what supplies the centripetal force Well, I didn't know you'd actually be listening you guys trick me and that's That's the normal force the normal force doing the track surface and the car Because there are no other forces if we're not including then drag and Frost which would be well anyway if it's at constant speed around the circle. So there's our centripetal force and normal direction centripetal force the end of the z-direction the z-component so it's that inward Directed component of the normal force that gives us the centripetal force for this corner And so there shouldn't be anything left other than to solve for theta just like it was on the first problem You got something you got something It's like you guys agree That's either comforting or scary Which one of you you are you careful when you draw that so long It's too long to be the component of that you'll forget that that's that is the centripetal force that component of the normal force on your drawings make sure that this normal component the centripetal force is a Legitimate component of and don't drive a lot longer or when you go to look at this in a couple weeks You'll forget where the centripetal force came from from the normal force to come the horizontal component of the normal force Joey got something well, you can now solve the solve the kinetics equations and the normal force that's theta then This is also I believe They're right so this would be n sine theta equals m v squared over rho because that is the centripetal force and and we can then find From the fact that there's no acceleration in the z direction which will take to be the up direction so and cosine theta equals mg and once again n cancels and m cancels doesn't it So this is not dependent upon the mass of the car Which isn't as much a concern anymore, but nascar rider drivers used to be kind of hefty All that barbecue ribs and stuff before life beer came up All right, putting those two together you get a theta bank banking angle of Check that 27 for 27.4 degrees Okay, no different than the problem. We just did so here's the different part Here's the different part if the car goes a lot faster At some time it's going to start skidding up the track What is the maximum speed before it starts to skid up the track to do that one? I'll have to give you a coefficient of friction of Some kind will take it to be 90 so remember that's the point at which the static friction is Gives way to the kinetic friction kinetic friction is one of the two surfaces are sliding over each other and the the car then Starts to skid so fine V max there's also though a concern if it goes too slow Then it could slide down the slope. So there's also a minimum velocity Two separate problems, but how does this problem change then? So it's two parts these can't be answered at the same time So for V max, how does the free-body diagram change? Well, the weight doesn't change that's still there. Is there still a normal force? There must be because it's still going in a circle. It's going in a different circle If it starts to skid up the slope, but it is still going in a certain effect We're looking for the point at which it starts to go up the slope What other forces how does it act on this free-body diagram? For V max if The speed is too high and it starts to slide up the track friction would counter that and be Down the track other than that Everything's the same it does change the normal Direction sum But that is still These squared over row. It's just we don't know what that V is. We're looking for this maximum velocity But this is before it starts to slide up the track. So the row is the same So let's see to get our angles, right? That's theta That's theta two is it not? Take the same angle now that we know the angle use that 27 for for the angle. So the Centripetal force is now a little bit different we've got n sine theta plus Mu s n because that's the friction force cosine theta Equals m V max over New x squared over all remember we're looking for that V max squared take theta as known now so that's two unknowns the Actually three because we have the mass the velocity and the normal force are all unknown here at this point some of the forces in the z direction and n cosine theta minus Mu s n remember we're looking for that maximum Static friction so we are right at that limit on that Dave is that okay? Are you raising your hand or just red? Okay, didn't know if I lost something there Actually, they're going to sum to zero So the up force is just put equals there and add on so I think that's two equations two unknowns then Because n and V are unknown so it's a bit of algebra now to solve it So you can go through that on your own on the weekend, but it comes out to be something like Well anybody have it Chris did you actually get that solved? 160 Yeah, I have 227 so I win the race because you're going to have to speed I am But that's feet per second so What do we have what we have is the velocity first? 100 Yeah, so there's still a lot of margin before Reaches any trouble, but it just it changed the free body diagram We're right at the maximum static friction That's why we can do this substitution me us and further friction But we've got two equations to announce all this left is geometry algebra It's not real clean algebra Because here we have sign plus cosine here we have cosine minus sine so it doesn't clean up real easily like it's doable especially since They just know so it's just those that just become numbers the more interesting question is What about the minimum velocity What's the minimum velocity where it goes too slow? Then actually slides down to the infield of the track. What changes now? There's the car Still has some weight to it Still has some normal force though as we saw the normal force does change between these problems What else the minimum velocity the dangers it's sliding down the hill So there must be some Part of the friction up the hill again. We're looking at that limit So we can take that to be mu s times the normal force whatever the normal force happens to be What changes with this equation in the normal direction anything? We get a minus sign right here is all because now the Frictions point in the other direction. So use the same equation in the end direct the normal direction and sine theta minus mu s and cosine theta Equals and and now we're looking for the minimum velocity Remember take theta be known same track 27.4 What changes in the z direction? With this equation just these are on the same side I believe it's the only difference and cosine theta plus mu s and Sine theta equals energy Does that look right? So again, it just comes down to be an algebra problem The interesting thing here Remember you get you know as you solve this as you put theta in Need to solve for and you can divide through or whatever you're gonna end up sometime with a velocity squared As your term you have something with the minimum velocity squared The trouble is when you do this one when you put in the theta solve through the equations as you've got them you get a negative number Which means the square root of that is imaginary. I Don't happen to have what that number is well, I have I have what the maximum sorry the minimum velocity is I Do have that I don't have what the square of it is You get something like seventy point nine I Remember I is the imaginary root feet for a second What's that mean if you get an imaginary number one? We don't have a real solution for What's that mean? That means it's not going to slide you you can come to a stop and Stay on the banking of the track now Typically in a race like this they don't come to speeds that low to the point where they actually stop on the track and worry about sliding down, however in Bicycle racing on a banked track This is actually a very shallow banking bicycle racing the typical banking is closer to 45 degrees for a small track and there is a race called the match race Where the riders are supposed to ride three laps? With one rider being in the lead for the first lap But then after that either rider can go into the lead the thing is neither rider wants the lead because that rider is Drafted by the second rider so what they will do is they'll come to a stop on the track and Do what's called a track stand they'll balance there on the track Daring the other guy to take the lead because the advantage always goes to the second rider And if the track is steep enough with low enough coefficient of friction One or both riders could actually fall off the track just because there's not enough friction to hold them on the bank and they'll start their track stands and then actually slide down the track and If the upper rider rider higher up the track does it you'll take out the lower rider If the lower rider falls generally the first rider cruises to the easy win So maybe you'll see that in the Olympics this summer from the left that matched It's called a match racing track racing and bicycle track on the ball Just for your interest it turns out you can prove this to be the minimum Coefficient of friction before this does become a concern is something like 0.52 now remember The 0.9 is pretty typical for dry tires on dry pavement But if these races are running the rain that coefficient of friction can't come down and it can't come down to that All right, let's see. Let's any questions on that before we clear the board You had just enough time to just start a new problem lay it out Give you something to do for the weekend. You've got a jail. That's about what I put it Alright, so here's a different problem Conveyor belt delivery package is kind of thing UPS would be concerned with Delivered to a circular ramp so we're looking at a side view of Package coming down a conveyor conveyor moves at some constant velocity and Neglecting friction on the ramp. So it might have little rollers or something on its own. There's a point at which The box will lose contact with the ramp because it's got enough speed and then Has a free fall trajectory of its own so find where the box loses contact with the ramp pay to max if you will because after that It's no longer in contact with the ramp. So if you were designing this and you wanted some kind of special coating or Rollers along here. There's no there's a point beyond which there's no sense of putting any more in especially if they're Costs Maybe they cost too much. All right, we can put some numbers to this if we'd like two kilograms a speed one meter per second and the radius of the curve ramp Half a meter so the equations of motion we can set up from a free-body diagram Just somewhere along the track anywhere along the track. It's got some weight mg and Some normal force if it's still in contact with the track at some certain point though It will lose contact with the track. So let's get enough time. We'll set up the equations of motion Let's see Maybe a normal and tangential components. There's the tangential direction There's the normal direction Force of these some of the forces in the tangential direction now remember Once it gets on to this ramp without friction, it's going to be picking up speed It will not still have this velocity that it had back here. It's going to be picking up some speed We don't know what that is So some of the forces in the normal direction with that backwards I'm some acceleration in the tangential direction and do the same thing for the normal direction And it should have components of acceleration in both of those directions Because it's changing direction as it goes around the path It's on a circular path when it's still in contact and at some point It it will lose contact. So in the in the normal direction we've got That angle is the theta Remember we don't know what it is. We're looking for the maximum theta. So we set up the constant the Equations of motion in terms of theta and then we can see what that becomes when we're at the limit of contact So we have n minus mg cosine theta that's This component of n in the normal direction and the component of the weight in the normal direction What's that equal to now? We're not at the point where we've lost contact. So it still has contact with the track What's that equal to? I guess the choices are zero or Or Some unknown acceleration or something else we might know a little more of This is some place where it's still in contact with the track We'll then drive these equations to the limit and find out what theta max is Yeah, you're right you're right very sorry Switch there that look better. Yes, the tangential one is just simply mg Sine theta But what's that equal to well the tangential one the tangential acceleration? It's picking up speed in the tangential direction Because it's falling even though it's in contact with the track to a certain point. So all we can do with that is Is put it as that the tangential acceleration is unknown But it does show That the mass is not material at least for that equation Which is fewer ups and you're designing this that's good because boxes come in all kinds of masses You don't want to have to sort them by weight before they even go anywhere What's the top one equal to go? mv squared over row mv, but this v is Not the same as this one. It's the tangential velocity And it's picking up a little bit of speed in that tangential direction But while it's in contact it's still in contact with it's still going in a circular path So it must have centripetal force now the deal here is to solve these Well, the normal force Goes to zero because once the normal force is zero It's no longer in contact with the ramp and that's the point that we were looking for Joe is that a hand up? So Oh Yeah, because it's in the normal direction, so That fix it because the normal direction is is in Yeah, in fact there it is right there So n drops out of this then once again that mass disappears. So again, this is a Solution that's independent of the mass and We've only got a few minutes here So I'll give you the part you need to finish it because we have three unknowns right now The is unknown Beta is unknown and is unknown I guess technically But m cancels out so at is the other so we have three unknowns two equations those right there again, we'll be looking for the solution when The normal force is out of the equation So to finish it up that's a remember that the acceleration in the tangential direction as it moves Tangentially so I'll call that s as distance along the along the path the arc length if you will equals V dv remember this this Non-constant acceleration equation where the acceleration is changing with position Which is exactly what's happening here? so the You can then take oh, there's one other little part and that's that DS equals row Rowdy theta so if you put those two things together Put that into here You can separate variables and integrate and then put it into there so I guess if we Number on the right number. This is equation one This is two These two together when you combine those would be three just were substitute DS for row to theta Then you can put three into two integrate and Then that goes into one and you can solve then you can solve for the maximum theta Because this will give you the velocity of the velocity in there, and then you solve for theta That's it, but I'll give you the answer so you can Check it not lose any sleep over the weekend 42.7 degrees gotta be careful that Integration is a little bit tricky. Just make sure you don't lose any of your your any of the values remember row is constant V and AT are not Theta's not well theta is constant. I guess at the limit. That's what you're looking for and then the mass doesn't matter