the following content is provided under a Creative Commons license your support will help MIT open courseware continue to offer highquality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT open courseware at ocw.mit.edu the technical topic for for today we start I rushed right at the end a little bit about tangent and normal unit vectors and I'm going just to recap that quick and then we're going to go on really a review which of by review this is the sort of stuff that for the most part I'm sure you've seen in 801 physics and other physics you've had before and that's impulse linear momentum and impulse so that'll be a quick review and then the third subject is one that's much deeper and this is angular momentum and angular momentum with respect to moving points which you probably haven't encountered before so those are the those are the three topics for today let's get started so last time the piece that I rushed a bit is this notion of tangent and normal coordinates so I gave this had this example you're driving down the road you drunk or whatever and when you're at this point this point and this point I'd like to know what the accelerations are so I'll call these one two 3 and this curve Y is of the form some y of some f ofx and in this case it's a sinine KX and K a is what's known as wave number is 2 pi over the wavelength 2 pi over Lambda and the wavelength then is for example from here to here now the velocity I'm going pick a point here the velocity at any point we know is just the tangent to the path so this is the path and this is horizontal so you're driving driving down the road like this right that's what we're trying to do here the gravity is down into the board doesn't really come into into the problem so the velocity at any at any time the vector we could describe as a magnitude and a unit Vector we'll call UT which is the tangent unit vector and at any instant in time it's just aligned with the tangent to to the curve and its perpendicular uh partner is a normal unit Vector which it points inward on the curve and this would be unu n now we're interested in accelerations so we'll need to take a derivative of this but uh so let I was going to say put taking aside but I won't so the acceleration vector we have to take a derivative of this well it's a v do UT plus a v UT dot it's a unit vector and we've encountered this problem before because if this rotates the unit Vector has a nonzero derivative because of its rotation and the this derivative of the unit Vector is given by Theta dot u n hat now what's Theta dot well on this curve at any instant in time here's U can't draw arrows today here's your UT at any instant when you're traveling Along on a curve you are going around a circle of some radius we'll call row so there's some radius of curvature at any instant in time and that radius of as you drive along in little time delta T you go forward and amount row Theta dot is the velocity at which you're traveling tangent to the curve so this is an there's an angle in here so in here there's some Delta you advance some Delta Theta so r r Omega is a velocity right so row Theta dot is a velocity and that gives you this velocity magnitude B okay and we've gone through I'm not going to go through that little deriv that little argument before but the the change in direction of this unit Vector UT when you go forward a little bit is actually inward un n by an amount uh row Theta dot or by an amount Theta dot one times Theta dot so we've I'm not going to do that piece of the derivative before but here is this derivative of the unit Vector it's Theta do unu n so if we substitute those back in here we can get an expression for our acceleration this is just the acceleration along the path plus v Theta dot u n and then we need to take into account the fact that we know that V the magnitude of the Velocity the speed in other words is row Theta Dot and that means Theta Dot is V over row so if we plug that in up here then we get an A final expression for our acceleration V do UT plus v^2 over row u n and that v^2 over row this is a this is a centripetal acceleration term you're going around a curve there's an acceleration inward it's just like as we did from Polar coordinates from it comes from rotating things well as soon as you go around a curve you're going to generate an acceleration that is of the same kind but now we because you tend to know speed it's easier to express it this way when you're doing these tangent normal problems and this is just the acceleration along the path the usual hit the gas pedal and speed up that's this term okay so the the piece that I didn't have time to put up last time which is in the book as is this little derivative the derivation of that is one page in the one of the early chapters of the book how do you get row and row when you have y as a function of X there's just this formula from mathematics from calculus that says row is dy DX 1 + dydx 2ar to the 3es power all over the magnitude of d2y dx2 and you just calculate these quantities and plug them in so for y = a sin KX X then dydx is a k cosine KX and D2 Y dx^ 2 is minus a k 2 sin KX so we now have these two quantities we pick a value of x at which we want to know the answer like 0.1 so at 0.1 KX is pi over two right because s's a maximum X is Lambda over 4 KX is Pi / 2 sin KX is 1 cosine KX zero and so we can calculate row 1+ and dydx the derivative is cosine that's zero so this term in here is zero this is to the three halves that's pretty easy to calculate this term down here D2 Y dx^ 2 well s of KX is 1 so that's uh minus a^ 2 but this is an absolute value sign so this just turns out to be one over a k 2 and we plug in some numbers I'm going to let uh my Lambda be 150 m M my amplitude here that I'm swerving back and forth let's make that 5 m uh we need a that's all we need to get row so Row in this case works out to be 114 m so the radius of curvature of this road that I'm driving down at that that point right at the peak in the curve is 114 m so the acceleration that I'm looking for a well it's V Dot UT and if I'm not accelera I'm not hitting the gas pedal constant speed I'll let that one be zero and the other one then is v^2 over row in the unu direction and let's let V equal um I have here in my example 20 meters per second which is about 40 knots and a knot is 15% more than a mile per hour so it's somewhere about 45 miles an hour so a typical Road speed driving down the road 20 m/ second now you can plug in here you can plug in here and you can find that the acceleration I work out here is 3.51 m/s squar and in what direction is it what direction is that acceleration the normal it's in the normal direction is it to the inside of the curve or the outside of the curve inside always to the inside of the curve so these tangent this tangent normal coordinates are really simple coordinates they're they're meant for a particular kind of simple problem when you know the path and it's just defined the normal unu is positive always inward to the center of the curve pointing toward where the center of rotation where that radius of where your radius of curve is always pointing at the origin of that radius of your curve okay so G is order of 10 m/s squared so that's about a third of a g would you third of a g enough to notice yeah absolutely you get you get pushed to the side of the car if you do that okay all right so let's we'll any questions about that I'm going to move on next if not I'm going to move on to linear impulse and momentum and this will be a quick review because this is is consists of the physics the kind of physics that you've done lots of times so I'm going to hit it quickly and then move on to angular momentum so we know for a particle this is what Newton told us you've got a particle here that's some mass m we know for a part it has some external forces on it we know for a particle that the sum of these external forces a vector sum is equal to the mass times the acceleration for the particle Newton Second Law and that's the mass times the derivative of the Velocity with respect to time that's where we get acceleration so this formula we can just rearrange the DTs a little bit so the summation of the forces DT is M DV and I want to integrate these so if I integrate this over time from some T1 to T2 this equality says we're going to find some change in velocity from V1 to V2 and this is our this is kind of the beginning of our impulse momentum relationship we tend to call the integration of forces over time impulse and if you have a nonzero impulse that leads to a change in the linear momentum okay so when you carry out so so they're writing this a little bit different way you can do summation because these are vectors and we have rules about vectors you can bring the summation outside of the integral so that this says the summation of the integral from T1 to T2 of these external forces is just m V2 minus mv1 because the integral on the right hand side is really simple and we'll finally State this in a uh way that we most commonly use it moving this term to the left so you start off with an initial momentum mv1 you add to it the summation of the impulses that you that occur over time T1 to T2 of the external forces and you get the final momentum mv2 and this is the way you usually use the formula so what happens if there's no external forces that's that's where we get the law of conservation of momentum if there are no external forces on the particle the momentum doesn't change and mv1 is got to be equal to mv2 whoops so just a a really trivial example drawing for something that we've done before you get the block sliding down the hill draw your free body diagram got friction got gravity got a normal force and we'll set ourselves up with a coordinat system aligned with the motion we're interested in so this is an inertial system XY and we've done this problem before so we said the summation we found out that the summation of the forces in the X Direction and ah this is a good moment to take an aside for a second this was a vector expression but you can break it you can implement it you can break it down into its individ idual Vector components so that equation gives you for particles gives you three sub equations one in the X one in the Y one in the Z and you can use them each independently so in this case we're going to use the only need the X component in order to do the problem so the sum of the forces in the X Direction now and this problem we've we which we have done before is mg sin Theta that component of gravity pulling it down the hip minus mu mg cosine Theta and that's the friction and none of these are functions of time so that whole thing is just some constant I'll just call it K right so the total forces on this system in the X direction is just some constant and that makes this integral up here pretty trivial to uh Implement so now we can say that m V1 I'll just make it v1x to emphasize that integral from zero or T1 to T2 but I'm going to let T1 be zero usually makes it problems easier so from Zer to t 0 to T of kdt equals mv2 I can't write this morning in the X Direction and if V1 X is zero starts off at zero time and zero velocity when you let it go to start with then you this then this term will go away and we can just implement this integral and the integral of kdt is KT right so this says then that KT evaluated from zero to some time that we want to know the answer is m V2 in the X Direction and so let's let T = 3 seconds you find out that v2x is 3 K Over M right and in this problem then that looks like 3 G sin Theta minus mu cosine Theta so this really a trivial almost trivial example it's not a hard example but it emphasizes all of the key points in the problem start off with a vector equation you can apply it in any one of the three Vector d three Vector component directions you integrate the forces the sum of the forces on the object in that direction Direction over time and you apply the impulse momentum formula and you get the answer you can do that so that basic step-by-step process is how you basically most impulse momentum problems are done and if there's no forces then you have conservation of momentum okay so let's do uh there's this was for particle just a quick reminder of we need to ask ourselves does this apply to groups of particles systems of particles well remember that we said if you've got a bunch of particles a system all of these are the M eyes we've already figured out that the total mass of the system times the velocity of the center of mass this is G is my center of mass with respect to my inertial frame so this is the momentum of the system that was just the summation of the individual Mi VI I with respect to O and further more we took the derivative of this so this is the momentum of the system the time derivative of the momentum should be the external forces so that we were able to by taking that derivative you get Mt the total masstimes v g with respect to O DOT and that had better give you be equal to the summation of the external forces on all the particles because you can do it one particle at a time and they all add each each of these particles has forces you sum them all up you get these forces and this is this allows you to say that the sum of all of the external forces on the system is equal to the total mass of the system times the acceleration of the center of mass so this formula now can it looks very similar to where we started with this little derivation for a particle you can now say that the summation this all leads to the summation of the integral from T1 to T2 of all of these forces over time gives you the change in the total linear momentum of the system system v g with respect to o1 two rather minus VG with respect to o 2 the second one so this is exactly the same kind of formulation the change in the linear momentum of the system is equal to the integral of the forces on the system over time so the impulse to the system so it all this allows you to do problems that you might not have thought about before so I've think about uh something like this I've got an old uh back re you know Revolutionary War earlier period Canon I can't I'm trying to draw so here's my Cannon barrel and back in those days draw a little better job here so we got this cannon sitting here and in the barrel you've got a bunch of what's known as in the old days known as grape shot so sometimes for kind for anti Personnel kind of stuff they would just throw a whole bunch of metal and junk into it but just imagine a bunch of iron balls or lead balls or rocks or whatever you want to put in there and you put a whole mess of them in here and you have a charge you set it off and it shoots them out the end of the gun I want to know what's the reaction what's the reaction force on the gun now how much what's when that gun when that charge goes off if you ever if you go down to the Constitution which is the oldest commissioned warship in the US Navy still afloat it's that I don't know if any of you have seen it down here in the dock in Chelsea was built around 1799 but when one of those when they shot one of those guns the gun would roll back several feet and get dragged to a stop by a bunch of restraining lines and then dragged back up for another for another load so the reaction force the force pushing back these things are used to sitting on Wheels could be pretty large let let's take an estimate of it so this is going to be a relatively small gun the uh mass of the total mass of the shot here time G is 10 lb so 10 lbs a shot a gallon of milk weighs eight pounds we're not this isn't a very big uh very big load and let's say that this Barrel here from here to here the acceleration length is 7 feet so that shots accelerated out the barrel over a distance of 7t and it has a muzzle velocity an exit velocity of 700 feet per second are most guns subsonic supersonic projectiles when they come out of guns thought about that I'm just asking seeing if you have a feeling for Speed how close to the speed of sound is that so speed of sound is about 1100 ft per second 340 m/ second so this is order of mack6 or something like that something like 6 not 06 0.6 so about 60% of the speed is down that's kind of slow actually as guns go so it the uh initial velocity the shots just sitting there is zero the final velocity is 700 feet per second the average velocity which I'm going to need for a second because I'm just making some estimates here the average velocity average of 0 + 700 is 350 ft per second to do the reason I need this is I need to estimate the delta T how long does it take to get the shot out of the barrel so distance equals rate times time right so the uh uh V average time delta T equals 7 ft at 350 ft per second you find out delta T is about 02 seconds so it gets out the barrel pretty quickly 0.02 seconds 20 milliseconds now that powder when it goes off it's putting a lot of forces on the shot so the total force on the shots 5,000 PBS what's the reaction force on the cannon so we're going to going to apply a principle here that we talked about earlier Newton Newton had three laws this is a group this is this massive balls in there the hot exploding gases are pushing them out the barrel what must be the push of the same gas on the cannon that's containing it equal opposite equal and opposite Newton's third law right so the reaction the reaction force that we're looking for is minus the force that it takes to get the shot out the barrel okay so force on the balls on the shot minus the reaction force and that's Newton's third law okay so we're almost there but just applying now this concept of impulse and momentum so now we can say the integral and this is the summation you got a whole bunch of balls in there they're all got forces on them but we're treating the group of balls as a system so we can we can use this notion of the ma total mass times velocity of the center of gravity to do this problem so this is the force on all of these little balls integrated over time it's going to be the mass total of the balls times the change in the velocity of the center of gravity the final velocity minus the initial velocity this is zero okay we know this we know this so This Mt is the 10 lbs over G we know this is 700 feet per second we know that all of this mounts up to the force external on the group delta T and this is the weight of the ball W over G and this is 700 feet per second that's 10 PBS divided by gravity 700 ft per second and if you work out those numbers you get 10,870 lb now is this is an average force right because we at delta T is the length of time it took assuming we had some one we assumed an average velocity going down the barrel so the peak force is probably higher than this and the pressure of the barrel probably isn't constant this actually isn't a bad estimate if you know the muzzle velocity which you could figure out probably from the distance it goes and things like that you can make this estimate so 10,000 pounds of reaction force is quite a lot this is this is a dinky gun 10 this is 10 lbs of shot is not much so that's this is this is the force to get the balls out the barrel if it's positive what's the reaction force minus 10,870 and all through this I've been doing this in a single Vector Direction so I would have set up my coordinate system like here's o this is the X Direction y direction so this is obviously in the positive X direction is what I've lined up my positive velocities and positive forces to be so just a really simple way one of the kind of ways in which you use this notion of impulse and momentum but you can use it to make estimates too okay and if you ever fired a shotgun not yeah any kick right so what gauge shotgun did you fire you know know okay okay 12 gaug shotgun you know the the the cartridge is about 3/4 of an inch diameter and it probably has a little amount of powder in there it amounts up to about that much stuff but it can give you a bruise in the shoulder if you don't hold it right so that's what that's what this is about all right so this kind of a quick review uh you've done lots of conservational momentum problems in your time the homework set this this time has two or three problems on it that are conservation or momentum linear momentum but now I want to move on to talking about angular momentum and angular momentum in a way in which you probably haven't done angular momentum problems before so anything about La last thing on this so mostly this I want you to read the read the chapter I think it's chapter 15 first few sections of it are on linear momentum the last few sections on angular momentum go read them and just work the problems that's where they that's where you'll get most of your refresher is work is doing the practice problems okay angular momentum so we'll start with particle we're very rapidly going to get to rigid bodies now Mo many times you've done angular momentum problems before mostly rotations about fixed axes so here's our inertial frame fixed XY and you have a particle out here and and it just has some mass m and it has some total force on it the force and this is the particle is located at B here so the force at B total Force at B just some Vector it also is traveling with some velocity at that instant in time so it has momentum okay and the momentum is p of this particle at B with respect to 2 O and I'd like to the standard expression then for the angular momentum of this particle at B with respect to O we have a position Vector here RB which we've used lots of times by now is just the cross product of the position Vector with the linear momentum and that's just that's the definition of angular momentum I'm using a lowercase H here because I'm going to do that to indicate single particles and I use a capital H later on when we're referring to rigid bodies so the angular momentum of this particle with respect to this point is given by that and you've where you've used this before then is the derivative of HB with respect to O DT is what is that and remember what does this give you the time rate of change of the angular momentum is the torque right and it's the torque it's a sum of the torqus applied on that object with respect to the coordinate system in which you're Computing the angular momentum so that's you've used this formula many times and in planer motion where you only have one axis of rotation usually the z-axis of rotation you usually usually would write this as I Theta double dot right in the simplest form if it's a rigid body it has the mass moment of inertia times Theta double dot is equal to the sum of the external torque so that's where you've met this before but for the moment this is just a particle let's stick with a particle so the piece that's new here probably new for you is that um what if I want to know the angular momentum with respect to another point so here's a point a and I'd like to compute h of B with respect to a well that's r b with respect to a cross p b with respect to O and this is really easy to forget the momentum is always calculated with respect to your inertial frame and that's why I keeping this with respect to and telling you what the frame name is is pretty important but we're out here at some arbitrary Point Computing the C the cross product of the position Vector from this arbitrary point to this moving mass and we're defining this is just a definition defining the angular momentum with respect to this point a as RBA cross p p with respect to O all right and what I want to get to is now is the torque on this system on this part around this particle B with respect to a is the time rate of change of HBA with respect to T but now it's a little more complicated plus the velocity of a with respect to O all vectors cross p b with respect to O messy term I was trying to think of an example where you uh you might want to do this so imagine that you've got an arm which can rotate you know might be on a robot or something like that and it has attached to it another arm with a mass on it and you've got a motor here which can make this rotate and you're trying to design the motor has has to be able to put out a certain amount of torque so this is O this is B this is a and you actually I want to know the torque required in this motor to drive this thing around but the motor is here and it only cares about what it feels so the torque at this point to drive this thing but this whole system is now in motion you'd have to use this formula okay so there are practical times when you'd like to be able to be able to calculate something like this so there's a I'm going to show you the a very brief derivation of this and but just so you get a feeling for where this comes from because there's a couple of outcomes that are very important to us so the sum of the forces there at those forces at B give you the time rate of change of the momentum at B with respect to O this is that moment momentum Vector of our particle okay and this FB this is a total external forces acting on the particle we know that's m a single particle times the velocity of uh that certainly could be right times that's our familiar formula for fals ma right so the time derivative of the linear momentum gives you this and the torque of B with respect to a is r b a cross I'm this total I'm going let this just be a total Vector I'll call it f B it's a vector the torque with respect to a is just RBA cross this total external Force okay and that's R ba cross the time derivative of PBO because the forces give us R which is from the time rate of change of the linear momentum of that particle so I can say it like this but now this there's just a little Vector identity for products of vectors that I'm going to take advantage of and I'll call this Q so Q so Q is of the form and I'm going to say m this is a Quant a vector a and this is a quantity a vector time derivative of a vector B it's a cross dbdt so there's a little identity that you can use it says a cross DB DT you can you can alternatively write that as the time derivative of A cross B minus time derivative of A cross B and if we're we're going to take advantage of this and just re construct this formula using this expression so that says torque with of B with respect to a the time derivative of RBA cross PBO minus the derivative of RBA cross PBF so we've just made this substitution down here in terms of RBA and PBO all vectors but we know that RBA from all the previous work we've done is just rbo minus r AO so the derivative of RBA with respect to time is VB minus V AO all right we're almost there we're almost there I'm going to need another board though this quantity here this is just h b a this is the angular momentum of the particle with respect to a it's just R cross B if you recall so then we can rewrite this expression for the torque as the time derivative of H of B with respect to a that's because of what I pointed out there and now this is minus VB o minus B AO cross P Bo o so P of B with respect to O is just m velocity of B with respect to O right so vbo cross PBO gives you what nothing get you zero because they're parallel to going they're exact they're in the same direction so this you only get a nonzero piece out of this minus time minus gives you a plus and you end up with and that's what we set out to find I said this is where I was trying to get and now we're there now the import importantly there's a couple of special cases of this this can be kind of a nuisance term to have to deal with lots of times you'd like to be able to get rid of it and just be able to go back to that Old Reliable formula torque is time rate of change of angular momentum so there are two obvious almost maybe obvious conditions in which this will go away what's the most obvious one of terms Zer something what did you say is zero one of the terms well yeah this this this term presumably not this guy if this is zero is just back to our old familiar formula that's one case the case one but now actually this this is a really important result because a can be anywhere as long as it's not moving and so this allows you to do things talk about rotations about fixed axes that aren't at the center of mass right so if you have a fixed axis of rotation and something going around it that's what this allows you to do okay that's one case so this is typ this we'll soon get the rigid bodies rigid bodies obey exactly the same formulas and you can have a rigid body now that it's not rotating about its Center but rotating maybe about its end like this that formul applies if the velocity of that axis about which it's rotating about which you're Computing this is not moving then you can just use you don't have to deal with that messy term so this is one case in which this term goes away the other case is if this velocity is parallel to the direction of the momentum and there's a really useful time that that happens so also this formula is true case two is when vao is parallel to PBO the direction they're going in the same direction now that happens when a is guaranteed to be true if a is at the center of mass because momentum is defined as the mass of the object object times the velocity of its Center of mass right even for rigid bodies so this is true when a is and I'll call it g at the center of mass so this gives us another really important generalization that we'll use that we make great use of in Dynamics and that says that the torque with respect to the center of mass is time rate of change of H with respect to G now I'm not going to go I did the proof this went through this little proof just for a particle but by summing a bunch of particles and kind of going through all the summations as we did to prove the center of mass formula you can show this allows you to very quickly show that this formulation is also true for rigid bodies so your way you say it for rigid bodies is that the sum of the torqus about with respect to some point for rigid body is H now Capital H H dot with respect to a plus the velocity of a with respect to O cross p and now I'm going to say G it's Center of Mass center of gravity with respect to O So the same statement for a rigid body is it's a torque with respect to some point a which can be moving now even accelerating is the time rate of change of the angular momentum with respect to a of that rigid body plus vao cross PG okay and again when would this te what would this messy second term go to zero when the velocity of a is zero fixed axis rotation or when this is parallel to that which is true for the center of mass always so the same same two special cases apply for rigid bodies when velocity of a with respect to O equals zero or when the velocity of a with respect to O is parallel to the uh p and this the the most important case of that is all always [Music] true that's always the case and in these when you can say this then the torque with respect to a is d h respect to a DT no second terms and it's those kinds of this you've applied generally in your physics like in 801 you've used formulas like this a lot done problems either with respect to the center of mass so objects run doing things like that you'll do the torque formulas with respect to the center of mass or when you have things that are pinned to points and rotate about fixed axes then you use the other formulation where it's with respect to a [Music] non-moving point so this is fixed axis rotation okay all right so that's kind of the dry derivation part of it I'm going to see if I can find an example [Applause] here okay this is quite a bit like a number of the homework problems so I've got a carnival ride got a bar and it's got a seat out here you're riding in it right so you're taking this ride it's rotating this is some fixed axis here and they you have a fixed coordinate system your o XYZ and then you'd probably have some rotating coordinate system with the point a fixed here at the axis of rotation and this is going the what's what's unusual about this ride not only can go around and round but the arm can go in and out so it might take a path inwards like if you pull the arm in as you're going around what are the forces that you feel in the ride okay so this is my point B out here it's where the person's at have some mass m and let's let for now first case Theta dot let that be constant a constant angular rate here here's Theta and r dot all right now you already know quite a bit about things like this what forces if you're riding in that bucket what forces do you think you would feel or I should say it in more carefully there will be accelerations that you feel in that bucket you'll feel like forces on you but what are the accelerations that you will feel you expect to be if you're riding in it I hear one here Cal acceleration so there'll be centripetal acceleration right everybody agree with that how anything else acceler it's going to be some coris because our DOT is not zero right it's changing in position and when the length of the arm of something rotating at constant speed changes what momentum changes angular for sure how about linear momentum yeah it's changing too because R cross p is angular momentum so both linear and angular momentum are changing as this radius gets longer or shorter and if the if that angular momentum changes it takes torque to drive it and so we ought to be able to use the formulas that we've just derived to calculate something about the torqus required to make this happen and the forces on the rider okay so we'll treat this one as a particle so H the angular momentum of that of the rider out here at B with respect to O it's going to be rvo cross p and in this problem I'll use polar coordinates pretty easy this is a planer motion problem it's confined to the XY plane and rotation in Z so polar coordinates are pretty convenient so this should look like radius r r Hat Cross and the the linear momentum of this is the masstimes r dot R hat that's the extension rate but it also has velocity in the Theta hat Direction R Theta hat so this is p with respect to o m v and this is the rad this is the r crossed into it now r r Hat Cross R hat gives you zero so you only get a single term out of this and you get an R cross R Hat Cross Theta hat positive K so this looks like plus m r 2 Theta dot k hat that's my H Bo o now I would that's my first piece that I wanted to get that's the a part that's find the angular momentum so this is a B I want to know the torque okay well which formula can we use are we allowed to use do we have to account for that second term no why yeah the velocity of the point about which we're Computing the angular momentum and therefore the torque is not moving so you only have to deal with the first term so the torque required to move that particle at B with respect to O is just the time rate of change d by DT of h b with respect to O and that's d by DT of m r 2 Theta dot k hat so what are the constants in this expression so we don't have to worry about their derivatives K hat change length no change direction no Theta dot does it change change in this problem no we fixed it we said it arbitrarily started off saying that we'll just let Theta dot be constant R though is changing so the time derivative of this particular one only we have to only have to deal with the r actually I'll I'll forget that for a second because I want to get both terms and then we'll let it be zero so this time derivative then when you work it out all right the derivative of the r term gives you 2 m r r do theta dot k hat and I'll just go ahead and forget for a minute that I said that's constant let's get the other term that might be there that term gives us uh m r 2 Theta double dot k hat this term comes from what we'll we'll find is the cholis one and this is from what we'd call the oil Arian acceleration this torque must the torque get this up a little higher we should be able to write as some R cross F right and so the r cross F terms this will be from the coris force and this would be from that or oian Force so this will look like r r hat crossed I'm just factoring this back out into its cross products RR Hat Cross 2 m r dot Theta dot in the Theta hat Direction plus r Theta double dot in the Theta hat Direction so there's two terms in this torque expression they come from R cross two Force terms the first Force term is what we know to be the cholis force and the second Force term I'm missing an M here RM Theta double do theta hat that's the force that takes to if the thing were're accelerating just to accelerate that ball Theta double dot takes that takes a force you do these cross products R cross Theta hat gives you k r cross Theta hat gives you K it all comes out in the right directions and now to do the problem we had we said oh yeah but this one's zero so we'll let this term go to zero we're left with a single term open and if we plugged in some numbers let's just see if how this works out and this we know it's just R cross the coris well let's let M you let M be 100 kilogram and R be 5 M and R dot 0.4 m/ second all kind of perfectly reasonable dimensions and Theta dot equals 3 radians per second so 2 pi radians per second means it goes around once once a second right say a little less than half a REV per second yeah probably yeah good catch got to have an r in it because that R comes from here right so I had the coris force written down but not the r you have to multiply it by so now we got if you plug in all of these numbers let's see if it see if it's ride you could survive I actually computed the Coriolis Force first it's just a I think it's just interesting to get a physical feeling for how whether or not you can feel these forces so the Coriolis force is just everything but the r 2m r. Theta dot so if you calculate that you get 240 Newtons and R times that the torque five times that about uh 1,200 Newton M and the acceleration how do we gets acceleration well F core Corola force is some mass times an acceleration so we can solve from that 240 Newtons the ma the acceleration of the system acceleration of a b with respect to O 240 Newtons divided 100 kilogram 2.4 m/ second s square and what's that in G's yeah about quarter of a g and again would you feel that so now you're riding so now you're riding in the bucket right let's say it's just spinning around the arm's not going in and out at all so you're riding in it what what force would you feel would you feel any forces pushing you into your seat constant speed r dot zero somebody out there would you feel any Force if you're going around and around in this thing sitting in the seat okay what's it come from centripetal acceleration right all right so and that is in what direction the acceleration is inward what would you actually feel you you'd feel if this but if this seat could swing out so you're facing in you'd feel like you're being thrown out in your seat right because you have to have a force on you to make that acceleration happen the acceleration is inward the force it better be pushing you inward so it's pushing on your back if you're looking in going around so that's the you know the uh forgotten the term for but you know they put the astronauts in a centrifuge right spin them around to see if they can take High G's okay that's the that's the G the inward high G acceleration due to the centripetal acceleration make you going in a circle but now not only are you going to feel that but now you start changing the length of the arm and if you change and if you change our DOT and make it positive so the arm is getting longer and we made this4 m/ second so it's moving out you know about like that it's not real fast but it's going to create a quarter of a g acceleration on you and and if R Dot's positive which direction is that acceleration I was careful we walked all the way through this the acceleration is positive but in what direction for coris Theta hat positive Theta hat so it's in direction of increasing Theta so this one is perpendicular to the arm right and this is this is the uh you know when my pingpong ball in here the force that actually causes it to speed up is partly cholis and partly orer and if I can make this go at constant speed the force that drives that the force that actually speeds that up is it going out is the coris force the normal one so if you were on this ride you'd be feeling about a quarter of a g perpendicular to the arm you'd be feeling another Force inward that's the Cent due to caused by the centripetal acceleration so you'd be feeling both of them how big is the centripetal acceleration sort of like uh R Omega squ right yeah R is 5 Omega is 3 3^ 2 is 9 * 5 9 * 5 45 divided by 10 4 and a half gez you wouldn't notice the cor you wouldn't notice the cholis very much it'd be kind of this' be kind of a tough tough ride on the uh on that side of things Okay so we've now we've done that uh you've seen a simple application pretty straightforward application of using time derivative of angular momentum to calculate torqus so homework again there's a couple problems you know very similar to the one I just did things going around Circus rides that kind of stuff so so have a good weekend see you on Tuesday