this is lecture 6b and now we're coming back to the idea of acceleration in we talked last time about tangential or linear acceleration and angular acceleration and how they're related and now we're going to focus on motion that is accelerated in both of these ways and so what we have here are the same two pictures we had on the previous page and we have here on the left in this table we have these tangential or linear which we often also might call rotation or translational quantities and here on the right we have angular or rotational quantities but these equations on the left should look pretty familiar to you they are for example if we go back to chapter two if you replace this with x instead of s right here x equals v naught t plus one half a t squared x equals v naught plus v over 2 t and so on these four equations are those equations for constant or average tangential or linear acceleration and now we're going to take those and just think about what's going on in this motion so again here we have at an initial moment of velocity v naught which would be the tangential velocity at a point on the rim and some final velocity up here where maybe it's going faster v and so there must be some tangential acceleration in the direction of that velocity so it's making it speed up and there's this arc length that goes through during this motion s and there's an amount of time t and those equations uh those these equations relate these quantities to one another in the same way they do in this picture where we would have a velocity v naught and a final velocity v where it's going faster and some displacement s which would be identical to this arc length here and some acceleration a if it's gaining speed and then there's some time that passes and so these five quantities here also the reason these are applicable in both situations is because this rolling the rolling motion includes spinning with the translation but it spins in such a way that this arc length is equal to the displacement because it rolls without slipping and so s here on the arc length is the same as s here and that's why we can say these equations are the same as these equations because this same would be true of the tangential velocity relative to the center here is equal to the velocity at whichever moment we're looking at and the tangential acceleration relative to the center is going to be the same as the translational acceleration so that's why we can say the equations are the same for this motion and this motion all right and then we add to that so we have all of these four equations and then we add to that the fact that we're we have motions not only in these translational and linear quantities or tangential quantities but we also have angular or rotational quantities we have [Music] angular acceleration if it's gaining speed rotationally and so we talked about that previously but we have omega naught which now omega naught is related to v naught by this equation right here and there's a final omega that we're going to write two separate equations so v naught is related to omega naught and v is related to omega at this moment or at this moment and that comes from just these equations right here from the same picture that vehicles are omega these were average or constant velocities but this equation v equals or omega is going to be true at any moment as well and the arc length is equal to r theta and the acceleration is equal to r alpha where alpha is the angular acceleration so let's go back down there and that's where we get these four equations now these might look similar to the ones on the left and i'll show you where we get them how we get them these come from these four but i'll go through those with you in a minute but let's finish just writing out what's all happening here so there's some angular acceleration alpha occurring for this whole time and there's a delta theta or a theta that takes place and this time is also this time is not um attached to either translational or rotational it's just time and so that time could be either it goes with either set of quantities and same idea here we could have an initial angular velocity here and a final angular velocity here that is related to the fact that it's angularly accelerating and these quantities are related v naught and omega naught are related to each other right here v and omega are related to each other with this equation v equals r times omega a is related to alpha by this equation down here and the same in this case this a tangential acceleration is related to this alpha same equation and then the last one is that s this arc length is equal to r times theta that's this one up here it's the top equation here and that's also true here this theta that it has rolled through is related to that arc length s which is the same as that distance s and so we can relate all of these quantities with these equations here in the middle that relate the angular quantities over here on the right to the tangential or linear quantities here on the left i wanted to show you where one of these comes from let's just take a look at this one here this first equation that looks pretty familiar except it has an s instead of an x and so let's just look over here at this one s equals v naught t plus one half a t squared so that would apply to this tangential motion or this translational motion and now we know though that these equations relate all the quantities that are translational only so we know that s equals r theta and we know that v naught is equal to r omega naught and we know that alpha is equal to or sorry a is equal to r alpha so i'm just using these substitutions right here we didn't use this one v equals r omega just because v is not in this equation only v naught all right and then if we just complete that equation then we have s equals v naught t plus one half a t squared but you notice that there's an r in every term and that r gets cancelled and we're left with this orange equation theta equals omega naught t plus one half alpha t squared and so it looks just like this purple equation except we've replaced translational displacement with rotational displacement or we've replaced linear displacement or tangential displacement with the angular displacement and same we've replaced the velocity with the angular velocity and the acceleration with angular acceleration and that's how we get this equation for theta from this equation for f and this equation gives us this equation with a few substitutions and the same thing is true for this equation uh s equals v naught plus v over 2 multiplied by t if we do few substitutions into this equation we get this equation over here on the right for theta and the same idea for these two bottom equations and they should look really similar to you to the equations over here on the left all right so with that let's just take a look at an example problem c number um c number seven and the calculations are simple but i do want to just show you at least one one problem here where we use these ideas this is on page seven the beginning of 6b so this is a typical music cd which not too many people use these anymore but it spins for about 74 minutes from beginning to end that's if the whole cd is used all this all the space on it is used so 74 minutes i'm to just start labeling these things that is some sort of an amount of time and at the beginning it says the cd rotates at an angular speed of 480 revolutions per minute that's angular speed or angular velocity would this would be the magnitude of the angular velocity so angular speed of 480 revolutions per minute or rpm that's at the beginning so it's implying that would be omega not angular speed or angular velocity of 480 revolutions per minute the cd rotates at the end the cd rotates at 210 rpm so that is omega final and we want to know what is the cd's average angular acceleration in radians per second squared so what we want to know is alpha and so you can see here we just have a few quantities and we just need to relate them to one another we have omega naught alpha omega and t and if you wanted to even sketch like a circle and you could say okay let's say that's kind of a flat looking circle but it'll do so it says starts here at some initial moment and it might go around and around and around and around many many many times because it's a 74 minute cd and it does this many revolutions per minute so it's going around and around a lot but let's just say at one moment it has some initial omega knot and then after it's gone around and around and around at some final moment here it's going slower some final omega which is less and so that's the final omega and so we know there's an angular acceleration that's slowing it down which means i'm expecting the angular acceleration to come out negative because it's opposite the direction of rotation if it's going to slow it down very similar to how if we have a velocity that's uh that's one direction and it's slowing down the acceleration is going to be the opposite direction and then we have a time that's given here now let's just look at the equation we note that relates these four quantities here if we go back here right here is an equation right here is an equation that relates those four quantities the one unknown is alpha so we don't even need theta we could solve for theta we could actually figure out how many revolutions it went through or radians it went through but all we need is the time alpha omega and omega naught and we can relate them using this equation it looks very similar to the equation v equals v naught plus 80 but we're solving this for alpha and so alpha equals omega minus omega naught over t and here's a question do we need to convert units here and the answer is we want this alpha in radians per second squared so we could convert units either at the beginning of the problem or at the end of the problem i'm going to do that at the end because i already have everything written out here so i'm just going to go ahead and put these numbers in and just show you what we get final omega is 210 don't get these mixed up the order and rpm is revolutions per minute so i'm going to write it this way revolutions per minute minus that's omega minus omega naught is 480 revolutions per minute and i'm going to divide by t which is 74 minutes now don't get too excited to cancel those minutes because they don't cancel they multiply because it's like multiplying by one over minutes this dividing by a minute and so this ends up being revolutions per minute squared and the number is just 210 minus 480 take that and divide by 74 and so what you get is negative approximately 3.65 revolutions per minute squared and again if you're having trouble seeing that why the minutes don't cancel just pull this what 74 minutes over here and write it as 74 uh 1 over 74 minutes quantity 74 minutes and then the minutes you see multiply in the denominator now we're going to convert revolutions to radians and so we want to get rid of revolutions and convert to radians and we know that for every one revolution there are two pi radians and so that will do the revolution unit conversion and then we want to convert minutes squared to second squared so we're going to do this minute up here seconds down here to cancel the minutes and then we're going to square the whole conversion factor and we know that in one minute there are 60 seconds and we're going to square it and so the the revolutions cancel with the revolutions and the minutes squared cancel with the minutes squared and we're left with seconds squared and so we're going to multiply by 2 pi that number that we got there 3.65 times 2 pi divided by 1 over 60 squared the 1 gets squared the 60 gets squared the seconds get squared everything in here gets squared and so if you do that carefully um you end up with alpha oh and i forgot a negative sign there's a negative there i might have said that but i didn't write it anyway this comes out to be negative and i'm going to round this to two sig figs six point four times ten to the negative three radians per second squared and that is our angular acceleration and the negative just means it's opposite omega it's in the opposite direction of omega and so we're losing speed all right and so that's number seven let's look at one more problem number 10 a car has tires of radius 0.3 meters and it initially moves at 20 meters per second along a straight road then the car speeds up with an acceleration of 1.5 meters per second squared for 0.48 seconds determine an angle through which the tire on the car rotates during this interval so let's just uh draw a little picture here all right so that's a road and then let's draw a little let's borrow another picture here there's a tire rolling down the road okay and there it is a little later so this is just one tire of the car um and it says that the car has tires of radius 0.3 meters so that radius is given right here that's r and that would be just that radius all right and then we have some translational quantities initially it moves at 20 meters per second along the road along a straight road so that would be its initial velocity so we'll say initially it's moving at v naught of 20 meters per second and then it speeds up with an acceleration of 1.5 meters per second squared for 8 seconds and so it has an acceleration [Music] a of 1.5 meters per second squared for a time of 8 seconds and we know that there are a couple other quantities that we could write down here like the final v we know it's going to be bigger we don't have a number but i'm going to just write it here we don't know it and the other thing we don't know is the displacement s we don't know that but we could find both of those if we want but let's before we find anything let's just see what else we have here we want to know what's the angle through which the tire rotates so imagine here that this tire we know it's going to rotate many times but let's just suppose that it rotates you know through several revolutions and it ends up here and so when it started here it's going to rotate through several revolutions and it's going to end up going through and we could even write it like this if we wanted like that so that's our theta or you could just draw it going there once so that's that's what we want to know what's the angle theta through which the tire rotates so there's there's a theta there's also an initial velocity an initial angular velocity omega naught there's a final angular velocity omega we don't know what it is but we know that it does have a final angular velocity and we know there's some angular acceleration alpha that's making it speed up we don't know that theta we don't know and none of these we actually know but we do know the time and so these are the same quantities then i'll just write time equals 8 seconds it's the same time but you notice for every linear translational quantity v naught i have an angular quantity omega naught v i have omega s i have theta because that arc length s would be the same arc length that the wheel rolls through that would be that same distance s here and so we know relationships as well we know that s is equal to r theta we know that v naught equals r omega naught we know that v equals r omega and we know that a equals r alpha those are the equations relating all the linear translational quantities to the rotational quantities through the radius and then we have some equations relating [Music] these quantities to one another so just for example if we take a look at this equation i want theta and this is the one that has it so i may not use these i might but i might i might not i want theta and let's just see what we have here theta is unknown and s is unknown but r is known so if i could get s i could get theta and we'll we'll do that but let me show you here v naught we know r we don't we do know but omega naught we don't but look we have an equation right here with only one unknown we could find omega naught v we don't know and omega we don't know um a we know are we know alpha we don't know but you might see here these there are a couple of equations we could get omega naught and alpha immediately from these two and then we have two other equations i'm going to write down one is this one i can use this equation s equals v naught t plus one half a t squared that's using all of these linear or translational quantities here but we could also write a very similar equation for the rotational quantities theta equals omega naught t plus one half alpha t squared and that equation is very similar to this equation and now let's just look at the unknowns in this equation both of these equations s is unknown but we know everything else we have v not t a and t so we could find s so there's a solution right there we can find s here substitute it into s equals r theta and then we can solve for theta so that's one solution but there's another one if you maybe saw it this way instead theta is unknown and omega naught is unknown and alpha is unknown but you might notice here oh i could solve for omega naught using this equation v naught equals r omega naught i could find omega naught and i could find alpha a equals r alpha so i could find both of those and substitute them into this equation and get theta that way so we actually have two possible ways to solve this problem and that's often the case in these problems where we could start with one set of equation or the other and we can use these green equations to relate the quantities all right so i'm going to let you guys do the math so you do the math and what you should end up with is theta is equal to v naught t plus one half a t squared that over r and that over r and it doesn't matter which way you do it you could do it the first way i showed you or the second way but play around with those and do that and then if you solve this as long as everything is in the right units and let's just check these are in meters per second and meters per second squared in seconds and the radius is in meters that's 0.3 meters and so theta should come out in radians and you can just double check that but this comes out in to be 693 radians so again here i'm showing you this problem not to just get the answer but i want to show you how to approach problems like this knowing that you have a set of quantities for angular motion and a set of equations for that angular motion and a set of quantities for in this case translational motion or it could be tangential motion and a set of equations that go with those quantities and we have a set of equations the green ones here that relate the two types of motion the angular to the linear all right so that's it for number ten okay so now we have um number seven and i'm going to add to the list number 10 as an example of motion with constant or average tangential or linear acceleration and angular acceleration in the same problem so we have two examples of that uniform circular motion is our next topic a couple of terms one is period of rotation and that would be the amount of time in seconds that it takes to go through one revolution so this would be seconds or seconds per revolution so that's what we call the period or the period of rotation and then the tangential speed would be v sub t and uh you could just leave the t off we've been talking about it like in this context as just v without a subscript t but here i'm putting the subscript just to emphasize that it's tangential so here we have a tangential speed and it's two pi r divided by the period if it's uniform or meaning if we have a constant tangential speed so when we say uniform we mean that that tangential speed is constant that's what we mean when we say uniform and so it means it's going around at a constant speed although the velocity is not constant and so um it's a this is the magnitude of the tangential velocity vector so at every point like at this point in this circular motion this velocity vector is going up on the paper or on the sheet here on the screen here and then over at this point it would be going to the right and over here would be going down and over here be going to the left and so on and so that vector is constantly changing direction which means the velocity vector is not constant and whenever the velocity vector is not constant we know we have acceleration i want to show you what i mean by that so here this is a demonstration that i showed you back in chapter two again looking a little too serious here um but you might remember this example this illustration and what this is meaning to show you is if we go back here when i'm speeding up i'm speeding up to the right this ball lags behind because of its inertia and so you might remember that shows us that it's accelerating the velocity is changing if the velocity is constant it would just hang straight down like you probably noticed this in your car if you have something hanging from your rear view mirror if you're just driving along at a constant speed it's just gonna or a constant velocity it's just going to hang there but if you speed up or slow down it's going to deflect one way or the other and so that's kind of what's happening here is when i'm speeding up it's kind of lagging behind and now when i slow down it keeps it tries to keep going forward so it deflects the opposite way and this ball inside this container of water it's actually suspended up because it's it's floating inside the water there if we continue and watch that when i first start speeding it up notice how the ball deflects forward right when i first start speeding it up let's go back so this is at rest and then now it's speeding up the ball deflects forward and that tells us there's acceleration whenever the velocity changes there's acceleration right and so that is speeding up and then it's kind of bobbling around it's kind of going like this just from the water sloshing around inside but when i bring it to rest right there at the end notice how as i slow it down and bring it to rest the ball deflects back the other way and that means the acceleration is back in the direction opposite the velocity when it's slowing down and so up to now we've talked about acceleration as speeding up or slowing down but here what we're talking about is not either of those now in this chapter because we're not speeding up or slowing down the tangential speed is constant but i have here a demonstration with the same two pieces of equipment but now i'm going around in a circular circle and watch what happens and i slowed it down so you can see it really easily that ball is deflected just like it was when we were speeding up or slowing down and here this ball gets deflected toward me that hanging ball was deflected out away from me but this one gets deflected toward me you see it from the side there you can see it real nice and clear how it's deflected right toward me and so in the sense that the ball gets deflected in the case of speeding up slowing down or in the case of now circular motion we could say that all of these are accelerated motions there's acceleration and the acceleration you might remember when it was speeding up we said the acceleration was in the direction that that ball was uh the acceleration was this way because the ball was deflected in that direction and here the ball is deflected backwards and so we say the acceleration vector is pointing back in the opposite direction same idea here we're going to say there's an acceleration vector toward the center of the circle which is where i am my head is kind of at the center of the circle so let's watch that one more time this ball gets deflected away from the direction of acceleration and this ball because the water is behaving like the hanging ball then this cork that's inside this orange cork inside gets deflected um opposite the water and so it gets deflected toward the center of the circle and so we call that centripetal acceleration or acceleration that's directed toward the center and i want to show you here um some equations the centripetal acceleration is directed toward the center its magnitude we call it a sub c for centripetal and centripetal is a word now you might be thinking stuart you're saying it wrong you're saying isn't it centrifugal and there is a word centrifugal but it's not this same word this word is centripetal and so it's spelled here for you centripetal and i'm going to highlight these in purple because it has the same units as this kind of acceleration or this kind tangential acceleration or this kind of linear acceleration it has the same units meters per second squared but we call it centripetal acceleration and centripetal is a word that's very similar to the word tangential it's a directional word a tangential velocity vector what we're saying is it's along a tangent line what we mean when we say tangential velocity is it's in that direction tangential is a directional word that's all that word means same thing with the word centripetal it's a directional word it comes from i believe a latin root two latin word root words meaning center and seeking so it's center seeking so it's a vector that points toward the center and i actually have that labeled here this would be the centripetal acceleration right here it's toward the center and we could draw that on this vector on this picture too a sub c meaning it's toward the center of the circle and so i'm repeating myself a lot just to drill this idea in um to emphasize that it's only a directional word there's another word that you're maybe familiar with called centrifugal that word is similar to centripetal but it has a different meaning it means center fleeing and so it's away from the center there are certain vector quantities that are centrifugal that point away from the center but this is not one of them this is acceleration vector points toward the center and we're going to get into that a little more as we go later in this chapter in section 6 c but for now it's toward the center and you can see that hopefully you're convinced of that by that little demonstration that i showed you that here we know when the car is speeding up that cork points in the direction of acceleration so here the cork is pointing to the right as the car speeds up to the right and when the car slows down like right there the cork points to the left even though the car's moving to the right it's slowing down and finally it stops but that cork deflects or points in the direction of acceleration and so when we see this motion this circular motion that cork deflects toward the center and so hopefully that helps convince you that's the direction of the acceleration in this case and we still call it acceleration because there's a change in velocity i want to remind you of something way back from chapter two well i won't go there but acceleration is always defined to be the rate of change of velocity and that could be a change in just the direction so this velocity actually has a constant magnitude if it's uniform circular motion but its direction is continuously changing and that's why this quark deflects and that's why we call that acceleration all right i'm not going to explain where this equation comes from but the main equation for centripetal acceleration is right here it's the tangential speed squared divided by the radius and i'm not going to go into a derivation of that but that is one way that we calculate centripetal acceleration and another way we calculate it is we use the fact that v sub t is equal to r times omega if we substitute e sub t equals r times omega and if we substitute that in here right there we would get r squared omega squared over r and we'd be left with this equation right here so that's another equation for centripetal acceleration if we know omega the angular speed and then there's another third equation that we can use as long as it is uniform circular motion and that is we just use this equation for v sub t we plug in the equation here v sub t b sub t equals 2 pi r the circumference divided by big t the period and when we substitute that in and square it you can see we square that we get 4 pi squared r squared over t squared but we would divide by r and we'd lose that squared and so that's another equation so these we will use each of these depending on the situation and what other quantities we know we use each of those at different times let me show you one more idea and that is in this uh box down here and we'll do one more example for you and that is what happens if we have both of these so this would be we might have tangential acceleration when the magnitude of tangential velocity changes or in other words when the tangential speed is not constant so we would have tangential acceleration only when it's not uniform circular motion so uniform circular motion another way we could describe that is we could say so that tangential acceleration is zero meaning the tangential speed doesn't change so the tangential speed uh when it changes we have what we call tangential acceleration that's speeding up or slowing down gaining or losing speed so that's due to a change in the magnitude of the velocity that's gaining losing speed magnitude so that's one type of acceleration the other type is that what we call the centripetal acceleration which i wrote down here and i'm just going to write 2 i wrote two versions of the equation here a sub c equals vt squared over r this type of acceleration so there's tangential or linear acceleration then there's centripetal acceleration that is always happening in circular motion because we always have the direction of the velocity changing so this is a change in the direction of the velocity whereas up here we said this was a change in the magnitude of velocity so this has to do tangential or linear acceleration has to do with speeding up and slowing down centripetal acceleration just has to do with changing direction that's it not speeding up or slowing down but it still makes that cork deflect and so it's still the same type of thing going on as when you speed up or slow down the velocity vector is not constant and notice that both of these quantities have units of meters per second squared so they both get the purple ink here and they could be combined to find the total acceleration a sub c can be combined with a sub t to find a total acceleration in meters per second squared and i'll draw that down here so for example if a sub c is this way and a sub t is this way you could think of those as like x and y components of a vector but these are actually two vectors that add together to give us a total combined acceleration which would be that way now there's another kind of acceleration we've talked about in this chapter i'm just drawing a dividing line here the other kind of acceleration is angular acceleration and it is here this is angular or rotational acceleration and that's the quantity in radians per second squared and that is due to changes in angular velocity and that would be gaining or losing speed rotationally and so alpha the angular acceleration has to do with just the tangential acceleration these two are related and let me draw this here these two all those equations we were talking about earlier in the chapter before we got to centripetal acceleration were just these two kinds of acceleration and how they're related using this equation that now i just put the subscript t there for tangential acceleration just to make sure it's distinguished from centripetal acceleration but you notice your centripetal acceleration is not related to alpha it's related directly to the tangential speed and then we do have this other equation that i'm sort of separating here again this r omega and that's only because we do we can just do a substitution because the tangential speed is equal to r omega so we can relate the centripetal acceleration to omega but not to alpha which is interesting anyway so keep in mind the different types of accelerations and how they're related and not related tangential and angular accelerations are related they're due to speeding up and slowing down centripetal acceleration is related to tangential speed in an equation and it can also be related to angular speed in an equation or to period in an equation here's the other part of the equation up here and it is like tangential acceleration in that it has the same units and it's the same kind of physical quantity it's through the change in direction of the velocity but tangentials do the change in the magnitude of the velocity and they can be combined because they're the same kinds of quantity then they can be combined to find a total acceleration that we could find the magnitude and direction of this total acceleration vector all right let's look at one problem where we're going to do that c number 15. all right so the blades of a large fan have a radius of 0.38 meters and just before the switch is turned to a higher setting the blades rotate about a fixed axis with angular velocity of 1.5 radians per second then they speed up with angular acceleration 2 radians per second squared and at a moment 0.5 seconds after they begin to speed up we want to know what is the total acceleration magnitude of a point on the tip of a blade and what is the angle between the total acceleration and the centripetal acceleration all right so let's draw here first of all a circle so there's a lot in that but i think if we break it down and label everything and set it up in a way that these equations kind of make sense here okay then that will hopefully make sense of this for you so this would be the circular path of the point on the tip of a blade of a ceiling fan so here's a like the if we're looking up at a ceiling fan let's draw this in the darker color here so let's say we're looking up at the ceiling fan and here's um let's say a blade at one moment let's see it's right there and then let's say later it's over here and then later it might be over here and later it might be over here so that might be where all the blades are at any one moment but this blade is the one we're looking at we're just looking at the point on the tip right out here and we want to know the point on the tip of a blade we want to know what is the total acceleration magnitude and to find that what we're going to do first is let's just go back and look at what are we even going to do to get there well the total acceleration we need both the tangential and the centripetal accelerations and we have equations that we can get those from tangential we can relate it directly to angular acceleration centripetal acceleration we can relate directly to tangential speed and radius and this also has radius or we can relate it to angular speed so i'm going to write those equations down in a bit but once we find both of these the tangential and centripetal accelerations we can use this vector diagram to relate them and that's what we're going to do so let's go back here so we're lying on our back on the ground looking up at the ceiling fan and it says that they it gives us the radius of the blades 0.38 meters so let's write that down that's the radius r and we can label that here so that is r .38 meters just before the switch has turned to a higher setting so you know how your ceiling fan maybe if you have one at home there might be a switch on the wall that you can change the setting to make it spin faster or slower all right so we're going to call this moment here at the bottom we'll call that initial and that's where we have an angular velocity just before the switch so this would be our initial angular velocity um so that will be the initial oh and let me just label this let's say this is our final moment and so it's going to speed up for a little while and it doesn't matter really what the angle it goes through is here but let's just say it's going to speed up a little bit and so now we have a final angular velocity that we don't necessarily know but let's go ahead and label the quantities that we know and don't know so initially the initial angular velocity is 1.5 radians per second and then it goes so let's say it's moving this way clockwise and then later it's going to speed up and be going faster with some final angular velocity that we don't know but this is just to kind of help us understand what's happening here and then if they tell us that the angular acceleration so this was the initial angular velocity and now we have the angular acceleration which is alpha angular velocity angular acceleration so that's alpha for this whole time that's causing it to speed up is 2 radians per second squared and then it says at a moment 0.5 seconds after they begin to speed up what is the total acceleration magnitude so we're looking at this time at a moment 0.5 seconds after they begin to speed up so initially that's where they begin to speed up and finally that's that point five seconds later and so it's here at the final moment so let's write here t is 0.5 seconds and at that final moment that's when we want to know what is the um total acceleration magnitude in meters per second squared of a point on the tip of the blade so in other words what is the acceleration right here at this point and what is the angle between the total acceleration and centripetal acceleration vectors and so let's just go back here for a minute and look at the summary sheet one more time so we're looking and i oriented it so it's pretty much the same we're looking at these two quantities at that moment when that blade is horizontal and we want to know an angle between two of these vectors and we'll read the problem again but so it's these two vectors at that final moment that we want to know and just real quick let's review these equations that we have to work with this one right here between the tangential and the angular so that's going to help us because we know the angular and this one right here because we know [Music] the radius and we need the centripetal acceleration somehow and so we may have to find the final angular velocity at that point because that's going to tell us the centripetal acceleration at that point but so these are a couple of equations that we'll use and let's go back here now and so what we're going to find for part a is what is the acceleration centripetal acceleration here and what is the tangential acceleration here and then what is the total acceleration is what we want to know so there's the total acceleration magnitude that's the first part we want to know what's the total acceleration magnitude in other words what's the hypotenuse of this vector triangle if we add these two vectors together and then what is the angle between total and centripetal so it would be the total would be the hypotenuse and the centripetal would be the lower one so we want this angle right here and i'm going to call this angle instead of theta i'm going to use phi or something like that because that angle is not the same as the theta that the fan goes through so there's lots of quantities floating around here all right well let's write down equation for tangential acceleration we need that first we know that's just equal to r alpha and we know both r and alpha and so we can calculate that number it comes out to be 0.76 meters per second squared if we just multiply 0.38 times 2 0.38 meters times 2 radians per second squared gives us 0.76 meters per second squared the centripetal acceleration is tangential squared tangential speed squared over the radius or it's equal to the radius times the angular speed squared now we already have initial angular speed to find this angular accel this centripetal acceleration here at the final moment we need to know this angular acceleration at the same moment and so we'll use this equation because we have an easy way to find omega using this set of equations omega equals omega naught plus alpha t they're the set of equations being all the ones for constant angular acceleration which we're told this is the rate at which they speed up so we can assume that's constant while it goes through this arc length here and so we can find omega final because we know omega initial and alpha and t so this is 1.5 radians per second plus 2 radians per second squared for a half a second and you can see that adds one radian per second and so that gives us two and a half radians per second and so that we can then use here to calculate the centripetal acceleration and we're going to get 2.375 meters per second squared and i'm keeping extra sig figs here and now i'm going to redraw the triangle over here just so you can see what quantities we're getting so we have a triangle with centripetal acceleration here which is quite a bit larger we see so there's the centripetal acceleration and the tangential acceleration i'm going to draw them more close to scale here and so we want the total acceleration which would be the hypotenuse of this triangle and that's just going to be because this side over here is equal in magnitude to the tangential acceleration and so there's a right triangle so this is just going to be the square root of a sub t squared plus a sub c squared and so it's just adding those two it's as if this is like the x and the y components of a vector it's kind of what's happening here and that comes out to be 2.49 if we round that off to three sig figs so that's the total acceleration that counts the part the 0.76 for speeding up and the 2.375 that is always there for going around the circular path and then theta is the theta between the total acceleration and the centripetal is what it says and so or not theta we'll call it phi and so the tangent of phi is equal to a tangential over a centripetal and i wrote out tangential but that's a sub t over a sub c and so phi is the inverse tangent of a sub t over a sub c and so we just get a value for phi of 17.7 degrees and those are our answers now you could call this theta if you want i just called it something other than theta just so that you're not confusing it with the theta that would be the rotational quantity and radians that the fan itself goes through i want to show you one more thing that illustrates this happening and that's back to this video clip and we'll wrap up with this is that this is right where i start speeding this up this cork in this water tank and i want to i want you to see what's happening right at this one moment when it's still speeding up so i'm just speeding it up and you can see it right here you can see the cork is actually deflecting to the left here which is in the direction that it's being sped up so it's in the direction of tangential acceleration so there's some tangential acceleration but there's also centripetal you can't quite see it at this angle but it's also deflecting toward me so it's doing both at the same time and that's what we're looking at in this problem there would be one deflection in the tangential direction due to the speeding up and another deflection toward the center if this were instead of a cork hanging down or pointing up inside a tank of water if it were just a ball hanging down from the tip of that fan you would notice two things one is that ball is swinging out like this because of the going around in the circle that's in response to the centripetal acceleration the ball seems to lag behind the fan and so it seems to swing out and then there's also the ball lagging behind this way because the fan is speeding up and so the corks doing the opposite of that it's pointing in and forward in the direction that it's speeding up and and again you can see that here and if we let this go a little longer you can sort of see it's doing it's not quite pointing at me and it's not quite pointing out this way in the tangential direction it's pointing kind of diagonally until it's fully sped up which is somewhere around now in the video okay so that wraps up six b and all types of acceleration i'd say go through this and review this and make sure you just understand the terms and the units and which equation applies when just to keep all these different kinds of accelerations straight