Transcript for:
Newton's Laws and Circular Motion Overview

this is lecture six c and we're now going to be applying newton's laws to circular motion and when we use newton's laws in circular motion we're going to use the same rules that we used when we applied newton's laws to straight line motion and i want to go back over a couple of things one thing real quickly is you remember when we have circular motion the reason we can apply newton's law specifically newton's second law is one we'll use is the fact that there is acceleration that's the reason that we would use f equals m a with acceleration and this is the acceleration that's now in circular motion usually we're going to be looking at uniform circular motion or at least we won't be dealing with tangential acceleration in the problems that we're looking at only centripetal acceleration we won't be looking at a point in a motion where it's speeding up or slowing down in this part of the chapter anyway in 6c so just a real quick review newton's second law is here and in components we do the same thing that we've been doing all along as we use the x components of the force we add them all up and set them equal to the mass times the acceleration x component same idea with the y components of the force and the mass times the acceleration y component newton's third law for every action force there's an equal and opposite reaction force and this one is really important because it's important that we know which for which object a force is acting on some forces act on one object some forces act on the other object and we want to know when we draw a free body diagram for example and use newton's second law we want to know only the forces acting on that one specific object so it's really important we understand which force is acting on which object and that's a little easy to get confused in circular motion i'll show you what we mean as we go one other thing to remember when we're evaluating circular motion problems with newton's laws is that there are no weird new forces um there's just the same old forces that we've been dealing with up to now basically these ones here on this list these are the types of forces that go in the free body diagram there's the weight vector which is caused by the earth there's tension caused by a string or a rope there's a normal force caused by a surface and maybe a friction force also caused by surface and we know the directions of those and so there's no weird strange forces and um you might have to really put aside your preconceived notions in this section because some things i'm going to say and explain at first you're going to be like wait a second that doesn't sound right but give it a chance and i think you'll see how it all makes sense when we evaluate it using these just basic standard rules but it is going to go against some of the ideas that maybe you've held up until now about circular motion and what you feel or what you've experienced in circular motion but i think we'll be able to make sense of all of it including those the the way circular motion feels and relate it to what's actually happening with the forces all right so with circular motion there's two sort of versions of this we're going to look at one is we're going to look at horizontal circular motion or horizontal circular paths in other words the the circle itself is just a horizontal circle and so those are the the first ones we're going to look at and we're just going to do a few of these we're going to do number 21. we're going to do quite a few examples in this section we're going to do number 23. we're going to do number 25 then we're going to look at vertical circular motion a little bit after and that would be vertical circular paths and we're going to look at just one of those as a full example but i'm going to talk about a couple other ones so i'll just mention those here so we're going to look at number 28 and we're going to discuss that's an and sign and discuss numbers 29 30 and 31 i'm gonna just discuss various aspects of those problems so we're going to be looking at quite a few problems only these we're going to do a few examples with horizontal motion and we'll do one example with vertical circular motion and there is actually one other example we're going to do when we talk about apparent weight down here and that one is number 27 i believe and so we're going to mainly do a bunch of examples here and we'll come back and talk about this and this when the time comes but for now let's just go look at number 21. all right number 21 is a situation that's a child standing on the rim of a merry-go-round of radius 2.5 meters and it's spinning at the fastest speed possible without slipping up so the child is about to slip off and so we're dealing with maximum static friction just about to slip and so we're given the static friction the coefficient of static friction here between the child's shoes and the floor of the merry-go-round that number right there we're going to draw a free body diagram of the child and then determine the amount of time it takes the child to go around once and the acceleration of the child first i want to show you a little clip that this is a situation very similar to what's described in this problem so that that little metal piece is kind of like the child and this uh is kind of like the merry-go-round and i'm gonna spin it up until this uh little piece of metal this little copper metal piece flies off i'm going to keep speeding it up and speeding it up and you know there's some speed at which it's going to start to slip and so we want to know what's that speed or we want to know what's the period that's the amount of time that it takes to go around so that's what question b is asking is the amount of time it takes the child to go around once that is what is the period t and the acceleration in particular because it's a circular motion it's we know it's circular motion so we want a sub c because it's circular motion and um so let's see here we have the radius that's r and it's spinning at the fastest speed possible without slipping off so just about to slip and so we have 0.4 is mu sub s 0.4 and because we're just about to slip we know that we have the friction force is fs max so this is a little review from chapter five and we're going to use that here that's mu sub s times a normal force so we know those are two of the forces the static friction force and the normal force so i'm going to draw two pictures um one picture i'm going to draw is this kind of this view from the side and uh i'm going to draw it like right here when the when the child is right there but i want to draw it when it's spinning so let's find that so right there so the child is spinning but i froze it and right there in the screen that the velocity vector is moving away from you and so the velocity vector is actually going into the video screen that you're looking at and it's right there and the radius would be this radius right here if the kid is at the rim it would be that radius but let's assume the rim is right where this is if we were to draw this for the kid it would the kid would be way out here so i'm going to draw an edge view so that just the edge of the merry-go-round and this kid standing here at the edge so this is an edge view and so here is the merry-go-round and it has some shaft and it's stuck in the ground and so it's uh it's spinning spinning i'll just say it spins and then there's this kid standing here at the edge and there's a radius r there's the center of the circular path and that's where the kid is standing right at the edge so let's just say they're right on the edge of the merry-go-round and that's that value r right there and so the velocity vector is into the page here um just a little thing that we could make a note of here is for vectors that are going in we sometimes use this notation that would be in for vector directions in and out of the page um vector directions in or out of the page if you're writing on a page of a piece of paper or in and out of the screen in this case and so what we're looking at if you think about an arrow with like say the fletchings on the end like this if you're looking at the tail of the arrow going away like if you put your eyeball right here and you looked at that arrow going away from you i know sorry my eyeball's disgusting there but um if you saw that arrow going away from you you would see these tail feathers crossed something like that and so that represents an arrow going away from us and if you put your eyeball right here and don't don't do this not safe but that vector coming towards you that would look like a tip of an arrow coming out at you so that's what it would look like if it were coming out that's just a little notation thing of how we might represent a vector that's either coming out coming out or going in depending on the direction and so this velocity vector we could say is that direction now it doesn't mention velocity but we know it's moving so i'm going to go ahead and label that and so that velocity vector is going into the page so that's the edge view there's also a top view let's look at that too all right so the top view is going to look something like this where we have a big circle and maybe you can do a better job than i did i'm free handing that and then this is the top of the kid's head and so there's the radius right there the radius is from the center out to where the kid is and then the velocity vector is going that way um and so that's a top view and so now that's a tangential velocity vector and because it's tangent to the curve if we could put the kid right at the edge it would be tangent to the rim right there i'm just drawing them slightly in from the edge just for the diagram all right so anyway so now we have those two vectors now we also know we have centripetal acceleration we have acceleration toward the center of the circular path let's go back here for just a minute to remember that's always going to go toward the center so if this is be like in the top view picture we have the tangential velocity and the centripetal acceleration so we we could label that v sub t or just leave it as v the way i have it already in the problem all right so that acceleration if we draw it on the picture here then this centripetal acceleration a sub c is going to be going toward the center whenever we have circular motion we always have acceleration toward the center as long as it's uniform circular motion we only have centripetal acceleration we don't have tangential or angular acceleration it's not speeding up or slowing down and so here also we could draw the centripetal acceleration toward the center and if it helped you could even label very clearly the center like this would be for the kids body the the center of the merry-go-round is right down here but the kid's body is going around a circular path somewhere up here like the center of mass of the kid is going around and around and so that would be the center of the circular path all right so i'm being really specific about all these ideas of where the center is where the velocity vector is which way it points where this centripetal acceleration vectors which way it points all those kinds of things so i want that to be really really clear as we go through this okay so now the first part of the question is draw a free body diagram and remember for the free body diagram and i'm i'm drawing so much on this problem and have to do that over here for part a the free body diagram we're not going to make up any weird forces okay so we're only going to look for forces that we know are on that list let's just go look at that list one more time right here this is just review is there a weight vector answer yeah there's the earth is pulling vertically down on the kid there's a weight vector is there a string or a rope no is there a surface yes is there a normal force and a friction force yes those are the only three forces that are going to be in our free body diagram and then also remember newton's third law because one of these forces you're going to have a little bit of a hard time with i think the direction that i give it and let's go look at that again okay so the free body diagram is going to have three forces the weight vector the normal force and the friction force and i'm going to draw it here at this point right here so let's just take the kid right there that's the kid and we're going to do the edge view the top view the weight vector would be going down away from us into the screen and the normal force would be coming out away toward us out of the screen so i'm not going to try and draw that i'm going to draw it from the side here so there's a weight vector going down a normal force going up so we know both of those so this is the weight the sorry that's the normal force and there's the weight vector so this is on the child by the earth this is on the child by the surface so on the child by the merry-go-round and this is on the child by the earth now the friction force which way do you think the friction force is i want you to think hard about this and i want you to make a commitment sort of but draw it in pencil because you might have to erase it okay so try that out try to draw the static friction force we know it's a maximum static friction force but pause the video draw that and then we're going to talk about it okay so i hope you've drawn one here for yourself and i hope you use pencil because this one's a little tricky so when you stand on that merry-go-round you feel a force pulling out on you right it feels like something's pulling out on you and there is a force that is outward if you think about which way you're um standing and how you would stand in order to stay on this merry-go-round you would be kind of leaning toward the center and that's because your feet would be pushing out this way your feet would be pushing to the left on the merry-go-round surface at this point in the picture or at this point in the picture same thing you'd be leaning that way and your feet would be pushing out away from the center but think about this that's the force on the merry-go-round by you so there is a force that is out there's a force and that would be a friction force there is a force there's a i'll call it fs max that is on the merry-go-round by the child but that is not the force that goes in this free body diagram does not go in this free body diagram and the reason is because this is a free body diagram of the child and all the forces that are acting on the child are the forces that go in this free body diagram so if this is you standing there you know you're pushing kind of outward with friction on the surface in order to keep yourself from flying outwards but that means by newton's third law if this fs max is the force that you push on the merry-go-round with then the merry-go-round is pushing back the other way on you and i'll tell you a couple of reasons that has to be but i'm going to draw it here so this is f s max that's on the child by the merry-go-round okay two reasons it has to be that way one reason is is if you just think about physically what if there were no friction what would happen and and you kind of saw it in this video once the static friction goes away you you see what happens is that block slides now watch what it does though it doesn't slide out meaning it starts sliding about here so let's just go back and now we'll go forward and it's i think it's about to slide right about here but notice what it's doing see how it's sliding but it's sliding it's still moving tangentially so right here it's coming toward us and it doesn't notice it's not sliding that way it doesn't there's no force pushing or pulling it that way it's still coming around this way and it's going to eventually completely lose contact and fly off but it's still going to go in a tangential direction when it flies off it goes tangentially you see that right there so when it leaves the surface it would go tangentially and that's what would happen if the kid if there were no friction the kid would just go that way so think about this if the kid is going to go that way but without friction but with friction the kid curves to the right and doesn't it make sense that there must be a force to the right that's pulling the kid to the right all right so that's one reason that that force has to be to the right because without it the kid would go straight but with it the kid curves to the right in this picture here's another reason is look at the direction of the acceleration if you have acceleration to the right at this moment then there has to be a net force to the right you can't have a force to the left causing an acceleration to the right the force on the child is to the right the force on the merry-go-round is to the left the friction force and so the force that goes in our free body diagram is the one to the right so that's a couple of reasons why it has to be to the right and now we're going to then just solve the problem using our basic procedure i'm going to say with my free body diagram i'm going to go like this f net x equals m a x and f net y equals m a y and remember the rule that we had in chapter 4 and 5 and that is i'm going to pick some other color here for my axes i'm going to choose an axis parallel to the acceleration so i'm going to make that my x-axis and i'm going to make this my y-axis in this picture the y-axis would be coming out of the screen toward us and the x-axis would be going toward the center but now i'm going to go here and i'm going to draw this as my x-axis and this is my y-axis and remember why i choose those axes and the reason is because i want my acceleration x component to be plus a and i want the y component to be 0 and that's what i get when i choose an x-axis along the acceleration vector and a y-axis perpendicular to it and now that we have centripetal acceleration this is centripetal and i have two equations that i'm going to i actually have three equations sorry let's go back here and just remember what those are and this is how i'm going to solve for the period is because i have a centripetal acceleration equation that involves the period right here and so i have a choice for the magnitude of centripetal acceleration i can say it's the speed tangential speed squared over the radius the angular speed squared times the radius or 4 pi squared times the radius divided by the period squared this question is asking me for the period so this is the one i'm going to use so here i'm going to say this is plus 4 pi squared r over t squared where t is that period that i'm trying to calculate and so now we're just going to do some real simple substitution and i'm going to write out my force vector components they're really easy here right i hope you see that because all the forces are on the x and y axes and so for example n x is just zero but n y is plus n those are the components of the normal force vector you don't necessarily have to write this out but i just want to make sure you're really clear what i'm doing here weight x is also zero and weight y is minus the weight or minus mg and you might be saying i don't have a mass in this problem that's okay don't worry it'll it'll work out i'm still going to use this equation and then my friction force has only an x component so f s max x is positive f s max and i'm going to use this equation over here mu sub s times n so it's positive u sub s times n and f s max y equals zero now you might want to just go through and do this for every single force every time if you're having trouble getting these equations right because it's these little picky details that are going to determine whether you get the question whether you get the right answer or not and so here i'm going to just add up remember this f net x it's the sum of all the x components well that's 0 plus mu sub s n plus zero so i get zero plus mu sub s n plus zero remember that's the sum and that's equal to this is a sub x right here that guy is a sub x and so we're just going to plug that in here that's equal to m times 4 pi squared r over t squared so that is i didn't give myself quite enough room there so that is the net forcing the x direction we have the three x components of the forces we're adding them up and so what i end up with is mu sub s n equals m times 4 pi squared r over t squared that's one equation that i get the other equation i get is this one which is really simple because it's just plus n plus zero minus mg equals zero so plus n zero minus mg because remember this is a sum of all the y components now you don't have to write it out this way you can just ignore the zeros you could have just written mu sub s n because you know that's the only one in the x and the y you could just write plus n minus mg that's fine too equals zero because m a y is zero and what that gives us is that the normal force is just equal to the weight that's only true when those are the only two forces with vertical components and so now i have a pretty simple solution to solve for t remember my first question is what's t so i'm going to find t now and i don't know the normal force it's not given and i don't know the mass but i do know mu sub s and i do know r and here i don't know the normal force and i don't know the mass but i'm going to just take that normal force and substitute it over here and some nice things happen we get mu sub s times mg equals m 4 pi squared r over t squared and the m's go away you see that and so the m the mass doesn't it's not uh a factor it cancels out um the reason it cancels out is because if you had a big kid versus a small kid well a big kid would have more friction force but it also requires more force to accelerate that larger mass so the larger normal force is going to mean a larger friction force but it also is just going to take a larger force to accelerate that larger mass all right and so now we just solve this for t and so i'm going to bring the t squared up and i'm going to bring the mu sub s and the g down so we're basically just making those two swap places and then we take the square root of both sides and we get t equals and if we plug in all those numbers you could take the 4 pi squared out and make it a 2 pi out here but either way if you plug in all those numbers what you get is 5.02 and the units are seconds if we have meters here for the radius and we have meters per second squared here for g the per second squared here comes up and becomes second squared and so that's the period after we take the square root and then so that's part b part a is actually then really simple because the acceleration we already know it if we have an equation for it already if we have the period then the acceleration is just 4 pi squared r over t squared and so for part b we can just take that number 5.02 seconds and plug it in here to answer part b and if we plug in those numbers we get 3.92 meters per second squared equals a sub c and that's the answer for part b if we just do that all right so i took up way too much space doing that problem so that's uh one example of how to deal with circular motion so let's go back to the summary sheet let's review now what we just did in problem number 21. so we drew a free body diagram for the child that's the one that was actually moving at a constant speed and in circular motion and so that's an object of interest we chose x and y axis based on this we made one axis toward the center parallel to the centripetal acceleration because now we're dealing with circular motion and so a sub x is plus a sub c and that could be plus v squared over r or it could be omega squared r or it could be what we used was 4 pi squared r over t squared in this section of the chapter we most often use this first equation or the last because we're more often interested in the period or the speed tangential speed so that is um those are equations for the magnitude of the acceleration and then the other axis the y axis is perpendicular to that so that a sub y is zero and that would that was in the vertical direction in that case and then we just substituted our components into these two equations and solved for whatever we were solving for so that is um how do we how we apply newton's laws to circular motion and i'm just going to do this in every one of these problems so now let's go look at number 23. okay so number 23 is a banked curve and if you've ever been driving on an on-ramp on a freeway you've been on a banked curve you know how it goes around like that uh when you're driving and entering or exiting the freeway it's banked and so the the picture looks something like this i'll do my best to try to sketch an edge view a curved track has a radius of 120 meters it's banked at an angle of 18 degrees from the horizontal if there's no friction what speed must a car have to stay on the track at this radius that's a real simple question there's not much given is there but let's draw an edge view so let's say this is kind of the bottom of everything and then there's a banked curve and it's banked at 18 degrees from the horizontal that's our 18 degree angle we could call that theta if we wanted to and so this is an edge view or you could think of it as a rear view of the car so in other words i'm looking at the back of the car so there's one tire here's the other tire and there's like the car and there's a driver right there if they're going again here we'll say the velocity vector is in uh so we could say it's like that that's the velocity vector so that car is going in away from you in the screen and then that curve goes around like so i'm trying to now draw sort of a three-dimensional version and then it comes back out way over here so this would be the other side of that banked curve something like that i'm trying to kind of draw sort of a three-dimensional thing and maybe you're more artistic than i am and um so that would be sort of a cutaway view i guess so that would be the car is tilted this way over here and the car would be tilted this way over here so on this side the car would be v would be coming out and the path that the car follows is somewhere along here so this would be the path that the car follows this so it's a horizontal circular path this would be the circle that let's say we track the center of mass of the car so it's not even on the road it's some circular path above the road and the center of that where that circle is horizontal so even though the track is banked at an angle the circle itself is horizontal that's half the circle and you can think of it as we're looking sort of three dimensionally at a half of circle the side that the half that's farthest from us and the center of that circle is going to be up here and i'm going to draw the radius right about there that would be the radius r because the center is right here center of the car's path and we could also draw a top view of this situation if it might help you see it so there's a sort of an inner i'm going to draw this a little smaller so i can squeeze it on here the banked part of the track is this band looks kind of like a donut from above that's a terrible picture but anyway the car is approximately here that's like the front and back of the car and the velocity vector is that way and the path of the car is this way and that's where the car would be in that other side over here going this way the velocity vector there would be going that way and then the car would continue so i don't have this part of the circle this part down here drawn in this picture we're only drawing sort of the half above this these two cars in the picture so that's like this sort of cutaway view that i have drawn here all right so now hopefully you can kind of picture that and i have a little demo clip video clip of something that's kind of similar to this this is maybe you've seen one of these toys where you you know some stores have this as a donation something like this and so that circular path i i was showing you this part here if you look at the circular paths on the side view here the circular path as you watch those pennies go around from the side you can sort of see the shadows and you see how that penny is going across horizontally so it's it's so we're looking edge on now and so that's kind of what we're drawing there the car is over here in this in the picture right here you can sort of see it that's what the car would be right there but it's actually going around in a horizontal circular path so that's also maybe helpful for you to see the whole thing and then so you could have it going either in or out in this picture in this picture so hopefully that helps you hopefully that one helped so that's the uh picture and the radius we're told is 120 meters and that angle is 18 and those are the only two numbers we have in the whole problem and so it wants to know what speed the car must have so we want to know what is v or that's that tangential speed i'll just call it v or you could call it v sub t so that's what we want to know we don't know the mass of the car um we don't know very much at all do we but we do know a couple things and so what we can do here when we have circular motion and especially if we have v and r that gives us acceleration you notice here that we could say there's an acceleration toward the center we could draw it on both pictures there's going to be an acceleration toward the center it's not going down the incline it's going that way toward the center and here also toward the center the center and this picture is here and remember that center is even with that horizontal circle the car's path is making it's not down here at the bottom that would only be the case if the car were right there at the very bottom so it's the center of the car's path and um all right so that's the centripetal acceleration and that defines our x and y axes for a free body diagram and so a the reason again that i bring up acceleration is we have circular motion and you notice here that we have r and we want to know v and so we could say oh okay so a sub c is v squared divided by r and so if i choose x and y axes we can write this here so if i can find the acceleration though then i can find the speed because i already know the radius and so let's see if we can find the acceleration or we'll sort of set it up using newton's laws and so i'm going to make an x-axis toward the center and a y-axis perpendicular to that or vertical and let's do a free body diagram based on this picture all right so we're going to do uh two guidelines here now this problem doesn't say make a free body diagram but i'm going to make one because i know the acceleration of the car and i have an angle for force vectors so that's why i'm going to do a free body diagram because i have at least some information that i'll maybe i can get out of this free body diagram of the car so let's draw these some different guidelines here meaning horizontal that's our x-axis because that's the direction of acceleration and the y-axis is vertical and we also though have axes parallel and perpendicular to the ramp and so we'll pick some other color for those so like this this direction parallel and that's at an angle of 18 degrees from the horizontal so this would be parallel and you can see that 18 degree angle there are parallel to the surface it's not really a ramp now it's just a surface that's tilted we're not sliding up and down that surface now we're just trying to go around that track and then there's another uh perpendicular to the surface is right here that's perpendicular to the surface and i draw both of those because we kind of need to know those directions in order to draw all the forces accurately so based on this edge view here we have let's go through our list of forces again do we have a weight vector the answer is yeah we do we just have a weight vector that goes vertically so we know the weight vector goes down that's the weight vector and we also have a surface so there's no tension there's no string or rope or anything but we have a surface and we're told in the problem that we don't want any friction so there's no friction and so we only want the force perpendicular to the surface which is the normal force so it's going to be going this way so there's a normal force going that way and that's it those are the forces just the weight vector and the normal force vector and and very similar to problems back in the other chapters chapter four and chapter five we then have to break up these forces into components but let's look at this um let's look at a couple of things here there's one thing that's basically the same and that is that there's this 18 degree angle and you you might be able to see it here i'll draw one more little orange line here this is also horizontal so that 18 degree angle is between the horizontal and the parallel and so that 18 degree angle right here that's 18 degrees that's 18 degrees that's 18 degrees and that is 18 degrees and we'll decide which one we need we need this one with the weight vector maybe or the normal vector we'll see which one we need um i want to remind you of a situation back in previous chapters where we had a free body diagram that looks just like this and the x and y axes that we chose though were not these ones we actually chose the other ones and i want to go back there just for a second and remind you why we did that so this is way back in number 12 and chapter four and i just chose this one because i wanted to remind you this one is similar it's not exactly the same but the free body diagram looks very similar because we're on a tilted surface like this but there's a tension here because there's also a cable but if we can ignore the tension then the normal and the weight vector look basically just like they do in that other problem that we were just now looking at in chapter six however in this problem the weight vector we found x and y components of the weight vector on this this x-axis here which was the axis parallel and the axis perpendicular the reason we chose that is look right here which way was this crate accelerating the crate was accelerating up the incline is actually speeding up so that's why we chose those x and y axis so that a sub x was plus a and a sub y is 0. now in the problem we're looking at back in chapter 6 let's go back there now here the acceleration is toward the center because the car is not sliding up or down it's not going this way at all it's moving around in a circular path and so the acceleration is completely different remember that um bottle that had that cork in it the acceleration would be the direction that cork would deflect in this picture and it would be that way toward the center of the circle even if the car is tilted like this it's only based on the direction of the center of the circle so the cork would just tilt that way it would lean toward that center whichever way that is for you from your point of view and so i'm choosing the acceleration vector i'm choosing my x and y axes to be parallel and perpendicular to that acceleration vector because what i want is i want a sub x to be plus a sub c which is plus v squared over r because it's centripetal acceleration and i want a sub y to be zero that is why i choose these axes because it makes my math simpler if one of my acceleration components is zero and so that's why i chose these axes so in this problem instead of finding components of the weight vector i'm actually going to find components of the normal force vector because the weight vector is already along the y axis and so if i wanted to write down my weight vector components weight x is zero and weight y is just minus the weight or minus mg so here my weight vector looks like it's completely along the y axis so it only has a y component the normal force vector though now has a y that would be n y right there and an x component which you could draw down here or i'll write it up here and so there are let's just draw this triangle here and so now you can see where we want that 18 degree angle it's right here so that is our 18 degree angle this is a right triangle and this is our normal force vector the hypotenuse is the magnitude this is n sub x this is n sub y and so n sub x over n is the sine of the angle because the angle is measured from the y axis not the x axis here and so n sub x is positive it's going in the positive x direction n sine of 18 degrees and n sub y is positive and cosine of 18 degrees and so we have those components and now we have all the components of the vectors force vectors and acceleration vectors and so now we can do this f net x is equal to m a x and f net y equals m a y and those are the only two forces is the weight and the normal and so i'm just going to add those two x components n x is n sine of 18 plus zero those are the weight x component and the normal force x component and on the right side is m and here i'm going to put v squared over r that's positive and this is positive so i get n sine 18 is m v squared over r and this force is the sum of the y components so i have plus n cosine 18 plus the y component of the weight is minus mg and that is equal to zero because the acceleration in the y direction is zero so i get n cosine 18 equals m g now we have a few unknowns here we don't know very many numbers in this problem at all the mass is unknown the normal force is unknown and the speed is unknown the mass is unknown the normal force is unknown but this problem is very similar to the last one it seems like we have too many unknowns but they we get some things cancelling if we solve this for n and then we're going to substitute so we get n is equal to mg divided by the cosine of 18. and here we're going to substitute that and so we get mg over the cosine of 18 sine of 18 equals m v squared over r and now you notice what happens is the masses cancel that was an unknown but it doesn't matter it doesn't matter what the mass is in this problem either and the speed is our only unknown you might notice sine 18 and cosine 18 sine over cosine is tangent so we have g times the tangent of 18 degrees and i'm going to multiply by r and that's equal to v squared and so v is the square root of that now remember g is just 9.8 here we already took into account the negative the downwardness of it right here and we did that here and then when we do our algebra the negative was it went away when we brought that over here so it's just 9.8 when we plug it in and if you plug in these numbers and those are the only two numbers we were given was the radius and the angle and that's all we need here besides g and so if you plug in the number there the numbers there what you get for your answer is two well to two sig figs it's 20 and so i'm going to write it this way 2.0 times 10 to the 1 meters per second because it's that's all i have written there but it might be like um 19.6 or something like that but i wanted to stress that we actually can say it to two sig figs it's 20. we know the zero is a sig fig so i'm just writing it this way all right so that is um number 23. now keep your hand or your thumb or something on this one so you can get back to this one because the next one is very similar to this and let's go look at number 25 now here this is a block hangs from a string attached to the ceiling of a bus when the bus moves at a constant speed of 28 meters per second around an unbanked curve of radius 150 meters the block hangs at an angle theta from the vertical determine theta now you notice this this problem has the same numbers different unknown but in this problem we're given the speed um we are given so that's the speed that's given in the last problem the speed was the unknown so that's v [Music] theta is our unknown theta the angle is our unknown but we're given the radius so it's the same three numbers last problem we were given theta and r and we found v here we're given v and r and we're going to find theta it's really similar to the last problem but here is the rear view of this picture this would be the edge or the rear view oops so here is this bus and we're looking at the bus driving away from us and there's something hanging from the ceiling of the bus but the the bus is going around an unbanked curve so it's on a flat horizontal surface and let's say it goes and comes back toward us over here so it's going around a circular path so here v would be coming out but here v is going in we don't need to write out an n if we have those little vector symbols but and so it's circular path this is like a rear view of us its circular path actually just kind of looks like a horizontal line is going away from us and it's over here the bus would be moving to the right and then it would be coming back around and finally at this point coming toward us and then it would go back around so it's it's going in a horizontal circle and a top view might look something like this i'll draw a minute a mini top view so here's the bus and there's we'll draw the block hanging from it in a minute but there's the velocity vector is going that way that's when it's going in and then over here the bus velocity vector is going the opposite direction [Music] and it's following a circular path something like this maybe you could do a better job than i'm doing of drawing a circle but that's the picture and then there's a center right here of the circle that would be this point right here in this picture here we're just looking edge on at that circle and so we know there's an acceleration vector toward that center and what we're going to be looking at the acceleration is actually for a block that's hanging from the ceiling of the bus so that's the thing we're interested in this problem and so the way that block would look if you were uh watching this block and i i've shown you this actually before let's look at this video clip here there it is right there now it's just going the opposite direction so here the velocity is coming out at us but you see how it's banked away from the center but the acceleration is toward the center and that we know that from our experience in this one with that if it were the cork in the bottle then it would be angled toward the center and that gives us the direction of the acceleration is right here it's toward the center which would be to the right in this picture all right so back to this picture so this block is hanging from the ceiling of this bus and maybe it's hanging here and it's like right there and the radius that we're talking about is the let's say that radius just corresponds to the radius of the circular path of the block itself and it just says that's the radius of the buses path and let's just assume that's right where the block is is right at that 150 meter mark and so um this one uh take a look here at the picture and now what we're going to do for similar reasons to last time is we're going to draw a free body diagram and we're going to use these same x and y axes so that our acceleration vector which our acceleration vector a sub c is v squared over r and we have a sub x is plus a is plus v squared over r and our a sub y is zero and i'm going to draw a free body diagram now based on those axes i'm going to draw a free body diagram of this little block that's hanging from the rear of the car so let's draw this free body diagram here and so think about what what are the forces acting on the block all right there's does the block have a weight vector yes it's down what about now there is is there a string or rope and the answer is yeah in this problem there is there's a string and it's pulling the block that way it's pulling along the string tension is always parallel to the string pulling on things so the tension vector is that way and there's no surface in this problem not on the block there would be for the bust but that's not what we're looking at here we're looking at the block and so the free body diagram would look like this it would have a weight vector and a tension vector and actually let's call that f sub t just to keep from confusing it with period t if we ever wanted to use the period that would be the tension because the period sometimes in circular motion problems i might use period t and i don't want to confuse it with tension so i'm going to call this f sub t here and now i'm going to set up my x and y axes and try to figure out what angles i have here so there's my x-axis there's my y-axis and now let's just take a look at the angles here inside the bus it says it hangs at an angle theta from the vertical so let's just draw that on here and i'm going to zoom way in here to be real clear about what theta is and let's do theta in blue here or something okay so that is theta right there and you guys might remember this rule from geometry or trigonometry but i think these are called alternate interior angles because that's a vertical line and this y-axis is also vertical and this one the string basically is a line between these two parallel lines that intersects both of them and so this is the same angle theta and i point that out because that goes in our free body diagram as the angle between the tension vector which is along the string and the vertical so that is the angle theta that we're trying to find okay so now i'm going to just kind of outline the rest of this problem and explain a little bit to you and then i'm going to have you guys finish it so now we're going to do this f net x equals m a x f net y equals m a y and you finish this one and i'm going to tell you sort of what you're going to get before i do that though i want to compare this to the previous problem that we just did i want you to study this free body diagram look what we have we have a tension vector here and an angle theta from the vertical and a weight vector look back here what did we have in this problem for this car that was on this bank track we had a normal force that was at this angle theta let's draw that as theta there the normal force was at a direction theta from the vertical and the weight vector that's all we had here and that's all we have in the other problem number 25. and so this problem looks almost exactly like number 25 except one difference is what do you think that difference is it's just that this is a tension vector in number 25 and it's a normal force vector in this problem but the mass otherwise is identical and i wanted you to see that i want you to think about what this normal force is doing the normal force is doing two things it has two functions here one is it's canceling out the weight vector that's the y component of the normal force that cancels out the weight vector you can see it right here in this equation and you can see it right here on the diagram that y component is just as big as the weight vector that's holding the car up the x component of the normal force is what is providing the acceleration toward the center of the curve that x component is what's keeping the car the car has inertia and if it weren't for that x component of the normal force the car would go in a straight line remember newton's first law objects tend to go in a straight line unless there's an external force pushing them to the right then they'll accelerate to the right they'll curve to the right so if there weren't any force to the right due to that banking of the curve there wouldn't be any acceleration to the right but the fact that there is a component of the normal force to the right because the curve is banked that gives us an x component of the normal force and that allows us to accelerate the car to the right and so that x component of the normal force is causing the acceleration that's what's causing the car to go in a curved path that's why the banked curve makes your car go uh in that curved path and this is the speed that we calculated here is the speed that you could go around that curve at 20 meters per second approximately without friction so even if the road were icy you could go at that speed around the curve without any friction um i'll just tell you a little story i used to live in eastern washington and i don't know if you've ever experienced freezing rain but freezing rain is something that occurs in the midwest states and some of the northern states it's not the same as hail so freezing rain is rain that the air is actually not freezing temperatures but the ground froze hard overnight and the air is kind of warmed up to the point that the the water droplets falling through it are not freezing they're still liquid water but when they hit the ground the ground is still frozen from being frozen hard overnight and when the water drops hit the ground they go splat and they they flatten out and they freeze into a sheet of ice and it is the the worst driving conditions i've ever driven in my life i used to live in denver colorado for quite a few years and i lived in the mountains in new mexico where there was a lot of snow and um so i was pretty good at driving and ice and snow but this freezing rain was next to impossible to drive in anyway i had a banked curve that i had to drive around to enter a freeway when i uh on my way to work and it was early in the morning i had an early morning class and so i was getting on this bank curve and i'm trying to get on the freeway i was actually going kind of this way on an on-ramp anyway so this bank curve what i would do is i would put this right tire way up here where the road was sanded so there was actually some friction because what was happening was cars that were going around this bank curve they were going kind of slow because they couldn't even get up speed and they would slide down they would just slide right down and there was a bunch of cars just sort of stuck on this part of the on-ramp and on the highway there was also part of the highway it was banked like that and but i managed to never do that because i went way up here as i got onto that bank section and i put my tire way over on the right where it was sandy and i could go around without sliding down but if i could get up to if these numbers were the same if it were 18 degrees if i could get up to 20 meters per second i could actually keep going at that speed even without that friction that's what this problem is telling us if i went any slower than that i would slide down toward the middle i would veer in toward the middle and if i went any faster than that i would veer out and eventually slide outward but right at 20 meters per second it's just the right speed to keep me on that circular path all right so back to the other problem number 25 so remember this normal force has two functions it holds the car up and it accelerates it that's exactly what the tension force is doing in this problem this tension force has two components one of those components is holding the block up one component of the tension the y component is just canceling out the weight that's f t y and it also has an x component that is what is accelerating that is f t x okay so i'm going to say you guys go through the math here and what you're going to be able to show is that the tangent of theta is equal to v squared over rg do this math actually there's a little bit of physics in getting this equation built and then theta is the inverse tangent of that equals 28 degrees keep this equation in your head for just a second tangent theta equals v squared over rg let's go back one more time here look at this this is exactly the same equation tangent of theta is v squared over rg if we brought the rg down there it's the exact same equation all right guys that's number 25. now at this point you might want to take a break from this video because this video is a little long and go and work on all the horizontally curved problems up through number 26 and then come back to this point in the video and next we're going to be looking at vertical circular motion okay so i hope you took a break and went and tried to solve everything up through number 26 and now we're switching gears a little bit still in the theme of circular motion but we're going to do one kind of related problem here number 27 and these figures that i've drawn here for you which i'm really proud of because they're really cute um they are either on earth or in deep space is the setting of each of these scenarios and it tells you which scenario they are and the difference is is that on earth there's gravity the acceleration due to gravity is 9.8 meters per second squared in these two pictures and in these three pictures there's no gravity and so we're out in deep space without any gravity so here in uh this first picture we're remaining at rest we're just standing in this elevator or this box that's hanging from this cable and we're remaining at rest and so um the first thing we're being asked to do is draw the person's acceleration vector or write a equals zero if it's zero so let's just do that for for all of these so in this case there is no acceleration a equals zero here though somebody cut the cable and so this person is accelerating at free fall so at 9.8 meters per second squared downward so the acceleration is that way and we could say it has a magnitude of 9.8 meters per second squared down is there on earth here this person is in deep space there's no gravity and they're just remaining at rest so this person also has no acceleration and i put a vector symbol on this one and this one uh you could leave it off it's the same thing the magnitude would be zero so the whole vector is zero in deep space here i added a comma here this comma shouldn't be here if we're accelerating at 9.8 meters per second squared you can see that this is like a rocket blasting uh fuel out that way which pushes the rocket upward this way so the acceleration is this way 9.8 meters per second squared and that's the acceleration and then here is a little tiny version of this person right here and so this is kind of a zoomed way out picture so we could see this whole rotating space habitat there's still no gravity but it has centripetal acceleration a that is mean meaning the person has centripetal acceleration a that way of 9.8 meters per second squared and so we have a circular motion and an acceleration that way and so now we're going to draw all the force vectors uh acting on the person and the way we determine what the force vectors are is we ask the question um is there a weight vector yes uh for the people on earth anyway so these two both have weight vectors so the force vectors there's going to be a weight vector for these two because they are on earth there is no weight vector for any of these three because they're out in outer space there's no gravity out there and then the other thing that could be happening is there there are no ropes touching the person so there's no tension but there could be a normal force if the person is touching the floor and that floor is going to push up on the person like in this picture there's going to be a normal force that cancels the weight vector because the person's touching the floor and the normal force cancels out that wave vector and makes them not accelerate so that kind of all should make sense together here there is no normal force so we could say n equals zero there is no normal force in this case there's no contact in other words they're not touching anything so there's no normal force and that's why this person accelerates is because the floor is not keeping them from falling anymore and so they're just falling because of the weight vector so they're falling in free fall at 9.8 meters per second squared here we have no weight and we have no normal force so there's no normal force and there's no weight here there's still no weight because we're still in outer space no weight and here there's still no weight uh we're out in space and but there is right here they are touching the floor and that floor is pushing on them and that is what makes them accelerate upward in our picture here because just like this person the floor is pushing up on their feet this floor is pushing up on their feet in this case it's just canceling the weight and making them not accelerate in this case there is no weight to cancel and so they are accelerating and finally in the last picture well what makes this person if we remove this floor here and this person from this person's situation they have a tangential velocity to the right at this moment because they're going around counter-clockwise the way it's drawn here and so they would if we remove the floor they would just keep going with their tangential velocity forever and ever because that they would obey newton's first law objects in motion keep going in a straight line motion unless they're acted on by an external force and the external force in this case is from the floor and that's what's causing them to accelerate it's a normal force and i'll draw it over here it's just like this one it's pointing in the direction of acceleration and that's what's causing the acceleration all right so we're going to circle the choices here and we want to know does the person have true weight and does the person have a parent let's just do the true weight first true eight is the ones that are on earth they actually weigh something they have a true weight vector there they have a gravitational pull so these do not apparently is when you feel like you weigh something whether you do actually weigh something or not and if they don't have a parent weight then that means they would feel weightless or they would have apparent weightlessness possibly true weightlessness but apparently this is when you don't feel like you weigh anything so in this first one yeah they feel like they weigh something in fact if you put a bathroom scale under them that scale would just read their weight and so they do have an apparent weight also but in this one if you put a bathroom scale under them it would just float like they're floating and they would not have any apparent weight they feel weightless even though they're not weightless they feel weightless there's a uh an airplane you can look this up on youtube it's called the vomit comet and that's kind of a disgusting name but it's a an airplane that they you can take rides in and you have to pay a lot of money and they do science experiments in this uh airplane as well and they they do is they take this plane in a like a parabolic trajectory and they use uh this guided path this flight path uh with it's all programmed in to make it feel like they're in a weightless environment they have true weight but they make it so that it's basically behaving as if it's in free fall the whole time but it's it's as if it's in free fall going up peaking and then coming back down for for quite an extended period of time and you feel weightless and so your acceleration is always 9.8 meters per second squared the reason they have to use the engines to do that is because air resistance would keep you from accelerating at this rate and so on the way up they actually have to um adjust the thrusting from the airplane in order to make the acceleration to make this plane slow down at this rate and then on the way back down they have to adjust the thrusting to make it speed up at this rate so that it feels weightless but anyway it's there's truly weight and that's why they're falling and accelerating but they don't feel any weight they they don't feel any force of their weight they they feel weightless and that's why they call it the vomit comment is that feeling of weightlessness makes you feel uh different and kind of makes your stomach do weird things all right so what about this person same thing they feel weightless so they have no apparently either this person this person though if you put a bathroom scale like imagine the bathroom scale here that's a way of understanding the bathroom skill would just float right along with them they couldn't step on it and make it read anything but here if they put a bathroom scale between their feet and the floor it would actually read something and that would be their apparent weight and they feel like they weigh something because that floor is pushing against them and making them accelerate and they actually feel like they weigh something even though they don't and this person too this is a an artificial gravity simulation is what this is and this would this is another way to simulate artificial gravity in a space habitat maybe you've seen sci-fi movies where they have a rotating space habitat and so they also if they put a scale beneath their feet and the floor it would actually waste they would have an apparent weight it wouldn't be a true weight but it would be an apparent weight all right so for each question below circle the correct choice at the right so go ahead pause the video for a minute and answer all these questions all right i hope you've tried to answer these questions yourself so can a person have non-zero apparent weight while having zero true weight and the answer is sure we see a few examples or a couple of examples non-zero apparent weight right here zero true weight right here so yes that is a yes these two cases right here can a person have second question non-zero true weight while having zero apparent weight apparent weightlessness and the answer is yes there's one case of it right here when you're in free fall that's when you would experience that so take a look at all these situations that have apparent weight and notice what is common in these free body diagrams these ones that have apparent weight they're the ones where the people's the person's feet are touching the floor and if they put a scale between their feet and the floor it would read something and they have a normal force that scale would push up on their feet with the normal force that's what the floor is doing and that is why they feel apparent weight the force that you feel right now if you feel like you weigh something if you're sitting down like i am there's a chair pushing up on you that's the force you feel the earth is pulling down on you you don't actually feel that force the force you feel like if you were in this situation if someone removed the chair you would feel this weightlessness you might feel kind of sick but you don't feel like there's a force especially if you were in a closed environment you would just feel weightless you'd feel like you're floating around even though you wouldn't be weightless the force that you feel the reason your rear end is tired after sitting for a long time is not the earth pulling down on you it's the chair pushing up on you or your feet at the end of the day if you're standing it's not the earth pulling down on you it's the floor pushing up on your feet that's the force that you actually feel and so that's what we call the apparent weight that's the force you actually feel is not your weight it's the normal force um if there were a tension force if you were hanging from a cable instead of being supported from the force below that tension would be your apparent weight and so it's the contact force it's the normal force or the tension force if it was if you're hanging from a cable and that contact force is the one that you actually feel you don't actually feel this long range weight vector this long range non-contact force of the earth pulling on you you don't feel that you just feel the ground pushing back up on you when it stops you from accelerating all right so hopefully that makes some sense you might want to go back through that again yourself but that's going to really help you understand some things going on in vertical circular motion like this situation here we're going to compare that all right so let's go look at one example of vertical circular motion and actually i have a couple of demonstrations here as well but let's read this one first a 9.5 kilogram ball hangs from a cable and swings in a vertical circle of radius 0.85 meters at the lowest point in the circle the ball speed is 25 2.8 meters per second at the lowest point draw a free-body diagram of the ball determine the magnitude of the centripetal force acting on the ball and determine the magnitude of the tension in the cable and we want to know what is the apparent weight at the lowest point okay so here is i'm going to draw a circle here's the center of the circle and i'm going to draw a circular path do my best to make this circular and the um the ball hangs from a cable and it swings in a vertical circle so we're going to look at the lowest point so here's the ball and it's hanging it's not just hanging there it's actually moving tangentially let's draw the tangential velocity vector there and it's going around in this path and we know that v is 2.8 meters per second and the radius is 0.85 meters and that's the mass of the ball i'll just write it here the mass is 9.5 kilograms and so at this point we want to know the what the free body diagram is so that's part a before we actually do the free body diagram though let's draw the acceleration vector remember this is a centripetal acceleration vector so it's going to go toward the center because it's circular motion and so that means we know that we have to have a net force in that direction and so think about what that means with the forces in the free body diagram so the forces in the free body diagram think about what's going on the ball does it have a weight vector yes it has a weight vector the earth is pulling on it it's pulling down and does it have anything else well there's a string pulling it and think about the the string if the ball is just hanging there then that tension would be just as long as the weight vector but because the ball is going around in the circular path and because right at this moment it's accelerating that way it's not moving that way it's actually moving this way but the acceleration is toward the center and so it has to have more force toward the center than it has away from the center and so there's a tension force that's larger than the weight vector let me show you a similar situation here in another video this video i call it um vertical circles here but it's not really i didn't do a vertical circular motion here so what's happening here is first of all the spring scale is just um let's go back a little bit here this spring scale is just uh sitting there and it's just holding this block so there on the block here there's a weight vector of about 10 newtons and a force up that's the tension in this hook that is about 10 newtons and that's what the scale reads is the tension in the hook now when i accelerate it upward here in the video let me just play it and then i'll go back and show you kind of the part that i want to talk about [Music] okay so just right there is the main part that i want to focus on is right right when i first start accelerating it upward here we go it goes and you see there's a rapid force increase here now the weight didn't change it still weighs about 10 newtons just under 10 newtons but now there's an upward force of 17 newtons so imagine those two vectors 10 newtons down 17 newtons up and so you know there's going to be a net force of seven extra newtons up so it's cancelling the weight but giving seven more additional more units of force up and that's what's causing it to accelerate is that net seven newtons of force that's what's happening here except this ex it's not speeding up in that direction but it's accelerating in that direction and that the velocity vector is getting pulled to the left if you were if you called that straight then it would be pulled to the left and that force larger force upward than downward is what makes that happen and that's why we call that acceleration is because there's a net force making that happen and so here the free body diagram we've done part a determine the magnitude of the centripetal force now what we mean by that is the centripetal force b the centripetal force is just the net force toward the center or in other words if we make a y-axis let's just make a y-axis toward the center and we'll make that our y-axis here which i guess i already did draw a y-axis here but i'll color it orange here so it's just the net force in the y direction is what we mean by that so that's just more of a kind of a funny question f net y well we know that's just equal to m a y and what is a y well in this case the centripetal acceleration is also in the y direction so a sub c is v squared over r and a sub y is just plus a sub c because it's going in the positive y direction is plus v squared over r and so i could just say oh okay so that centripetal force is just the net force toward the center or the net force in the y direction and one way i can find the net force is because i don't necessarily know both of these forces is i can just take the mass and multiply it by the centripetal acceleration which is m times v squared over r and i know all those numbers m v and r and so i can just calculate that value so that's what we mean by centripetal force it's just the net force in the centripetal or the net force toward the center or in this case the y direction and when we calculate that number it's just 87.6 or about 88 newtons so that's the centripetal force now we want to know part c what is the magnitude of the tension in the cable so part c we want to know what is f sub t part b was what is the net force part c is what is just the tension force and now we're going to do what we've been more accustomed to doing is we're going to go like this f net y equals m a y i'm just going to start over and say okay this is the sum and this is pretty simple i only have two forces one with a positive y magnitude one with a negative so it's plus f t plus the negative of the weight and i'm just going to say negative mg this is this is a y component of plus f t the weight has a y component of negative weight or negative mg and that's equal to m and this is v squared over r and now i can just solve this and say okay f sub t is just mg plus m b squared over r and we could even factor the m out if we wanted g plus v squared over r and that comes out to be 180.7 approximately or about 181 newtons or we could round it to 1.8 times 10 squared if we wanted to be to keep it to 2 sig figs we could write it as 1.8 times 10 squared i'm not super picky as long as you keep it to two or three in the end we could write it as that instead okay so um finally the last question part d is what is the apparent weight and that is what if we were holding it with a scale like like what we had there what in other words what would the scale read or what would it feel like if that were you hanging onto that string instead of the ball what would it feel like and you would be pulling just as hard as that string is in other words it would feel like you weigh this much the tension force and so that is for part d the apparent weight is equal to f sub t which is that 1.8 times 10 squared newtons it's the magnitude of the contact force it's the magnitude of the force that's that you actually feel and that's typically going to be either a normal force or a tension force depending on which one is are you feeling and how you're connected to whatever it is that's making you go on that curved path all right so that is number 28. now that's the last full example but i do want to talk about a couple of other problems i want to talk about we'll come back and talk about 29 but let's go talk about number 30 so i have another little demonstration i want to show you um this one says a ball in number 30 is twirled on the end of a string in a vertical circle of radius 0.9 meters at a constant unknown speed at the top of the circle the tension of the string has magnitude equal to half the weight of the ball so it's telling you here's how we could write this f t at the top has a magnitude of one-half the weight or one-half mg that's what that's saying and so here we have a let's draw this that'd be like the center and then there's this um it's just a ball and there it is at the bottom so this actually looks a lot like the problem we just did let me see if i can draw a reasonably good circle here so that would be at the bottom pardon my my very crooked circle all right so here at the bottom let's say it's moving that way that would be its tangential speed or just we could just say v and up here at the top it would be up here and there's a string connecting it here or a string connecting it up here at the top it's moving that way and we could call that v sub t or just v and so and and it's moving at the same speed here so you've maybe done that if you've twirled something i have a little video clip i wanted to show you here where i look very serious so that's kind of what's going on i did this is a little different in that what i did here was i put a little mass inside the cup it's that little copper mask that i used in another demonstration because i just wanted to talk about that as well how does that mask stay in the cup and so when it's getting twirled around ah right there you kind of heard the mask flop around because what i'm doing is i'm slowing it down listen again and watch slowing down slowing down slowing down thunk you heard that so that's where things are starting to break down at the top here that's where the string is going slack and that little mass is starting to fall out so there's like a minimum speed that i need in order to keep that string tight or taut and i need to keep that mass in there if it's like something um stuck to the bottom of that cup as it's twirling around so that is there that's plenty fast but then when i slow it down there's that minimum speed and you can kind of see the string it's a little hard to see in the video clip here but and i can't necessarily see the string very well but right here that string is going slightly slack and that's when the mass everything's kind of in free fall right there the strings not doing anything and it's just under gravity's effect only and so it's that's when the mass loses contact and feels weightless and uh so anyway but at the bottom uh the ball looks just like what it did in the previous problem that we just did number 28 but at the top it looks a little different so i just want to talk about the top so i'll let you do part a part one let's do part two up here at the top free body diagram and you can do part one at the bottom the free body diagram um so at the top think about what's happening is the string is now below the cup the cup was up here and or the ball would be up here and the string is below it pulling down so the ball is going over the top and the string is pulling down on it the string would be pulling and so there's a tension force that way and the weight vector they're both that way and um and so here there's going to be a tension force that way at the top and we were told that it's half the size of the weight vector so we know that the weight vector which is also pulling down is bigger twice that size and we can actually figure out what the magnitude is because that's the mass and the weight is equal to mg and that tension at the top is half mg so we could figure out what both of those are if we wanted to so that's at the top of the path at the bottom of the path it's going to look just like what we did in the previous problem so i'm going to let you do that part but i want to talk about so what happens at the minimum speed because it talks about this part b it says determine the speed of the ball and then determine the tension in the string at the top so or sorry at the bottom so do this once you draw on this then do f net y equals m a y and you have a couple of choices here like if we go back here let's just look at what we did here we made the y direction up positive and so what we did here is we had plus ft minus mg equals plus ma that was a plus right there and uh that because the acceleration was also up toward the center now let's look at this other situation that we just looked at here in this problem is at the top there's an acceleration vector but it's going down it's toward the center so the acceleration here is that's the acceleration at the top and the acceleration at the bottom is going up toward the center because the acceleration is always toward the center so if you're up here it's pointing down if you're down here it's pointing up so that's the acceleration at the bottom they're different they're both going to be they have the same magnitude a equals v squared over r and since it tells us that it's a constant unknown speed that v is the same and so v at the top and v at the bottom are the same and so this is going to be the acceleration at the top magnitude and the acceleration at the bottom magnitude and so that's the idea um in terms of how we set this up but so this acceleration is down so you have a couple of choices here what you could do you could stick with the positive direction being up make a positive y-axis going up and i'll just do that here and you can choose how you want to do this but if you make the positive y-axis going up then notice that this is now going to be negative this acceleration y component will be negative and so this would be minus v squared over r i'm just going to say it i'm going to let you guys write it out this would be minus v squared over r and both of these would be this would be minus this magnitude and the weight would be minus mg so all three of them would be negative you'd have minus the weight minus the tension equals m times minus a minus v squared over r in other words or if you wanted you could switch it and just make the y axis for this free body diagram and this actually makes things a little simpler if you want to do it this way you could make the y axis down for this part of the free body diagram and that way your acceleration is positive here and both of these forces are positive then it's just plus this force plus this force equals m times plus a and that's simpler because it just avoids the three negative signs but it doesn't really matter because even if you have a negative negative and a negative if you multiply that whole equation by negative one it becomes all positive so as long as they're all the same as the important thing because they're all three down the acceleration the weight and the tension are all down now down here at this for this free body diagram you can have a different positive direction for this free body diagram you can do it just like you did in the previous problems you can make your positive direction up for this one as long as you're consistent in this equation and then you'll do this free body diagram and you do this f net y equals m a y at the bottom and here you're doing this at the top and so as long as you're consistent in each one it doesn't matter which way you choose the y-axis for each one as long as you're consistent within that part in other words this is negative this is negative and this is negative the acceleration and the two force vector y components are all negative or they're all positive but the result is going to be the same and down here same idea just be consistent with whatever your y-axis choice is in building this equation all right so that's a little bit that i wanted to mention um number 31 let's talk about number 31. this one is says the track is at it is a vertical circle with radius 3.5 meters and the stunt driver goes past the point one at the bottom at the speed of 18 meters per second what speed must she have at point 2 at the top so the normal force there has the same magnitude at the bottom so in this problem v1 is 18 meters per second that's v1 and what we want to know is is what is v2 so these are different speeds but the normal forces are the same it says the normal force at the top is the same magnitude as it did at the bottom so the normal force one equals normal force two we could call it n we don't know what it is but we know they're equal to each other this problem is really similar it's a practice problem but i really recommend that you do it because i think it adds some things that are a little different it's very similar though to number 30 in that at the top think about the forces here there's a track above the car so the track is pushing down on the car which is very similar to what we saw in number 30 where we had a string below the ball pulling down on the ball so there was a tension pulling down on the ball in number 31 we have a normal force pushing down on the ball but the free body diagram is going to look almost just like this except this is a normal force instead of a tension force so that's one difference with number 30 and 31 and then also just kind of what you're solving for is different okay and then one last one i want to talk about is number 29. um a skier of mass 65 kilograms has a constant speed of 15 meters per second and she passes over the top of a hill so let's just draw this i want to kind of help you with just drawing this one so there's a hill which has a radius of curvature 75 meters so this is a hill that she's passing over and so if we could think of this as part of a circular path and she's skiing over the top of this hill she's very happy she's skiing okay and anyway there's a radius of curvature of 75 meters and the center of curvature if she goes over the top of this hill is going to be down here somewhere that would be the center and the radius and let's make the radius to her center of mass that would be the radius and let's call that r1 is 75 meters and now think about what's different than from this one compared to the one we just looked at number this one number 31. here the car is under the track and so think about this the directions of the forces there's the weight vector pointing down and the normal force from above is pushing down so there's two forces down just like in number 30 there's a tension force down and a weight vector down when we're at the top but here in number 29 here she's on top of the curve not below the curve like in number 31 the car is below underneath the track but here she's on top of the track that's curving like so and so there's normal forces actually up but think about what her free body diagram is going to look like it's going to have a normal force up and a weight vector down because the normal force is going to push up on her because she's on top of that snow but the acceleration is toward the center so think about what of those two forces has to be bigger is the weight vector bigger or is the normal force vector bigger to make her acceleration go down she has to have more weight than normal force if that makes sense because we need more force down than up to cause an acceleration down that's why when you go over a curve really fast like in your car if you've ever driven over a curve really fast it makes your stomach feel kind of weird because you're feeling less weight less apparent weight and that's what it's asking you to solve for is you want to find what is her normal force at the top you want to find what is that normal force n1 we'll call it and actually just call it n i we don't need to call this r1 i was thinking there was another part to this problem but there's not so this is just the normal force we want to solve for what's her normal force or her apparently that's what she feels and then in part d of the problem suppose the skier goes over again at higher speed at what speed will she just barely feel weightless in other words her apparent weight is zero or in other words that means she's gonna lose contact you've seen this in the movie where the car goes over the hill too fast and it loses contact with the road that's what's happening here is she's going so fast that she just barely loses contact with the snow and the normal force becomes zero so the normal force in part d is zero in other words that's what we mean that her apparent weight is zero she loses contact and so there is no normal force like what we saw back in this case when you lose contact that's when there's no normal force and no apparent weight in these two pictures all right everybody that was a very long section um let's just go do a quick summary here of everything so again we just follow the same process we always follow when applying newton's second law we draw a free body diagram don't make up any weird forces that aren't there apply newton's second law making the acceleration the y-axis or the x-axis along the acceleration vector and that's going to be the magnitude is v squared over r or usually that or 4 pi squared r over t squared are the two that we normally use here and then if there is another direction to consider then that component would be zero if it's a vertical circle typically we're going to make a sub y equal to the acceleration because our acceleration is going to be along the y axis and then we substitute in and solve for whatever it is we're trying to find a apparent weight is this idea of the weight that you feel and that's the magnitude of the contact force all right so go back through those example problems and if you haven't done the horizontal circular paths i'd suggest go through those examples and just solve those ones up through number 26 first then go back through example number 28 and the ones i discussed with you here and then go and try and solve those problems all right and that is it for this section