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Video transcript

people find centripetal force problems much more challenging than regular force problems so we should go over at least a few more examples and while we're doing them will point out some common misconceptions that people make along the way so let's start with this example and it's a classic let's say you start it with a yo-yo and you were lit around vertically and I think this is called the around the world if you want to look it up on YouTube looks pretty slick they whirled it around goes high and then it goes low so this is a vertical circle not a horizontal circle we're not rotating this ball around on a horizontal surface this ball is actually getting higher in the air and then lower in the air but for our purposes we just need to know that it's a mass tied to a string let's say the mass of the yo-yo is about 0.25 kilograms let's say the length of the string is about 0.5 meters let's say this ball is going about 4 meters per second when it's at the top of its motion and something you might want to know if you're a yo-yo manufacturer is how much tension should this rope be able to support how strong does your string need to be so let's figure out for this example what is the tension in the string when this yo-yo is at its maximum height going 4 meters per second and if it's a force you want to find the first step always is to draw a quality force diagram so let's do that here let's ask what forces are on this yo-yo well if we're near earth and we're assuming we're going to be near the surface of the earth playing with our yo-yo there's going to be a force of gravity and that force of gravity is going to point straight downward so the magnitude of that force of gravity is going to be M times G where G is positive 9.8 G represents the magnitude of the acceleration due to gravity and this expression here represents the magnitude of the force of gravity but there's another force the string is tied to the mass so this string can pull on the mass strings pull the exert a force of tension which way does that tension go a lot of people want to draw that tension going upward that's not good ropes can't push if you don't believe me go get a rope try to push on something you'll realize oh yeah it can't push but it can pull so that's what this rope is going to do this rope is going to pull how much I don't know that's what we're going to try to find out this is going to be the force of tension right here and we'll label it with a capital T we could have used F with a sub T there's different ways to label the tension but no matter how you label it that tension points in toward the center of the circle because this rope is pulling on the mass so after you draw a force diagram if you want to find a force typically you're just going to use Newton's second law and we're going to use this formula as always in one dimension at a time so vertically horizontally centripetal e one dimension at a time to make the calculations as simple as possible and since we have a centripetal motion problem we have an object going in a circle and we want to find one of those forces that are directed into the circle we're going to use Newton's second law for the centripetal direction so we'll use centripetal acceleration here and net force centripetal e here so in other words we're going to write down that the centripetal acceleration is going to be equal to you the net centripetal force exerted on the mass that's going around in that circle so because we chose the centripetal direction we're going to be able to replace the centripetal acceleration with the formula for centripetal acceleration the centripetal acceleration is always equivalent to V squared over R the speed of the object squared divided by the radius of the circle that the object is traveling in so we set that equal to the net centripetal force over the mass and the trickiest part here the part where the failure is probably going to happen is trying to figure out what do you plug in for the centripetal force so now we've got to decide what is acting as our centripetal force and plug those in here with the correct signs so let's just see what forces we have on our object there's a force of tension and a force of gravity so when you go try to figure out what to plug in here people start thinking they start looking all over know you drew your force diagram look right there our force diagram holds all the information about all the forces that we've got as long as we drew it well and we did draw well we included all the forces so we'll just go one by one should we include should we even include the force of gravity in this centripetal force calculation we should because we're going to include all forces that point centripetal e and remember the word centripetal is just a fancy word for pointing toward the center of the circle and this force of gravity does point toward the center of the circle so we're going to include this force and our centripetal force its contributing in other words to the centripetal force it's one of the forces that is causing this ball to go in a circle so we included in this formula so mg is the magnitude of the force of gravity we have to decide do we include that as a positive or a negative many people want to include it as a negative because it points down but when we're dealing with the centripetal direction its inward that's going to be positive not necessarily up that's going to be positive if up happens to point in then we'd consider it positive so if we were down here up is positive but up here down is positive because it points in toward the center of the circle and if we're over here diagonally up and left would be positive because any force that would be pointing toward the center of the circle is going to be included as a positive sign and there's a reason for that the reason we're including toward the center of the circle as positive is because we chose to write our centripetal acceleration is positive and since we know that the centripetal acceleration points toward the center of the circle if we make the centripetal acceleration positive we've committed to toward the center of the circle as being positive as well in other words we could have decided that out of the circle is positive but if we did that this centripetal acceleration that points inward would have had to be included with a negative sign over here and that's just weird nobody does that so we choose into the circle is positive that makes our centripetal acceleration positive but it also makes every force that points inward positive as well and long story short this force of gravity is going to be counted as a positive centripetal force since it points inward toward the center of the circle and we keep going we've got another force here we've got a force of tension do we include it in here yes we do because it points toward the center of the circle and do we include it as a positive or negative since it also points toward the center of the circle we're going to include it as a positive centripetal force it is also one of the forces it causes this ball to move around in a circle in other words the combined force of both gravity and the force of tension are making up the net centripetal force in this case so now if we want to solve for the tension we just do our algebra we'll multiply both sides by the mass then we'll subtract mg from both sides and if we do that we'll end up with the tension equals MV squared over R minus mg which if we plug in numbers we get that the tension in the rope is five point five five Newtons so this is to be expected we subtract the force of gravity the magnitude of it from this net centripetal force so this term here represents the total amount of centripetal force we need in order to cause this yo-yo to go in a circle but the amount of tension we need is that amount minus the force of gravity and the reason is the force of gravity and tension together are both acting as the centripetal force so neither one of them have to add up to the total centripetal force it's just both together that have to add up to the centripetal force and because of that the tension does not have to be as large as it might have been however if we consider the case where the yo-yo rotates down to the bottom of its path down here once the yo-yo rotates down to this point our force diagram is going to look different the force of gravity still points downward the force of gravity is always going to be straight down and the magnitude is always going to be given by M times G but this time the tension points up because the string is always pulling on the mass ropes can only pull ropes can never push so this rope is still pulling the mass the yo-yo toward the center of the circle so now when we plug in over here one of these forces is going to be negative before they were both positive and they were both positive because both forces pointed toward the center of the circle now only one force is pointing toward the center of the circle and we can see that that's tension tensions pointing toward the center of the circle gravity is pointing away from the center radially away from the center that means tension still remains a positive force but the force of gravity now for this case down here would have to be considered a negative centripetal force since it's directed away from the center of the circle so if we were to calculate the tension at the bottom of the path the left-hand side would still be V squared over R because that's still the centripetal acceleration the mass on the bottom would still be M because that's the mass of the yo-yo going in circle but instead of T plus mg we'd have t minus mg since gravity is pointing radially out of the center of the circle and if we solve this expression for the tension in the string we'd get that the tension equals we'd have to multiply both sides by M and then add mg to both sides and we'd get that the tension is going to equal M V squared over R plus mg this time we add mg to this MV squared over R term whereas over here we had to subtract it and that should make sense conceptually since before up here both tension and gravity were working together to add up to the total centripetal force so neither one had to be as big as they might have been otherwise but down here not only is gravity not helping the tension gravity's hurting the centripetal caused by pulling this mass out of the center of the circle so the poor tension in this case not only has to equal the net centripetal force it has to add up to more than the net centripetal force just to cancel off this negative effect from the force of gravity and if we plugged in numbers we'd see that the tension would end up being bigger we'd actually get the same exact term here except that instead of subtracting gravity we have to add gravity to this net centripetal force expression and we get the tension would be ten point four five Newtons so recapping when solving centripetal force problems we typically write the V squared over R on the left hand side as a positive acceleration and by doing that we've selected in toward the center of the circle is positive since that's the direction that centripetal acceleration points which means that all forces that are directed in toward the center of the circle also have to be positive and you have to be careful because that means downward forces can count as a positive centripetal force as long as down corresponds to toward the center of the circle and just because the force was positive during one portion of the trip like gravity was at the top of this motion that does not necessarily mean that that same force is going to be positive at some other point during the motion