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I want to show you how to do a slightly more sophisticated centripetal force problem and this one's a classic this is the one where there's a mass tied to a string and that string is secured to the ceiling and the mass has been given an initial velocity so that it swings around in a horizontal circle so this mass is going to maintain a constant height it's not moving upward down but it revolves in a horizontal circle so if you were to view this thing from above or below it looks something like this you'd see the ball tracing out a perfect circle if you're looking at this from right below and the question I want to ask is two things what is the tension in the rope and what has to be the speed of the mass and these are given variables we know the mass is three kilograms we know the length of the rope is two meters and this rope is making an angle of 30 degrees with respect to this vertical line right here now this problem is a classic for a good reason if you don't have a clean conceptual understanding of what we really mean by the terms when dealing with centripetal forces and if you don't have a problem solving strategy to use to tackle problems this problem will expose you so that's why we should go over this to make sure you have a clean understanding of exactly what we mean by all the different terms and to show you that there's a strategy you could use to solve any centripetal force problem and that strategy is this first draw a quality force diagram for the object or objects in your problem so let's do that first the forces that are acting on this three kilogram sphere are the force of gravity you'll have a force of gravity straight down that's going to be M times G so the way we find the force of gravity is with the formula mass times 9.8 and the only other object that's touching this mass is the rope so the only other force on this mass is the force of tension so I'm going to label that force with a capital T this will be the total force of tension from the rope and really the only other step in this problem solving strategy it's really only two steps use Newton's second law but only use it for one direction at a time and we typically only have two directions I mean we've got this vertical direction or we've got this horizontal direction we got a 50/50 shot I mean we got to pick one of the two it's not like you can pick the wrong direction if you pick the wrong direction it'll just mean that you can't solve because there'll be too many variables but what'd you write down in that process it's probably gonna be useful later in the problem anyway so don't erase it if you pick a direction that you can't solve since you have too many variables just go to the other direction and pick that direction I know what direction to pick because I've done this one before but if you don't know the worst thing you do is freeze up you got to try something and if you try the wrong direction no big deal there's only one other direction to pick so I'm going to choose to analyze these forces in the vertical direction so I'm going to say that the acceleration in the vertical direction this Y Direction is equal to the net force in the Y Direction divided by the mass and now I simply ask myself what is the acceleration in the vertical direction it's not 9.8 a lot of people want to say negative 9.8 but that's only if this ball were free-falling and this ball is not free-falling in fact this ball is not even changing its vertical height its remaining at the same constant height and that means not only is there no vertical velocity there's no vertical acceleration since the ball is not even moving vertically so on the left-hand side here this is great we could put a zero zeros are wonderful they make calculations easier so this is going to equal the net force in the vertical direction so one force is going to be the force of gravity so I'll put that in here you'll have M times G that force is definitely a vertical force but you got to be careful if we're going to consider upward is positive this mg is a negative force so in terms of Y directed forces downward typically the convention is that you choose that to be negative so I'm going to say that mg is negative and now we ask ourselves are there any other vertical forces well we just look at our force diagram so our force diagram will tell us if there's any other vertical forces we look over here the only other force is this tension force and part of this tension force is vertical we can't put in the entire tension into this formula we could only do that if this tension was directed vertically upward but it's not part of its vertically upward and part of its horizontal so part of this tension force is directed this way I'm going to call that the X component of the tension and then part of this tension is directed vertically I'll call that the Y component of the tension so we can only plug in this Y component of tension into this formula over here since we're only plugging in Y directed forces into this equation so I could just write plus T Y but I don't want to do that I want to solve for what T is not what why is so I want to write this T Y in terms of T and I can do that look at the triangle this is making this tension and this T Y is going to make an angle right here and that angle has to be the same angle that the rope makes with this vertical line so the tension is not the rope the tension is a force exerted by the rope and the tension lies along the same direction as the rope but the tension is not the rope this is a really common misconception sometimes people think oh the tension is two meters right no that's not even a force that's the length of the rope and yes the length of the rope lies along the same direction as the tension force but the tension is different from the length of the rope however this angle between the tension and this vertical component of the tension is the same as the angle between this rope and the vertical direction so that means we can label this as 30 degrees right here and now I can figure out what is the vertical component of the tension in terms of T and in terms of theta I got a right triangle here's a right angle this side that I want to find out T Y is adjacent to this 30 degrees since it's adjacent we're going to use cosine in other words we're going to say that cosine of theta equals adjacent over hypotenuse people get confused sometimes they're like wait I thought vertical was always the opposite side it'd be the opposite side if we knew this angle we don't know that angle we know this vertical angle and that means this vertical side is adjacent to that angle we know the angle we could write the cosine of 30 degrees is going to equal the adjacent side and the adjacent side is T Y and the hypotenuse is going to be T people get freaked out here they're like I've got too many variables that's okay we can label this total hypotenuse is T even though we don't know it it's alright we're going to manipulate symbols and we're going to solve for T Y if I multiply both sides by T I get to the T Y the vertical component of the tension is equal to the total tension times cosine of 30 degrees this is the vertical component of the tension and this is the force that I can plug into my vertical net force so I'm going to add because this points upward T the total tension times cosine of 30 and then I have to divide by my mass so I can solve this for T now if I multiply both sides by M M times zero is going to be zero I'll move the mg over and then I divide by cosine of 30 that the tension in the rope is going to equal mg the force of gravity divided by cosine of 30 and if we plug in numbers we'll get the T equals the mass was three kilograms G is always 9.8 divided by cosine of 30 gives us a tension of about 30 3.9 Newtons I'll just say that that's 34 Newtons so that's the tension in the rope that's the first thing we wanted to find we just found it over here that is the tension in the rope so let's do the next part let's try to find the speed people get a little concerned now they're like what do I do don't deviate from the plan we drew our force diagram we used Newton's second law for one of the directions you still got work to do then do Newton's second law for another direction we're just going to do this for the x-direction so we'll do the x-direction I'll see that the acceleration in the x-direction equals the net force in the X Direction divided by the mass and I'll ask myself the same question is there any acceleration in the X direction there wasn't any acceleration in the Y direction you might think there's not any in the X direction but there is this mass is going in a circle that means there's going to be centripetal acceleration in this direction so this horizontal Direction is essentially just the centripetal direction so to make that more clear I'm just going to put AC and FC and whenever you have centripetal acceleration we can replace that with V squared the speed squared divided by the radius and that's going to equal the net centripetal force over the mass what force is acting in the centripetal direction well you can figure that out it's just the force that was acting in the X direction because this is the X direction the X Direction is the direction that happens to be pointing toward the center of the circle that's why the X Direction here is just the centripetal direction so to figure out which forces are centripetal I just look at my force diagram I drew this for a reason I drew this so when I look at it I can figure out what forces are vertical to put in over here and what forces are centripetal ie horizontal to put in over here the only force that's horizontal is the horizontal component of the tension that's this TX but again instead of just plugging in TX we'll plug in what this component is in terms of the angle and the total tension just like we did over here T why was t cosine 30 so it might not be that big of a surprise that TX the horizontal component is just going to be T sine 30 and if you don't believe that you can prove it to yourself think about it this TX component is the opposite to this angle and for opposite we'll use sine so we could say that sine of 30 would be the X component which is the opposite side over the hypotenuse which is T and if you solve this for T X you multiply both sides by T you indeed get that TX is just T the total tension times sine 30 so we could plug that back in over here we know that the only component that's acting as a centripetal force ie that's pointing toward the center of the circle is this X component which we just found is T sine 30 degrees now you see why picking this direction first wouldn't have allowed us to solve because we wouldn't have known the speed V and we wouldn't have known the tension t only because we chose the Y Direction first we were able to find the tension and now that we know this tension being 34 Newton's we could plug that in over here and solve for our speed but there's a really common mistake that people make here people really want to plug in R as 2 meters because they're like hey you gave me 2 meters over here I'm going to use it that's an R right isn't that art no that is not our R here always in the centripetal formula for the acceleration this R represents the radius of the circle that the object is moving in and this object doesn't move in a circle with a radius of 2 the radius of the circle this objects traveling in looks like this that's the radius of the circle and that's not 2 meters how do we find that will again use trigonometry we're just going to say that we've got a right triangle this time though we're going to make a right triangle out of the length not of the force not of this tension we're making the right triangle out of the length but again we know that that side is a right angle we know that this side is 30 degrees so we can say that this radius is the opposite side of that 30 degrees so we're going to use sine theta we're going to say that sine of theta which is 30 equals the opposite side and that's R divided by the total length of the string L and so if I solve this for the radius I get the radius of the circle that this ball is traveling in would be L times sine of 30 where L would be this 2 meters we'll plug that in back over here we'll say that V squared divided by r which is L sine 30 is going to equal T sine 30 divided by the mass and now we can solve we can multiply both sides by L sine 30 and we'd get that V squared is going to equal T sine 30 times L sine 30 divided by the mass of the sphere and since I want to find V naught V squared I'll take a square root of both sides and when I take a square root of both sides end up with V is the square root of T sine 30 L sine 30 over the mass so if we plug in our numbers we get that V is the square root of T which is 34 Newton's times sine of 30 times L and this L is referring to this total length is 2 meters times sine of 30 all divided by the mass which was three kilograms which if you solve gives you a speed of about two point three eight so I'll just say two point four meters per second which is the speed that we were trying to find so recapping when you're solving a sophisticated centripetal force problem be sure to draw a quality force diagram then use Newton's second law for a single direction and only plug forces in that direction into the net force if the direction you choose happens to lie along the centripetal direction ie it points toward the center of the circle then you could use V squared over R for your centripetal acceleration but again only plug in forces that lie along that direction for the centripetal force and make sure that you understand when we say the radius we're talking about the radius of the circle that the object is traveling in