so let's say if you have a ball and the ball is moving to the right and there's a force that's acting on the ball and it's parallel to the velocity vector what's going to happen to the ball if the force and velocity vectors are parallel the ball is going to speed up now what if the force and the velocity vectors are anti-parallel if they're opposite to each other what's going to happen to the speed the ball is going to slow down now what about if the velocity vector is perpendicular to the force vector what happens if these two are perpendicular then the ball is going to turn at constant speed it's going to change direction this force is known as a centripetal force provided that that force remains perpendicular to the velocity vector if that happens then we have uniform circular motion now the centripetal force is not really a new kind of force rather it's created by another force so let's say if you have a ball attached to a rope and you're swinging this ball in a horizontal circle what provides the centripetal force in this problem you provide the centripetal force and that force is transmitted through the rope which is known as a tension force so in a horizontal circle the tension force provides the centripetal force in this example now what about when the sun orbits the earth i mean the earth orbits the sun rather what provides the centripetal force in this case because we have circular motion in this example so what keeps the earth in orbit around the sun in this case it's a gravity gravity provides the centripetal force gravity pulls the earth towards the sun but the earth wants to fly outward towards outer space so instead it turns so gravity keeps the earth in orbit around the sun so as you can see the centripetal force is not really a new kind of force it's a force that's provided by another kind of force so any force that causes an object to move in a circle can be labeled as a centripetal force so another example is let's say if you have a car turn in in this case the static frictional force provides the centripetal force it allows the car to make that turn on the road another example that you won't see in this video but you'll learn more about it in the next semester physics is the magnetic force let's say if you have a proton or any positive charge and this positive charge is moving towards the right and the magnetic field let's say it's going into the page in physics 2 you'll see it's represented by an x symbol a dot means out of the page so the magnetic field is going into the page and this charge is moving towards the right it's going to field a magnetic field i mean a magnetic force that's going to cause the proton or the positive charge to turn and so whenever you have a moving charge in a magnetic field that charge is going to turn it feels a magnetic force exerted on it by the magnetic field and so this magnetic force can behave as a centripetal force because it's causing the object to turn so those are the main four examples that you'll see in physics in the case of a rope tension can provide the centripetal force if an object is moving the circle in the case of planets it's gravity in the case of a car turn in static friction and if you have a moving charge in a magnetic field it's going to be the magnetic force that provides the centripetal force so how can we find the equation of the centripetal force now according to newton's second law the net force is equal to the mass times the acceleration so the centripetal force is mass times centripetal acceleration and the centripetal acceleration is v squared over r so the centripetal force is m v squared of r it's directly proportional to the mass and it's proportional to the square of the speed and it's inversely related to the radius of the circle now there are some other things that you need to know about this equation for instance whenever the mass increases you need to know that the centripetal force will increase as well if the speed increases the centripetal force will increase and if you increase the radius of the circle the centripetal force will decrease so imagine if you're driving a car and if you make a sharp turn if you make a sharp turn you're going to experience a very large centripetal force you're gonna feel like you're flying out of the window if you decrease the radius the centripetal force will increase but if you make a wider turn you won't feel like you're flying out of the window as much because the centripetal force is a lot less if you increase the radius of curvature now if you try to make a wide turn at a very high speed you're going to feel a much larger centripetal force and if you make a sharp turn with a very high speed you might actually fly out the window so that feeling that you get when you make a turn you can increase or decreasing so if you want to decrease that feeling when you're turning just decrease the speed and increase the turn the radius of the turn and you won't feel that centripetal force effect as you would when you make that turn so just make a wider turn at a slower speed now if you want to fly out the window drive fast and make a sharp turn i would not recommend that though now here's the question for you if you double the mass what effect will that have on the centripetal force if you double the mass the centripetal force will double if you triple the mass the centripetal force will triple to answer questions like that just plug it in into the formula everything else that doesn't change replace it with a once you get three likewise if you double the speed and keep everything constant the centripetal force will increase by a factor of four now what if you double the radius of the circle what's going to happen the centripetal force will be one half of its original value and if you reduce the radius to one half the centripetal force will be one over one half which is two it's going to double now what if you triple the mass double the speed and decrease the radius by a factor of three what effect will that have on the centripetal force so if you triple the mass double the speed and reduce the radius to one third of its value two squared is four and three times four is twelve twelve divided by one third is the same as twelve times three so it's thirty the centripetal force will increase by a factor of 36. now i'll move another example let's say if you quadruple the mass triple the speed and reduce the radius to one-fourth of its value go ahead and find the effect that it will have on the centripetal force so we quadruple the mass triple the speed and the radius is one-fourth so it's four times three squared three squared is nine and four divided by one fourth that's like four times four that's 16. so we got sixteen times nine ten times nine is ninety six times nine is fifty four so this adds up to one forty four so if you quadruple the mass triple the speed and reduce the radius to one fourth of its value the centripetal force will be 144 times greater than its uh current value now let's move on to some practice problems a 0.25 kilogram ball moves in a horizontal circle of radius 1.5 meters at a speed of 30 meters per second what is the centripetal force acting on the ball so here's the ball it's moving in a circle the centripetal force is always directed towards the center of the circle and it's equal to m v squared divided by r so we have everything that we need we have the mass of the ball which is 0.25 kilograms and we have the speed which is 30 meters per second and the radius is 1.5 30 squared is 900 and 900 times 0.25 is 225. if you divide that by 1.5 you're going to get 150. so that's the centripetal force acting on this ball it's 150 units now let's move on to this example a 500 kilogram car makes a circular turn at a speed of 12 meters per second if the centripetal force acted on the car is 9 600 newtons what is the radius of the circular turn so we have the mass which is 500 kilograms we have the speed which is 12 meters per second and we have the centripetal force which is 9 600 newtons all we need to find is r the radius so let's start with this equation and let's rearrange it to get r so i'm going to multiply both sides by r and so these will cancel so i have the centripetal force times the radius is equal to mv squared next i'm going to divide both sides by the centripetal force so the radius of the circle is going to be the mass times the square of the speed divided by the centripetal force so this is the formula that we need for this problem so let's go ahead and get the answer it's going to be the mass which is 500 kilograms multiplied by a speed of 12 meters per second squared divided by the centripetal force of 9 600 newtons 12 squared times 500 is 72 000 and if you divide that by 9 600 you're going to get a radius of 7.5 meters and that's it for this problem a 0.35 kilogram ball is attached to a rope moving in a horizontal circle of radius 85 centimeters if the rope can withstand a maximum tension force of 1500 newtons what is the maximum speed that the ball can travel in this circle without breaking the rope so in this problem we need to calculate v the maximum speed so let's solve for v in this formula so what i'm going to do is i'm going to multiply both sides by r over m now the reason why i'm doing that is to cancel the radius on the right side and also to cancel the mass so all i have left over on the right side of the equation is v so now i need to take the square root of both sides so v is equal to the radius times the centripetal force divided by the mass with a square root symbol so now let's plug in everything that we have into this formula the radius is 85 centimeters and we need to convert that to meters one meter is 100 centimeters so to convert it take 85 divided by 100 and that will give you 0.85 meters the centripetal force is going to be the maximum tension force which is 1500 newtons whenever you have an object moving in a horizontal circle and if it's attached to a rope the tension force is approximately equal to the centripetal force if the ball is moving fast enough if not you're going to have what is known as a tetherball experiment and it's not going to be the same but if it's moving fast enough you don't have to worry about the weight force of the ball it's going to be insignificant but i'm going to go into more detail into that topic in another video right now let's plug in the mass which is 0.35 0.85 times 1500 that's 1275 and if you divide that by 0.35 you should get like 36 43 if you round it to nearest whole number and then take the square root of that number and then the maximum speed is going to be 60.4 meters per second so now you know how to calculate it if needed but now let's go over some free body diagrams now imagine if you're holding a rope in your hand that's attached to a ball and if you spin it fast enough in a horizontal circle it's going to look something like this if it's going fast enough it's going to appear nearly horizontal if it's not going fast enough it might be slightly at an angle and if it's going slow the angle is going to be larger so it's going to look like this now the tension force is the force in the rope so that's going to be the tension force that tension force has an x component and a y component which i'm going to call f t x and f t y now the ball also has a weight force mg this if it's moving relatively slow this angle is going to be significant now the centripetal force is the force that's going to keep it moving in the circle so it's the horizontal component of the tension force so when the ball is moving slow ftx is equal to the centripetal force this is always true but i'll talk more about when it's moving fast fty is equal to the weight force this component balances the weight force now if you want to find a tangent force it's going to be the square root of f t x squared plus f t y squared to find f t x you need to find the centripetal force which is mv squared over r and then fty is mg so if you want to get the exact answer for the tension force here's the equation you need to use ftx is mg right no i take that back f t x is uh mv squared of r so you have m v squared over r squared plus f t y squared which is mg squared so for any situation this will be the exact answer for the tension force now if the ball is moving fast enough it's going to look like this now we can't really see anything so i'm going to use this picture this is still going to be ft and you're still going to have ftx and fty but when it's nearly horizontal ft is approximately equal to ftx fty will be negligible so if the ball is moving fast you could say that ft is approximately ftx which is the centripetal force so that's going to be mv squared over r this part becomes insignificant now let's use this equation to see if we can get the same answer so i'm going to square both sides so i'm going to have ft squared is equal to mv squared over r squared plus mg squared so the tension force is 1500 the mass is 0.35 we don't have the value of v r is going to be 0.85 and this is squared and then we have mg squared so that's 0.35 times 9.8 squared now 1500 squared that's a big number that's 2 million 250 000. 0.35 divided by 0.85 that's 0.411 well let me square it too so i could just make this a lot easier it's like .41176 and then it repeats once you square it you're going to get 0.16955 and then v squared squared that's going to be v to the fourth power 0.35 times 9.8 that's 3.43 and then once you square it it's 11.7649 now if you take this big number and subtract it by that small number it's going to be insignificant but we'll do it anyway let's subtract both sides by 11.7649 so this is going to be about 2 million 249 988.235 so now let's divide both sides by 0.16955 so you should have one three two seven zero three five two point three two is equal to v to the fourth so now i'm going to raise both sides to the 1 4 power to get v so take the 4 root of both sides and this will give you 60.356 which rounds to 60.4 meters per second so as you can see the answer doesn't change much if the ball is moving pretty fast and if the mass is small then we could say that the tension force is approximately equal to the centripetal force so it's going to be mv squared over r if it's a horizontal circle so if you have an object that looks like this if it's moving the horizontal circle then the tension force is just mv squared of r if it's moving fast enough if it's not if it's at an angle then it's different then you need to use this equation it's going to be the square root of mv squared over r squared plus mg squared if it's at an angle but that will give you the exact answer it's just a lot more work here's the last problem asteroid x orbits moon x y z with a speed of 500 meters per second and a centripetal force of 1500 newtons and asteroids x y and z all have the same mass now if asteroid y orbits the same moon with the same orbital radius as asteroid x but at a speed of a thousand meters per second what is the centripetal force acting on asteroid y so let's say this is a moon a moon that's orbiting some planet and then we have an asteroid orbiting that moon so this is asteroid x now asteroid y is in the same orbital radius as x now x is moving at 500 meters per second currently and asteroid y is moving at twice the speed a thousand meters per second and let's go ahead and put asteroid z in part b asteroid z is at half of the orbital radius relative to x so somewhere here we're going to have asteroid z and it's going at 1500 meters per second now the centripetal force on x is 1500 newtons what is the centripetal force acting on y how can we find the answer well we can't use this equation directly we have to use it indirectly now if we focus on x and y everything is the same they have the same mass and the same orbital radius the only thing that's different is the speed now what happens to the centripetal force if the speed is doubled so the mass doesn't change let's put a one the speed doubles and the radius doesn't change two squared is four so the centripetal force will increase by a factor of four if we double the speed so if x has a centripetal force of 1500 then y is going to be 1500 times four so the centripetal force is going to be 6 000 newtons now if you want to come up with an equation that relates centripetal force to speed given that everything else stays the same like the mass and radius here's the formula that you could use so we're going to say f2 over f1 where these two represents the centripetal force of two different objects that's going to be m v2 squared over r divided by m v 1 squared over r now i didn't use any subscripts for m and r because they're the same so therefore they're going to cancel so f2 over f1 is equal to v2 squared over v1 squared this is the form that you want to use now if you want to solve for f2 multiply both sides by f1 so f2 is going to be f1 times v2 squared over v1 squared or you could just write it like this if you want v2 over v1 and then square so let's use that formula so f1 where one is the first object asteroid x is 1 500 the speed of asteroid x is 500 and the speed of asteroid y is a thousand so a thousand divided by five hundred is two and two squared is four four times fifteen hundred is six thousand so this is the formula that you could use if you need to relate the centripetal force with the speed together now let's move on to the next part part b if asteroid z orbits moon xyz with a speed of 1500 meters per second but an orbital radius that is half of asteroid x what is the centripetal force acting on asteroid z so let's think about it conceptually the speed relative to x increases by a factor of three it goes from 500 to 1500 and the radius is one half of its value so what happens to the centripetal force if we triple the speed and cut the radius in half so let's use the equation the mass doesn't change so we're going to replace it with a one the speed triple so let's put a three and the radius is reduced to one half of its value three squared is nine nine divided by a half or nine times two that's 18. so the centripetal force will be 18 times its value so it's going to be this number 1500 times 18. so asteroid z is going to have a centripetal force of 27 000 newtons now let's confirm this answer with an equation so i'm going to use f2 over f1 again which is equal to m v2 squared over r2 divided by m v1 squared over r1 so the only thing that's the same in this example is the mass that's the only thing i can cancel now using the keep change flip principle if you keep this the same change division to multiplication and flip this fraction you're going to get this expression it's going to be v2 squared times r1 over v1 squared times r2 and then let's multiply both sides by f1 so f2 is going to be f1 times r1 over r2 times v2 over v1 squared so now let's plug in the values that we have so f1 that's the original force for asteroid x that's 1500 r1 we don't have a value so we're just going to plug in one r2 is half of r1 so it's one half or you could say if if r1 had a radius of two r2 is going to have a radius of one if r1 has a radius of four r2 is going to have radius of two either case the ratio of r1 to r2 is going to be 2. v2 is 1500 v1 is 500 and then that's squared so we have fifteen hundred times two and then fifteen hundred over five hundredths three three squared is nine and two times nine is eighteen so then we have 1500 times 18 which is going to be 27 000 newtons so that's how you can get the answer now let's see if all of this makes sense so asteroid x is moving slow and it's making the y turn so therefore the centripetal force is very small asteroid y makes the y turn like x but it's moving faster so it has a much larger centripetal force asteroid z makes a sharper turn and it's moving very fast so it has an extremely high centripetal force make sure you understand this concept so if you're in a car and if you make a wide turn driving slow the centripetal force that you're going to feel is very very low you may not even feel it at all now if you make a sharp turn going very fast you're going to feel it and the faster you go and the sharper the turn is the more you're going to feel so just make sure you understand that and that's it for this video hopefully it gives you a a good understanding of centrifugal force and how to solve problems associated with it so thanks for watching you