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# AP Physics 1 review of Centripetal Forces

## Video transcript

what does period and frequency mean the period is the number of seconds it takes for a process to complete an entire cycle circle or revolution so if there's some repeating process the time it takes that process to reset is the period and it's measured in seconds the frequency is the number of cycles or circles or revolutions completed in one second so if there's some process that's repeating the number of times the process repeats in one second would be the frequency this means as units of one over second which is just called the Hertz and because the period and frequency are defined in this inverse way as seconds per cycle or cycles per second each one is just the inverse of the other in other words the period is just one over the frequency and the frequency is equal to one over the period one example of a repeating process is an object going in a circle at a constant speed if this is the case you can relate the speed the radius of the circle and the period of the motion since speed is just a distance per time and the distance the object travels in one cycle is two PI R the circumference the speed would just be 2 PI R / the period or since one over the period is the frequency you could write the speed as 2 PI R times the frequency since time is not a vector these quantities are not vectors and they cannot be negative so it's an example involving period and frequency look like let's say a moon travels around a planet in a circular orbit of radius R at a constant speed s we want to know what the period and frequency are in terms of given quantities and fundamental constants so we use the relationship between the speed the period and the frequency we know that for objects in circular motion the speed is 2 pi R over the period and that means the period here would be equal to 2 pi R over the speed and since frequency is 1 over the period if we take 1 over this quantity we just flip the top and bottom and we get that this is the speed over 2 pi R but we can't leave our answer in terms of V we had to express this in terms of given quantities we were given s so our answer for the period has to be 2 pi R over s and for frequency it would be s over 2 PI R which is C what is centripetal acceleration the centripetal acceleration of an object is the acceleration it's causing that object to go in a circle and it's important to note that this centripetal acceleration always points toward the center of the circle the formula to find the centripetal acceleration is speed squared divided by the radius of the circle the object is traveling in even though this has a bit of an exotic formula for the acceleration it's still an acceleration so it still has units of meters per second squared and it is a vector which means it does have a direction ie toward the center of the circle but this centripetal acceleration does not cause the object to speed up or slow down this centripetal acceleration is only changing the direction of the velocity if the object going in the circle is also speeding up or slowing down there's also got to be a component of the acceleration that's tangential to the circle in other words if the object is going in a circle and speeding up there's got to be a component of acceleration in the direction of the velocity and if the object is slowing down there's got to be a component of acceleration in the opposite direction to the velocity so centripetal acceleration changes the direction of the velocity and tangential acceleration changes the magnitude or size of the velocity but this formula V squared over R is only giving you the magnitude of the centripetal acceleration this does not account for any tangential acceleration so what's an example problem involving centripetal acceleration look like let's say particle a is traveling in a circle with a constant speed s and a radius R if particle B is traveling in a circle with twice the speed of a and twice the radius of a what's the ratio of the acceleration of particle a compared to particle B so particle a is going to have a centripetal acceleration of the speed squared over the radius and particle B is also going to have an acceleration of the speed squared but this speed is twice as much as the speed of particle a and it's traveling in a circle with twice the radius of particle a when we square the two we'll get four over two gives us a factor of two times the speed of a squared over the radius of a so the ratio of the acceleration of particle a compared to particle B is going to be one-half since the acceleration of particle a is half the acceleration of particle B centripetal forces are not a new type of force centripetal forces are just one of the any other forces that we've already met that happened to be pointing toward the center of the circle making an object to travel in a circle so for a moon going around the earth gravity is the centripetal force for a yo-yo going around on a string the tension is the centripetal force for a skateboarder doing a loop-de-loop the normal force is the centripetal force and for a car going around a round about the static frictional force is the centripetal force and these forces still follow Newton's second law but using centripetal forces means you're also going to have to use the expression for the centripetal acceleration now if a force is directed radially inward toward the center of the circle you would count that force as positive since it points in the same direction as the centripetal acceleration and if a force points radially out from the center of the circle you would count that as a negative force and if a force is directed tangential to the circle you wouldn't include it in this calculation at all you could include those forces in their own Newton's second law equation but you wouldn't be using V squared over R for that acceleration those tangential forces change the speed of the object but the centripetal force changes the direction of the object so what's an example problem involving centripetal forces look like imagine a ball of mass M rolling over the top of the hill of radius R at a speed s and we want to know at the top of the hill what's the magnitude of the normal force exerted on the ball by the road so we'll draw our force diagram there's going to be an upward normal force on the ball from the road and there's going to be a downward force of gravity on the ball from the earth and these two forces are not going to be equal and opposite if they were equal and opposite they'd balance and if the forces are balanced the object would maintain its velocity and keep traveling in a straight line but this ball doesn't travel on a straight line it starts accelerating downward so this normal force is going to have to be less than the force of gravity to figure out how much less we could use Newton's second law with the formula for centripetal acceleration the speed is s the radius is R the force of gravity is going to be a positive centripetal force since it's directed toward the center of the circle the normal force is going to be a negative centripetal force since it's directed radially away from the center of the circle if we divide by the mass which if you solve this for normal force gives you the force of gravity - M s squared over R which makes sense because this normal force has to be less than the force of gravity Newtons you versal law of gravity states that all masses in the universe pole ie attract every other mass in the universe with gravitational force and this force is proportional to each mass and inversely proportional to the square of the center to Center distance between the two masses in mathematical form it just says that the force of gravity is equal to big G a constant which is six point six seven times 10 to the negative 11th multiplied by each mass in kilograms and then divided by the centre to centre distance between the two masses in other words not the surface-to-surface distance but the centre to centre distance and even if these two objects have different masses the magnitude of the force they exert on each other is going to be the same this is illustrated by the formula since you could swap these two masses and you get the same number and it's also something we know from Newton's third law this force of gravity is a vector and it has a direction the direction is always such that it attracts every other mass and since this is a force that unit is in Newtons so what's an example problem involving Newton's universal law of gravity look like let's say two masses both of mass M exert a gravitational force F on each other if one of the masses is exchanged for a mass 3m and the center-to-center distance between the masses is tripled what would the new gravitational force be we know the gravitational force is always big G times one of the masses multiplied by the other mass divided by the center-to-center distance squared so the initial force between the two masses would be big G M times M over R squared but the new force with the exchanged values would be big G times three M times M divided by three times the radius squared the factor of three squared on the bottom gives nine and three divided by 9 is 1 over three times big GMM over R squared so we can see that the force with the new values is one-third of the force with the old values what's gravitational field mean the gravitational field is just another word for the acceleration due to gravity near an object you can visualize a gravitational field as vectors pointing radially in toward a mass all masses create a gravitational field that points radially in toward them and dies off like 1 over R squared the farther you get away from them so the formula for the gravitational field created by a mass M is big G times the mass creating the field divided by the distance from the center of the mass to the point where you're trying to determine the value of the field and again this value for the gravitational field is going to be equal to the value for the acceleration due to gravity of an object placed at that point the gravitational field is a vector since it has a direction ie toward the center of the object creating it and since gravitational field is equivalent to acceleration due to gravity the units are meters per second squared but you could also write that as Newton's per kilogram which is another way of thinking about what gravitational field means not only is it the acceleration due to gravity of an object placed at that point but it's the amount of the gravitational force exerted on a mass M placed at that point so you can think of the gravitational field as measuring the amount of gravitational force per kilogram at a point in space which when rearranged gives you the familiar formula that the force of gravity is just M times G so it's an example problem involving gravitational field look like let's say a hypothetical Planet X had three times the mass of Earth and half the radius of Earth what would be the acceleration due to gravity on Planet X ie the gravitational field on Planet X in terms of the acceleration due to gravity on earth which is GE so we know that the gravitational field on earth has to be big G times mass of the earth over the radius of Earth squared which we're calling G sub e and the gravitational field on Planet X would be big G times three times the mass of the earth divided by half the radius of the earth squared and when we square this factor of 1/2 we'll get one-fourth which is in the denominator so 3 divided by 1/4 is 12 times big G mass of the earth over radius of the earth squared and since this entire term here is the acceleration due to gravity on earth the acceleration due to gravity on Planet X is going to be 12 times the acceleration due to gravity on earth sometimes when you're solving gravitational problems you'll be given the density instead of the mass the density is the amount of mass per volume for a given material the symbol for density is the Greek letter Rho and you can find it by taking the mass divided by the volume so that units of density are kg/m^3 and it's not a vector since it has no direction but it does let you solve for mass if you know the density you could say that the mass is the density times the volume so what's an example problem involving density look like let's try the hypothetical planet problem again but this time instead of being told that Planet X has three times the mass of Earth let's say Planet X has three times the density of Earth and again half the radius of Earth what would be the acceleration due to gravity on Planet X in terms of the acceleration due to gravity on earth GE we could write down the formula for gravitational acceleration or gravitational field which is big G M over R squared but this time we don't know the mass we just know the density so we want to rewrite this formula in terms of density which we can do by rewriting the M as Rho times V since density is mass per volume and mass is density times volume but we don't know the volume of this planet we just know the radius so we need to rewrite volume in terms of radius which we could do since planets are spherical and the volume of a sphere is 4/3 PI R cubed we can substitute this expression in for the volume and finally an expression for the acceleration due to gravity of big G times Rho 4/3 PI R cubed divided by R squared and we can cancel out an R squared on the top and the bottom which leaves us little G as equaling big G Rho 4/3 PI R so the gravitational acceleration on earth would be big G Rho of earth 4/3 pi times the radius of Earth and the gravitational acceleration on Planet X would be big G times the density of Planet X which is 3 times the density of Earth times 4/3 pi times the radius of Planet X which is 1/2 the radius of Earth which when we pull out the 3 and factor of 1/2 gives us three-halves times the expression for the acceleration due to gravity on earth so the gravitational acceleration on Planet X is going to be 3 halves the gravitational acceleration on planet Earth gravitational orbits are just a special case of centripetal acceleration where some object is orbiting another object due to the gravitational force and if that orbit is a circle we can relate the speed the radius of the orbit and the larger mass to each other using Newton's second law and centripetal acceleration you just in the acceleration as the centripetal acceleration v squared over R and since the centripetal force is the force of gravity you can plug in the expression for the force of gravity as the centripetal force which is big GMM over the distance between them squared and since the mass of the orbiting object cancels we get an expression that relates the speed of the orbiting object the larger mass that's pulling that object in and the center-to-center distance between the objects which if we solve this for V gives us the square root of big G times the mass pulling in the object divided by the center-to-center distance between the objects note that this formula does not depend on the mass that's in orbit since that mass cancelled out in the calculation so what's an example problem involving gravitational orbits look like well imagine a space station of mass M s is orbiting at an altitude of 3r above a planet of mass M P and radius R as seen in this diagram and then imagine a different space station of mass 3m s is orbiting at an altitude of 2r above a planet of mass 4m p and a radius of 2 R as seen in this diagram and we want to know if the speed of the space station of mass M s is V then in terms of V what's the speed of the space station of mass 3m s well we just show that the speed of an orbiting object is going to be equal to the square root of big G times the mass of the larger object pulling in the smaller object divided by the center-to-center distance between the objects and since this formula doesn't involve the mass of the orbiting object it doesn't matter that the objects have different masses but the mass of the planet can make a difference so to get the speed of the space station to M s we could say that it's the square root of big G the mass of Planet P over the center to Center distance which is not going to be the radius of the planet or the altitude it's going to be the radius of the planet plus the altitude since this has to be the center to Center distance which in this case will be 3r plus R which is 4 R and now to get the speed of the space station of mass 3m s we use the same formula which is big G mass of the planet which in this case is 4 MP divided by the center to Center distance which in this case would be 2 R plus 2 R and again that's for R it we compare the only difference between these expressions is that there's an extra factor of four within this square root so if we take that factor out the square root of four is two we'd get two times the expression for the speed of the spacestation m/s so the space station 3ms is traveling two times the speed of the space station m/s
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