- AP Physics 1 review of 1D motion
- AP Physics 1 review of 2D motion and vectors
- AP Physics 1 review of Forces and Newton's Laws
- AP Physics 1 review of Centripetal Forces
- AP Physics 1 review of Energy and Work
- AP Physics 1 review of Momentum and Impulse
- AP Physics 1 review of Torque and Angular momentum
- AP Physics 1 review of Waves and Harmonic motion
- AP Physics 1 Review of Charge and Circuits
In this video David explains each concept for centripetal motion and solves an example problem for each concept. Created by David SantoPietro.
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- Hi all. Here's a dropbox link to download the 2D motion notes if you want them. Good luck on the exam!
- For the example question on centripetal force, why is it Mg - M(S^2/R) rather than Mg + M(S^2/R)? My thought process was that if normal force equals to the forces in the opposite direction and that since the weight of the ball and the centripetal force both point inward, the normal force would equal the weight + centripetal force rather than minus. What is the problem to my thinking?(3 votes)
- that would hold true if the ball remained at a constant height, however because the ball is moving downwards N<Mg so N = Mg - M(s^2/R) not Mg + M(S^2/R) as Mg - M(s^2/R)< Mg(1 vote)
- How do we know that all planets are spherical?(1 vote)
- we dont. And they are not all spherical.
However, it makes sense that most things in the universe would be spherical. At least when they were intially formed. Why? a) because most things would be 'fluid' in nature when formed. Either because they are very hot or because they are formed from gas, dust etc that would be able to flow.
b) gravitational fields are radial. This means they tend to pull those fluids towards a center. All the bits of fluid are trying to get close to that centre so the lowest energy arrangement of the whole thing is spherical.
I guess if bodies are not spherical, it because they have had their shape changed after they have been formed. By impact or strong gravitational fields (eg moons of Jupiter)(2 votes)
- Acceleration = (Velocity)' = (position)''. Typically using the power rule, subsequent derivatives lead to lower exponents. However, this is just how the numbers work out. Another method is to divide by dt (A=(dv/dt). Given these trends, why does it increase when we talk about rotating objects. Here, ||P(t)|| = r, ||V(t)|| = rw, ||A(t)|| = r(w^2). Why do the subsequent derivatives get an extra w? NOTE: rw/(dt) = rα = A (but r/dt is nothing). I am a little hazy on the difference between Ac and A. However, nevertheless, why is there that correlation (multiplying by w) as seen above.(1 vote)
- You need to use the chain rule in taking that derivative. Watch sal's vids about it in the calculus section.(1 vote)
- At6:08he says that the normal force has to be less than the force due to gravity.
But how is that possible? If the object is at rest, the object is obviously not sinking thus the normal force is equal to the gravitational force. Are we just saying that when it is moving then the normal force isn't necessarily the same?
I thought like this;
1. Object is moving past the top part of the circle - Normal Force is keeping it from moving down.
2. Object moved past the top part of the circle and is not colliding with it anymore, so gravity moves it down.
3. Depending on how fast the object is moving and how much acceleration is due from gravity, the object will move along the circle, if not actually colliding with it. If it does collide with it then the normal force will actually prevent it from moving down farther at that moment.
Where is my logic wrong?
- Well, first of all, the object is not at rest. It is moving at a constant speed along the top of the hill. If an object undergoes circular motion, that means that the net force must point towards the center of the circular path, this means that the normal force must be less than the weight of the car. As for your logic: There is no point where the car loses contact with the hill. If the car is to momentarily lose contact, its speed must be greater than a certain threshold. We do not have any information about the car being at this threshold, therefore, we can infer that the car does not lose contact with the hill. Also, the normal force doesn't always point straight up, it points perpendicularly out of the hill. So the normal force stops the car from going into the hill but it does not stop the car from traveling down the hill. Hope this helps!(1 vote)
- At12:20, he cancels out the R^3 in numerator with the R^2 in denominator, leaving R. But aren't these R's actually different variables? Isn't the R^3 in the numerator the radius of the planet (from the volume of sphere), and isn't the R^2 in denominator the distance of the object from the mass in the gravitational field? If I am not mistaken, then these two can't cancel each other out....(1 vote)
- At6:00, we say the object accelerate down because the net force is downward (The Mg is more than Fn). What about the moment in which the object is going upward? I am asking that because I believe the Fn can't be more than Mg. So what is causing the upward motion?(1 vote)
- if there are other forces accelerating the object downward, such as an elevator accelerating upwards, Fn will be Mg + Fo where Fo is other forces other than Mg acting along the same axis. Fn always = Mg+Fo but in many cases Fo=0 so we ignore it(1 vote)
- Wouldn't it be better to refer to "v" as velocity instead o speed?(0 votes)
- [Voiceover] 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 it has 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 distance per time, and the distance the object travels in one cycle is two pi R the circumference, the speed would just be two pi R per the period, or since one over the period is the frequency, you could write the speed as two pi R times the frequency. Since time is not a vector these quantities are not vectors and they cannot be negative. So, what's an example involving period and frequency look like? Let's say moon travels around a planet in a circular orbit of radius R at a constant speed S. And we wanna know what the period and frequency are in terms of given quantities and fundamental constants, so we'll use the relationship between the speed, the period, and the frequency. We know that for object in circular motion the speed is two pi R over the period. And that means the period here would be equal to two pi R over the speed. And since frequency is one over the period, if we take one over this quantity we just flip the top and bottom and we get that this is the speed over two 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 two pi R over S, and for frequency it would be S over two pi R, which is C. What is centripetal acceleration? The centripetal acceleration of an object is the acceleration that'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 travelling 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, i.e 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 gotta 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 gotta be a component of acceleration in direction of the velocity, and if the object is slowing down, there's gotta 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 of 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 travelling in a circle with a constant speed S and a radius R. If particle B is travelling 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 gonna have a centripetal acceleration of the speed squared over the radius, and particle B is also gonna have an acceleration of the speed squared, but this speed is twice as much of the speed of particle A, and is travelling 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 gonna 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 happen to be pointing towards the center of the circle making an object 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 loopty loop, the normal force is the centripetal force. And for a car going around a roundabout, 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 force is directed tangential to the circle, you wouldn't include it in this calculation at all. You can 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 wanna know, at a 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 gonna be an upward normal force on the ball from the road, and there's gonna 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 would balance, and if the forces are balanced the object would maintain its velocity and keep travelling in a straight line. But this ball doesn't travel in a straight line, it starts accelerating downward. So, this normal force is gonna have to be less than the force of gravity. To figure out how much less, we can use Newton's second law with the formula for centripetal acceleration. The speed is S, the radius is R, the force of gravity is gonna be a positive centripetal force since it's directed toward the center of the circle. The normal force is gonna be a negative centripetal force since it's directed radially away from the center of the circle. Then we divide by the mass, which, if you solve this for normal force, gives you the force of gravity minus M, S squared over R, which makes sense 'cause this normal force has to be less than the force of gravity. Newton's universal law of gravity states that all masses in the universe pull, i.e 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 6.67 times 10 to the negative 11th multiplied by each mass in kilograms, and then divided by the center to center distance between the two masses, in other words, not the surface to surface distance, but the center to center distance. And even if these two objects have different masses, the magnitude of the force they exert on each other is gonna 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, the 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 3M times M divided by three times the radius squared. The factor of three squared on the bottom gives nine, and three divided by nine is one over three times big G M M over R squared. So, we can see that the force with the new values 1/3 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 one over R squared the farther you get away from them. So, the formula for the gravitational field little G 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, i.e. 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 newtons 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 could 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, what'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, i.e. 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 a half we'll get 1/4, which is in the denominator, so three divided by a 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 gonna 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, the units of density are kilograms per meter queued. 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 that 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 four thirds pi R cubed. We can substitute this expression in for the volume and finally get an expression for the acceleration due to gravity of big G times rho four thirds pi R cubed divided by R squared. And we can cancel an R squared on the top and the bottom, which leaves this little G as equaling big G rho four thirds pi R. So, the gravitational acceleration on Earth would be big G rho of Earth four thirds 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 three times the density of Earth times four thirds pi times the radius of planet X, which is one half the radius of Earth, which when we pull out the three and factor of a half, gives us three halves times the expression for the acceleration due to gravity on Earth. So, the gravitational acceleration on planet X is gonna be three 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 plugin the acceleration as the centripetal acceleration, V squared over R, and since the centripetal force is the force of gravity, you can plugin the expression for the force of gravity as the centripetal force, which is big G M M over the distance between them squared. And since the mass of the orbiting object cancels we get an expression that relates to 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 canceled out in the calculation. So, what's an example problem involving gravitational orbits look like? Well, imagine a space station of mass MS is orbiting at an altitude of 3R above a planet of mass MP and radius R as seen in this diagram, and then imagine a different space station of mass 3MS is orbiting at an altitude of 2R above a planet of mass 4MP and a radius of 2R as seen in this diagram, and we wanna know, if the speed of the space station of mass MS is V, then in terms of V, what's the speed of the space station of mass 3MS? Well, we just showed that the speed of an orbiting object is gonna 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 MS, 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 gonna be the radius of the planet or the altitude, it's gonna 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 4R. And now to get the speed of the space station of mass 3MS, we'll use the same formula, which is big G mass of the planet, which in this case is 4MP divided by the center to center distance, which in this case would be 2R plus 2R, and again that's 4R, and if 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 space station MS. So, the space station 3MS is travelling two times the speed of the space station MS.