AP Physics 1 review of Centripetal Forces | Physics | Khan Academy

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, 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.

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. The distance the object travels in one cycle is 2πr, the circumference. The speed would just be 2πr per the period, or since one over the period is the frequency, you could write the speed as 2πr * 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? Let's say a moon travels around a planet in a circular orbit of radius r at a constant speed s, and we want to know what the period and frequency are in terms of given quantities and fundamental constants.

We'll use the relationship between the speed, the period, and the frequency. We know that for objects in circular motion, the speed is 2πr over the period, and that means the period here would be equal to 2πr over the speed. 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 2πr.

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 2πr/s, and for frequency, it would be s/2πr.

What is centripetal acceleration?

The centripetal acceleration of an object is the acceleration that's causing that object to go in a circle. 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 m/s 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 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 v²/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?

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 4 over 2, giving us a factor of 2 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 1/2 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 other forces that we've already met that happen to be pointing toward 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 loop-the-loop, the normal force is the centripetal force, and for a car going around a roundabout, the static frictional force is the centripetal force.

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.

If a force points radially out from the center of the circle, you would count that as a negative force. 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 v²/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 a 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.

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. 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 in 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, and we divide by the mass.

If you solve this for normal force, it gives you the force of gravity minus m s²/R, which makes sense because 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 * 10⁻¹¹ 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. 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, the unit is Newton.

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 * 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 * m/R², but the new force with the exchanged values would be Big G * 3m * m/(3R)². The factor of 3² on the bottom gives 9, and 3/9 is 1/3 * Big G m m/R².

So we can see that the force with the new values is 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 1/R² the farther you get away from them.

So the formula for the gravitational field g created by a mass m is Big G * 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. 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. Since the gravitational field is equivalent to acceleration due to gravity, the units are m/s², 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 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 * g.

So what's an example problem involving gravitational field look like? Let's say a hypothetical planet X had 3 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?

We know that the gravitational field on Earth has to be Big G * mass of the Earth over the radius of the Earth squared, which we're calling g sub E. The gravitational field on Planet X would be Big G * 3 * the mass of the Earth divided by half the radius of the Earth squared. When we square this factor of 1/2, we'll get 1/4, which is in the denominator. So, 3/1/4 is 12 times Big G * mass of the Earth over radius of the Earth squared.

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 the units of density are kilograms per m³, 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 3 times the mass of Earth, let's say Planet X has 3 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/R², 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 m as ρ * V since density is mass per volume, and mass is density times volume.

But we don't know the volume. 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)πR³. We can substitute this expression in for the volume and finally get an expression for the acceleration due to gravity of Big G * ρ * (4/3)πR³/R².

We can cancel out an R² on the top and the bottom, which leaves us g as equaling Big G * ρ * (4/3)πR. So the gravitational acceleration on Earth would be Big G * ρ of Earth * (4/3)π * the radius of Earth. The gravitational acceleration on Planet X would be Big G * the density of Planet X, which is 3 * the density of Earth * (4/3)π * the radius of Planet X, which is 1/2 the radius of Earth.

When we pull out the three and the factor of a half, it 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 3/2 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. 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 plug in the acceleration as the centripetal acceleration v²/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 G mm over the distance between them squared.

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 canceled out in the calculation.

So, what's an example problem involving gravitational orbits look like? Well, imagine a space station of mass Ms orbiting at an altitude of 3R above a planet of mass Mp and radius R, as seen in this diagram. Then imagine a different space station of mass 3Ms orbiting at an altitude of 2R above a planet of mass 4Mp and a radius of 2R, as seen in this diagram. We want to 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?

We just showed 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. 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 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 + R, which is 4R.

Now, to get the speed of the space station of mass 3Ms, we use the same formula, which is Big G * the mass of the planet, which in this case is 4Mp divided by the center-to-center distance, which in this case would be 2R + 2R, and again that's 4R.

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 4 is 2; we'd get 2 times the expression for the speed of the space station Ms. So, the space station 3Ms is traveling two times the speed of the space station Ms.