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More solving problems with Newton's Laws. Free body diagrams. Friction effects.	Circular motion. Centripetal acceleration. Universal gravitation and motion of the planets.	Work, kinetic energy, potential energy. Conservation of energy.

Circular Motion

Now that we have learned about Newton's Laws, we need practice in using them to understand examples of motion. You have seen many examples now where the motion is in some straight line, or where the motion in vertical and in the horizontal directions, are easily separated. Now, we consider a new type of motion that is quite common, but not so easily separated. Namely, we are headed towards discussion of circular motion. In particular, we will discuss uniform circular motion, motion around a circular path and at constant speed.

In understanding any type of motion, we always want to ask two questions:

1. Is the system accelerating?
2. What are the forces and the net force?

In the case of circular motion, THE DIRECTION OF THE VELOCITY VECTOR IS CHANGING ALL THE TIME. Therefore, we definitely have an acceleration. For uniform circular motion, the entire change in velocity is just a change in direction, because the magnitude (the speed) is fixed.

Let's try to determine the size and direction of the acceleration:

Centripetal acceleration

Here, I follow the same derivation as you find in Giancoli. First, draw a picture to show the motion of the system. Let's take a ball on a rope and swing it in a circular path at constant speed:

Here, I'm showing the circular path, the object, and the velocity vectors at two particular times. You can see that the velocity vector changes with time (points in different directions) so that there is acceleration. We refer to it as centripetal acceleration.

Anytime there is acceleration, we can use it to understand forces and motion. In this case, we need to determine the actual size and direction of the acceleration. Giancoli uses the idea of similar triangles to figure it out. Let's look at the triangle formed by the two radii when the object moves a small distance, D l, around the circle. It looks like this:

Next, think about the velocity vectors at each of these positions and the triangle you form from them when you take the vector difference:

Recall that the acceleration is defined as:

We have not said anything about the time required for this change in position and velocity (that would depend upon the speed). However, the direction of the acceleration is determined entirely by the change in velocity vector! Our diagram shows that the change in velocity, Dv, points inwards, toward the center of the circle. Therefore, our first major result is that the centripetal acceleration points inwards, toward the center of the circle.

Next, let's determine the magnitude of the acceleration. Notice that the triangle formed by the two radii is isosceles, with some angle between the radii. Similarly, the triangle formed by the two velocity vectors is also isosceles (same speed) with the same angle. Therefore, these triangles are similar. By similar triangles, we know that the ratios of the various side lengths are equal. Therefore,

or

The magnitude of the centripetal acceleration is then:

Overall, the centripetal acceleration has this magnitude and points to the center of the circle.

Example: You whirl a 1 kg mass attached to a string around a 1 meter radius circle at 10m/s. What is the tension in the string necessary to this motion? Let's just for the moment ignore gravity and say that the only force acting on the mass is the string tension (it's simpler this way). As usual, start with a picture:

The only force acting on the mass is the string tension, T. Therefore, the net force is T. We know by Newton's 2^nd Law that net force is equal to mass multiplied by acceleration. In this case, the acceleration is the centripetal acceleration associated with going around the circle. Therefore,

Notice that we reason using exactly the same tools as usual. We need Newton's Laws and we look at the accelerations. Make a diagram. Find the forces. Now let's work some concept tests:

Now that you have some experience with the free body diagrams, let's work another example to show the type of results you can extract:

Example: You whirl a rock on a string at uniform speed around a circle. The circle is vertical and the force of gravity is included. How do the tensions in the string compare at the top of the circle and the bottom of the circle? Start with the free body diagram:

Notice that the force of gravity is always pointing down. However, the tension on the string is always a pulling force, so it points along the string. Further, these forces result in a net force that, by Newton's 2^nd Law, must give the same magnitude centripetal acceleration. I emphasize here that we ASSUME that the object is following the circle at uniform speed. If it does not, then we need a different acceleration. Further, if the forces fail to balance correctly, say the string breaks, then the object will leave the circle. In any case, assuming uniform circular motion, we must have:

Here, you can see that the tension at the bottom must cancel the negative force of gravity, and then have the additional amount needed to get the positive centripetal acceleration. However, at the top, the force of gravity is already in the correct direction to get the correct sign of centripetal acceleration. Therefore, the tension at the top is there to add whatever small amount (if any) that you need. Therefore, the tension at the top is less than the tension at the bottom in magnitude. For any given acceleration, you can find the tensions by solving the equations above.

Connection to motion of planets and satellites

Circular motion is a close approximation to the way planets move or to the way satellites move around the Earth.

Some net force, pointing in towards the center of the circle, is required to cause this type of motion. The net force could be due to a rope, track, etc. In the case of planets, the net force comes from gravity. Newton was the first person to successfully explain how the force of gravity causes circular motion. Here is his insight:

Imagine doing a problem on projectile motion. You already have done a number of these. Say that we have a cannon that can fire a projectile at some initial speed. The picture looks like this:

The motion of the cannon ball is described by uniform velocity in the horizontal direction and uniform acceleration (at the acceleration of gravity, g) in the vertical direction. Now imagine that we look at a sequence of cannon shots at higher and higher initial velocities. They would look like this:

As the shots go further and further, the shape of the Earth (it's a sphere roughly) causes the surface to bend away from the curve of the projectile. Now, suppose that the curve of the projectile and the curve of the Earth's surface are the same.

The fact that gravity could lead to circular motion of satellites around planets, planets around suns, suns around galaxies, etc. was an inspiration. It lead Newton to develop a complete theory of gravity.

Newton’s Law of Universal Gravitation

At this point, we have developed several force laws, by looking at accelerations and figuring out the forces needed to match the observed motion. For example, near the surface of Earth, all objects, regardless of mass, fall with the same acceleration, g. Therefore, we decided that the force of gravity near the Earth must be described by the force law:

To find force laws, you look at the acceleration and determine the required net force.

Newton had an insight about the circular motion of objects, like planets around stars. Therefore, he decided to look at examples to see if he could learn the force law for gravity NOT necessarily close to the Earth.

For example, the Moon is an object that orbits the Earth in nearly a circle. We know the radius of the circle, and the speed of the Moon (it goes around the circle once each month), so we can determine the centripetal acceleration. After trying various forms, Newton proposed is Law of Universal Gravitation. Imagine two spherical objects, one of mass M and the other of mass m, separated center-to-center, by a distance, r. Then, the force of attraction between them is:

All objects attract each other with a force that depends upon both their masses and that gets weaker as they get further apart.

Example: What is the force of attraction between the Earth and a spherical ball of mass, m, near the surface of the Earth? Of course, you already know that the answer for the force of gravity on a ball near the surface of Earth must be -mg, where g is roughly 10 meters/s^2. However, we can also do it with the universal gravity law.

The mass of the Earth is roughly 6E24 kg. The radius of the Earth is 6E6 meters. Therefore, if you are near the surface of the Earth, your center to center spacing with the Earth is roughly the radius of the Earth. Therefore, the force of gravity near Earth is:

Universal Gravitation returns the result that we expect.

Although we won't have time to cover it further in lecture, try out the concept tests and, of course, the homework, to look further at the gravitational force between objects.

Important points

• An object in uniform circular motion feels an acceleration, pointing towards the center of the circle. This acceleration is referred to as centripetal acceleration. It has a magnitude of
• Circular motion problems are solved using Newton's Laws.
• By studying the nearly circular motion of the Moon and planets, Newton was able to propose that the gravity force between two point objects of mass M and m, respectively, and separated by distance, r, is given by . Here, the constant

. This equation is referred to as Universal Gravitation.