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Angular Work and Power Since work in the linear sense equals force times distance, angular work is similarily found by,  Since power is work over time, the rotational power is found by,  Since omega equals theta over time, the equation for power becomes,  Again, this is the rotational analog to P = Fv. The rotational power equation says that when a certain amount of energy is needed at a certain rate by means of a rotating shaft, as the angular speed increases, the torque needed decreases and vice versa. Consider the problem below to tie these two concepts together. An alternator on a truck engine is connected to the engine by a V-belt. As show below.  The torque created on the V-belt is due to the difference in tensions from the top and bottom parts of the belt, which is why the bottom is slacked a bit and the top is tight.  It may seem odd at first that the bottom part of the belt is pulling in the opposite direction of rotation. It is only because the top part of the belt is pulling harder is there a net force in the direction of rotation.  Let's say the alternator produces 225 Watts of electric power when it rotates at 3200 rpm, which is 335 rad/s. If the alternator's pulley has a radius of 40.0 mm, and has a 95 percent efficiency, find the difference between the top and bottom tensions in the V-belt. Recall that efficiency is given by,  Therefore, power in is equal to Power out divided by the efficiency. The torque is then found by taking Power in divided by omega, Therefore, the force is easily found by dividing the torque by the radius, since torque equals force times lever arm, which is the radius of the pulley. This is approximately 4 pounds of force, which is the difference between the top and bottom parts of the belt. Angular Momentum Angular Momentum is one of my personal favorite physics concept. Ever wonder why you can easily balance a bicyle when you are moving, but it is terribly difficult to do so when you are stopped? The reason: angular momentum. But before we go into why, let's take a look at the basics. Again, angular momentum is very similar to its linear counterpart, only this time, instead of inertia in motion, it is moment of inertia in angular motion. The equation therefore depends on the moment of inertia, and omega given by:  Angular momentum is a vector quantity, and points in the same direction as omega. Therefore, you have to use the Right-Hand Rule to determine the direction it points. Now let's get back to why it helps you balance a moving bicycle.
Is is much easier to show the effect of angular momentum then to explain it. Unfortunately, that is not
possible due to the fact this is a book. Therefore, I will attempt to describe the effect by explaining
the demonstration I use in the classroom. Consider the picture below.
 This is a picture of a person holding onto a bike tire that has handles extending from the middle of the spokes where the bike frame would normally connect. The arrows show the direction of rotation. Here is the cool effect: when one tries to turn the wheel, like you would when trying to turn the handle bars of a bike, the tire actually resists this motion.  In fact, this resistance actually causes a person to spin around if he was sitting or standing on something that was free to rotate, and he tilted the spinning tire to the left or right!  The fact that spinning objects resist being rotated is used in so many different applications from something as simple balancing a basketball to as complicated as orbiting satellites. Everytime you see a magician balance plates on a tall rod, they are always spinning. It isn't so much of a trick as it is an application of angular momentum. When the plates are spinning, they resist falling off because they resist that tipping motion.  As mentioned above, often times astonauts will spin the satellite before releasing it into orbit in order to stabilize its motion.
Another amazing demonstration I like to show in class is the spinning tire hanging from a rope.
If you connect the rope to one of the ends of the handles, and hold the rope out, obviously the
tire will hang flat as shown below.
 However, when you spin the tire and then let go and hold on by the rope, the tire stays upright as shown below.  This is exactly how a gyropscope works. The angular momentum of the spinning tire or gyroscope keeps it upright. Conservation of Angular Momentum As with linear momentum, there is also a conservation of angular momentum, which states the initial momentum of a system will always equal the final momentum of the system.  This principle is beautifully illustrated with an ice-skater. Recall how an ice-skater will begin a spin with her arms out, and then she will slowly bring them into her body until they are over her head. What happens to the rate of her spinning? (Which is her omega). The rate increases. The reason has everything to do with the conservation of angular momentum.
When her arms are far from her body, she has a larger moment of inertia, so her body spins
at a slower rate. As she brings the arms in closer, the moment of inertia decreases. Since
the initial angular momentum must equal the final angular momentum, the omega must increase
with the decreasing moment of inertia, so that the product of the two remains the same. This
is illustrated below.
 I want to reiterate what I mean by the product of the moment of inertia, I, and omega, w, must be constant. Let's say the ice-skater began with I = 20, and omega = 5. If after the arms are brought in and now I = 10, omega must increase to 10 as well so that the product stays at 100. Let's take a look at another example with actual numbers. A merry-go-round is spinning .50 rev/s, and has a mass of 500-kg and a radius of 2.0 meters. A 50-kg man jumps on the very outer edge of the ride. Find the new rate of rotation of the merry-go-round. Since the man is not rotating initially, there is only one term on the initial angular momentum side. When he does jump on though, we have to consider him like a point particle that is rotating with a radius of 2.0 meters, which is the radius of the ride itself, because he is on the outer edge. (It is like the man is a ball attached to a string going in a circle). So, the equation becomes,  Since the man and the ride are rotating together, they share the same final omega. Solving for omega gives .12 rev/s. Proving Kepler's second Law (This is not a crucial piece of information and gets quite technical. So, one can skip this section without missing anything vital. However, it is a neat little proof that ties into previously learned material). Recall that Kepler's Second law is that planets sweep out equal areas in equal time intervals. This can actually be proven by using the conservation of angular momentum. Angular momentum must be conserved because the sun's force on the planet acts along the line that joins the centers of the sun and that planet. As a result, there is no net torque on the planet. Consider the picture below.  This picture shows a planet at two different times. We call the entire time of the trip delta t, and we consider it to be a very short time interval. During this time, the planet sweeps an angle theta, which equals omega times delta t.  The omega is the planet's omega during this short time interval, because remember omega changes according to where the planet is with respect to the sun. If the planet is close to the sun, omega is larger. If the planet is far from the sun, omega is smaller. (This is Kepler's Second Law). As the planet is making that angle, since we are considering the fact the orbit is elleptical and does not have a constant radius, we are going to assume here that the radius increases by a small distance labeled delta r in the diagram. So, the base of the triangle created by the planet's orbit is r + delta r, and the altitude, which is the height, is r times theta. Therefore, since the area of a triangle can be expresses as 1/2(base)(altitude), we can write,  Since the time interval was considered to be short, delta r is actually quite small (it is exaggerated in the picture for clarity), we can consider r + delta r to be basically equal to just r itself and neglect delta r altogether. (If this bothers you, remember we are dealing with distances from the moon to planets here, so even ten-thousand miles makes little difference!) Therefore, we can write the area as simply,  If we divide by the time interval delta t, we have,  Since omega = theta/time, we have,  Recall from previous section that the moment of inertia for an object that is like a ball attached to a string, such as a planet in this situation is mr^2. In this case, m is the mass of the planet. In order to put an m in the top part of the equation above, we need to put one in the denominater as well to keep things legal. Also, the two in the 1/2 will be moved to the denominater.  And there is the desired result! This last equation says that the change in area over time equals the angular momentum of the planet divided by two times the planet's mass. Since we already established that the angular momentum must be a constant for orbiting planets, and the two and the mass in the denominater are obviously constant, it means the change in area over any time interval is a constant as well! Thus, Kepler's Second Law of planetary motion has been proven.
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