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Nonlinear Motion The first five chapters dealt exclusively with linear motion, which means motion in a straight line. We are now going to deal with nonlinear motion or motion not in a straight line. For our discussion, we will only consider motion in a circle. Consider a ball rolling across a flat table as shown below.  It is traveling straight with a certain velocity denoted by the vector. Now, imagine a string that suddenly appears attached to the ball and fixed at a point on the table. The ball will no longer travel in a straight line; it will travel in a circle.  The velocity vectors now change direction as the ball travels around the circular path and they stay tangent to the circle, which means that they are lines that touch only one point on the circle, and are at right angles to the string. We call the velocity of an object in a circle the Tangential Velocity. The velocity vector also tells you the direction the object would travel if the string were cut that instant. Centripetal Acceleration and Force The ball in the picture is traveling at a constant speed around the circle. (This is evident because the vectors are the same length). So, this brings a good question. Is the ball accelerating? At first thought you might say no because the speed is constant. However, you have to remember that acceleration by definition is change in velocity over time, and velocity includes both speed and direction. If you change only one of those, you change velocity. So, it seems that even though the ball is traveling at a constant speed, the direction is contsantly changing. Thus, the ball is indeed accelerating. What does the object have to have in order for it to accelerate? The answer is force. According to Newton's second law, f = ma. So, if you have a mass and acceleration you have to have a force. So, the question of the day is what direction is that force acting on the object. In other words, if you were to draw in the force vector on the diagram below, in what direction would it point?  A lot of people will draw in the force shown below. After all, when you round a curve in a car, you always lean out away from the circle right? This is what is known as the centrifigual force. It means "outer seeking force."  Let's talk about rounding a curve in a car. Imagine you are a passenger in a car that is traveling at a certain speed in a straight line. All of a sudden the driver turns left and you immediately begin to lean to the right. What is making you lean to the right? You could say that it was the centrifugal force pushing you out. If you say that, my question to you is what is supplying that force? Don't think to hard because the answer is absolutely nothing! So how can a force exist when there is nothing there to exert it? It simply cannot exist. Therefore, the famous centrifugal force vanishes into thin air and we are left with only one other option. When do you stop leaning to the right? When you hit the door. According to Newton's Third Law if you hit the door and push it out, because you want to go straight, the door exerts a force on you that is opposite your force which would be in towards the circle. That force is called the centripetal force, which means "inner seeking force." This is the only force that is acting on you once you cancel out the force due to gravity and the upward force from the chair. Therefore, the force that actually makes you go in a circle is a force that pushes you toward the center of the circle!  You might be asking yourself a few questions right about now. One, if the force is inward, why don't I move to the center? And, what makes me lean outward? Those are great questions and knowing centripetal force does not sufficiently answer them. To answer both of them, I need to start with what makes you lean outward. Before you turned left in your car, you were going in a straight line. When the car turns left, what does your body want to keep doing? According to Newton's First Law, all objects will continue to move in a straight line and constant speed unless acted upon by an outside force. So, your body wants to keep moving in a straight line, which is why you lean out. The leaning is your bodies attempt at keeping its straight line. You continue to lean until you hit the door which is the outside force that makes you go in a circle. If there was no door and seatbelt to keep you in, your body would be successful at keeping its straight line motion and you would fall out of the car. You can look at this another way. When you are at rest in a car, your body wants to stay at rest. (Newton's First Law again). When you step on the gas what does your body want to keep doing? It wants to stay at rest which is why you lurch backward into your seat. Nobody says that there is some mysterious force pushing you back. This is the exact same effect that your body experiences when traveling in a circle, only this time your body wants to stay in straight line motion. O.K., you might still be saying that this does not explain why you don't move towards the center of the circle. Oh, but it does. Any object that is going in a circle constantly desires to move in a straight line, which is why the object would continue to move in the direction of the tangential velocity (a straight line) if the inward force was removed. So, you have to things in constant conflict. The object wants to go straight and the force wants you to go towards the center of the circle. The result: you travel in the circle.  It turns out that the centripetal acceleration is related to the tangential velocity and radius of the circle by the equation,  Therefore, since according to Newton's Second Law F = ma, the centripetal force becomes,  There are a couple of things to note about the centripetal force equation. First, notice that the force is directly proportional to the velocity squared. (where the mass is the constant of proportionality). This should ring a bell because that is the exact same relationship that work has with velocity. So, when you double the velocity that you have going in a circle, you have to quadruple the force to keep you in that circle. If you triple the velocity, you have to have nine times the force! The point is, increasing the velocity makes the force increase rapidly. If you have ever been on a merry-go-round and hung on with your arms, when the velocity is increased by even a little, your arms have to work pretty hard. This is why it is a good idea to not have a small child hanging from the very edge, because if you double the speed, the child has to exert four times the force from his or her little arms. Force also has a relationship with radius but it is not direct. Since radius is at the bottom of the equation or denominator, force has an inverse relationship with radius. This means that if one increases the other decreases by that same amount and vice versa. So, if you double the radius of the merry-go-round, you actually have to hold on with half the force. If you decrease the radius by a half, you have to hold on with double the force. Frequency and Period You are probably familiar with the phrase rotations per minute or rpm's. This is the frequency of the rotating object. It tells you how many revolutions it completes in the given amount of time. So, it could be rotations per minute, hour, or day. It doesn't matter what time unit frequency is in. However, when you use it in equations, it has to be in seconds! Period is how long it takes to make one complete revolution. So, the units are in time such as seconds, minutes, or hours, but it has to be in seconds when you use it in equations as well. Period and frequency have an inverse relationship given by, Where t is the period in seconds, and f is frequency in revolutions per second. This means that if your frequency is 2 rev/s, your period is 1/2 second. Think about it. If an object makes 2 complete revolutions every second, it takes 1/2 seconds to complete just one revolution. Tangential Velocity We know that speed is distance over time. The distance traveled in one complete revolution is the circumference of the circle given by, Where (pi) is 3.14..., and r is the radius. So, tangential velocity can be found by taking distance over time, only now we need to use 2(pi)r for distance and the period for time. Let's consider an example that ties this all together. The finish fling at Worlds of Fun is basically a big barrell that spins around fast enough so that you literally stick to the walls. In fact, the floor drops out from under you, and you still stay on the wall. I am not sure about the actual radius and frequency of the barrell, so I will make some up to give a rough idea of what is going on. Let's say the radius is 2.43 m (about 8 feet), and the frequency is 1.0 rev/s. From this information, we can find the centripetal acceleration and centripetal force. In order to find those, we need to know the speed which we can by using the radius and frequency. Because the frequency is 1.0 rev/s, the period is 1/1 or 1 second. Therefore, the speed is found by, Which yields 15.3 m/s. So, the centripetal acceleration would be, In order to make this answer more relatable, we can put the answer in terms of g, the acceleration due to gravity. So, divide 95.9 by 9.8 and you get 9.79 or 10. This means that you would be feeling 10g's on that ride. (I am pretty sure that is bigger than the actual amount). Here is what 10g's means. Right now you are experiencing one g (as long as you are not reading this while accelerating of course). One g means you feel like you weigh exactly what you do weigh. If you were experiencing 2g's, you would feel like you weigh twice as much. So, 10g's means you feel 10 times heavier! That is why it is so hard to lift your arms and head from the wall of that ride. It feels no different if you were to lay on the floor and suddenly be 10 times heavier. Getting up would not be an easy task. Well, neither is getting off of the wall on finish fling. Let's now consider the centripetal force with somebody who has a mass of 80 kg. (about 177 pounds) That answer is in Newtons, so dividing by 4.448 converts the answer to 1732 pounds! Is that actually 10 times the weight? Not really because remember I rounded the g's to 10. It was actually 9.79. So, taking 9.79 and multiplying by 177 you do indeed get 1732 pounds! We can also consider another problem that involves friction. I am not going to do any problems here because a few are posted in the questions archive section. I am going to discuss the implications of this type of problem. If say a car is rounding a curve, the only force present that can supply the centripetal force needed to make that curve is friction. Therefore centripetal force equals the friction force. If you recall, the frictional force is given by,  Where the funny symbol is the static coefficient of friction. Setting that equal to centripetal force yields,  Notice how both sides have mass. This means we can just cancel them out of the equation which leaves,  We can now solve for v which will give us the maximum speed that is allowed before the car will begin to skid. This of course depends on the conditions of the road, (which affects the static coefficient of friction), the radius of the turn, and g.  It is interesting to note here that the mass of the car does not affect the maximum speed around a curve! Remember, mass canceled out of the equation. This is why there is only one posted speed limit for an approaching curve. It is a good thing because if mass did make a difference, we would have some long signs that had to post a different speed for each weight division of cars. Speaking of those posted signs. These equations are basically what are used to determine what the posted speed should be. It does get a little more complicated because I did not include the banking of the road for example. Anyway, the designers will calculate the maximum speed and post a speed that is a lot less because they know there are people that do not take them very seriously. I for one always slow down when I go around a curve. (wink wink). In general, when dealing with friction as the centripetal force, the following inequality must be true.  This says that the frictional force (without the mass) must be greater than or equal to the centripetal force in order to stay in a circle. The centripetal force is the force that is needed to make a circle with the given the speed and radius. The frictional force is the available force. It is the only source that supplies the centripetal force. So, if the supply is smaller than the demand, the object will not stay in a circle.  In the example above, we solved for the maximum speed that can be achieved before the supply becomes to small for the demand. So, for any speed greater than the maximum, the inequality is not true and the object will leave the circle and travel in a straight line. For any speed smaller than the maximum, the inequality is true, and the object will continue in the circle. Vertical Circle We now turn our attention to motion in a vertical circle which means we have to consider gravity in the equations. Consider the picture below.  At what point do you think the velocity of the ball is the greatest? The smallest? Since the ball in the picture is falling on the right side of the path, it stands to reason that the ball is traveling the fastest at the very bottom of the circle. Furthermore, since the ball is traveling agaisnt gravity on the left side of the path, the velocity is the smallest at the top of the circle. Therefore, the force that the string would exert on the ball would be greatest at the bottom and smallest at the top. Now let's consider the forces acting on the ball so that we can come up with an equation that describes the force of the string on the ball at the top and bottom. Top of Circle We know that there is a centripetal force acting on the ball because it is traveling in a circle. There are two forces that make up that centripetal force at the top as shown in the picture below,  The Ft is the force of the string at the top of the circle, and mg is the ball's weight. They both act downward. Since the centripetal force is also down (because that is where the center of the circle is located), and we make down negative we have the equation,  Since all terms in this equation are negative, we can simply multiply through by negative one and have,  This equation says that the centripetal force acting on the ball at the top is a combination of the force of the string and the ball's own weight. It is both the string and the weight that are supplying the centripetal force. Plugging in the equation for centripetal force and solving for the force of the string we get,  Bottom of Circle There are still two forces acting on the ball at the bottom, but this time the force by the string (Fb) is up instead of down.  Therefore, the centripetal force equation becomes,  Solving for Fb and plugging in the centripetal force equation yields,  We just proved mathematically that the force of the string at the bottom is greater than the force at the top. Notice that for the Ft equation the mg is subtracted from the centripetal force, and for the Fb equation it is added. Even though we found these two forces using a string and ball, they will work for any situation that deal with a vertical circle. If you want to know the force on a pilot while pulling out of a dive, use the Fb equation. In this case, instead of a string supplying the force, the pilot's seat exerts the Fb. Let's consider an example of a pilot pulling out of a dive as shown below.  Suppose the radius of the circle is 900 meters, and the speed at the bottom of the circle is 250 m/s, and the pilot's mass is 65-kg (143.65 pounds). We can use the Fb equation to find the force that the seat pushes up on the pilot. This corresponds to 1147 pounds! This means that the pilot "feels" like he weighs 1147 pounds. You can imagine how hard it is to even breathe in that situation. Another thing that they have to deal with is the blood rushing from their head to their feet. That is why they wear pressurized suits to help keep the blood in the upper half of the body. Anway, we can determine the g-force on that pilot if we take the weight he feels and divide by the his actual weight. 1147/143.65 is roughly equal to 8. This means that he feels like he weighs 8 times more than his normal weight in a one-g environment. We can find the g-forces another way as well. We can use the method utilized in the finish fling problem above, which involves the centripetal acceleration equation. Plugging in the numbers gives, If we divide that answer by 9.8 we actually get approximately 7. You might be wondering why the two equations did not yield the same g-force. Here's the deal. The Ac equation only tells you the extra g's needed to make that circle. It did not include the 1-g that is always present due to the earth. So, you have to add one to the seven to get the correct g-force which is 8. Critical Speed Let's now turn our attention to the top of the circle. Remember the Ft equation, One way to use this equation is to find the speed in which the ball must go so that the string stays taught. When the ball is traveling in a vertical circle, it has to be going fast enough so that it does not fall when it reaches the top of the circle. There is a certain speed for the object so that if it travels below that speed the string will be loose, and if it travels above that speed, the string will be taught. That speed is called the Critical Speed. We can find that critical speed if we set Ft equal to zero and solve the equation.  If you solve that equation for v, you get,  Which is the critical speed. The cool thing about this equation is you can apply it to any situation that involves traveling at the top of a circle. For example, consider the roller coaster below.  When the coaster reaches the first hill after the big drop, if it is traveling above the the critical speed, you will "lift" off of your seat, which means you will need a seat belt. If it is traveling below the critical speed, you will stay in your seat. For example, let's say the radius is 20 meters.  the critical speed will be given by, So, the car will have to be moving at a speed greater than 14 m/s in order for the riders to experience air time. Here is a good question. Does it matter how heavy you are in this last example? In other words, would a person weighing 300 pounds experience the same air time or "lift" off of the seat at the same time a 100 pound person would? The answer is a surprising yes! To explain why we need to turn to Newton's First Law once again. The instant the car is at the top of the hill, both you and the car have a tangential speed that is pointing horizontally. (See picture below under "before"). As soon as the car begins to drop, due to the track, your body still wants to follow that straight horizontal line so it does. That is when you "lift" off of the seat. So, you are not really being lifted up.  The point is, it has nothing to do with the masses of the people. It has to do with the seat literally dropping away from the people. As soon as the people are air born, you are now under the influence of gravity which still means mass does not matter. This is cool because every person will experience the same fun from a roller coaster regardless how heavy they are. You can also apply this to a roller coaster that turns upside down such as the Orient Express. Critical speed still applies, only this time if you travel above the critical speed you actually don't need a harness to keep you in the car! It is like the ball and string. What keeps the ball from falling at the top of the circle once it passes the critical speed? You guessed it: Newton's First Law. The ball wants to go straight but the string and its own weight are pulling it straight down, so it travels in a circle. This can be applied to the person in a coaster going upside down. At the very top of the circle, as long as the coaster is traveling above the critical speed, the person will want keep going straight and the coaster will be curving back down. Because of that, the person is "stuck" to the seat. However, if the coaster is traveling below the critical speed, a person would fall out of the seat if there was not a harness. To understand this let's look at the ball and string again. If the ball is traveling below the critical speed at the top of the circle, it still "wants" to go straight, but not enough to fight against the string and its own weight so to speak. So, the two forces win out and the ball falls toward the center. |