1. 1. 5/20/2015 The Coriolis Effect The Coriolis Effect Zachary Waller Bridgewater College 1 | P a g e
2. 2. 5/20/2015 The Coriolis Effect Table of Contents I. Abstract ..3 II. Introduction .4 III. Theoretical Calculations9 IV. Methods and Procedures..19 V. Results and Discussions.....20 VI. Conclusion...29 VII. References35 VIII. Tables..36 2 | P a g e
3. 3. 5/20/2015 The Coriolis Effect Abstract Pseudo forces are not real forces. They are imaginary forces that help to describe how things work in the physical world. These include both the Coriolis Force and Centrifugal Force. In this experiment we will be analyzing the Coriolis Effect, deflection, Coriolis Force, and Centrifugal Force of a projectile being fired. This experiment will be done by allowing a projectile to be fired from a rotating table. To understand the concepts better we will be working through theoretical calculations. Through experiments we collected a deflection, velocities, and time in order to calculate the Coriolis Force and Centrifugal Force. The average percent error between the two deflections calculated and observed, excluding outliers, is shown to be 5.688%. In the end, we discovered that most all the variables, including deflection, Coriolis Force, Centrifugal Force, and the radial velocity, would be dependent on the rotational velocity. 3 | P a g e
4. 4. 5/20/2015 The Coriolis Effect Introduction Every day people are affected by the rotation of the earth. It allows people to experience light from the sun during the day and the stars at night. This project will analyze another factor that deals with rotation and is often something that is overlooked on earth. In many introductory physics courses, the frame work and laws of physics are laid out. From this, we plan to delve deeper and apply these concepts of force, energy, and motion in a real world example. The purpose of the study is to explain the Coriolis Effect and how it affects people all around the world daily. Every time a plane takes off, a missile is fired, a rocket is launched, a hurricane develops, or even when you throw a ball at your friend on the other side of the merry-go-round, the Coriolis Effect is experienced. One challenge of observing the Coriolis Effect is that a picture cannot be taken or seen that proves the existence of it. This is partially because we experience it without knowing it is happening. In order to have physical proof, a video of pictures over time would have to be taken from a fixed point outside of the frame that the Coriolis Effect is occurring in. It is a displacement that must be shown or proven through experiment. This project will answer what the Coriolis Effect is and how it impacts our lives by designing and building a model that will serve as a physical representation of the Coriolis Effect. This design will represent the challenges that are faced on earth when launching projectiles in the rotating frame. The Coriolis Effect was first discovered in the early 20th century by Gaspard Coriolis, a French engineer1 . In this paper, the force due to the Coriolis Effect will be referred to as the Coriolis Force. This is because, like the Centrifugal Force, it has been introduced in an artificial manner as a result of our arbitrary requirement to satisfy Newtons second law in a noninertial reference frame.2 4 | P a g e
5. 5. 5/20/2015 The Coriolis Effect The Coriolis Effect occurs because a point on the surface of an object moves at a faster linear velocity where it has a greater radius r from the axis of rotation. For example, the linear velocity of a point on the surface of the earth, is much greater in Mxico because it is closer to the equator in comparison to the tip of Canada. This is because the distance, or radius r, is greater. In Figure 1, it can be seen that the stars move only 500 miles, while below the smiles move twice that. This is because the radius r2 is twice that of r1. The earth is rotating at one full rotation in 24 hours, and since the radius of the top is 1200 miles linearly, it only travels 500 miles in one hour. Likewise, the smiles will move 1000 miles in one hour respectively. This is shown by , (1) , (2) , (3) where v represents the velocity of each particle, dr the change in radius, and dt the change in time. This shows how far they each travel in an hour in respect to where they are located. This is not disregarding the fact that each will still make one full rotation in 24 hours; although, the one with the larger radius, r2, will travel a greater linear distance in respect to the entire frame. 5 | P a g e
6. 6. 5/20/2015 The Coriolis Effect Figure 1: a visual example for linear velocities dependence on radius. This means that when you are trying to move something closer or farther from the equator to a point of a smaller or greater radius, you must take into account the Coriolis Effect. In Figure 2, you can see an example of the particle moving more at a greater radius. A particle, represented by a smile, is being shot from the equator to the upper line. In this example, we must take into account the rotational change and the Coriolis Effect. If it takes one hour for a package to get from the middle line with the smiles to the upper line with the stars being shot directly upwards, we can see that the actual path will put the smile onto the stars line exactly 500 miles ahead of the star. Because the goal is to make the smile land where the star was, there will have to be adjustments in the angle of the initial launch. 6 | P a g e
7. 7. 5/20/2015 The Coriolis Effect Figure 2, a visual example of the Coriolis Effect for a projectile changing its radius. Another thing to keep in mind is that Coriolis is not a force; it is rather an effect due to the rotation. It is often referred to as a force because it causes a projectile in the rotating frame to move a certain way because of the direction of rotation. The angle the projectile is projected to, the direction, and the amount of movement are all dependent on the initial conditions. These include the rotational velocity , the force the projectile is launched at, and any outside forces, which in this experiment is gravity. For this experiment we will also be ignoring the air resistance. We can also note that the faster the rotation around the origin of the projectile, the greater the deflection, from the Coriolis Effect, will be. This is because the axis of rotation spins with no linear velocity, while the twist on the equator or outermost region can be considered all linear. In other words, the equator rotates, but does not spin. If there is no twist or spin, there is no Coriolis Effect. This means that in this experiment we must launch the projectile at some point between the outermost and innermost regions. 7 | P a g e
8. 8. 5/20/2015 The Coriolis Effect Figure 3, a visual example of the Coriolis Effect and the direction of deflection.5 In Figure 3, the rotational velocity is in the counter-clockwise direction. This causes the curve of the Coriolis Effect, each curve being relative to its point of launch (indicated by the dot). If the rotation is counter- clockwise, then the direction of the Coriolis will be to the right of the linear velocity and left for clockwise respectively. Coriolis Forces are noninertial. This means that the Coriolis Effect will only be present in a rotating frame. This experiment will be in a non-inertial frame. This means that the velocity of the frame will be non- constant. An inertial frame is one with a constant velocity. Think of playing ping pong on a train. If the train is moving at a constant velocity, meaning it is in an inertial frame, then the game can be played the same as playing in a garage. This can only be true if air resistance is ignored, so the train car would have to be enclosed. The ball will go exactly where it is expected to. This is because in both situations the velocity is constant and the frame is inertial. Now, if the train began to have a change in acceleration, the ping pong ball would not go where you expect it to. The non-inertial frame will have an effect on the ball. Theoretical Calculations 8 | P a g e
9. 9. 5/20/2015 The Coriolis Effect Before examining the math behind the Coriolis Effect, variables must be defined and analyzed. The following is a set of variables and their definitions to clarify their use. For these calculations we will be using the following: is the velocity of the particle with respect to the rotating system; is the velocity relative to the fixed axis; is the linear velocity of the moving origin; is the angular velocity vector which is equal to the rotation rate and is directed along the axis of rotation of the rotating reference frame; is the Coriolis acceleration; is all of the physical, non-pseudo, forces; is the net force in the rotating frame including pseudo forces; is the Coriolis Force; is the Centrifugal Force; is the acceleration of the rotating system with respect to the inertial frame; is mass of the projectile; is the velocity of the projectile due to the rotation of the moving axis; h is the height of the projectile above the surface; g is acceleration due to gravity. We can see that all the forces acting on the particle, when launched, are 9 | P a g e
10. 10. 5/20/2015 The Coriolis Effect . (4) Since the entire frame is not moving and the projectile is non-inertial, (5) from equation (4) is not a relevant factor. This is because they both occur when the entire system is moving. These would have to be taken into account if we were performing the experiment on a train or rolling down a ramp, for example. In this experiment, we will be looking at a nonmoving reference frame. This means we would be looking at the earth as its own frame. The next part of Equation (4) is the Centrifugal force written as . (6) This is something that will occur in this frame, although it is not part of finding the Coriolis Effect. Here the fundamental force equation , (7) is used in order to accurately find the unknowns in this experiment. Newtons simple force equation is then taken and manipulated in order to find the impact of the Coriolis Effect on the system. When manipulated, it can be written as or , (8) for the Coriolis Force. The vector formula for the magnitude and direction of the Coriolis acceleration is , (9) which incorporates the rotational and linear velocity in all vectors. 10 | P a g e
11. 11. 5/20/2015 The Coriolis Effect The matrix, , (10) shows how is calculated.2 Example 1 In this example we will be proving the deflection of a particle projected vertically upward to a height h above a point on the earths surface. The particle is at a northern latitude . We will show that the projectile strikes the ground at point to the west.2 For this problem we can also neglect air resistance and consider only small vertical heights. In order to start, first analyze the forces acting on the particle of mass m. Use (11) because it is in a fixed frame, meaning the entire frame is not moving. Then, dividing the mass, Equation (4) becomes . (12) When analyzing the first term on the right, , (13) it is the same as Equation (5) without the mass. The second term, , (14) 11 | P a g e
12. 12. 5/20/2015 The Coriolis Effect respectively replaces the second part of Equation (4) not including the mass. The third term, , (15) is the final part of Equation (4) not including the mass. When these are all laid out together, in format, we can see , (16) which then simplifies to . (17) This is the same as Equation (4). This is because we are observing the particle in the rotating frame versus the fixed frame from the beginning. For this problem can be voided because is constant, meaning the angular acceleration is zero. This is because there are no other forces relative to the moving coordinate system. They can be declared irrelevant to what is being looked for because it is bound to a moving frame, while this problem is for a fixed frame. We interpret, , (18) to be the centrifugal force term and , (19) to be the Coriolis force term. 12 | P a g e
13. 13. 5/20/2015 The Coriolis Effect Now that the equation for force has been established we can manipulate it in order to solve for the height h. First, let us look back to the introductory physics energy equation. This will come into play as we move forth in the calculations. The initial kinetic energy will be equivalent to the final potential energy as shown: . (20) This problem can be manipulated as, , (21) in order to find the initial velocity which will later be used. Equation (21) is focusing on the direction because both g and h are variables in that direction. Now we solve for the Coriolis force. If Equation (19) is broken down, the cross product can be solved as: , (22) and after, the constant -2 is added back in, in a later step. Equation (22) shows that there is only one force, due to the Coriolis Effect, in the direction. The velocities in and direction will be ignored because they are not of value to solving the deflection in the direction. They are also small in comparison to the velocity in the direction. The value in this direction is . (23) When the equation is rewritten, plugging the Coriolis acceleration back in, it is shown in vector form as: (24) 13 | P a g e
14. 14. 5/20/2015 The Coriolis Effect Now take each acceleration, from Equation (22), and integrate in order to solve for the velocity and the position in each dimension. For , , (25) , (26) , (27) defines the values of velocity and position. Equation (25) comes straight from the value of Equation (22). From there is integrated in Equation (26). The initial velocity in the direction is zero because there is no initial force giving it an initial velocity. Then, position is found in Equation (27) by integrating the velocity found in Equation (26). The initial position is also found to be zero because the origin is declared to be zero. For , , (28) , (29) , (30) defines the values of velocity and position. Equation (28) comes straight from the value of Equation (22). From there, is integrated in Equation (29). The initial velocity in the direction is zero because of the noninertial frame. Then position is found in Equation (30) by integrating the velocity found in Equation (29). The initial position is also found to be zero because the origin is declared to be zero. Finally for 14 | P a g e
15. 15. 5/20/2015 The Coriolis Effect , (31) , (32) , (33) defines the values of velocity and position. Equation (31) is the acceleration in the direction. It was solved in Equation (22) to be zero, but there is an initial acceleration g. This explains where -g comes from in Equation (31). From there, is integrated in Equation (32). The initial velocity in the direction is because of Equation (21). Then, position is found in Equation (33) by integrating the velocity found in Equation (32). The initial position is also found to be zero because the origin is declared to be zero. Now, in order to find the positions to the west we go back to Equation (30). What was found in Equation (33) can be disregarded because it is simply telling the vertical height the projectile lands. Now we go to the position found, , (34) in the direction. This is what is used to find the position to the west. Now go back to, , (35) the initial velocity. This will be substituted in as the initial velocity in Equation (34) to find, . (36) There is one last step in solving this and that is finding the time equivalent for this equation. Refer back to Equation (32). Note that, , (37) 15 | P a g e
16. 16. 5/20/2015 The Coriolis Effect is velocity in the direction. This is equal to zero at height h; therefore, in this problem, we can make equal to zero and solve for t as: . (38) The final step is to plug in t to Equation (36). The result, , (39) can be simplified further to our answer, . (40) Example 2 If a projectile is fired due east from a point on the surface of the earth with a northern latitude , a velocity of magnitude , and at an angle of inclination to the horizontal of , show that the lateral deflection when the projectile strikes is: . (41) Start off by first revisiting the Equation (8) , (8) which defines the equation of the Coriolis Force. To solve for acceleration the mass can be pulled out, since it is a constant, and from there we will solve for the accelerations in the ,, and directions. This is done by referring back to Equation (22), which for the subject part of the acceleration is subject to change. Here we are taking into account both angles and , representing the angle of northern latitude on the earths surface and 16 | P a g e
17. 17. 5/20/2015 The Coriolis Effect the angle of inclination to the horizon respectively. When solving for displacement, the acceleration must be integrated. Here we start with the acceleration, , (42) and find the corresponding components. From this it can be noted that the, and directions are not the direction of the deflection. Using the right hand coordinate system and what is given in the problem, we denote that is south, is east, and is the up or outwards direction. This attests that the deflection will be in the southern direction so both and can be ignored. The direction can be used in order to find the total time it takes the particle to hit the ground. First, we must assume that the projectile will hit the ground at . (43) Knowing this, we must go back to the equation for acceleration, Equation (4), and use the acceleration found in the direction. If we integrate twice, , (44) , (45) , (46) the equation for distance in the direction is shown. From this point refer back to Equation (43), plugging it into Equation (46), , (47) , (48) , (49) 17 | P a g e
18. 18. 5/20/2015 The Coriolis Effect in order to find time. The part from Equation (46) will go to zero and be irrelevant to Equation (47) because at Z(t) = 0 both and are zero. This is the time that the projectile takes from being shot to hitting the ground. Since the displacement is in the or south direction, we can integrate the acceleration in that direction, , (50) , (51) , (52) in order to find the value of the displacement. After doing this, the time can be plugged into Equation (52), , (53) , (54) , (55) which allows the displacement to be simplified. This shows that Equation (55) is the deflection of Example 2. Methods and Procedures This experiment will demonstrate the Coriolis Effect by creating a physical experiment that will allow us to find the pseudo-force and deflection of a projectile that will be launched. In this experiment a cart will be fixed onto a track. The track will be rotating and the cart will initially be at a small radius. As the track rotates, the distance of the cart from the origin of rotation will increase because of the centrifugal force. When the cart reaches a radius of 30 cm from the origin of rotation, a projectile will be launched. From this we will observe the deflection caused by the Coriolis Force. 18 | P a g e
19. 19. 5/20/2015 The Coriolis Effect The goal of this experiment is to find the variables to calculate the Coriolis Force and deflection of a projectile. To find these we will use the equation from the Coriolis Force, , (8) which can be derived from Newtons second law, . (16) Results and Discussion In this experiment we will be calculating and finding the angular velocity in all directions, the linear velocity in all directions, the time t it takes from the start of the rotation to the launch of the projectile, and the mass of the projectile being launched. We will be ignoring all air friction. The goal of this experiment is to find the Coriolis Force, the deflection because of the Coriolis Effect, the Centrifugal Force, identify the independent factors that these Forces rely on, and discuss how they relate to one another. Identifying the Angular Velocity The surface that the projectile is rotating on is level to the horizontal, illustrating the rotation in a disk- like manner. This means that the launch point is not above the horizontal; rather, the launch point will be located on the horizontal. Since the projectile is launched from a level surface to the horizontal and vertically upwards, the value of will be 90. This implies that the values, , (56) for the rotational velocity will be different than if they were calculated at an increased . This means that can be found by simply observing the amount of rotations per change in time. This is observed through a video that is taken of the experiment. The initial time is the moment that the track begins to rotate and the final 19 | P a g e
20. 20. 5/20/2015 The Coriolis Effect time is the moment the cart is launched. The amount of rotations is measured by the rotational distance the cart travels within the initial and final times. From this, , (57) the angular velocity is found.3 Identifying the Directions Terms In order to find the velocity of the projectile, the three dimensional vectors must be broken down. In the direction a photo gate, placed at the launch point of the projectile launcher, was used to record initial and final times of when the projectile enters and leaves the gate. The only force acting on the projectile, other than the launcher, is the acceleration due to gravity. This can be integrated, , (58) , (59) , (60) to show the relation between the height of the projectile and time it takes to get to that height. The time it takes the projectile to enter and leave the gate will be determined by a photo gate and used to calculate the change in time. There were ten trials taken in order to account for any error. Initial Time (s) Final Time (s) Time (s) 1.976 1.981 0.005 0.827 0.832 0.005 2.089 2.095 0.006 1.194 1.200 0.006 2.594 2.600 0.006 0.341 0.347 0.006 20 | P a g e
21. 21. 5/20/2015 The Coriolis Effect 0.582 0.588 0.006 1.011 1.016 0.006 1.032 1.038 0.006 0.432 0.437 0.006 Table 1: The change in time when the projectile entered and exited the gate. After finding the time it takes the projectile to move through the gate, the diameter of the spherical projectile was needed. In order to find the change in distance traveled by the ball between the two times, we took a rectangular slice of tape and made a mark as shown in Figure 4. Figure 4, an initial marker on the projectile. After this, the tape is wrapped around the projectile, and a second mark was made at the overlay (shown in Figure 5). 21 | P a g e
22. 22. 5/20/2015 The Coriolis Effect Figure 5, an illistration of the two marks overlayed on the piece of tape. Finally, the tape was unrolled, layed out flat, and measured, as shown in Figure 6, giving us the circumference of the object. Figure 6, the circumference of the projectile being measured. Next, the circumference was used to solve for the diameter. This is done by, (61) (62) 22 | P a g e
23. 23. 5/20/2015 The Coriolis Effect taking the circumference and dividing by pi. Once this distance is acquired, it can be used with the gravitational acceleration and time, previously found to find the initial velocity. Equation (63) can be broken down into two parts, (63) when solving for the initial velocity in the direction, the first being the diameter or change in distance of the ball per unit change in time and the second being the contribution of acceleration term on the velocity. Table 2 shows the overall velocity, the change in distance per time, and the velocity relative to the acceleration. We can see that the contribution of acceleration term can almost be considered irrelevant. This is because the amount of time that it takes the projectile to move through the gate is small. If the value of time is small, then the time the acceleration has to contribute the velocity is also small. Hence the value of the Acceleration term is small. Table 2 shows the average of the data collected during the set of ten tests. For each trial there were constants, including the acceleration due to gravity, -9.81 m/s2 , and the change in distance, 0.025 m, from when the gate collects the initial and final time. The time varied because it was collected at the moment the ball entered and left the gate. This caused the velocity to also be an inconsistent variable. Time (s) Diameter per time (m/s) Acceleration Term (m/s) Voz (m/s) Average 0.006 4.455 0.028 4.483 Table 2: The velocity calculated based on the ball being launched through a photo-gate. Equation (60) is used to find the velocity of the diameter per time and the acceleration term. The diameter- dependent velocity term is 99.937% and acceleration-dependent velocitys contribution is 0.063% of the total velocity. This shows that that the acceleration-dependent velocitys impact is small. This is because acceleration is only acting on the projectile for an extremely small about of time. Calculating the Coriolis Force 23 | P a g e
24. 24. 5/20/2015 The Coriolis Effect After finding the initial velocity in the upward or direction, it must also be found in the horizontal and vertical vectors. The initial velocity in the direction is found experimentally by using a motion detector that is mounted onto the track. As the cart moves 0.3 m down the track the motion detector collects its velocity. The velocity we will be using is the one right as the projectile is launched. This is found by collecting the instantaneous velocity at 0.3 m. This was done for ten trials to ensure the data varied. The velocities collected differ because the force on the cart, pushing it the 0.3 m, varies. This is because the centrifugal force acting on the cart, pushing it outward, increases as the rotational velocity increases. With these found values we can input them, 2, (64) into Equation (10) to find the acceleration. This results in, , (65) an acceleration caused by the Coriolis Effect only in the direction. The values of , each depend on the trial; while the mass is constant. These values are used in the following equation, , (66) to find the Coriolis Force. If the centrifugal force increases, then the amount of time it takes the cart to reach the end will decrease, and the velocity in the direction will increase as shown in Figure 8. The graph is linear, ignoring air friction, because the Coriolis Force is the dominating force; which is what would be expected. To insure we have this exact value, a barrier is placed on the track to cause the cart to instantly stop where the projectile is launched. There is no initial velocity in the direction. Calculation of the Deflection 24 | P a g e
25. 25. 5/20/2015 The Coriolis Effect To find the deflection of the projectile in the direction the value of time t must be found. This is the time that it takes from the moment the projectile is launched until the moment it hits the ground. This can be found by taking the time it takes the ball to hit the ground in the direction. This is because the projectile has a constant height, time, and force in the direction. We can use, , (60) for the displacement in the direction in order to find the time. Since the height that the projectile launches to is a constant 0.783 m, we will use this equation to solve for the time. The average time it takes the projectile from the point it is launched to the point that it hits the ground is 0.671 s. This value can be checked using Equation (60). When solving for the deflection we can use the acceleration . (65) By integrating twice, , (67) , (68) We find an equation for displacement in the direction. Using what is found in Equation (68) the deflection in the direction can be identified as . (69) Using the values previously found for we can solve for the deflection in the direction. Equation (69) was used to find the theoretical in Table 3. rad/sec Velocity (m/s) Theoretical Deflection (m) Deflection (m) Percent Error 25 | P a g e
26. 26. 5/20/2015 The Coriolis Effect 2.182 0.311 0.306 0.290 5.395 2.886 0.390 0.507 0.465 9.023 2.886 0.397 0.516 0.478 7.962 2.942 0.437 0.579 0.565 2.509 2.980 0.437 0.587 0.575 2.040 3.222 0.465 0.675 0.650 3.837 3.491 0.477 0.750 0.740 1.359 3.427 0.556 0.858 0.840 2.189 4.398 0.574 1.137 0.973 16.88 9.861 1.391 6.179 1.250 394.3 Table 3: The calculated deflection for ten trials. Calculation of the Centrifugal Force The Centrifugal Force is also a non-inertial, pseudo force that is caused by the rotation. The direction of this force is radially outward from the origin or center of rotation.4 This is depicted in Figure 7: Figure 7, A visualization of Centrifugal Force.5 26 | P a g e
27. 27. 5/20/2015 The Coriolis Effect The equation for the centrifugal force, , (70) comes from Newtons second law and can be written as, . (71) This equation can be rewritten using . (72) This is because the linear velocity is dependent on the rotational velocity. When rewritten, , (73) we use Equation (72) in place of the rotational velocity, allowing us to write the centrifugal force in terms of velocity. When the previous found velocity, mass 0.01 kg, and radius 0.03 m are plugged in, we find that our centrifugal force is: (rad/sec) Velocity (m/s) Fcent (N) 2.182 0.654 0.014 2.886 0.866 0.024 2.886 0.866 0.024 2.942 0.883 0.025 2.980 0.894 0.026 3.222 0.967 0.030 3.491 1.047 0.035 3.427 1.028 0.034 4.398 1.319 0.056 9.861 2.958 0.283 Table 4: The calculated Centrifugal Force of the cart at the point where the ball is launched. We can see that the Centrifugal Force increases as increases, which is expected. 27 | P a g e
28. 28. 5/20/2015 The Coriolis Effect Conclusion and References This experiment has shown that both the radial velocity, Coriolis Force, and Centrifugal Force depends on and can be written in terms of the rotational velocity. In Figure 8 the angular velocity is linearly proportional to the velocity in the direction. This would be expected because it is dependent on the angular velocity as shown in Equation (71). Figure 8: The increase in linear velocity and angular velocity because of the angular velocity. This increase in velocity also goes to show that there is an increase in the Coriolis Force, allowing for greater deflections. As shown in Table 3, the theoretical and actual deflections are close to each other with a low percent error other than the last data point, which could be because of air friction, which has been ignored. This concludes that the calculations of the deflection holds true in this situation. There were also a lot of assumptions that we were able to make in this project. They included ignoring air friction, the initial velocities, values for the rotational velocity, and the use of final data as initial data. We were able to ignore air friction because it was not a force being analyzed in this experiment. It also has a minimal effect on calculating things such as the deflection seen in Table 3. We were able to assume the initial velocity to be zero, before the cart started moving down the track, because the cart started from rest. This was relevant to solving the Centrifugal Force. The angular velocity was unchanging and constant because the rotating table rotated at a constant rotational velocity until the projectile was launched. When it came to calculating the deflection because of the Coriolis Effect, we used some final terms as initial terms. This includes the final velocity in the direction, the rotational velocity, and the distance 0.3 m. We were able to do this 28 | P a g e
29. 29. 5/20/2015 The Coriolis Effect because the projectile was launched at this point, and we marked it as the initial point for calculating the deflection. The final assumption that can be made is that the time it took for the ball to hit the ground, in the direction, was the same as the time it took from the moment the ball was launched till the moment it hit the ground, which was able to be assumed because time is constant in all vectors. Both the Centrifugal Force and Coriolis Force are dependent on the angular velocity. The Centrifugal Force is dependent on second power of shown in Equation (70), likewise the Coriolis Force, shown in Equation (66) has a similar dependence which can be shown through the velocity in the direction. The Velocity can be derived from the Centrifugal force Equation: . (74) We know that the force is in the radial outward direction From this, Equation (74) can be rewritten as . (75) The acceleration term that derives from the force and is in the direction is . (76) From this differential equation, the solution, , (77) will be used to solve for the velocity. This can be derived to find the equation for velocity . (78) The initial velocity is zero because the cart is starting from rest. This means that the constants A and B can be solved for as shown: 29 | P a g e
30. 30. 5/20/2015 The Coriolis Effect , (79) . (80) In this situation both constants are the same. This means that both equations for velocity and distance can be simplified using the value of the constant , (81) , (82) , (83) (84) where l is, 0.17 m, the length of the cart. Because this is the velocity of the cart moving down the track, the time at the 0.3 m is not consistent for all ten trials and must be solved for. This is done by taking the equation, , (85) where we found radius, and solving for time: , (86) . (87) This is the time when the projectile is launched at the radius 0.3 m. With this time we can resolve the Coriolis Force by plugging in the differential equation for the velocity as shown: , (88) , (89) , (90) 30 | P a g e
31. 31. 5/20/2015 The Coriolis Effect . (91) This shows how the Coriolis Force is a function of the angular velocity like the Centrifugal Force, although it also has a angular velocity term in the Sinh function. Time also relies on because the greater the value of the smaller the value of time t will be. Figure 9: A graph of the rotational velocity squared and force. Figure 9 shows how both the Centrifugal Force and Coriolis Force are related to . Both forces increase as the rotational velocity does. The forces would increase quadratically if it were compared to . Normally the Coriolis Force would be seen as linear in omega and Centrifugal Force seen quadratic in omega, but in this case it is different because the only force pushing the cart down the track is the Centrifugal Force, which is dependent on . This causes the velocity in the Coriolis Force to also be dependent on . This proves that the Forces are dependent on the rotational velocity. 31 | P a g e
32. 32. 5/20/2015 The Coriolis Effect Citation key 1 Lackner, J. R., and P. Dizio. "Rapid Adaptation to Coriolis Force Perturbations of Arm Trajectory." American Physiology Society. Journal of Neurophysiology, 1 July 1994. Web. Apr. 2015. 2 Marion, Jerry B., and Stephen T. Thornton. Classical Dynamics of Particles and Systems. Third ed. Florida: Harcourt Brace Jovanovich, 1988. Print. 3 NCS Pearson. "Angular Velocity Formula." Angular Velocity Formula. NCS Pearson, 2014. Web. Apr. 2015 http://formulas.tutorvista.com/physics/angular-velocity-formula.html 4 "Uniform Circular Motion (centrifugal Force) Calculator." High Accuracy Calculation for Life or Science. CASIO COMPUTER CO., 2015. Web. Apr. 2015 http://keisan.casio.com/exec/system/1271292951 5 Nave, R. "Coriolis Force." Centripetal Force. CASIO COMPUTER CO., 2015. Web. Apr. 2015 http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html 32 | P a g e
33. 33. 5/20/2015 The Coriolis Effect Table 5: Values for ten trials in the direction. Trial Initial Time z (s) Final Time z (s) Time (s) Distance (m) acceleration z (m/s2 ) Vo z (m/s) Diamiter per Time (m/s) Acceleration Term (m/s2) 1 2.594 2.600 0.006 0.025 -9.810 4.263 4.262 0.029 2 0.341 0.347 0.006 0.025 -9.810 4.353 4.353 0.028 3 1.194 1.200 0.006 0.025 -9.810 4.393 4.393 0.028 4 1.011 1.016 0.006 0.025 -9.810 4.395 4.395 0.028 5 1.032 1.038 0.006 0.025 -9.810 4.440 4.440 0.028 6 2.089 2.095 0.006 0.025 -9.810 4.445 4.445 0.028 7 0.582 0.588 0.006 0.025 -9.810 4.464 4.464 0.028 8 0.432 0.437 0.006 0.025 -9.810 4.572 4.571 0.027 9 0.827 0.832 0.005 0.025 -9.810 4.573 4.573 0.027 10 1.976 1.981 0.005 0.025 -9.810 4.683 4.656 0.026 Average 1.208 1.213 0.006 0.025 -9.810 4.458 4.455 0.028 Table 6: Values for ten trials in the direction. Trial Initial time (s) final time (s) Time (s) position y (m) Velocity y (m/s) 1 2.400 3.400 1.000 0.300 0.311 2 2.050 2.800 0.750 0.300 0.390 3 2.150 2.900 0.750 0.300 0.397 4 2.550 3.250 0.700 0.300 0.437 5 1.700 2.350 0.650 0.300 0.437 33 | P a g e
34. 34. 5/20/2015 The Coriolis Effect 6 1.400 2.050 0.650 0.300 0.465 7 1.950 2.600 0.650 0.300 0.477 8 1.800 2.350 0.550 0.300 0.556 9 1.850 2.350 0.500 0.300 0.574 10 3.450 3.650 0.200 0.300 1.391 Table 7: Solving for the Coriolis Force. Calculated Velocity (m/s) angular distance (deg) angular distance (rad) z (rad/sec) mass (kg) ac (m/s2) Fc (N) 0.300 125.000 2.182 2.182 0.010 1.357 0.013 0.400 124.000 2.164 2.886 0.010 2.251 0.022 0.400 124.000 2.164 2.886 0.010 2.291 0.022 0.429 118.000 2.059 2.942 0.010 2.571 0.025 0.462 111.000 1.937 2.980 0.010 2.605 0.025 0.462 120.000 2.094 3.222 0.010 2.997 0.029 0.462 130.000 2.269 3.491 0.010 3.330 0.032 0.545 108.000 1.885 3.427 0.010 3.811 0.037 0.600 126.000 2.199 4.398 0.010 5.049 0.049 1.500 113.000 1.972 9.861 0.010 27.433 0.266 Table 8: Solving the velocity for ten trials in the direction. distance z (m) Vo z (m/s) 1 0.774 4.263 2 0.779 4.353 3 0.781 4.393 4 0.782 4.395 5 0.783 4.440 34 | P a g e
35. 35. 5/20/2015 The Coriolis Effect 6 0.783 4.445 7 0.784 4.464 8 0.786 4.572 9 0.786 4.573 10 0.787 4.683 0.783 4.458 Table 9: Deflection and percent error for ten trials. Total time z (s) z (rad/sec) Velocity y (m/s) Deflection (m) Deflection (m) Percent Error 0.671 2.182 0.311 0.306 0.29 5.395 0.671 2.886 0.390 0.507 0.465 9.023 0.671 2.886 0.397 0.516 0.478 7.962 0.671 2.942 0.437 0.579 0.565 2.509 0.671 2.980 0.437 0.587 0.575 2.040 0.671 3.222 0.465 0.675 0.65 3.837 0.671 3.491 0.477 0.750 0.74 1.359 0.671 3.427 0.556 0.858 0.84 2.189 0.671 4.398 0.574 1.137 0.973 16.88 0.671 9.861 1.391 6.179 1.25 394.3 Table 10: Solving the velocity over ten trials in the direction. a (m/s2) z (rad/sec) w^2 possision y (m) Velocity y (m/s) Vx=wr (m/s) -9.81 2.182 4.759446856 0.300 0.311 0.654 -9.81 2.886 8.326400553 0.300 0.390 0.866 -9.81 2.886 8.326400553 0.300 0.397 0.866 -9.81 2.942 8.655743743 0.300 0.437 0.883 -9.81 2.980 8.882918963 0.300 0.437 0.894 -9.81 3.222 10.38178988 0.300 0.465 0.967 35 | P a g e
36. 36. 5/20/2015 The Coriolis Effect -9.81 3.491 12.18418395 0.300 0.477 1.047 -9.81 3.427 11.74515055 0.300 0.556 1.028 -9.81 4.398 19.34361044 0.300 0.574 1.319 -9.81 9.861 97.23740306 0.300 1.391 2.958 Table 11: Comparison of the Centrifugal Force and Coriolis Force mass kg Fcent (N) Fcor (N) 2y/l Time (s) Collected Time (s) Fcor (N) 0.010 0.014 0.013 3.529 0.886 1.000 0.013 0.010 0.024 0.022 3.529 0.670 0.750 0.023 0.010 0.024 0.022 3.529 0.670 0.750 0.023 0.010 0.025 0.025 3.529 0.657 0.700 0.024 0.010 0.026 0.025 3.529 0.649 0.650 0.025 0.010 0.030 0.029 3.529 0.600 0.650 0.029 0.010 0.035 0.032 3.529 0.554 0.650 0.034 0.010 0.034 0.037 3.529 0.564 0.550 0.033 0.010 0.056 0.049 3.529 0.440 0.500 0.054 0.010 0.283 0.266 3.529 0.196 0.200 0.271 36 | P a g e