1. 5/20/2015 The Coriolis Effect The Coriolis Effect Zachary
Waller Bridgewater College 1 | P a g e
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. 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. 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. 5/20/2015 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. 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. 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. 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
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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 |
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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. 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
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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. 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. 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
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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. 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. 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. 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. 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
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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. 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. 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. 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
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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. 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. 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. 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. 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. 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
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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. 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
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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. 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. 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. 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. 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