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Modeling a Thrown Tomahawk
Flathead Community College
Steve Waddle
May 13, 2015
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1
Abstract
Tomahawk throwing historically was a sport participated in at rendezvous
by trappers, Native Americans, and now by reenactors and sportsman. The
parabolic trajectory of a thrown tomahawk is analyzed in this system of first
order differential equations. Matlab and the numerical solver ode45 was used
to solve and plot the path of the centroid and a point on the blade.
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Contents
1 Abstract
2
2 Introduction
4
3 Assumptions
4
4 Initial Conditions
4.1 Initial Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Center of Mass and Moment of Inertia . . . . . . . . . . . . . . .
4.3 Distance for one Revolution . . . . . . . . . . . . . . . . . . . . .
5
5
6
6
5 Projectile Motion
5.1 Deriving Equations . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 System of Equations . . . . . . . . . . . . . . . . . . . . . . . . .
7
8
11
6 Results
12
7 Conclusion
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3
2
Introduction
This project models the motion of a thrown tomahawk. This motion is a
form of projectile motion with both rotation and twisting. The twisting motion
on a well thrown tomahawk should be minimal or else it would not stick into
the target. So This project will not look at the twisting motion that may occur
with a bad throw.
To be considered is whether the number of rotations a tomahawk makes is
dependent on how fast it is thrown.
To understand the initial forces that act on the tomahawk A short description of a properly thrown tomahawk is in order . First, the thrower rotates their
arm in a foreword circular motion till the arm is in line with the target. then
pressure is released on the handle to allow it to slide out of the hand. With this
release there is no rotation of the wrist. The tomahawks centroid moves in a
parabolic arc towards the target. As it is following this path the head and the
handle rotate around this centroid.
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Assumptions
The flight of the tomahawk starts when the thrower releases the handle.
When he releases there is an initial velocity but no forward acceleration. There is
acceleration due to gravity. When released the tomahawk travels in a parabolic
flight and at the same time rotates around its centroid in a forward direction.
It will be assumed that the throw is a good one and will not have a twisting
motion. Also, the ground will be assumed to be flat and level. If the ground
were to slope upwards or downwards an adjustment to the distance equations
would need to be made.
There are many different styles and shapes of a tomahawk.It will be assumed
that this does not change the flight characteristics and as such a very basic shape
will be modeled. Also the density of both the head and handle are assumed to
be uniform throughout.
After the tomahawk is released there is no longer a driving force. The only
forces that act on it while in flight are gravity and air resistance. The throw
is assumed to be in ideal weather conditions, meaning there is no wind, rain,
or snow. For this project impulsive forces shall not be looked at. two of these
force are wind and the impact into the target.
Newton’s 3rd law states that for every action there is an equal and opposite reaction. That means that air resistance will always act in the opposite
direction as the tomahawks velocity.This being said air resistance is not going
to be considered in these equations. Due to the short distance and the shape
of the tomahawk Air resistance is assumed to have no significant effect on the
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tomahawks trajectory.
4
Initial Conditions
The mass of the head and handle were measured. The hickory handle has a
mass of 0.424(lb) and the steel head a mass of 1.068(lb). The initial height is 6.5
(f t), the initial velocity will be determined, the initial distance is in the x-axis
is zero. All three of these would have a small effect on the tomahawk flight but
not enough to warrant putting them into this projects calculations.
4.1
Initial Velocity
The initial velocity is broken into two forms. There is a linear velocity (v)
and an angular velocity (ω). Velocity is a function time. the definition of velocity is the change in distance (∆s) divided by the change in time (∆t).
ds
∆s
=
= s˙
(1)
∆t
dt
Angular velocity is the rate of change in the radial angle per time rate of
change. This can also be defined as the linear velocity divided by the radius of
curvature.
v=
dφ
dt
A scalar equation for angular velocity can be written as
ω=
ω=
|v|
.
r
(2)
(3)
e
The initial linear velocity of the thrown tomahawk can be shown by solving
ω = |v|
r (3) for v and combining with equation (1) to get
ds
dφ
=
r.
(4)
dt
dt
Several throws were made and the times were recorded. The average of
these throws is 0.38 (s) and the distance away from the target was 13 (f t) so
the velocity of 34.21 ( fst ). With no way of calculating the initial velocity or the
angular velocity (ω) a velocity of 34.30 ( fst ) will be used for initial velocity and
using equation (2), with a radius of 2 (f t), an initial angular velocity of 17.15
( rads
s ) is calculated.
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4.2
Center of Mass and Moment of Inertia
The center of mass and moment of inertia can be found using calculus.
An attempt at locating these values was made but the time required and the
complex nature of the tomahawk made it impractical to continue. Since these
calculations were not completed an assumed center of mass will be used. To
find this point an attempt was made to balance the tomahawk and measure
this point. The center of mass is approximately 15(in) from the bottom of the
handle towards the top. The point on the edge of the blade will be 2 12 (in) in
the y-axis direction and 5(in) in the x-axis direction. Using the measurements
mentioned previously the distance and angle from the center of mass and the
point is calculated using trigonometry so β = 63.43◦ and r = 5.59(in).
4.3
Distance for one Revolution
The distance an object travels is based on the velocity and the length of
time traveled at that velocity.
s=v∗t
(5)
To find the arc length of any circle this equation can be used. S = θ ∗ r,
where (θ) is in radians an (r) is the radius. One full revolution is 2π so the arc
length of a circle is S = 2πr. The units for ω = ( rads
t ). For the time it takes to
rotate one revolution can be rewritten as:
t=
By solving equation ω =
(5) we see that
dφ
dt
=
|v|
r
2π
ω
(6)
(2) for velocity (v) and combining with
s=ω∗r∗t
(7)
Now substitute equation (6) and you get
s=ω∗r∗
2π
ω
(8)
simplify equation(8) and the equation for the distance it takes to get one
revolution is found
s = r ∗ 2π
(9)
This derivation shows that the distance for one full revolution is not dependent on time or velocity but on the radius from the shoulder to the centroid.
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5
Projectile Motion
Newton’s second law of motion will be used to for this section.[3, p.136-137]
This law states that the sum of the forces (F ) acting on a body equals its mass
(m) times acceleration (a). In this case the forces acting on the thrown tomahawk are gravity and air resistance.
X
F = ma
(10)
Acceleration is the second derivative of position so (??) will be rewritten as
(11).
X
F = m¨
r
(11)
Since, gravity (g) is the only acceleration in the y-axis direction and it always
acts in a downward direction and up is the positive direction the force of gravity
(g) is negative in equation (??)
X
F = −mg.
(12)
Combining equations (??), and (12) the equation of force can be written as
(16)
X
Fy
=
may
(d2 s)y
= −mg
dt
(dv)y
m
= −mg
dt
(dv)y
= −g
dt
The force equations for the x-axis and the z-axis are
X
Fx = max
m
(13)
(14)
(15)
(16)
(17)
2
m
(d s)x
dt
(dv)x
Xdt
Fz
(dv)x
dt
(d2 s)z
m
dt
(dv)x
dt
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=
0
(18)
=
0
(19)
= maz
(20)
=
0
(21)
=
0
(22)
=
0
(23)
5.1
Deriving Equations
Acceleration is the second derivative of position This can be written as
d2 s
a = dv
dt = dt . In order to find the height and distance equations it is necessary to integrate twice. The only direction that has any acceleration is in the
vertical direction, y-axis direction. This acceleration is due to gravity which is
g = 32.2( fs2t ).The integration below shows the steps needed to find the height
and distance equations. It is important to notice that the time in these equations are from time the tomahawk is released and not from the start of motion.
ax = 0, ay = −g, az = 0
Z
v
dv
v0 dt
Z v
dv
= −g
Z
(24)
t
−gdt
=
v0
(25)
0
v − v0
v
= −gt
(26)
= −gt + v0
(27)
(28)
v0 is the initial velocity after release of handle. Velocity is v =
integrated to get the position equation.
Z
y
y0
Z y
dy
dt
=
−gt + v0
Z
dy
=
y0
ds
dt
which can be
(29)
t
(−gt + v0 )dt
(30)
0
y − y0
=
y
=
−gt2
+ v0 t
2
−gt2
+ v0 t + y0
2
(31)
(32)
(33)
−gt2
+ v0 t + y 0
(34)
2
The acceleration of both the x-axis and z-axis are zero. Since they both equal
zero the integration will be the same for both. The x-axis will be integrated here.
y(t) =
8
v
Z
dv
dt
=
0
(35)
dv
=
0
(36)
v − v0
=
0
(37)
v
dx
dt
=
v0
(38)
=
v0
Z t
(39)
v
Z 0v
v0
Z
x
dx
=
x0
v0 dt
(40)
0
x − x0
=
x =
v0 t
(41)
v0 t + x 0
(42)
(43)
This function represents the distance traveled in the x-axis direction.
x(t) = v0 t + x0
(44)
The function for the z-axis can be solved by substituting z in for x.
z(t) = v0 t + z0
(45)
The equations for the z-axis were derived here but will not be used in the system
of differential equations. The reason for this is equation not being used is that
there is no velocity in the z-axis direction. If the tomahawk were to be rotated
about the x-axis some angle and and then thrown it would be necessary to use
the z-axis in the system of equations.
To show the relationship between the distance traveled and height a new
2
function must be found. To find this function lets look at y(t) = −gt
2 + v0 t + y0
and x(t) = v0 t + x0 . By solving for t in the x(t) and then substituting this
equation into the y(t) a new equation will now relate distance and height.
x =
x − x0
=
t =
v0 t + x 0
(46)
v0 t
x − x0
v0
(47)
(48)
Since the tomahawk’s initial velocity may not originate horizontally it is now
necessary to figure in an initial velocity angle (θ). This angle (θ) is the angle of
the initial velocity, as shown in this graph.
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Vy = 0
Vx
V
Vy
ttotal
2
Vx
Vy V
V
y
Vy
V
Vx
θ
x
Vy
V
Vx
This graph was found at http://tex.stackexchange.com/questions/215773/projectilemotion-diagram-using-pgfplots-tikz and it shows projectile motion similar to the
motion the tomahawk travels.
The velocity in the x-axis direction is found to be v0 cos(θ) and in the
0
y-axis directionv0 sin(θ). Making this substitution you get t = v0x−x
cos(θ) and
y=
−gt2
2
+ v0 sin(θ)t + y0 .
Now that that t is solved for it can be inserted into y(t).
y
y
y
−gt2
+ v0 sin(θ)t + y0
2
x − x0
−g(x − x0 )2
=
+ v0 sin(θ)
+ y0
2v0 (θ)
v0 cos(θ)
g(x − x0 )2
= tan(θ)(x − x0 ) −
+ y0
2v0 cos(θ)
=
(49)
(50)
(51)
This second order equation shows that the trajectory is a parabolic path.
In order to track a specific point on a body it must be related to a another
point of know velocity. In this case the center of mass will be used. A point on
the face of the blade will be used. This point is at the midpoint of the blade
and on the edge of it.It can be shown that the velocity of a point (A) on a rigid
body is equal to the velocity of the centroid (G) plus the velocity of (A) with
respect of (G).[4, p.548-550]
vA = vG + vA/G
(52)
Using a free body and mass diagram this equation is found.
¯ − mgi
vG j = Iαk
(53)
The vector equation for velocity is
v =ω×r
10
(54)
Combing equations (52), (53) and (54) a system of equations is found.
¯ − mg) + (ω × rA/G )
vA = (Iα
(55)
This equation can be separated into component form. The angle β represents the angle formed by the center of mass the x-axis and the point being
modeled. This point can be any point on the tomahawk but for this project it
is a point on the edge of the blade.
dx
dt
dy
dt
dz
dt
5.2
= −ωrcos(β)
(56)
= −mg + ωrsin(β)
(57)
¯
= Iα
(58)
System of Equations
A function was written in Matlab to solve the system of equations derived in
this project. The first order equations bellow model the position of the center
of mass and a point relative to it. The equations for the z-axis were not used in
this system even though they were derived. The reason for this is simple. The
time it was taking to solve for the moment of inertia to much.
x(1)0
=
x(2)0
=
x(3)0
=
x(4)0
=
The list of definitions bellow
sents.
dx
= −ωrsin(β)
(59)
dt
dy
= −mg + ωrsin(β)
(60)
dt
dy
= −gt + v0
(61)
dt
dx
= v0
(62)
dt
describes what each first order equation repre-
x(1)
=
Position in the x direction
x(2)
=
Position in the y direction
x(3)
=
Position of the center of mass in the y direction
x(4)
=
Position of the center of mass in the x direction
0
=
Velocity of a point in the x direction
0
x(2)
=
Velocity of a point in the y direction
x(3)0
=
Velocity of the center of mass in the y direction
0
=
Velocity of the center of mass in the x direction
x(1)
x(4)
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Results
This paper shows that the number of rotations a tomahawk takes before it
reaches a certain distance is dependent on the distance from the shoulder to the
center of mass. This means that no matter how fast or how hard the tomahawk
is thrown the distance remains the same for one revolution. Using equation (9)
and a distance of 2 12 (f t) for the radial distance it is calculated for one revolution
it takes 15.71 (f t).
This figures are graphs of the phase plane for this system of equations
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Conclusion
This project ended up being harder than expected. This project proved
that the distance needed for a full revolution is based on the distance from
the shoulder to the center of mass and not on the power and velocity of the
throw. It was shown that the center of mass travels in a parabolic motion.
An point on a rigid body such as the tomahawk can be related to the center of
mass and thus find this position relative to the center of mass or any other point.
A graph showing the phase plane of this system of equations was produced
along with graphs showing position versus time. These graphs can be used to
predict the path of the tomahawk and the given point. With more time and
research this project should produce a graph showing the circular motion of the
point on the blade relative to the path of the center of mass and the ground.
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References
[1] Ashbaugh, Mark S., Carmen C. Chicone, and Richard H. Cushman. ”The
Twisting Tennis Racket.” Journal of Dynamics and differential Equations
3.1(1991):67-85. Print.
[2] Kawano Daniel T., Alyssa Novelia, and Oliver O’Reilly M. ”Tossing a Book
in the Air” University of California Berkeley. 24 March 2015.
[3] Arnold David, Albert Boggess, and John Polking.Differential Equations
With Boundary Value Problems. 2nd ed. Upper Saddle River: Pearson
Prentice Hall
[4] Meriam, James L. and L. Glenn Kraige.Engineering Mechanics Dynamics.
7th ed. Hoboken: Wiley
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