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The Mathematics
of
Ice Skating
soniaburney
tanyahou
sybillam
andreaolarig
Background Information
o a double axel and a split
parabolic
path
jump create
a
o the moment of inertia
while spinning changes
as arms are brought
towards the body
o maximum
potential
energy is reached at the
maximum vertical height
of a jump
maximum kinetic energy
o is
at reached
the beginning and end
of a jump,
Kinematics
no air resistance or friction
o assuming
s
o skaters must be concerned with
maximum height of jumps, rotational
speed, and horizontal speed
o skaters try to achieve maximum
displacement during their jumps
horizontal
and
vertical
o unlike speed skaters and ski jumpers, ice skaters generally are not
moving fast enough to have their jumps be affected by air
resistance
center of gravity always follows a parabolic shape
o the
Y
X
Kinematics and Ice Skating
o the take-off
take-off angle
angle take-off velocity,
take-off velocity height of takeoff height of take-off
are the three factors which determine
the figure skater's trajectory during a
jump
o (it is very important to separate the
object's vertical take-off velocity from
their horizontal take-off velocity)
o once gravity has slowed the skater's
upward vertical velocity to zero,
gravity then accelerates the skater
velocity
back to is
theZERO
earth
o velocity is ZERO at the peak of the
jump/vertical displacement is of
Kinematics Equations
v  vo  at
1 2
s  vo t  ( )at
2
2
2
(v  vo )
s
2a
1
s  ( )(v  vo )t
2
s position
vo
initial
velocity
v final
velocity
aconstant
acceleration
Calculus of Kinematics
ds DS
v

dt DT
dv Dv
a 
dt Dt
a(t )(dt )  d (v(t ))
 v(t )( dt )  ds

 v(t )(dt )   d (s(t ))
 a(t )(dt ) v(t )
 v(t )(dt )  s(t )
 a(t )(dt )   d (v(t ))
Application of Kinematics
o projectile motion
1 2
s  vo t  ( )at
2
ds
 v  vo  at
dt
o Donald Duck is about to perform a double axel. He takes off at an
angle of 45° and his vertical displacement is 1 meter and his
his flight timeisis2.5
0.5meters.
seconds, what is the initial velocity of
horizontal If
displacement
his jump? [ignore rotation]
g
1m
  45
continuedpartI
x component
y component
v  vo  vo cos 
v  vo  gt  vo sin   gt
a0
t t
xx
ag
t t
yy

v  vo  vo cos( 45)

v  vo  gt  vo sin( 45)  gt
a0
t  0.5 sec
x  2 .5 m
ag
t  0.5 sec
x  1m
continuedpartII
1 2
x  vox (cos  )t  at
2
2.5m  vox (cos 45)(0.5 sec)
m
vox  7.071
sec
dx
 vox (cos  )  at
dt
m
vx  5
sec
1 2
y  voyt  gt
2
1
1m  voy (sin 45)(0.5 sec)  g (0.5) 2
2
voy  0
o so, using kinematics we can find the initial and final velocity in the x
direction of the figure skater given specific quantities
Rotation [Moment of Inertia]
o assuming no air resistance or friction
o rotational
inertiaisisthethe
resistance
rotational inertia
resistance
to to
change
change in
in rotation
rotation
o scratch spins: the axis of rotation is the
body of the skater.
o the rotational inertia of a given mass is
given
I by
mr 2
o inertia also equals
I  lim r 2 m  r 2 dm
m0

asas
thethe
o the
the rotational
rotationalinertia
inertiaincreases
increases
square of
of the
the distance
distance to
to the
the axis
axis
square
o for example, if distance is doubled,
then inertia is quadrupled. Even a
small change in position can affect the
rotational inertia
The Moment of Inertia
o the moment of inertia of a spin is increased as arms are brought out
since the radius of the spin increases
o the
the radius
radius and
and the
the moment
moment of
of inertia
inertia are
are directly
directly proportional
proportional
radius of arms is large and so
inertia is also large
since radius is small, inertia is
small and so the skater is less
likely to resist change in rotation
Application of Inertia
o moment of inertia for a skater
o the mass of our skater will be 50 kg.
Her torso and standing leg are about
40 kg. The radius of the skater will
be 0.1meter. Her torso and one leg
will be represented by a cylinder so
the moment of1 inertia
will be
2
I  mR
2
o this is the moment of inertia around her torso
and standing leg
1
IA  (40kg)(0.1m) 2
2
 0.2kg  m 2
continued
o now, we have to find the moment of inertia for her arms and other
leg as they move closer to the body
o in this caseI  MR 2
o this is the moment of inertia of her arm and leg
IB  (50kg  40kg)(0.1m) 2
 0.1kg  m2
o this diagram represents the basic inertia of the skater
(central axis) and the circle she creates
o to find the total moment of inertia, we can add the inertia of the
torso/leg with the inertia of the arms/leg
IA  IB  0.2  0.1  0.3kg  m2
Rotation [Angular Momentum]
o angular momentumanisobject’s
an
resistance
object’s toresistance
change in rotation
to
change in rotation
conserved
o angular
momentum
if there is no
force presentis
conserved if there is no
force present
o the force needed to create
angular
momentum
is
produced when the ice
skater pushes against the
L  I
ice
o angular
momentum
is
any change in
rotation is due to a
force
Calculus of Angular Momentum
o since angular momentum is conserved with noIforce, if
changes
must
stays constant at
 then
Lchange so
various time intervals
o as the moment of inertia decreases, angular velocity

increases
  o  t
o rotational kinematics for spins as long as is constant
d
1 2
  o  t
  o  t  
2
dt
2
 2  o  2 ( )

o angular
o angular
d
dt
d d 2

 2
velocity
dt
dt
d
L

I

(
acceleration dt )
Application of Angular Momentum
o we can use rotational kinematics to find angular momentum
1
2
o position  ot  t 2
o since L  I  (
d
)
dt
we can take the derivative of position at a given time to find angular
momentum.
d
dt
  o  t  
o for example, if initial velocity of a spinning skater is 20 rad/sec with a
constant acceleration 10 rad/sec², at time = 0.1 sec, what is the angular
momentum of the spinning skater? (using inertia
found from pervious
d
L

I

(
)
d
application)
dt
  o  t  
 I 
dt
rad
rad
 10 2 (0.1sec)
sec
sec
rad
 21
sec
 20
 0.3kg  m 2  21
kg  m 2
 6.3
sec
rad
sec
Rotation [Calculus of Torque]
o torque is the type of force that makes something rotate
o there is no net torque when angular momentum is conserved
o the equation for torque is defined by
T  I
 Inertia  AngularAcceleration
o since during a scratch spin, angular momentum changes, so there
is a net torque acting on the skater
o also
d
d 2
Power  Torque  AngularVelocity
or
T  I ( )
T  I (
)
dt
dt 2
T  T (
d
)
dt
Torque  DerivativeofAngularMomentum
dL

dt
Application of Torque
o so, given a specific equation at
a specific time and inertia, one
could find the net torque
o for example, at t = 0.2 sec given
I  5kg  m 2
  5t 2  2t 3
d 2
 10  12t
dt 2
rad
  12.4 2
sec
d 2
T  2 I
dt
12.4
rad
 (5kg  m 2 )  310 Nm
2
sec
o And, we can find power
of the skater at t =0.2 sec
P T(
d
)
dt
d
 10t  6t 2
dt
rad
 2.24
sec 2
P  310 Nm  2.24
 694.4
Nm
sec 2
rad
sec 2
Energy
o to better understand how
potential and kinetic energy
works, rotation and torque will
be negligent
o the skill that also best fits this
type of criteria would be the
split jump
o something to always keep in
mind is that energy is always
conserved and neither created
nor destroyed
Starting with Potential
o as with any jump or skill, the skater
needs to build up or increase the
amount of potential energy they have
before executing a move
by definition potential energy
o since
by definition
potential energy is
is independent
of motion
independent of motion
o this means that the potential energy of
a skater is the amount of energy
stored in muscle power
o after a skater jumps, their muscle
power potential energy is converted to
kinetic energy, which is then converted
to gravitational potential energy (at the
very top of their flight), and then
converted back to kinetic once they
Kinetic Energy
o if potential energy is the amount
of energy when an object is not in
motion,
thenenergy
kineticisenergy
is the
kinetic
the energy
energy
object
has by
virtue
of
an
objectanhas
by virtue
of its
motion
its motion
o the sum of kinetic and potential
energy is mechanical energy
o assuming
there
are
no
nonconservative forces (such as
friction), then mechanical energy
is conserved
sinceof energy
is
Law
of Conservation
Total Energy
always conserved – Law of
Conservation of Total Energy
o this basically means that the
initial mechanical energy of the
skater is equal to the final
mechanical energy
E  K U
Ki  U i  K f  U f
E
mechanical energy
U
K
potential energy
kinetic energy
Further into Energy
o How does potential and
kinetic energy all relate to
the skater?
o First of all, potential is
U  mgh
mass  gravity  height
o This would be applied as
the mass of the skater
multiplied by gravity (a
constant) multiplied by
the skater’s height
continued
1 2
o next, kinetic energyKis mv
2
o where is this derived from?
F
a
o since the skater’s acceleration is
and will have
m
traveled a distance
of
then the final
can be
s
v speed,
proven with kinematic equations
F
v  v0  2as  2as  2
s
m
1
 Fs  mv 2
2
2
2
o one might notice though, that
W  Fs
o however, work done by the skater has transferred energy
to it, which comes in as kinetic energy
Application of Energy
o a skater is coming out of
o the strength of the friction
her split jump at a 20°
force on the skater is
angle, and lands skating
F f   k FN   (mg cos  )
on the ice a distance of
10m. If the coefficient of
kinetic friction of the
o so work done by friction is
calculate
the skater’s
skates and
ice is speed
0.2,
 F f d   k (mg cos  )  d
calculate
at
the end the skater’s
speed at the end
o the vertical height in the
end would be
h  d sin 
continued
o hence, with the Law of Conservation of Total Energy, this gives us
K i  U i  W friction  K f  U f
1 2
mv  0
2
1
mg (d sin  )  (  k mg cos   d )  mv2
2
1
gd (sin    k mg cos  )  v 2
2
v  2 gd (sin    k cos 
0  mgh  (  k mg cos   d ) 
v  2(10)(10)[sin 20  (0.2) cos 20 ]
v  30.8
m
s
Works Cited
o Hokin, Sam. “Figure Skater Spins.” Online. Available:
http://www.bsharp.org/physics/stuff/skater.html,
14 May 2006.
o King, Deborah. “The Science of Jumping and Rotating.” Online.
Available:
http://btc.montana.edu/olympics/physbio/biomechanics/biointro.html. 14 May 2006.
o Knierman, Karen and Rigby, Jane. “The Physics Ice Skating.”
Online.
Available:
http://satchmo.as.arizona.edu/~jrigby/skating/main.html,
14 May 2006.
o Nave, C.R. “HyperPhysics.” Online. Available:
http://hyperphysics.phy-astr.gsu.edu/HBASE/hframe.html,
14 May 2006.