the mathematics of
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The Mathematics of. Ice Skating. sonia burney tanya hou sybil lam andrea olarig. Background Information. a double axel and a split jump create a the moment of inertia while spinning changes as arms are brought towards the body - PowerPoint PPT PresentationTRANSCRIPT
The Mathematics of
soniaburneytanyahousybillamandreaolarig
Ice Skating
Background Information
o a double axel and a split 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
o at the beginning and end of a jump,
parabolic path
maximum kinetic energy is reached
Kinematics
o so skaters must be concerned with
o skaters try to achieve maximum horizontal and vertical displacement during their jumps
o unlike speed skaters and ski jumpers, ice skaters generally are not moving fast enough to have their jumps be affected by air resistance
o the
maximum height of jumps, rotational
speed, and horizontal speed
assuming no air resistance or friction
center of gravity always follows a parabolic shape
XY
Kinematics and Ice Skating
o the take-off angle take-off velocity,
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 back to the earth
o velocity is ZERO at the peak of the jump/vertical displacement is of maximum value
take-off angletake-off velocityheight of take-off
velocity is ZERO
Kinematics Equations
position
tvvs
a
vvs
attvs
atvv
o
o
o
o
))(2
1(
2
)(
)2
1(
22
2
initial velocity
final velocity
constant acceleration
s
ov
v
a
Calculus of Kinematics
)())(( tsdttv
))(())(( tsddttv
DT
DS
dt
dsv
dsdttv ))((
Dt
Dv
dt
dva
)())(( tvdtta
))(())(( tvddtta
))(())(( tvddtta
Application of Kinematics
o projectile motion
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 horizontal displacement is 2.5 meters.
45m1
g
2)2
1( attvs o atvv
dt
dso
If his flight time is 0.5 seconds, what is the initial velocity of his jump? [ignore rotation]
continuedpartI
mx
t
a
vvv oo
5.2
sec5.0
0
)45cos(
xx
tt
a
vvv oo
0
cosx component
yy
tt
ga
gtvgtvv oo
sin
mx
t
ga
gtvgtvv oo
1
sec5.0
)45sin(
y component
continuedpartII
o so, using kinematics we can find the initial and final velocity in the x direction of the figure skater given specific quantities
sec5
)(cos
sec071.7
sec)5.0)(45(cos5.22
1)(cos 2
mv
atvdt
dx
mv
vm
attvx
x
ox
ox
ox
ox
0
)5.0(2
1sec)5.0)(45(sin1
2
1
2
2
oy
oy
oy
v
gvm
gttvy
Rotation [Moment of Inertia]
o assuming no air resistance or frictiono rotational inertia is the resistance to
change in rotation o scratch spins: the axis of rotation is
the body of the skater.o the rotational inertia of a given mass
is given by o inertia also equals
o the rotational inertia increases as the square of the distance to the axis
o for example, if distance is doubled, then inertia is quadrupled. Even a small change in position can affect the rotational inertia
2mrI
dmrmrIm
22
0lim
rotational inertia is the resistance to change in rotation
the rotational inertia increases as the square of the distance to the axis
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 radius and the moment of inertia are directly 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
the radius and the moment of inertia are directly proportional
Application of Inertia
o moment of inertia for a skatero 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 of inertia will be
2
2
1mRI
2
2
2.0
)1.0)(40(2
1
mkg
mkgIA
o this is the moment of inertia around her torso and standing leg
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 case
o this is the moment of inertia of her arm and leg
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
2MRI
2
2
1.0
)1.0)(4050(
mkg
mkgkgIB
23.01.02.0 mkgIBIA
Rotation [Angular Momentum]
o angular momentum is an object’s resistance to change in rotation
o angular momentum is conserved if there is no force present
o the force needed to create angular momentum is produced when the ice skater pushes against the ice
o angular momentum is defined by IL
an object’s resistance to change in rotation
conserved
if there is no force present
any change in rotation is due to
a force
Calculus of Angular Momentum
o since angular momentum is conserved with no force, if changes then must change so stays constant at various time intervals
o as the moment of inertia decreases, angular velocity increases
o rotational kinematics for spins as long as is constant
o angular velocity
o angular acceleration
o angular momentum
)(2
2
1
22
2
o
o
o
t
t
t
dt
do
dt
d
2
2
dt
d
dt
d
)(dt
dIL
I L
Application of Angular Momentum
o we can use rotational kinematics to find angular momentumo position
o since
we can take the derivative of position at a given time to find angular momentum.
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 application)
2
2
1tto
)(dt
dIL
t
dt
do
sec21
sec)1.0(sec
10sec
202
rad
radrad
tdt
do
sec3.6
sec213.0
)(
2
2
mkg
radmkg
Idt
dIL
Rotation [Calculus of Torque]
o torque is the type of force that makes something rotateo there is no net torque when angular momentum is
conservedo the equation for torque is defined by
o since during a scratch spin, angular momentum changes, so there is a net torque acting on the skater
o also or
elerationAngularAccInertia
IT
)(dt
dIT
)(
2
2
dt
dIT
)(dt
dTT
ocityAngularVelTorquePower
dt
dL
omentumofAngularMDerivativeTorque
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 32
2
25
5
tt
mkgI
Nmmkgrad
Idt
dT
rad
tdt
d
310)5(sec
4.12
sec4.12
1210
22
2
2
2
2
2
2
2
2
2
sec4.694
sec24.2310
sec24.2
610
)(
Nm
radNmP
rad
ttdt
ddt
dTP
o And, we can find power of the skater at t =0.2 sec
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
o since by definition potential energy is 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 land
by definition potential energy is independent of motion
Kinetic Energy
o if potential energy is the amount of energy when an object is not in motion, then kinetic energy is the energy an object has by virtue of 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 since energy is 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
kinetic energy is the energy an object has by virtue of its motion
Law of Conservation of Total Energy
mechanical energypotential energy
E
U
ffii UKUK
UKE
kinetic energyK
Further into Energy
o How does potential and kinetic energy all relate to the skater?
o First of all, potential is
o This would be applied as the mass of the skater multiplied by gravity (a constant) multiplied by the skater’s height
mghU heightgravitymass
o next, kinetic energy is
o where is this derived from? o since the skater’s acceleration is and will
have traveled a distance of then the final speed, can be proven with kinematic equations
o one might notice though, that o however, work done by the skater has transferred
energy to it, which comes in as kinetic energy
2
2
1mvK
m
Fa
s v
sm
Fsasavv 2222
02
2
2
1mvsF
sFW
continued
Application of Energy
o a skater is coming out of her split jump at a 20° angle, and lands skating on the ice a distance of 10m. If the coefficient of kinetic friction of the skates and ice is 0.2, calculate the skater’s speed at the end
dmgdF kf )cos(
o the strength of the friction force on the skater is
o so work done by friction is
o the vertical height in the end would be
)cos( mgFF Nkf
sindh
calculate the skater’s speed at the end
o hence, with the Law of Conservation of Total Energy, this gives us
s
mv
v
gdv
vmggd
mvdmgdmg
mvdmgmgh
UKWUK
k
k
k
k
fffrictionii
8.30
]20cos)2.0(20)[sin10)(10(2
cos(sin2
2
1)cos(sin
2
1)cos()sin(
02
1)cos(0
2
2
2
continued
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/bio-intro.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.