Transcript Physics of Basketball

Physics of
Basketball
Kailey Medfisch, Sarah Chomos & Jordan Lake
How The Shot Works anatomically
● Isotonic Contractions
○ Concentric: muscles shorten (ex: biceps during prep phase)
○ Eccentric: muscles lengthen (ex: Triceps during follow through)
● Extensors: Bring body back to position
Flexors:
muscles pull on bone to bend at joint
--knee, hip, ankle, elbow
-shoulder and wrist
● Force behind contraction
o forces built up from flexion of knees and transfer to upper
body during shot, increasing acceleration
o Forces present in all joint; transfer power from quadriceps up
through all 17 muscles of the hand
lever mechanics of the forearm
i. 1st class: Extension of the triceps: load-ball in hand,
fulcrumelbow joint,
effort-triceps
ii. 3rd class: Load-ball in hand, effort - muscle nsertion
holes on
forearm,
fulcrum-elbow joint
iii. 3rd class: Load-ball in hand, effort-wrist muscles insert
how muscle contraction occurs &
energy transfer
● actin and myosin: two main filaments in muscle fiber
o atp (adenosine triphosphate) and Calcium ions needed for
contractions
o contractions are graded: each muscle fiber (cell) contracts
asynchronously
● atp is attached to myosin giving energy
o stimuli hits, causes a twitch
o continuous twitch = complete tetanus or contraction
o causes kinetic energy of entire muscle to be effected
● Filaments slide over one another to shorten fiber
● Potential energy comes from the muscles in legs wanting to pull
back
into
cartilage and coefficient of friction
● articular (hyaline) cartilage
(2.21mm thick in humans)
o porous and spongy - joint bends, provides pressure,
releases synovial fluid
o synovial fluid allows joints to have a negligible coefficient
of friction
 μs = 0.01
μk = 0.003
 reduced friction relates to the hydrodynamics (such
as a car hydroplaning)
● HYaluronic Acid
o lubricant and shock absorber
o Under high strain: hyaluronic molecules entangle -
Mechanics
projectile motion
-Starting from Rest, a force is applied to the basketball, and it
travels independently in the X and the Y directions.
-Forces acting in the Y direction: 1. Gravity (g)=(mass of ball)x(-9.8
m/s2)
2. Y-component of
Applied Force
3. Magnus Force
- The ball will travel upwards until it reaches a maximum heightat this instant, its y-velocity is equal to zero.
- v2final= v2initial + 2(acceleration=G)(Δdisplacement)
Mechanics Continued
Projectile Motion
-Completely independent of the Y components, a force is
applied to the Ball in the X-direction
-the only other force other than the applied force that
acts on the ball in the X DIRECTION is air resistancewhich will be considered negligible.
-because no forces act on the ball in the x-direction
through the course of its trajectory (ideally), the XVELOCITY REMAINS CONSTANT UNTIL THE BALL
MAKES CONTACT WITH THE HOOP.
MECHANICS CONTINUED
Rotational Motion
-As the ball projects through the air, it also rotates.
-Angular Velocity (ω)= ΔΘ / Δtime
-Angular Acceleration (α)= Δω / Δtime
The Rotation of the ball creates an upward Magnus force,
which will go against the force of gravity and help the ball
travel longer and further.
Mechanical Energy
-Mechanical Energy = Potential Energy + Kinetic
Energy
-potential energy= (mass of ball)x(G)x(height)
-Kinetic Energy= ½ (mass)(velocity)2 + ½ (I=
moment of
inertia)(ω=angular
velocity)2
-MOMENT
OF INERTIA= ⅔(MASS) X (RADIUS)2
- SO, KINETIC ENERGY= TRANSLATIONAL
Law of Conservation of Mechanical
Energy
The total mechanical energy of a
system (the sum of potential and
kinetic energies) remains constant,
as long as all of the forces acting
upon it are conservative forces.
-examples of nonconservative forces- friction, air
Conservation of Energy
-assuming that all energy is conserved, the ball
should have the same amount of energy at the
beginning and end of its trajectory
-because the ball is stopped at a point higher in
the trajectory than where it was launched, its
“final kinetic energy” will be less than its initial.
-as the speed of the ball increases, kinetic
energy increases.
Ideal Shot
Under ideal circumstances, the acceleration of
the ball should be -9.8 m/s/s because the only
vertical force should be the gravitational force;
also, there should be no horizontal acceleration
as the ball should remain at a constant
horizontal velocity. However, we know that in
the real world there are nonconservative forces
(such as air resistance), so it is likely that the
Three-Point Shot
Time in air: 1 second
Y-Components
Initial Position: 2.646
m
Maximum Position:
5.660 m
Final Position: 3.049
m
Initial Velocity: 5.872
m/s
Final Velocity: -3.743
m/s
Average
Angle of Shot: 49.38°
Energy
Initial Mechanical Energy:
KE=18.01J PE=15.07J ME=33.08J
ME at Top of Shot: KE=7.82J
PE=25.58J ME=33.4J
Free-THrow
Time in Air: .867
seconds
Y-Components
Initial Position:
2.789 m
Max Position:
3.889 m
Final Position:
3.192 m
Initial Velocity:
4.907 m/s
Final Velocity: 3.073 m/s
X-components:
Energy:
initial position= o.o mInitial point:
Final position= 3.905 PE=
m 15.88 j
initial velocity= 5.397 KE=15.46
m/s
J
final velocity= 3.899 m/s
Maximum: PE=
o J KE=
Angle of Launch= 42.44
22.14
Half-Court Shot
X COMPONENTS:
Y COMPONENTS:
INITIAL
INITIAL POSITIONPOSITION- 2.903
3.803 M
M
FINAL POSITIONFINAL POSITION3.048 M
10.66 M
Initial Velocity- 3.640
INITIAL
m/s
VELOCITY- 8.393
Final Velocity- -5.224
M/S
m/s
FINAL VELOCITYAcceleration- -9.757
7.895 M/S
m/s2
TIME- 0.0-1.0
SECONDS
Kinetic Energy:
initial point t= 0 sec
Vtotal=9.148 m/s
ke= 24.31 J
PE=21.65 J
maximum t=0.417
sec
vtotal=7.859 m/s
ke=17.94 j pe=
26.62 j
right before hoop
t=1.0 sec
vtotal=-9.467 m/s
Experimental sources of Error
-setting a scale in logger pro.
-tracking the ball through frames of video in
logger pro.
-due to equipment, we could not calculate the
components of rotational motion.
the end