Kinesiology 201 Solutions
Fluid and Sports Biomechanics
Tony Leyland
School of Kinesiology
Simon Fraser University
Fluid Biomechanics
1. Lift force is a force due to fluid flow around a body that acts perpendicular to the
direction of that fluid flow (whereas drag forces act in the same direction as the fluid
flow). Lift forces are important in such sports as swimming, discus, javelin, and ski
jumping. Refer to the various texts on reserve in the library for such specifics.
2. Primarily to generate propulsive lift forces. The hands are “bladed” through the water
creating an angle of attack with the relative flow of water past the hand. This
generates a low-pressure area behind the hand and a higher pressure area in front
of the hand thus creating a net propulsive lift force.
3. For most balls, as the relative speed with the air increases the drag increases until
turbulent flow is initiated in the boundary layer. This in turn moves the separation
point of the boundary layer further back and reduces the magnitude of the drag.
Drag
Force
Ball Velocity
4.
Yes a curve ball really curves? This was demonstrated many years ago when a
major league pitcher threw a baseball at three stakes to follow the path shown
below. The three stakes are in line (despite the optical illusion that this is not the
case).
O
O
O
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The curve is achieved by spinning the ball very fast. Different locations on a spinning
ball experiences different amounts of air friction and air pressure. This is due to the
relative differences in velocities on either side of the ball. The top of a top-spun ball for
example is moving through the air quicker than the bottom of the ball. This creates a
larger force on the top of the ball than the bottom (this force called a Magnus force). The
Magnus force causes the ball to drop quicker than it would without spin.
Sport Biomechanics
1.
a)
b)
c)
Detailed discussion of these concepts is in the Hay text.
Mass of bat. (Can control: use a more massive bat )
Mass of ball.
Velocity of bat. (Some control: swing hard and/or hold very low down to increase the
distance from axis of rotation and contact point [tangential velocity =Vt = ωr])
d) Velocity of ball. (Some control: wait for fastball)
e) Angle of incidence. (theoretically some control, but it is not practical to hit ball late so
as to increase this angle [ball would go out of play])
f) Coefficient of restitution. (You could argue that you have some control in that you
should make sure you hit the sweet spot [centre of percussion])
2.
Radial acceleration = vt2/r
Therefore centripetal force = mvt2/r (mass x acceleration)
So an increase in mass and tangential velocity, and a decrease in the radius of the
corner, will increase the amount of centripetal force required.
The centripetal force acts on the bike at the tires (force due to friction). As this acts
some distance from the centre of gravity of the bike/cyclist system a torque is generated
that tends to rotate the bike/cyclist system outwards. By leaning in the centre of gravity
of the bike/cyclist system moves outside of the base of support (towards the centre of
the corner) thus generating an opposing torque). The opposing torques balance out so
the cyclist does not fall inward or outward.
3. A rotating body will continue to turn about its axis of rotation with constant angular
velocity unless an external couple or eccentric force is exerted upon it.
or
If gravity is the only external force (hence no torque) acting upon a body it will continue
to turn about its axis of rotation with constant angular momentum
Angular momentum = H = Ιω
Therefore 10 = 3.5ω ω = 10/3.5 = 2.86 rad/s
Yes. As discussed in explaining manoeuvres like cat-rotation.
4. If the angular velocity of the bat is the same, then the linear velocity of the bat at the
point of contact would be less as the radius of rotation has been decreased and the
angular velocity of the swing is constant (vt = ωr). Therefore the speed of the hit ball
would be reduced. In terms of the direction of the hit ball (velocity is a vector!) this is
not really determinable from the information given. If the batter starts his/her swing
at the same time with the same angular impulse then the bat will have traveled
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further due to its lower moment of inertia (Ι). Therefore the ball will be pulled more to
the left (right-handed batter).
However, by choking up and reducing Ι the players can rotate the bat or racquet quicker,
In baseball this means you can wait longer before deciding where or whether to swing.
In squash the player can swing later which can add to deception or allow you to get the
racquet "around" on a ball you are scrambling to play.
5. The spin will not affect the vertical component of the velocity after contact with the
ground, this is determined by the coefficient of restitution between the ball and the
playing surface.
Assume a ball travelling from left to right for this discussion. Top-spin results in the
bottom of the ball moving backwards relative to the ball’s centre of gravity (that old
equation vt = ωr again!). This means that the bottom of the ball is moving slower relative
to the ground than if it had no spin or under-spin. This in turn results in less friction in
the negative direction and thus less linear impulse in this direction. This in turn results in
less change in the horizontal velocity of the ball. Indeed, the ball can often be spun so
much that the bottom of the ball is moving backwards relative to the ground. So if the
underside of the ball is traveling in a negative direction the friction will be positive, thus
increasing the horizontal velocity of the ball. Thus the ball will bounce out relatively
lower compared to the same ball coming in at the same angle with no spin or under-spin.
This is because the vertical component of the velocity after rebound is the same for the
top-spun and no-spin balls but the horizontal velocity is greater for the top-spun ball (see
below).
Top-spin
No-spin
However, to answer the next part of the question you have to understand something
about fluid forces. A top-spun ball will experience a downward force in addition to
gravity due to the fluid flow around it. This force is called a Magnus force. This results in
the ball deviating from its parabolic flight and striking the ground at a much steeper
angle. So although it kicks out low compared to the angle it comes in at, it comes in so
steep that it still bounces out relatively high. So the answer is to expect a higher relative
bounce when tracking down a top-spun ball, compared to no spin or under-spin.
6. Conservation of angular momentum. Once in the air no external torques can be applied
to the volleyball player (torques due to air resistance are negligible). Therefore the total
angular momentum of the system (volleyball player) is constant. The torques that
accelerate his arm clockwise as shown generate clockwise angular moment. An equal
in magnitude (but opposite in direction) amount of angular momentum is generated in
the legs to keep the change in angular momentum at zero. Therefore his legs must
swing anti-clockwise.
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7. a) Drag forces opposite to the direction of travel of the ball (specifically in the same
direction as the relative airflow past the ball).
b) Magnus forces perpendicular to the relative flow of air past the ball.
c) Gravity acts downwards towards the centre of the earth.
8. When the blade hits the ice there is an opposing force generated (mostly normal
reaction force and some friction). This generates a torque on the stick and if the
hands were together, this would severely retard the velocity of the stick. If the hands
are together, little torque can be generated by the player to offset this opposing
torque. Anyone who has hit the ball "fat" in golf or hit the ground first in field hockey
knows the truth of this description. By spreading the hands of the stick the player is
able to generate considerable torque (leverage) in the direction of intended motion of
the stick and keep it moving quickly through to puck contact.
9. Magnitude of force generated against the board and the angle of lean (how far is the
perpendicular distance from the line of action of the force to the gymnasts centre of
gravity.
The magnitude of the force the board pushes back with depends on the force exerted by
the gymnast and the amount of energy returned from the springboard.
10. To become unstable the line of action of gravity (from subject centre of gravity) must
fall outside the base of support. This will result in a moment turning the individual.
a) The mass of the subject. The more massive the greater the stability
b) The height of the centre of gravity. The higher the CG the greater the stability
c) The size of the base of support. The wider the base the greater the stability
d) The distance the centre of gravity is from the boundaries of the base of support. The
greater the distance the greater the stability
11. Long-jumpers take-off with forward angular momentum due to the direction of the
force vector at take-off (this is behind the body’s centre of gravity resulting in a
moment rotating the athlete forward). If a good jumper who is in the air a long time
does nothing about this she/he will rotate into a poor position for landing. Hence this
“hang” technique simply maximises the athlete’s moment of inertia about a
transverse axis and reduces their angular velocity (H = Ιω).
12. a) Some of the energy put into the initial jump is transferred into the board on
landing and returned during the second jump.
The landing and subsequent drive for the take-off initiates a stretch-shortening cycle
(similar to a plyometric exercise). This turns the muscles on (quadriceps for example),
initiates a forceful contraction and stores some energy in the muscle’s elastic tissues.
b) Strain energy
c) As only time is required all calculations are in the vertical plane. You could use the
quadratic this is the other methodology for projectiles, namely breaking the flight into
sections.
Vertical velocity = 6sin70 = 5.638 m/s
Vertical from take-off to peak height
vf = vi + at
0 = 5.638 -9.81t
t = 5.638 /9.81
t = 0.575 s
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Vertical displacement from take-off to peak height
but at peak height vf = 0
vf2 = vi2 + 2ad
0= (5.6382) + (2 x 9.81)d
d= (5.6382)/19.62
= 1.62 m
So total vertical displacement from peak height to water entry = -1.5 - 1.62 = -3.12 m
Vertical peak height to water entry
d = vit + 0.5at2
-3.12 = 0.5 x (-9.81) t2
as vi = 0 then vit = 0
t = √(6.24/9.81) = 0.78 s
Therefore: total time in air = 0.575 + 0.78 = 1.37 seconds
13. a) Horizontal velocity of ball’s centre of gravity = 10sin25 = 4.23m/s (the angle of
incidence is taken from the vertical)
Velocity due to spin.
vt=rω = -40π x 0.04 = -5.03 m/s
Resultant horizontal velocity. vh = 4.23 – 5.03 = -0.804 m/s
b) Note that the force given is the frictional force that acts in a horizontal direction.
Force and time are given in the horizontal so this part of the question likely requires
impulse-momentum relationship (Ft = ∆mv).
Horizontal the impulse would be Ft = 10 x 0.1 = 1 = ∆mv = 4v
Therefore ∆v =0.25m/s
So because of the large amount of spin that resulted in the ball’s underside travelling
backward, the frictional force acts forward and adds to the ball’s overall velocity.
Horizontal velocity after the bounce = 4.23 + 0.25 = 4.48 m/s
In the vertical the coefficient of restitution is given so Newton’s law of impact is to be
used. As the floor is stationary, the equation relating velocities can be rearranged to:
-e = va/vb
(va = velocity after impact, vb = velocity before impact)
Vertical velocity of ball’s centre of gravity = 10cos25 = 9.06m/s
So to calculate vertical velocity after bounce. 0.6 = va/9.06 va = 9.06 x 0.6 = 5.44 m/s
So the components of the rebound velocity are 4.48 and 5.44.
Resultant = √(4.482 + 5.442) = √(49.664) = 7.05 m/s θ = tan-1(5.44/4.48) = 50.5o
This makes sense as the ball kicks out compared to its steep angle of entry (the vertical
velocity was reduced while the horizontal added to. Remember that this angle is the
angle to the right horizontal. To stay with the convention of angle of incidence and angel
of reflection being given from the vertical we could right the answer as this.
Rebound speed = 7.05 m/s at an angle of reflection = 39.5o
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