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 1
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. Detailed discussion of these concepts is in the Hay text. a) Mass of bat. (Can control: use a more massive bat ) b) Mass of ball. c) 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 = v t 2 /r Therefore centripetal force = mv t 2 /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 (v t = 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 2
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 v t = 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. 3
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 v f = v i + at 0 = 5.638-9.81t t = 5.638 /9.81 t = 0.575 s 4
Vertical displacement from take-off to peak height v f 2 = v i 2 + 2ad but at peak height v f = 0 0= (5.638 2 ) + (2 x 9.81)d d= (5.638 2 )/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 = v i t + 0.5at 2 a s vi = 0 then v i t = 0-3.12 = 0.5 x (-9.81) t 2 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. v t =rω = -40π x 0.04 = -5.03 m/s Resultant horizontal velocity. v h = 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 = v a /v b (v a = velocity after impact, v b = velocity before impact) Vertical velocity of ball s centre of gravity = 10cos25 = 9.06m/s So to calculate vertical velocity after bounce. 0.6 = v a /9.06 v a = 9.06 x 0.6 = 5.44 m/s So the components of the rebound velocity are 4.48 and 5.44. Resultant = (4.48 2 + 5.44 2 ) = (49.664) = 7.05 m/s θ = tan -1 (5.44/4.48) = 50.5 o 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 = 39.5 o 5