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Linear Kinetics Objectives. Identify Newton’s laws of motion and gravitation and describe practical illustrations of the laws Explain what factors affect friction and discuss the role of friction in daily activities and sports
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Linear Kinetics Objectives • Identify Newton’s laws of motion and gravitation and describe practical illustrations of the laws • Explain what factors affect friction and discuss the role of friction in daily activities and sports • Define impulse and momentum and explain the relationship between them • Explain what factors govern the outcome of a collision between two bodies • Discuss the interrelationship among mechanical work, power, and energy • Solve quantitative problems related to kinetic concepts
Linear Kinetics Outline - The Relationship between force and motion • Read Chapter 12 in text • Classification of forces • Types of forces encountered by humans • Force and motion relationships – three ways to look at it: • Instantaneous effect – Newton’s law of acceleration (F=ma) • Force applied through time (Impulse-momentum)(Ft = mv) • Conservation of Momentum • Force applied through distance (work-energy) (Fd = 1/2mv2) • Conservation of Energy • Self-study problems • Sample problems: #2 p 392; #3 p 396, #4 p 397, #5 p 402, #6 p 405, #7 p 408 • Introductory problems, p 411: 1,3,5,7,8,10 • Homework problems (Due Thursday, April 20) • Additional problems, p 412: 6,8,9
Effect of forces on the system (can be total human body, or a part of the body) • Action vs reaction • Internal vs external • Motive vs resistive • Force resolution – horizontal and vertical components • Simultaneous application of forces – determining the net force through vector summation
External forces commonly encountered by humans • Gravitational force (weight = mg) • Ground Reaction Force (GRF)(Figure 12-4, p 386) • Vertical • Horizontal (frictional) • Frictional force (coefficient of friction) (pp 389-395) • Elastic force (coefficient of restitution) (pp 399-402) • Free body diagram - force graph (p 63)
Coefficient of friction, resistance to sliding: Cfr = Frf /Nof Sample Prob # 2, p 392
Coefficient of Restitution (COR) • COR is a measure of the liveliness of an object • When 2 objects collide: • When one object is stationary, this reduces to: • An alternative way to measure COR is to drop a ball and measure the ht bounced compared to ht dropped:
Coefficient of Restitution (COR) • COR of balls dropped or thrown at a rigid wooden surface is shown here. • COR increases directly with temperature and inversely with impact velocity.
Instantaneous Effect of Force on an Object • Remember the concept of net force? • Need to combine, or add forces, to determine net force • Newton’s third law of motion (F = ma) • Inverse dynamics – estimating net forces from the acceleration of an object • Illustrations from Kreighbaum: Figures F.4, F.5, and F.6 (pp 283-284)
Force Applied Through a Time: Impulse-Momentum Relationship (pp 295-399) • Force applied through a time • Impulse - the area under the force-time curve • Momentum - total amount of movement (mass x velocity) • An impulse applied to an object will cause a change in its momentum (Ft = mv) • Conservation of momentum (collisions, or impacts) • in a closed system, momentum will not change • what is a closed system? • It is a system where net forces are zero • Example – horizontal movement of airborne objects, or where frictional forces are negligible • Example: Sample problem #3, p . 396 • Second example – slide # 19 (football player jumping and catching a ball)
Impulse: area under force- time curve Net impulse (Ft) produces a change in momentum (mV) Sample problem #4, p 397
Vertical impulse While Running: Area under Force-time curve
Anterioposterior (frictional) component of GRF: impulse Is area under Force-time curve Positive and Negative impulse Are equal if Horizontal comp Of velocity is constant
Conservation of momentum: when net impulse is zero (i.e. the system is closed), momentum does not change Also, sample prob #3, p 396
Force Applied Through a Distance: Work, Power, Energy (pp 403-409) • Work - force X distance (Newton-meters, or Joules) • On a bicycle: Work = F (2r X N) • Running up stairs: Work = Weightd (slide 21) • On a treadmill: Work = Weightd X per cent grade (slide 22) • Power - work rate, or combination of strength and speed (Newton-meters/second, or watts) • On a treadmill: P = Weightd X per cent grade/ time • On a bicycle: P = F (2r X N) / time • Running up stairs: P = Weightd /time (See next slide) • Energy - capacity to do work • kinetic, the energy by virtue of movement (KE = 1/2 mv2 ) • gravitational potential, energy of position (PE = weight x height) • elastic potential, or strain, energy of condition (PE = Fd)
Sample prob #6, p 405 Power running up stairs: Work rate = (weight X vertical dist) ÷ time
Work while running on treadmill: Note that vertical distance equals the product of running speed, time, and %grade.
Calculating Power on a Treadmill • Problem: What is workload (power) of a 100 kg man running on a treadmill at 10% grade at 4 m/s? • Solution: • Power = force x velocity • Force is simply body weight, or 100 x 9.8 = 980 N • Velocity is vertical velocity, or rate of climbing • Rate of climbing = treadmill speed x percent grade = 4 m/s x .1 = .4 m/s • Workload, workrate, or power = 980N X .4 m/s = 392 Watts • Note: 4 m/s = 9 mph, or a 6 min, 40 sec mile • Calculate your workload if you are running on a treadmill set at 5% grade and 5 m/s. • Answer for 200 lb wt (91 kg) is: 223 Watts
Conservation of Energy • In some situations, total amount of mechanical energy (potential + kinetic) does not change • Stored elastic energy converted to kinetic energy • diving board • bow (archery) • bending of pole in pole vault • landing on an elastic object (trampoline) • Gravitational potential energy converted to kinetic energy • Falling objects • Videodisk on pole vault
Energy conservation – Case I : elastic potential (strain) and kinetic Potential energy (FD) + Kinetic energy (1/2mv2) remains constant
Energy conservation – Case II : gravitational potential and kinetic Potential energy (Wh) + kinetic energy (1/2mv2) remains constant
Conservation of energy: gravitational potential and kinetic Sample problem #7, p 408
Falling objects and work-energy relationship • Problem: • If a 2 kg object is dropped from a height of 1.5 meters, what will be its velocity and kinetic energy when it hits the ground? • Solution: • Kinetic energy at impact (mgh) equals the potential energy at drop height (½ mv2) • Potential energy at drop(mgh) = 29.43 Nm • Kinetic energy at impact = 29.43 Nm; v = 5.42 m/s
Three ways to minimize impact force of 2 colliding objects • Force-time, or impulse-momentum relationship (Ft = mv) • Increase time through which force is applied • Force-distance, or work-energy relationship (FD = ½ mv2) • Increase distance through which force is applied • Force-area, or pressure concept (P = F/a) • Increase area over which force is applied
Revisiting the problem from week 1 regarding man falling from ledge • A man fell from the railing of a walkway on a second-story apartment building. He was found lying unconscious on his back with his center of mass located 5 feet horizontally from a second story walkway and railing. The top of the railing was 21.6 ft above the ground. His blood alcohol content was found to be .30 (inebriated) and he has no memory of how he fell. In order to appraise liability for the accident, we need to determine if the victim walked into the railing or if he was sitting on the railing and fell off. Can this be done from the information given? How? (Hint: First, find time of flight, then find horizontal velocity, then try to figure out what forces were required to obtain this velocity by using Newton’s law of acceleration (F = ma)