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## Chapter 3

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**Chapter 3**Linear Kinetics Explaining the Causes of Linear Motion**Objectives**• Explain Newton’s three laws of motion • Apply Newton’s second law of motion to determine the acceleration of an object if the forces acting on the object are known • Apply Newton’s second law of motion to determine the net force acting on an object if the acceleration of the object is known**Objectives**• Define impulse • Define momentum • Explain the relationship between impulse and momentum • Describe the relationship between mass and weight**Newton’s First Law of Motion: Law of Inertia**• Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it • If no external force acts on an object, that object will not move, if it wasn’t moving to begin with, or it will continue moving at constant speed in a straight line if it was already moving**Newton’s First Law of Motion: Law of Inertia**• Newton’s first law of motion may be interpreted in several ways: • If an object is at rest and the net external force acting on it is zero, the object must remain at rest • If an object is in motion and the net external force acting on it is zero, the object must continue moving at constant velocity in a straight line • If an object is in motion at constant velocity in a straight line, the net external force acting on it must be zero**Conservation of Momentum**• Linear momentum—Product of the object’s mass and its linear velocity • The faster an object moves the more momentum it has • The larger a moving object’s mass, the more momentum it has • L = mv • L = linear momentum • m = mass • v = instantaneous velocity • Total momentum of a system of objects is constant if the net external force acting on the system is zero**Elastic Collisions**• Conservation of momentum principle can be used to predict post-collision movements of objects if we know their masses and their pre-collision velocities • Li = Σ(mu) = m1u1 + m2u2 = m1v1 + m2v2 = Σ(mv) = Lf • Velocity of penny A just before the collision equals the velocity of penny B just after the collision • mAuA = mBvB**Elastic Collisions**• Post-collision velocity of the nickel is equal to the pre-collision velocity of the penny times the ratio of the mass of the nickel to the mass of the penny—Penny’s mass (2.5g) is half the nickel’s mass (5g), so the nickel’s velocity is half the penny’s pre-collision velocity • mPuP = mNvN • vN = ( uPmP)/mN**Elastic Collisions**• In most situations, involving perfectly elastic collisions of two objects, each object transfers all of its momentum to the other object during the collision • If the collision involves a stationary object and a moving object, immediately after the collision, the previously stationary object will be moving, and the previously moving object will be stationary**Elastic Collisions**• What about head on collisions? • If the collision of the two pennies is perfectly elastic, then the precollison momentum of each is transferred to the other after the collision • Pre-collision momentum of penny A equals the post-collision momentum of penny B and vice versa**Elastic Collisions**• Third type of collision—Both objects moving in the same direction but at different velocities—Faster moving object collides with the slower moving object • Momentum of the faster moving object is completely transferred to the slower moving object • Immediately after the collision, the previously faster moving object stops, and the previously slower moving object now has the total momentum of the system**Inelastic Collisions**• In a perfectly inelastic (plastic) collision, momentum is still conserved, but rather than bouncing off of each other, the objects in the collision stay together after the collision and move together with the same velocity • m1u1 + m2u2 = (m1 + m2)v**Inelastic Collisions**• Most collisions occurring in American Football are examples of inelastic collisions—Two colliding players move together as one following the collision • Football puts a premium on momentum—mass and velocity are equally important—Faster and larger players are most successful**Inelastic Collisions**• Example: • Suppose an 80kg fullback collides in midair with a 120kg linebacker at the goal line during a goal line stand—Just before the collision, the fullback had a velocity of 6m/s toward the goal line, and the linebacker had a velocity of 5m/s in the opposite direction—Who will prevail?**Inelastic Collisions**• Towards goal line is positive direction • m1u1 + m2u2 = (m1 + m2)v • (80kg)(6m/s) + (120kg)(-5m/s) = (80kg +120kg)v • 480kg·m/s – 600kg·m/s = (200kg)v • -120kg·m/s/200kg = -.6m/s • Fullback will not score • Most collisions in sports are neither perfectly elastic or inelastic, but are somewhere in between**Coefficient of Restitution**• Means of quantifying how elastic the collisions of objects are • Defined as the absolute value of the ratio of the velocity of separation to the velocity of approach • Velocity of separation is the difference between the velocities of the two colliding objects just after the collision • Velocity of approach is the difference between the velocities of the two colliding objects just before the collision**Coefficient of Restitution**• e = |v1 – v2/u1 – u2| = |v2 – v1/u1 – u2| • E = coefficient of restitution • v1, v2 = post-impact velocities of objects one and two • u1, u2 = pre-impact velocities of objects one and two • The coefficient of restitution has no units—For perfectly elastic collisions, the coefficient of restitution is 1.0, its maximum value—For perfectly inelastic collisions, the coefficient of restitution is zero, its minimum value**Coefficient of Restitution**• Coefficient of restitution is affected by the nature of the objects in the collision • e = √bounce height/drop height • Critical measure in most ball sports since bounciness of ball or implement will greatly affect the outcome of a competition**Coefficient of Restitution**• USGA rules forbid drivers having a coefficient of restitution with the golf ball greater than .83 • NCAA men’s basketball rules require the ball to bounce to a height between 49 and 54in. (measured to the top of the ball) when dropped from a height of 6ft. • Rules of racquetball state that the ball must bounce to a height of 68 to 72in. if dropped from a height of 100in.**Coefficient of Restitution**• Baseball with a wooden bat is approximately .55 • Tennis ball on the court is about .73 • How the ball bounces is determined by its coefficient of restitution**Coefficient of Restitution**• Sample Problem 3.1 (text p. 86) • Golf ball is struck by a golf club—Mass of the ball is 46g and the mass of the club head is 210g—Club head’s velocity immediately prior to impact is 50m/s—Coefficient of restitution between the club head and the ball is .80 • How fast is the ball moving immediately after impact?**Coefficient of Restitution**• mball = 46g • mclub = 210g • uball = 0m/s • uclub = 50m/s • e = .80 • vball = ? (this is what we are ultimately solving for) • vclub = ?**Coefficient of Restitution**• m1u1 + m2u2 = m1v1 + m2v2 • mballuball + mclubuclub = mballvball + mclubvclub • e = |v2 – v1/u1 – u2| • e = vclub – vball/uball –uclub • vclub = e(uball – uclub) + vball • mballuball +mclubuclub = mballvball + mclub[e(uball – uclub) + vball]**Coefficient of Restitution**• (46g)(0) + (210g)(50m/s) = (46g)vball + (210g)[.80(0 – 50m/s) + vball] • (210g)(50m/s) = vball(46g +210g) – (210g)(.8)(50m/s) • (210g)(50m/s) + (210g)(.8)(50m/s) = vball(256g) • vball = (210g)(90m/s)/256g = 74m/s**Newton’s Second Law of Motion: Law of Acceleration**• If a net external force is exerted on an object, the object will accelerate in the direction of the net external force, and its acceleration will be directly proportional to the net external force and inversely proportional to its mass • ΣF = ma • ΣF = net external force • m = mass of the object • a = instantaneous acceleration of the object**Newton’s Second Law of Motion: Law of Acceleration**• Any time an object starts, stops, speeds up, slows down, or changes direction, it is accelerating and a net external force is acting to cause this acceleration • Force of gravity is equal to an objects weight (mass multiplied by the acceleration due to gravity) • W = mg**Newton’s Second Law of Motion: Law of Acceleration**• Standing in an elevator, initially, the only forces acting are gravity and the reaction force from the floor • These are vertical forces, so if we want to know acceleration and its direction: • ΣFy = may • ΣFy = R + (-W) = may • ΣFy = net external force in the vertical direction • m = mass • ay = vertical acceleration • W = weight • R = reaction force exerted on the feet by the elevator**Newton’s Second Law of Motion: Law of Acceleration**• If the reaction force, R, is larger than your weight, you feel heavier and the net force acts upward, resulting in an upward acceleration (e.g. speeding up in the upward direction) • If the reaction force, R, is equal to your weight, you feel neither heavier nor lighter, and the net force is zero, resulting in no acceleration (e.g. moving between floors) • If the reaction force, R, is less than your weight, you feel lighter and the net force acts downward, resulting in a downward acceleration (e.g. slowing down in the upward direction)**Newton’s Second Law of Motion: Law of Acceleration**• What about lifting a 10lb dumbbell? • The external forces are the pull of gravity acting downward and the reaction force from your hand acting upward • The net force is thus the difference between these two forces • To start the lift, you must accelerate the dumbbell upward—The force exerted on the dumbbell must be greater than 10lb.**Newton’s Second Law of Motion: Law of Acceleration**• To continue moving the dumbbell upward requires a net force of zero and must be equal to 10lb. • As you complete the lift, you need to slow down the upward movement of the dumbbell so the net force acting on the dumbbell is downward—The net force you exert on the dumbbell must be less than 10lb.**Newton’s Second Law of Motion: Law of Acceleration**• Accelerating something vertically requires much more force than accelerating something horizontally • Why?—Bowling ball example • ΣFy = Pulling force + (-W) = may • ΣFx = Pushing force + (-Ff) = max**Newton’s Second Law of Motion: Law of Acceleration**• A net force is needed to slow something down or speed it up or change directions • Sample Problem 3.2 (text p. 90) • 52kg runner is running forward 5m/s when foot strikes ground—Vertical ground reaction force is 1800N—Friction force is 300N—These are the only external forces acting other then gravity • What is the runners vertical acceleration as a result of these forces?**Newton’s Second Law of Motion: Law of Acceleration**• m = 52kg • Rx = 300N • Ry = 1800N • W = mg = (52kg)(9.81m/s) = 510N • ay = ? • ΣFy = (Ry – W) = may • 1800N – 510N = (52kg)(ay) • 1290N/52kg = 25m/s2 upward**Impulse and Momentum**• Except for gravity, most external forces change with time—So the acceleration of an object subjected to these forces also changes with time • Impulse is the product of force multiplied by the time that the force acts • Impulse-momentum relationship • ΣFΔt = m(vf – vi) • The average net force acting over some time interval will cause a change in momentum of an object—Because mass is constant, this usually means a change in velocity**Using Impulse to Increase Momentum**• The task in many sports skills is to cause a large change in the velocity of something • In throwing events, the ball has no velocity at the beginning of the throw, and the task is to give it a fast velocity by the end of the throw—We want to increase momentum of the ball • A large change in velocity is produced by a large average net force acting over a long time interval**Using Impulse to Increase Momentum**• Techniques in sports activities such as throwing or jumping are largely based on increasing the time of force application (technique modifications) to obtain a large impulse**Using Impulse to Decrease Momentum**• In certain other activities, an object may have a fast initial velocity, and we want to decrease this velocity to a slow or zero final velocity—We want to decrease the object’s momentum (e.g. landing from a jump, catching a ball, being struck by a punch)**Using Impulse to Decrease Momentum**• Sample Problem 3.3 (text p. 95) • Boxer is punching a heavy bag—Time of impact of the glove with the bag is .10s—Mass of the glove and hand is 3kg—Velocity of the glove just before impact is 25m/s • What is the average impact force exerted on the glove?**Newton’s Third Law of Motion: Law of Action-Reaction**• To every action there is always opposed an equal reaction • The effects of these forces are not canceled by each other because they act on different objects • It is the forces that are equal but opposite, not the effects of the forces**Newton’s Third Law of Motion: Law of Action-Reaction**• When you push or pull on something, what you feel is not the force that you are pushing or pulling with; it is the equal but opposite reaction force that is pushing or pulling on you • When you push against a wall, why don’t you accelerate as a result of the force the wall exerts on you?**Newton’s Third Law of Motion: Law of Action-Reaction**• The force you exert against the wall does not act on you, so it can’t counteract the effect of the force the wall exerts on you • Gravity pulls down on you with force equal to your weight • Friction between feet and floor • This frictional force opposes the pushing force from the wall and prevents you from accelerating as a result of the wall pushing against you