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:
1. If an object is at rest and the net external force
acting on it is zero, the object must remain at rest
2. 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
3. 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 precollision 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 precollision 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
Newton’s Third Law of Motion: Law
of Action-Reaction
• What about forces that cause
accelerations?
– Football example
– Newton’s third law helps explain how forces
act; it does not explain what the effects of the
forces will be (remember separate objects)
Newton’s Law of Universal
Gravitation
• All objects attract each other with a gravitational force
that is inversely proportional to the square of the
distance between the objects
• This force of gravity is proportional to the mass of each
of the two bodies being attracted to each other
• F = G(m1m2/r2)
• F = force of gravity
• G = universal constant of gravitation
• m1 and m2 = mass of the two objects involved
• r = distance between the centers of mass of the two
objects
Newton’s Law of Universal
Gravitation
• The gravitational forces between most of
the objects in sports are very small—so
small that we can ignore them
• However, the earth does produce
substantial gravitational force on other
objects
• The earths gravitational force acting on an
object is equal to the objects weight
Newton’s Law of Universal
Gravitation
• F = G(m1m2/r2)
– For objects close to the earth’s surface
several of the terms in the equation are
constant
– G = universal constant of gravitation
– m2 = mass of the earth
– r = distance from the center of the earth to its
surface
Newton’s Law of Universal
Gravitation
• If we introduce another constant:
– g = G(m2/r2)
• The equation then becomes:
– F = mg
– W = mg
– W = the force of the earth’s gravity acting on the
object, or the weight of the object
– m = mass of the object
– g = acceleration of the object caused by the earth’s
gravitational force
Summary
• Newton’s first law explains that objects do not
move or do not change their motion unless a net
external force acts on them
– Collisions and conservation of momentum
• Newton’s second law explains that if a net
external force does act on an object, it will
accelerate in the direction of the net external
force and its acceleration will be inversely
related to its mass
– Impulse-momentum relationship—Increasing the
duration of force application increases velocity
Summary
• Newton’s third law explains that forces act
in pairs—For every force there is another
equal force acting in the opposite
direction—Explains how forces act; it does
not explain what the effects of the forces
will be (remember separate objects)