Transcript Center of Mass and Linear Momentum

Chapter 9
Center of Mass and Linear
Momentum
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
9-1 Center of Mass

Motion of rotating objects is complicated

There is a special point for which motion is simple


Center of mass
of bat traces out a
parabola, just as a
tossed ball does
All other points rotate
around this point
Figure 9-1
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Goals for Chapter 9
To learn the meaning of the momentum of a particle and how an
impulse causes it to change
To learn how to use the principle of conservation of momentum
To learn how to solve problems
involving collisions
Goals for Chapter 8
To learn the definition of the center of
mass of a system and what determines
how it moves
To analyze situations, such as rocket
propulsion, in which the mass of a
moving body changes
Introduction
In many situations, such as a bullet hitting a carrot, we
cannot use Newton’s second law to solve problems
because we know very little about the complicated
forces involved.
In this chapter, we shall introduce momentum and
impulse, and the conservation of momentum, to solve
such problems.
9-1 Center of Mass

DEFINITION:
center of mass (com) of a system of particles:
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9-1 Center of Mass
For two particles separated by a distance d, where the
origin is chosen at the position of particle 1:
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9-1 Center of Mass


For two particles separated by a distance d, where the
origin is chosen at the position of particle 1:
For two particles, for an arbitrary choice of origin:
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9-1 Center of Mass

For many particles, we can generalize the equation,
where M = m1 + m2 + . . . + mn:
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9-1 Center of Mass

In three dimensions, we find the center of mass along
each axis separately:
More concisely, we can write in
terms of vectors:
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9-1 Center of Mass

For solid bodies, we take the limit of an infinite sum of
infinitely small particles → integration!

Coordinate-by-coordinate, we write:

Here M is the mass of the object
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9-1 Center of Mass

If objects have uniform density = ρ (“rho”)

Substituting, we find the center of mass simplifies:
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9-1 Center of Mass

The center of mass lies at a point of symmetry

It lies on the line or plane of symmetry)

It need not be on the object (consider a doughnut)
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9-1 Center of Mass
Answer: (a) at the origin (b) in Q4, along y=-x (c) along the -y axis
(d) at the origin (e) in Q3, along y=x (f) at the origin
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© 2014 John Wiley & Sons, Inc. All rights reserved.
9-1 Center of Mass
Example Subtracting
o
o
Task: find COM of a disk with another
disk taken out of it:
Find the COM of the two individual
COMs (one for each disk), treating the
cutout as having negative mass
Figure 9-4
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© 2014 John Wiley & Sons, Inc. All rights reserved.
9-1 Center of Mass
Example Subtracting
o
o
On the diagram,
comC is the center
of mass for Plate
P and Disk S
combined
comP is the center
of mass for the
composite plate
with Disk S
removed
Figure 9-4
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9-2 Newton's Second Law for a System of Particles

Center of mass motion continues unaffected by forces
internal to a system (collisions between billiard balls)

Motion of a system's center of mass:

Reminders:
1. Fnet is the sum of all external forces
2. M is the total, constant, mass of the closed system
3. acom is the center of mass acceleration
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6-2 Newton's Second Law for a System of Particles
Examples Using the center of mass motion equation:
o
Billiard collision: forces are only internal, F = 0 so a = 0
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6-2 Newton's Second Law for a System of Particles
Examples Using the center of mass motion equation:
o
Billiard collision: forces are only internal, F = 0 so a = 0
o
Baseball bat: a = g, so com follows gravitational trajectory
o
Or does it…
Figure 9-5
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6-2 Newton's Second Law for a System of Particles
Examples Using the center of mass motion equation:
o
Exploding rocket: explosion forces are internal, so only the
gravitational force acts on the system, and the COM follows
a gravitational trajectory
Figure 9-5
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9-2 Newton's Second Law for a System of Particles
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9-2 Newton's Second Law for a System of Particles
Answer: The system consists of Fred, Ethel and the pole. All forces are
internal. Therefore the com will remain in the same place.
Since the origin is the com, they will meet at the origin in all three cases!
(Of course the origin where the com is located is closer to Fred than to
Ethel.)
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Momentum and Newton’s second law
The momentum of a
particle is the
product of its mass
and its velocity:
What is the momentum of a 1000kg car going
25 m/s west?
Momentum and Newton’s second law
The momentum of a
particle is the
product of its mass
and its velocity:
What is the momentum of a 1000kg car going
25 m/s west?
p = mv = (1000 kg)(25 m/s) = 25,000 kgm/s west
9-3 Linear Momentum


Momentum:
o
Points in the same direction as the velocity
o
Can only be changed by a net external force
We can write Newton's second law thus:
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9-3 Linear Momentum

We can write Newton's second law thus:
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9-3 Linear Momentum
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9-3 Linear Momentum
Answer: (a) 1, 3, 2 & 4 (b) region 3
Eq. (9-25)
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9-3 Linear Momentum

We can sum momenta for a system of particles to find:
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9-3 Linear Momentum



Taking time derivative write Newton's second law for
system of particles as:
Net external force on system changes linear momentum
Without a net external force, the total linear momentum
of a system of particles cannot change
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9-5 Conservation of Linear Momentum


Without a net external force, the total linear
momentum of a system of particles cannot change
This is called the law of conservation of linear
momentum
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Impulse and momentum
Impulse of a force is product
of force & time interval
during which it acts.
Impulse is a vector!
On a graph of Fx versus
time, impulse equals area
under curve.
Impulse and momentum
Impulse-momentum theorem:
Impulse = Change in momentum J of particle during time
interval equals net force acting on particle during interval
J = Dp = pfinal – pinitial
J = Net Force x time = (F) x (Dt)
so…
Dp = (F) x (Dt)
Note!!
J , p, F
are all
VECTORS!)
F = Dp/(Dt)
9-4 Collision and Impulse


If F isn’t constant over time….
This means that the applied impulse is equal to the
change in momentum of the object during the collision:
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9-4 Collision and Impulse


Given Favg and duration:
We are integrating: we only
need to know the area under
the force curve
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9-4 Collision and Impulse
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9-4 Collision and Impulse
Answer: (a) unchanged (b) unchanged (c) decreased
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9-4 Collision and Impulse
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9-4 Collision and Impulse
Answer: (a) zero (b) positive (c) along the positive y-axis (normal force)
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9-5 Conservation of Linear Momentum

For an impulse of zero we find:

Which is another way to say momentum is conserved!

Law of conservation of linear momentum
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9-5 Conservation of Linear Momentum

Check components of net external force to determine if
you should apply conservation of momentum
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9-5 Conservation of Linear Momentum

Internal forces can change momenta of parts of the
system, but cannot change the linear momentum of
the entire system
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9-5 Conservation of Linear Momentum

Internal forces can change momenta of parts of the
system, but cannot change the linear momentum of
the entire system
Answer: (a) zero (b) no (c) the negative x direction
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9-5 Conservation of Linear Momentum

Do not confuse momentum and energy

Change in KE => SPEED changes (+ or – directions)

Change in P => direction may have changed, or
velocity may have changed, or
BOTH may have changed…
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Compare momentum and kinetic energy
Changes in momentum
depend on time over
which net force acts
But…
Changes in kinetic
energy depend on the
distance over which net
force acts.
Ice boats again
• Two iceboats race on a frictionless lake; one with mass m
& one with mass 2m.
• Wind exerts same force on both.
• Both boats start from rest, both travel same distance to
finish line.
• Questions!
• Which crosses finish line with more KE?
• Which crosses with more Momentum?
Ice boats again
• Two iceboats race on a frictionless lake; one with mass m and one
with mass 2m.
• The wind exerts the same force on both.
• Both boats start from rest, and both travel the same distance to the
finish line?
• Which crosses the finish line with more KE?
In terms of work done by wind?
W = DKE = Force x distance = same;
So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22
Ice boats again
• Two iceboats race on a frictionless lake; one with mass m and one
with mass 2m.
• The wind exerts the same force on both.
• Both boats start from rest, and both travel the same distance to the
finish line?
• Which crosses the finish line with more KE?
In terms of work done by wind?
W = DKE = Force x distance = same;
So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22
Since m2 > m1, v2 < v1 so more massive boat loses
Ice boats again
• Two iceboats race on a frictionless lake; one with mass m and one
with mass 2m.
• The wind exerts the same force on both.
• Both boats start from rest, and both travel the same distance to the
finish line?
• Which crosses the finish line with more p?
In terms of momentum?
Force on the boats is the same for each, but TIME that
force acts is different. The second boat accelerates slower,
and takes a longer time.
Since t2> t1, p2 > p1 so second boat has more momentum
Ice boats again
• Check with equations!
• Force of wind = same; distance = same
• Work done on boats is the same, gain in KE same.
• ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22
• And m1v1 = p1; m2v2 = p2
• ½ m1v12 = ½ (m1v1) v1 = ½ p1 v1 & same for ½ p2 v2
• ½ p1 v 1 = ½ p 2 v 2
• Since V1 > V2, P2 must be greater than P1!
p2 > p1
A ball hits a wall
A 0.40 kg ball moves at 30 m/s to the left, then
rebounds at 20 m/s to the right from a wall.
A ball hits a wall
• A 0.40 kg ball moves at 30 m/s to the left, then
rebounds at 20 m/s to the right from a wall. What is
the impulse of net force during the collision, and if it is
in contact for 0.01 s, what is the average force acting
from the wall on the ball?
Remember that momentum is a vector!
When applying
conservation of
momentum, remember
that momentum is a
vector quantity!
Remember that momentum is a vector!
When applying
conservation of
momentum, remember
that momentum is a
vector quantity!
Use vector addition to add
momenta in
COMPONENTS!
Kicking a soccer ball – Example 8.3
Soccer ball 0.40 kg moving left at 20 m/s, then kicked up & to
the right at 30 m/s at 45 degrees.
If collision time is 0.01 seconds, what is impulse?
9-6 Types of Collisions

Elastic collisions:
o
Total kinetic energy is unchanged (conserved)
o
A useful approximation for common situations
o
In real collisions, some energy is always transferred

Inelastic collisions: some energy is transferred

Completely inelastic collisions:
o
The objects stick together
o
Greatest loss of kinetic energy
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9-6 Momentum and Kinetic Energy in Collisions

For one dimension inelastic collision
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9-6 Momentum and Kinetic Energy in Collisions

Completely inelastic collision, for target at rest:
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9-6 Momentum and Kinetic Energy in Collisions
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9-6 Momentum and Kinetic Energy in Collisions
Answer: (a) 10 kg m/s (b) 14 kg m/s
(c) 6 kg m/s
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9-7 Elastic Collisions in One Dimension

Total kinetic energy is conserved in elastic collisions
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9-7 Elastic Collisions in One Dimension

For a stationary target, conservation laws give:
Figure 9-18
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9-7 Elastic Collisions in One Dimension

With some algebra we get:

Results
o
o
o
Equal masses: v1f = 0, v2f = v1i: the first object stops
Massive target, m2 >> m1: the first object just bounces
back, speed mostly unchanged
Massive projectile: v1f ≈ v1i, v2f ≈ 2v1i: the first object keeps
going, the target flies forward at about twice its speed
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9-7 Elastic Collisions in One Dimension

For a target that is also moving, we get:
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9-8 Collisions in Two Dimensions


Apply conservation of
momentum along each axis
Apply conservation of
energy for elastic collisions
Example For a stationary target:
o
Along x:
o
Along y:
o
Energy:
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A two-dimensional collision
Two robots collide
and go off at
different angles.
You must break
momenta into x & y
components and
deal with each
direction
separately
9-8 Collisions in Two Dimensions
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9-8 Collisions in Two Dimensions
Answer: (a) 2 kg m/s (b) 3 kg m/s
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Elastic collisions
•In elastic collision, total
momentum of system in a
direction is same after collision
as before… if no external forces
act in that direction:
•Pix = Pfx
•mavai +mbvbi =mavaf +mbvbf
But wait – there’s more!
Objects colliding along a straight line
Two gliders collide on an frictionless track.
What are changes in velocity and momenta?
Elastic collisions
•In an elastic collision, the total
kinetic energy of the system is
the same after the collision as
before.
•KEi = KEf
•½ mavai2 + ½ mbvbi2 =
½ mavaf2 + ½ mbvbf2 =
Elastic collisions
•In an elastic collision,
•Difference in velocities initially =
(-) difference in velocities finally
(vai - vbi) = - (vaf – vbf)
Elastic collisions
•In an elastic collision,
•Difference in velocities initially =
(-) difference in velocities finally
(vai - vbi) = - (vaf – vbf)
•Solve generally for final velocities:
vaf = (ma-mb)/(ma+mb)vai + 2mb/(ma+mb)vbi
vbf = (mb-ma)/(ma+mb)vbi + 2ma/(ma+mb)vai
Elastic collisions
Elastic collisions
Behavior of colliding objects is greatly affected by
relative masses.
Elastic collisions
Elastic collisions
Behavior of colliding objects is greatly affected by
relative masses.
Elastic collisions
Elastic collisions
Behavior of colliding objects is greatly affected by
relative masses.
Stationary Bowling Ball!
Inelastic collisions
• In any collision where
external forces can be
neglected, total momentum
conserved.
• Collision when bodies stick
together is completely
inelastic collision
• In inelastic collision, total
kinetic energy after collision
is less than before collision.
Some inelastic collisions
Cars are intended to have
inelastic collisions so the
car absorbs as much
energy as possible.
The ballistic pendulum
Ballistic pendulums are
used to measure bullet
speeds
A 2-dimensional automobile collision
Two cars traveling at right angles collide.
An elastic straight-line collision
Neutron collisions in a nuclear reactor
A two-dimensional elastic collision
Rocket propulsion
• Conservation of momentum holds for rockets, too!
F = dp/dt = d(mv)/dt = m(dv/dt) + v(dm/dt)
• As a rocket burns fuel, its mass decreases (dm< 0!)
Rocket propulsion
• Initial Values
m = initial mass of rocket
v = initial velocity of rocket
in x direction
mv
= initial x-momentum of rocket
vexh
= exhaust velocity from rocket motor
(This will be constant!)
(v - vexh) = relative velocity of exhaust gases
Rocket propulsion
• Changing values
dm
= decrease in mass of rocket from fuel
dv = increase in velocity of rocket in x-direction
dt = time interval over which dm and dt change
Rocket propulsion
• Final values
m + dm
= final mass of rocket less fuel ejected
v + dv
= increase in velocity of rocket
(m + dm) (v + dv) = final x-momentum of rocket (+x direction)
[- dm] (v - vexh) = final x-momentum of fuel (+x direction)
Rocket propulsion
• Final values
mv
= [(m + dm) (v + dv)] + [- dm] (v - vexh)
initial momentum
final momentum of
the lighter rocket
final momentum of
the ejected mass of
gas
Rocket propulsion
• Final values
mv
= [(m + dm) (v + dv)] + [- dm] (v - vexh)
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
Rocket propulsion
• Final values
mv
= [(m + dm) (v + dv)] + [- dm] (v - vexh)
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
Rocket propulsion
• Final values
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
Rocket propulsion
• Final values
0 = mdv + dmdv + dmvexh
Rocket propulsion
• Final values
0 = mdv + dmdv + dmvexh
Neglect the assumed small term:
m(dv) = -dmdv – (dm)vexh
Rocket propulsion
• Final values
m(dv)
= – (dm)vexh
Gain in momentum of original rocket is related to rate
of mass loss and exhaust velocity of gas!
Rocket propulsion
• Final values
m(dv)
= – (dm)vexh
Gain in momentum of original rocket is related to rate
of mass loss and exhaust velocity of gas!
Differentiate both sides with respect to time!
Rocket propulsion
• Final values
m(dv)
= – (dm)vexh
m(dv/dt) = – [(dm)/dt] vexh
Rocket propulsion
• Final values
m(dv)
= – (dm)vexh
m(dv/dt) = – [(dm)/dt] vexh
ma
= Force (Thrust!)
= – [(dm)/dt] (vexh)
Rocket propulsion
• Final values
m(dv)
= – (dm)vexh
m(dv/dt) = – [(dm)/dt] vexh
ma
= Force (Thrust!)
= – [(dm)/dt] (vexh)
rate of
change of
mass
Velocity of
gas
exhausted
Rocket propulsion
Thrust
= – [(dm)/dt] (vexh)
Example:
vexhaust = 1600 m/s
Mass loss rate = 50 grams/second
Thrust =?
Rocket propulsion
• Final values
Thrust
= – [(dm)/dt] (vexh)
Example
vexhaust = 1600 m/s
Mass loss rate = 50 grams/second
Thrust = -1600 m/s (-0.05 kg/1 sec) = +80N
Rocket propulsion
• Gain in speed?
m(dv)
dv
= – (dm)vexh
= – [(dm)/m] vexh
integrate both sides
Rocket propulsion
• Gain in speed?
m(dv)
dv
vf - vi
= – (dm)vexh
= – [(dm)/m] vexh
(integrate both sides)
= (vexh) ln(m0/m)
m0 = initial mass
gain in velocity
m0 > m, so ln > 1
Velocity of gas
exhausted
“mass ratio”
Rocket propulsion – Gain in Speed
vf - vi
= (vexh) ln(m0/m)
m0 = initial mass
gain in velocity
Velocity of gas
exhausted
m0 > m, so ln > 1
“mass ratio”
Faster exhaust, and greater difference in mass, means a
greater increase in speed!
Rocket propulsion
• Gain in speed?
vf - vi
= (vexh) ln(m0/m)
Example:
Rocket ejects gas at relative 2000 m/s. What fraction of
initial mass is not fuel if final speed is 3000 m/s?
Rocket propulsion
• Gain in speed?
vf - vi
= (vexh) ln(m0/m)
Example:
Rocket ejects gas at relative 2000 m/s. What fraction of
initial mass is not fuel if final speed is 3000 m/s?
ln(mo/m) = (3000/2000) = 1.5 so m/mo = e-1.5 = .223
Rocket propulsion
• Gain in speed?
vf - vi
= (vexh) ln(m0/m)
m0 = initial mass
gain in velocity
Velocity of gas
exhausted
m0 > m, so ln > 1
“mass ratio”
STAGE rockets to throw away mass as they use up fuel, so
that m0/m is even higher!