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Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Center of Mass and Linear Momentum
Chapter 9
© 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
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.
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-1 Center of Mass
DEFINITION:
center of mass (com) of a system of particles:
© 2014 John Wiley & Sons, Inc. All rights reserved.
For two particles separated by a distance d, where the origin is chosen at the position of particle 1:
9-1 Center of Mass
© 2014 John Wiley & Sons, Inc. All rights reserved.
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:
9-1 Center of Mass
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-1 Center of Mass
For many particles, we can generalize the equation, where M = m
1 + m
2 + . . . + m
n:
© 2014 John Wiley & Sons, Inc. All rights reserved.
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:
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-1 Center of Mass
If objects have uniform density = ρ (“rho”)
Substituting, we find the center of mass simplifies:
© 2014 John Wiley & Sons, Inc. All rights reserved.
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)
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-1 Center of Mass
Example Subtractingo Task: find COM of a disk with another
disk taken out of it:
o Find the COM of the two individual COMs (one for each disk), treating the cutout as having negative mass
Figure 9-4
© 2014 John Wiley & Sons, Inc. All rights reserved.
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-1 Center of Mass
Example Subtractingo On the diagram,
comC is the center
of mass for Plate P and Disk S combined
o comP is the center
of mass for the composite plate with Disk S removed
Figure 9-4
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
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 = 0o Baseball bat: a = g, so com follows gravitational trajectory
o Or does it…
Figure 9-5
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-2 Newton's Second Law for a System of Particles
© 2014 John Wiley & Sons, Inc. All rights reserved.
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.)
Momentum and Newton’s second law
The momentum of a particle is the product of its mass and its velocity:
m .p = v
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:
m .p = v
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
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:
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-3 Linear Momentum
We can write Newton's second law thus:
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-3 Linear Momentum
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-3 Linear Momentum
Answer: (a) 1, 3, 2 & 4 (b) region 3
Eq. (9-25)
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-3 Linear Momentum
We can sum momenta for a system of particles to find:
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
9-5 Conservation of Linear Momentum
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 = (SF) x (Dt)
so…
Dp = (SF) x (Dt)
SF = Dp/(Dt)
Note!!
J , p, F
are all
VECTORS!)
© 2014 John Wiley & Sons, Inc. All rights reserved.
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:
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-4 Collision and Impulse
Given Favg
and duration:
We are integrating: we only need to know the area under the force curve
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-4 Collision and Impulse
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-4 Collision and Impulse
Answer: (a) unchanged (b) unchanged (c) decreased
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-4 Collision and Impulse
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-4 Collision and Impulse
Answer: (a) zero (b) positive (c) along the positive y-axis (normal force)
© 2014 John Wiley & Sons, Inc. All rights reserved.
For an impulse of zero we find:
Which is another way to say momentum is conserved!
Law of conservation of linear momentum
9-5 Conservation of Linear Momentum
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-5 Conservation of Linear Momentum
Check components of net external force to determine if you should apply conservation of momentum
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
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…
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 m2v2
2
• And m1v1 = p1; m2v2 = p2
• ½ m1v12 = ½ (m1v1
) v1 = ½ p1 v1 & same for ½ p2 v2
• ½ p1 v1 = ½ p2 v2
• 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?
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-6 Types of Collisions
Elastic collisions:o Total kinetic energy is unchanged (conserved)o A useful approximation for common situationso In real collisions, some energy is always transferred
Inelastic collisions: some energy is transferred
Completely inelastic collisions:o The objects stick togethero Greatest loss of kinetic energy
© 2014 John Wiley & Sons, Inc. All rights reserved.
For one dimension inelastic collision
9-6 Momentum and Kinetic Energy in Collisions
© 2014 John Wiley & Sons, Inc. All rights reserved.
Completely inelastic collision, for target at rest:
9-6 Momentum and Kinetic Energy in Collisions
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-6 Momentum and Kinetic Energy in Collisions
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-6 Momentum and Kinetic Energy in Collisions
Answer: (a) 10 kg m/s (b) 14 kg m/s (c) 6 kg m/s
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-7 Elastic Collisions in One Dimension
Total kinetic energy is conserved in elastic collisions
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-7 Elastic Collisions in One Dimension
For a stationary target, conservation laws give:
Figure 9-18
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-7 Elastic Collisions in One Dimension
With some algebra we get:
Results
o Equal masses: v1f = 0, v
2f = v
1i: the first object stops
o Massive target, m2 >> m
1: the first object just bounces
back, speed mostly unchanged
o Massive projectile: v1f ≈ v
1i, v
2f ≈ 2v
1i: the first object
keeps going, the target flies forward at about twice its speed
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-7 Elastic Collisions in One Dimension
For a target that is also moving, we get:
© 2014 John Wiley & Sons, Inc. All rights reserved.
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:
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
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-8 Collisions in Two Dimensions
© 2014 John Wiley & Sons, Inc. All rights reserved.
9-8 Collisions in Two Dimensions
Answer: (a) 2 kg m/s (b) 3 kg m/s
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 mbvbi
2 = ½ mavaf
2 + 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
mv = initial x-momentum of rocket
vexh = exhaust velocity from rocket motor
(This will be constant!)
(v - vexh) = relative velocity of exhaust gases
m = initial mass of rocket
v = initial velocity of rocket
in x direction
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 momentumfinal 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) = – (dm)vexh
dv = – [(dm)/m] vexh
integrate both sides
Rocket propulsion
• Gain in speed?
m(dv) = – (dm)vexh
dv = – [(dm)/m] vexh (integrate both sides)
vf - vi = (vexh) ln(m0/m)
m0 = initial mass
m0 > m, so ln > 1
“mass ratio”
Velocity of gas exhaustedgain in velocity
Rocket propulsion – Gain in Speed
vf - vi = (vexh) ln(m0/m)
m0 = initial mass
m0 > m, so ln > 1
“mass ratio”
Velocity of gas exhaustedgain in velocity
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
m0 > m, so ln > 1
“mass ratio”
Velocity of gas exhaustedgain in velocity
STAGE rockets to throw away mass as they use up fuel, so that m0/m is even higher!