Download Monday, October 25, 2004

PHYS 1443 – Section 003
Lecture #17
Monday, Oct. 25, 2004
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
6.
Impulse and Momentum Change
Collisions
Two Dimensional Collision s
Center of Mass
CM and the Center of Gravity
Fundamentals on Rotational Motion
2nd Term Exam Monday, Nov. 1!! Covers CH 6 – 10.5!!
Monday, Oct. 25, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
1
Impulse and Linear Momentum
ur
Net force causes change of momentum 
Newton’s second law
By integrating the above
equation in a time interval ti to
tf, one can obtain impulse I.
So what do you
think an impulse is?

tf
ti
ur d p
F
dt
ur ur
d p  Fdt
ur ur ur
ur
r
t f ur
d p  p f  pi  D p  Fdt  I

ti
Impulse of the force F acting on a particle over the time
interval Dt=tf-ti is equal to the change of the momentum of
the particle caused by that force. Impulse is the degree of
which an external force changes momentum.
The above statement is called the impulse-momentum theorem and is equivalent to Newton’s second law.
What are the
dimension and
unit of Impulse?
What is the
direction of an
impulse vector?
Defining a time-averaged force
Monday, Oct. 25, 2004
ur 1
ur
F   F i Dt
Dt i
Impulse can be rewritten
If force is constant
I  FDt
I  F Dt
It is generally assumed that the impulse force acts on a
PHYS 1443-003, Fall 2004
short time
much
Dr. but
Jaehoon
Yu greater than any other forces present.
2
Another Example for Impulse
In a crash test, an automobile of mass 1500kg collides with a wall. The initial and
final velocities of the automobile are vi= -15.0i m/s and vf=2.60i m/s. If the collision
lasts for 0.150 seconds, what would be the impulse caused by the collision and the
average force exerted on the automobile?
Let’s assume that the force involved in the collision is a lot larger than any other
forces in the system during the collision. From the problem, the initial and final
momentum of the automobile before and after the collision is
p i  mvi  1500  15.0i  22500i kg  m / s
p f  mv f  1500  2.60i  3900i kg  m / s
Therefore the impulse on the
I  D p  p  pi  3900  22500 i kg  m / s
automobile due to the collision is
 26400i kg  m / s  2.64 104 i kg  m / s
f
The average force exerted on the
automobile during the collision is
Monday, Oct. 25, 2004
4
D p 2.64 10
F  Dt 
0.150
 1.76 105 i kg  m / s 2  1.76 105 i N
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
3
Collisions
Generalized collisions must cover not only the physical contact but also the collisions
without physical contact such as that of electromagnetic ones in a microscopic scale.
Consider a case of a collision
between a proton on a helium ion.
F
F12
t
F21
Using Newton’s
3rd
The collisions of these ions never involve no
physical contact because the electromagnetic
repulsive force between these two become great
as they get closer causing a collision.
Assuming no external forces, the force
exerted on particle 1 by particle 2, F21,
changes the momentum of particle 1 by
ur ur
d p1  F 21dt
Likewise for particle 2 by particle 1
ur ur
d p 2  F 12 dt
law we obtain
d p2
ur
 F 12 dt
So the momentum change of the system in the
collision is 0 and the momentum is conserved
Monday, Oct. 25, 2004
ur
  F 21dt  d p1
d p  d p1  d p 2  0
p system  p1  p 2
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
 constant
4
Elastic and Inelastic Collisions
Momentum is conserved in any collisions as long as external forces are negligible.
Collisions are classified as elastic or inelastic based on the conservation of
kinetic energy before and after the collisions.
Elastic
Collision
A collision in which the total kinetic energy and momentum
are the same before and after the collision.
Inelastic
Collision
A collision in which the total kinetic energy is not the same
before and after the collision, but momentum is.
Two types of inelastic collisions:Perfectly inelastic and inelastic
Perfectly Inelastic: Two objects stick together after the collision,
moving together at a certain velocity.
Inelastic: Colliding objects do not stick together after the collision but
some kinetic energy is lost.
Note: Momentum is constant in all collisions but kinetic energy is only in elastic collisions.
Monday, Oct. 25, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
5
Elastic and Perfectly Inelastic Collisions
In perfectly Inelastic collisions, the objects stick
together after the collision, moving together.
Momentum is conserved in this collision, so the
final velocity of the stuck system is
How about elastic collisions?
m1 v1i  m2 v 2i  (m1  m2 )v f
vf 
m1 v1i  m2 v 2i
(m1  m2 )
m1 v1i  m2 v 2i  m1 v1 f  m2 v 2 f
1
1
1
1
In elastic collisions, both the
m1v12i  m2 v22i  m1v12f  m2 v22 f
2
2
2
2
momentum and the kinetic energy
m1 v12i  v12f   m2 v22i  v22 f 
are conserved. Therefore, the
final speeds in an elastic collision
m1 v1i  v1 f v1i  v1 f   m2 v2i  v2 f v2i  v2 f 
can be obtained in terms of initial From momentum
m1 v1i  v1 f   m2 v2i  v2 f 
speeds as
conservation above
 m  m2 
 2m2 
v1i  
v2i
v1 f   1
 m1  m2 
 m1  m2 
Monday, Oct. 25, 2004
 2m1 
 m  m2 
v1i   1
v2i
v2 f  
 m1  m2 
 m1  m2 
PHYS 1443-003, Fall 2004
What happens when
the
two masses are the same?
Dr. Jaehoon Yu
6
Example for Collisions
A car of mass 1800kg stopped at a traffic light is rear-ended by a 900kg car, and the
two become entangled. If the lighter car was moving at 20.0m/s before the collision
what is the velocity of the entangled cars after the collision?
The momenta before and after the collision are
Before collision
p i  m1 v1i  m2 v 2i  0  m2 v 2i
m2
20.0m/s
m1
p f  m1 v1 f  m2 v 2 f  m1  m2 v f
After collision
Since momentum of the system must be conserved
m2
vf
m1
vf
What can we learn from these equations
on the direction and magnitude of the
velocity before and after the collision?
Monday, Oct. 25, 2004
m1  m2 v f
pi  p f

 m2 v 2i
m2 v 2i
900  20.0i

 6.67i m / s
m1  m2  900  1800
The cars are moving in the same direction as the lighter
car’s original direction to conserve momentum.
The magnitude is inversely proportional to its own mass.
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
7
Two dimensional Collisions
In two dimension, one can use components of momentum to apply
momentum conservation to solve physical problems.
m1
m1 v 1i  m2 v 2 i  m1 v1 f  m2 v 2 f
v1i
m2
q
f
x-comp.
m1v1ix  m2 v2ix  m1v1 fx  m2 v2 fx
y-comp.
m1v1iy  m2 v2iy  m1v1 fy  m2 v2 fy
Consider a system of two particle collisions and scattersin
two dimension as shown in the picture. (This is the case at
fixed target accelerator experiments.) The momentum
conservation tells us:
m1 v 1i  m2 v 2i  m1 v1i
m1v1ix  m1v1 fx  m2 v2 fx  m1v1 f cos q  m2v2 f cos f
m1v1iy  0  m1v1 fy  m2v2 fy  m1v1 f sin q  m2v2 f sin f
And for the elastic collisions, the
kinetic energy is conserved:
Monday, Oct. 25, 2004
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
What do you think
we can learn from
these relationships?
8
Example for Two Dimensional Collisions
Proton #1 with a speed 3.50x105 m/s collides elastically with proton #2 initially at
rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and
proton #2 deflects at an angle f to the same axis. Find the final speeds of the two
protons and the scattering angle of proton #2, f.
m1
v1i
m2
q
Since both the particles are protons m1=m2=mp.
Using momentum conservation, one obtains
x-comp. m p v1i  m p v1 f cos q  m p v2 f cos f
y-comp.
f
m p v1 f sin q  m p v2 f sin f  0
Canceling mp and put in all known quantities, one obtains
v1 f cos 37  v2 f cos f  3.50 105 (1)
From kinetic energy
conservation:
3.50 10 
5 2
 v v
2
1f
2
2f
Monday, Oct. 25, 2004
v1 f sin 37  v2 f sin f
(2)
v1 f  2.80 105 m / s
Solving Eqs. 1-3
5
(3) equations, one gets v2 f  2.1110 m / s
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
f  53.0
Do this at
home
9
Center of Mass
We’ve been solving physical problems treating objects as sizeless
points with masses, but in realistic situation objects have shapes
with masses distributed throughout the body.
Center of mass of a system is the average position of the system’s mass and
represents the motion of the system as if all the mass is on the point.
What does above
statement tell you
concerning forces being
exerted on the system?
m2
m1
x1
x2
xCM
Monday, Oct. 25, 2004
The total external force exerted on the system of
total mass M causes the center of mass to move at
an acceleration given by a   F / M as if all
the mass of the system is concentrated on the
center of mass.
Consider a massless rod with two balls attached at either end.
The position of the center of mass of this system is
the mass averaged position of the system
m x  m2 x2
CM is closer to the
xCM  1 1
heavier object
m1  m2
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
10
Motion of a Diver and the Center of Mass
Diver performs a simple dive.
The motion of the center of mass
follows a parabola since it is a
projectile motion.
Diver performs a complicated dive.
The motion of the center of mass
still follows the same parabola since
it still is a projectile motion.
Monday, Oct. 25, 2004
The motion of the center of mass
of the diver is always the same.
PHYS 1443-003, Fall 2004
11
Dr. Jaehoon Yu
Example 9-12
Thee people of roughly equivalent mass M on a lightweight (air-filled)
banana boat sit along the x axis at positions x1=1.0m, x2=5.0m, and
x3=6.0m. Find the position of CM.
Using the formula
for CM
m x

m
i
xCM
i
i
i
i
M 1.0 M  5.0 M  6.0  12.0 M
 4.0(m)

3M
M M M
Monday, Oct. 25, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
12
Center of Mass of a Rigid Object
The formula for CM can be expanded to Rigid Object or a
system of many particles
xCM
m x  m x      mn xn
 11 2 2
m1  m2      mn
m x

m
i i
m y

m
i
i
yCM
i
Dmi
ri
rCM
zCM
i
i
r CM  xCM i  yCM j  zCM k
r CM 
i
i

m x
i
i
i   mi yi j   mi zi k
i
i
 mi r i
m
i
i
i
i
M
A rigid body – an object with shape
and size with mass spread throughout
the body, ordinary objects – can be
considered as a group of particles with
mass mi densely spread throughout
the given shape of the object
Monday, Oct. 25, 2004
i i
i
i
i
The position vector of the
center of mass of a many
particle system is
m z

m
i
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
xCM 
 Dm x
i
i
i
M
xCM  lim
Dm 0
i
r CM 
 Dm x
i
i
M
i

1
rdm

M
1
M
 xdm
13
Example for Center of Mass in 2-D
A system consists of three particles as shown in the figure. Find the
position of the center of mass of this system.
Using the formula for CM for each
position vector component
y=2 m (0,2)
1
m x

m
i
(0.75,4)
xCM
rCM
One obtains r CM  x
CM
i i
1 1
2 2
1
i
yCM
2
3 3
3

m2  2m3
m1  m2  m3
i
i
i
i
i
m x m x m x m x
xCM  m  m  m  m

i
i
m y

m
i
i
(2,0)
m3
x=2
(1,0)
m2
x=1
i
r
r m2  2m3  i  2m1 j
i  yCM j 
m1  m2  m3
If m1  2kg; m2  m3  1kg
i
m y
yCM  m

i
i
i
i
i

m1 y1  m2 y2  m3 y3
2m1

m1  m2  m3
m1  m2  m3
Monday, Oct. 25, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
r CM 
3i  4 j
 0.75i  j
4
14
Example of Center of Mass; Rigid Body
Show that the center of mass of a rod of mass M and length L lies in midway
between its ends, assuming the rod has a uniform mass per unit length.
The formula for CM of a continuous object is
L
xCM 
x
dx
Therefore xCM
dm=ldx
1

M
1
M

xL
x 0
xdm
Since the density of the rod (l) is constant; l  M / L
The mass of a small segment dm  ldx
xL
1 1 2 1 1  L
1 1 2 

x0 lxdx  M  2 lx  M  2 lL   M  2 ML   2
x 0
xL
Find the CM when the density of the rod non-uniform but varies linearly as a function of x, la x
M

xL
x 0
ldx  
xL
x 0
xL
axdx
1
1

 aL2
  ax 2 
2
 x 0 2
Monday, Oct. 25, 2004
xCM
1

M

xL
x 0
1
lxdx 
M
xL
1
a
x
dx

x 0
M
2
xL
1 3 
 3 ax 

 x 0
1 1 3
1 2
 2L
 aL  
 ML  
M 3
3
 M 3

PHYS 1443-003, Fall 2004
xCM 
Dr. Jaehoon Yu
15
Center of Mass and Center of Gravity
The center of mass of any symmetric object lies on an
axis of symmetry and on any plane of symmetry, if
object’s mass is evenly distributed throughout the body.
How do you think you can
determine the CM of
objects that are not
symmetric?
Center of Gravity
Dmi
Axis of
One can use gravity to locate CM.
symmetry
1. Hang the object by one point and draw a vertical line
following a plum-bob.
2. Hang the object by another point and do the same.
3. The point where the two lines meet is the CM.
Since a rigid object can be considered as collection
of small masses, one can see the total gravitational
force exerted on the object as
F g   F i   Dmi g  M g
i
Dmig
CM
What does this
equation tell you?
Monday, Oct. 25, 2004
i
The net effect of these small gravitational
forces is equivalent to a single force acting on
a point (Center of Gravity) with mass M.
PHYS 1443-003, Fall 2004
16
The CoG is the point in an object
as
if
all
the
gravitational
force
is
acting
on!
Dr. Jaehoon Yu
Motion of a Group of Particles
We’ve learned that the CM of a system can represent the motion of a system.
Therefore, for an isolated system of many particles in which the total mass
M is preserved, the velocity, total momentum, acceleration of the system are
Velocity of the system
Total Momentum
of the system
Acceleration of
the system
External force exerting
on the system
If net external force is 0
Monday, Oct. 25, 2004
v CM
d r CM  d  1

dt  M
dt
 1
m
r
i
 i  
M
 mi
d r i  mi v i

dt
M
mv

 m v   p
p CM  M vCM  M
M
i i
a CM
d v CM  d  1

dt  M
dt
 1
m
v
i
 i  
M
 F ext  M aCM   mi a i  d dtptot
 F ext  0 
d ptot
dt
i i
p tot  const
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
i
 ptot
d v i  mi a i
 mi dt  M
What about the
internal forces?
System’s momentum
is conserved.
17