PHYS 1443 – Section 001
Lecture #13
Tuesday, June 28, 2011
Dr. Jaehoon Yu
•
•
•
•
•
Collisions
Center of Mass
Rotational Motion
Rotational Kinematics
Relationship Between Angular and
Linear quantities
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
1
Announcements
• Reading Assignment
– CH9.10
• Quiz #3 tomorrow, Wednesday, June 29
– Beginning of the class
– Covers CH8.1 through CH9.9
• Planetarium Show extra credit
– Must obtain the signature of the “Star Instructor” AFTER
watching the show on the ticket stub
– Tape one side of the ticket stubs on a sheet of paper with
your name on it
– Submit it on the last class Thursday, July 7
• Late submissions will not be accepted!!!
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
2
Valid Planetarium Shows
• Regular shows
– TX star gazing; Nanocam; Ice Worlds
• Private shows for a group of 15 or more
–
–
–
–
Bad Astronomy; Black Holes; IBEX; Magnificent Sun
Microcosm; Stars of the Pharaohs; Time Space
Two Small Pieces of Glass; SOFIA
Violent Universe; Wonders of the Universe
• Please watch the show and obtain the signature on
the back of the ticket stub
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
3
Extra-Credit Special Project
• Derive the formula for the final velocity of two objects which
underwent an elastic collision as a function of known
quantities m1, m2, v01 and v02 in page 8 of this lecture note in
a far greater detail than in the note.
– 20 points extra credit
• Show mathematically what happens to the final velocities if
m1=m2 and explain in detail in words the resulting motion.
– 5 point extra credit
• NO Credit will be given if the process is too close to the note!
• Due: Start of the class Tuesday, July 5
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
4
Extra Credit: Two Dimensional Collisions
•Proton #1 with a speed 5.0x106 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  to the same axis. Find the final
speeds of the two protons and the scattering angle of proton
#2,  . This must be done in much more detail than the book
or on page 13 of this lecture note.
•10 points
•Due beginning of the class Wednesday, July 6
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
5
Collisions
Generalized collisions must cover not only the physical contact but also the collisions
without physical contact such as that of electromagnetic ones on a microscopic scale.
Consider a case of a collision
between a proton on a helium ion.
F
F12
t
F21
The collisions of these ions never involve
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
r r
dp1  F21dt
Likewise for particle 2 by particle 1
r r
dp2  F12dt
ur
ur
r
r
Using Newton’s law we obtain
d p 2  F12dt   F21dt  d p1
ur
ur
ur
d p  d p1  d p2  0
So the momentum change of the system in a
ur
ur ur
collision is 0, and the momentum is conserved
p system  p1  p2  constant
3rd
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
6
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 whether the total kinetic
energy is conserved, meaning whether it is the same before and after the collision.
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 momentum is the same before and
after the collision but not the total kinetic energy .
Two types of inelastic collisions:Perfectly inelastic and inelastic
Perfectly Inelastic: Two objects stick together after the collision,
moving together with the same 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.
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
7
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?
r
r
r
m1 v1i  m2 v 2i  (m1  m2 )v f
r
r
r
m1 v1i  m2 v 2i
vf 
(m1  m2 )
r
r
r
r
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 
 2m1 
 m1  m2 
v1 f   1
v

v
v

v

v2i
1i
2i
2f
1i







m

m
m

m
m

m
m

m
 1
 1
 1
 1
2
2
2
2
What
Tuesday, June 28, 2011
happensPHYS
when1443-001,
the two
masses are the same?
Summer 2011 Dr.
Jaehoon Yu
8
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
pi  m1v1i  m2 v2i  0  m2 v2i
m2
20.0m/s
m1

After collision
Since momentum of the system must be conserved
m2
vf
m  m v
pi  p f
m1
vf

1
m2 v2i
m
1
What can we learn from these equations
on the direction and magnitude of the
velocity before and after the collision?
Tuesday, June 28, 2011

p f  m1v1 f  m2 v2 f  m1  m2 v f
 m2


2
f
 m2 v2i
900  20.0
 6.67m / s
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 1443-001, Summer 2011 Dr.
Jaehoon Yu
9
Ex.9 – 11: A Ballistic Pendulum
The mass of a block of wood is 2.50-kg and the
mass of the bullet is 0.0100-kg. The block swings
to a maximum height of 0.650 m above the initial
position. Find the initial speed of the bullet.
What kind of collision? Perfectly inelastic collision
No net external force  momentum conserved
m1v f 1  m2v f 2  m1v01  m2 v02
m1  m2 v f  m1v01

Solve for V01

v01 
 m1  m2  v f
m1
What do we not know? The final speed!!
How can we get it? Using the mechanical
energy conservation!
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
10
Ex. A Ballistic Pendulum, cnt’d
Now using the mechanical energy conservation
1
2

mv 2  mgh

m1  m2 gh f 
gh f 
1
2


1
2
v 2f
Solve for Vf
m1  m2 v 2f


v f  2gh  2 9.80 m s2  0.650 m 
f
Using the solution obtained previously, we obtain
v01
m  m v
m m 




1
2
f
m1
1
2
m1
2gh f
 0.0100 kg  2.50 kg 
2

2
9.80m
s
0.650 m

0.0100
kg





 896m s
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
11
Two dimensional Collisions
In two dimension, one needs to use components of momentum and
apply momentum conservation to solve physical problems.
m1
r
r
r
r
m1 v1i  m2 v 2i  m1 v1 f  m2 v 2 f
v1i
m2


x-comp.
m1v1ix  m2v2ix  m1v1 fx  m2v2 fx
y-comp.
m1v1iy  m2v2iy  m1v1 fy  m2v2 fy
Consider a system of two particle collisions and scatters in
two dimension as shown in the picture. (This is the case at
fixed target accelerator experiments.) The momentum
conservation tells us:
r
r
r
m1 v1i  m2 v 2i  m1 v1i
m1v1ix  m1v1 fx  m2 v2 fx  m1v1 f cos   m2 v2 f cos 
m1v1iy  0  m1v1 fy  m2 v2 fy  m1v1 f sin  m2 v2 f sin 
And for the elastic collisions, the
kinetic energy is conserved:
Tuesday, June 28, 2011
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
What do you think
we can learn from
these relationships?
12
Ex. 9 – 13: 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  to the same axis. Find the final speeds of the two
protons and the scattering angle of proton #2,  .
m1
v1i
m2

Since both the particles are protons m1=m2=mp.
Using momentum conservation, one obtains
x-comp. m p v1i  m p v1 f cos   m p v2 f cos 
y-comp.

m p v1 f sin   m p v2 f sin   0
Canceling mp and putting in all known quantities, one obtains
v1 f cos 37  v2 f cos   3.50 105 (1)
v1 f sin 37  v2 f sin 
From kinetic energy
conservation:
3.50  10   v
5 2
2
1f
(2)
v1 f  2.80 105 m / s
v
2
2f
Tuesday, June 28, 2011
Solving Eqs. 1-3
5
(3) equations, one gets v2 f  2.1110 m / s
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
  53.0
Do this at
home
13
Center of Mass
We’ve been solving physical problems treating objects as sizeless
points with masses, but in realistic situations 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 the
forces being exerted on the
system?
m2
m1
x1
x2
xCM
Tuesday, June 28, 2011
The total external force exerted on the system of
total mass M causes the center
ofumass
to move at
r
r
an acceleration given by a   F / Mas 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
xCM 
m1 x1  m2 x2
m1  m2
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
CM is closer to the
heavier object
14
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.
Tuesday, June 28, 2011
The motion of the center of mass
of the diver is always the same.
PHYS 1443-001, Summer 2011 Dr.
15
Jaehoon Yu
Example 9 – 14
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
Tuesday, June 28, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
16
Ex9 – 15: 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
(0.75,1)
xCM
rCM
i
r
i i
1 1
2 2
1
i
2
3 3
3

m2  2m3
m1  m2  m3
i
yCM 
 mi yi
i
m
i
i

m1 y1  m2 y2  m3 y3
2m1

m1  m2  m3
m1  m2  m3
Tuesday, June 28, 2011
m y

m
i
yCM
i
i
i
i
r
r
r
r  m2  2m3  i  2m1 j
i  yCM j 
m1  m2  m3
One obtains r CM  x
CM
m x m x m x m x
xCM  m  m  m  m

i
i
i
(2,0)
m3
x=2
(1,0)
m2
x=1
i
If m1  2kg; m2  m3  1kg
r r
r
r r
3i  4 j
r CM 
 0.75i  j
4
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
17
Center of Mass of a Rigid Object
The formula for CM can be extended to a system of many particles
or a Rigid Object
xCM 
m1x1  m2 x2    mn xn
m1  m2    mn
m x

m
m y

m
i
i i
i
yCM
Δmi
ri
rCM
zCM
i
i
i
r
r
r
r
r CM  xCM i  yCM j  zCM k

r
r
r
m
x
i

m
y
j

m
z
k
 i i  i i  ii
i
i
r
 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
Tuesday, June 28, 2011
i
i
i
r
r CM 
i i
i
i
The position vector of the
center of mass of a many
particle system is
m z

m
i
xCM 
 m x
i
M
xCM  lim
m 0
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
i
i
i
 m x
i
i
M
i

r
1 r
r CM 
rdm

M
1
M
 x dm
18
Ex 9 – 16: CM of a thin rod
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=λdx
1

M
1
M

xL
x 0
xdm
Since the density of the rod (λ) is constant;   M / L
The mass of a small segment dm  dx
xL
1 1 2 1 1  L
1 1 2 

x0 xdx  M  2 x  M  2 L   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, λ x
M

xL
x 0
dx  
xL
x 0
xL
xdx
1
1

 L2
  x 2 
2
 x 0 2
Tuesday, June 28, 2011
xCM
1

M
xCM
xL
1 xL 2
1 1 3 

xdx


x
dx

x 
x0

M x 0
M 3
 x 0
1 1 3
1 2
 2L

 L  
 ML  
M 3
3
 M 3

xL
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
19