Transcript PHYS 1443 – Section 501 Lecture #1

PHYS 1443 – Section 003
Lecture #11
Monday, Oct. 21, 2002
Dr. Jaehoon Yu
1.
2.
Collisions in Two Dimension
Center of Mass
•
•
•
3.
4.
5.
Definition
CM of a Rigid Object
Center of Mass and Center of Gravity
Motion of a Group of Particles
Rocket Propulsion
Fundamentals on Rotation
Today’s homework is homework #12, due 12:00pm, next Monday!!
Monday, Oct. 21, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
1
Announcements
• 2nd Term exam
–
–
–
–
Wednesday, Oct. 30, in the class
Covers chapters 6 – 10
Mixture of Multiple choice and Essay problems
Review on Monday, Oct. 28
• Magda Cortez, please come and talk to me after the class
Monday, Oct. 21, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
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 scatters in
two dimension as shown in the picture. (This is the case at
fixed target accelerator experiments.) The momentum
conservation tells us:
m1 v 1 f  m2 v 2 f  m1 v1i
m1v1ix  m1v1 fx  m2 v2 fx  m1v1 f cos q  m2 v2 f cos f
m1v1iy  0  m1v1 fy  m2 v2 fy  m1v1 f sin q  m2 v2 f sin f
And for the elastic conservation,
the kinetic energy is conserved:
Monday, Oct. 21, 2002
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
What do you think
we can learn from
these relationships?
3
Example 9.9
A 1500kg car traveling east with a speed of 25.0 m/s collides at an interaction with a 2500kg
van traveling north at a speed of 20.0 m/s. After the collision the two cars stuck to each
other, and the wreckage is moving together. Determine the velocity of the wreckage after the
collision, assuming the vehicles underwent a perfectly inelastic collision.
The initial momentum of the two car system before the collision is
p i  m1v1i i  m2v2i j  1500  25.0i  2500  20.0 j
 3.75 104 i  5.0 104 j
The final momentum of the two car system after the perfectly
inelastic collision is


p f  m1  m2  v fx i  v fy j  4.0  103 v fx i  4.0  103 v fy j
Using
X-comp. p fx  pix
momentum
conservation
p f  pi
Y-comp. p fy  piy
Monday, Oct. 21, 2002
m1  m2 v fx  m1v1x  0
m1  m2 v fy  0  m2v2 y


v fx
m v  0 
 1 1x
v fy
0  m v  

v f  v fx i  v fy j  9.38 i  12.5 j m / s
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
m1  m2
2 2y
m1  m2
3.75 104
 9.38m / s
1500  2500
5.0 104
 12.5m / s
1500  2500
4
Example 9.10
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
f
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.
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   v
5 2
2
1f
v
2
2f
Monday, Oct. 21, 2002
v1 f sin 37  v2 f sin f
Solving Eqs. 1-3
(3) equations, one gets
(2)
v1 f  2.80 105 m / s
v2 f  2.11105 m / s
f  53.0
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
5
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. 21, 2002
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
m1  m2
heavier object
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
6
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
i
i
yCM
i
mi
ri
rCM
i
zCM
i
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. 21, 2002
i i
i
r CM  xCM i  yCM j  zCM k
r CM 
m z

m
i
i
i
The position vector of the
center of mass of a many
particle system is
m y

m
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
xCM 
 m x
i
i
i
M
xCM  lim
m 0
i
r CM 
 m x
i
i
M
i

1
rdm

M
1
M
 xdm
7
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
mi
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   mi g  M g
i
mig
CM
What does this
equation tell you?
Monday, Oct. 21, 2002
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 2002
Dr. Jaehoon Yu
8
Example 9.12
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
2 2
1
i
2
3 3
3

m2  2m3
m1  m2  m3
i
i
i
i
One obtains r CM  x i  y j 
CM
CM
i i
1 1
yCM
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
m2  2m3  i  2m1 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. 21, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
r CM 
3i  4 j
 0.75i  j
4
9
Example 9.13
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. 21, 2002
xCM
1

M

xL
x 0
1
lxdx 
M
xL
1
a
x
dx

x 0
M
2
1 2
1 3
 2L
 aL  
 ML  
3
3
 M 3

PHYS 1443-003, Fall 2002
xCM 
1
M
Dr. Jaehoon Yu
xL
1 3 
 3 ax 

 x 0
10
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. 21, 2002
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
d ptot
F

0

 ext
dt
i i
p tot  const
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
i
 ptot
d v i  mi a i
 mi dt  M
What about the
internal forces?
System’s momentum
is conserved.
11