Download PHYS 1443 – Section 501 Lecture #1

PHYS 1443 – Section 002
Lecture #16
Wednesday, Oct. 31, 2007
Dr. Jae Yu
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Two Dimensional Collisions
Center of Mass
Fundamentals of Rotational Motions
Rotational Kinematics
Relationship between angular and linear quantities
Wednesday, Oct. 31, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
1
Announcements
• Make sure you come to the colloquium today
– 4pm in SH101
– Refreshment at 3:30pm in SH108
• A midterm grade discussion next Monday, Nov. 5
– Be sure not to miss this
Wednesday, Oct. 31, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
2
Wednesday, Oct. 31, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
3
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
q
f
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 cosq  m2 v2 f cos f
m1v1iy  0  m1v1 fy  m2 v2 fy  m1v1 f sin q  m2v2 f sin f
And for the elastic collisions, the
kinetic energy is conserved:
Wednesday, Oct. 31, 2007
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
What do you think
we can learn from
these relationships?
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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 putting 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
Wednesday, Oct. 31, 2007
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 1443-002, Fall 2007
Dr. Jaehoon Yu
f  53.0
Do this at
home
5
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
Wednesday, Oct. 31, 2007
The total external force exerted on the system of
total mass M causes the center
of umass
to move at
r
r
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
m1 x1  m2 x2
CM is closer to the
xCM 
heavier object
m1  m2
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
6
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.
Wednesday, Oct. 31, 2007
The motion of the center of mass
of the diver is always the same.
PHYS 1443-002, Fall 2007
7
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
Wednesday, Oct. 31, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
8
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
i
(2,0)
m3
x=2
(1,0)
m2
x=1
i
r
One obtains r CM  x
CM
m x m x m x m x
xCM  m  m  m  m

i i
1 1
i
2 2
1
i
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
Wednesday, Oct. 31, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
m y

m
i
i
yCM
i
i
i
i
r
r
r
r  m2  2m3  i  2m1 j
i  yCM j 
m1  m2  m3
If m1  2kg; m2  m3  1kg
r r
r
r r
3i  4 j
r CM 
 0.75i  j
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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
mi
ri
rCM
zCM
i
i

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
Wednesday, Oct. 31, 2007
i
i
r
r
r
r
r CM  xCM i  yCM j  zCM k
r
r CM 
i i
i
i
i
i
The position vector of the
center of mass of a many
particle system is
m z

m
i
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
xCM 
 m x
i
i
i
M
xCM  lim
m 0
i
 m x
i
i
M
i

r
1 r
r CM 
rdm

M
1
M
 xdm
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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
axdx
xL
1
1

 aL2
  ax 2 
2
 x 0 2
Wednesday, Oct. 31, 2007
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 1443-002, Fall 2007
xCM 
1
M
Dr. Jaehoon Yu
xL
1 3 
 3 ax 
x 0
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Center of Mass and Center of Gravity
The center of mass of any symmetric object lies on the
axis of symmetry and on any plane of symmetry, if the
object’s mass is evenly distributed throughout the body.
How do you think you can
determine the CM of the
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 a collection
of small masses, one can see the total gravitational
force exerted on the object as
ur
ur
ur
ur
F g   Fi   mi g  M g
i
mig
CM
What does this
equation tell you?
Wednesday, Oct. 31, 2007
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 1443-002, Fall 2007
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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
The external force
exerting on the system
If net external force is 0
Wednesday, Oct. 31, 2007
r
r
r
r
 1
d r i  mi vi
m
r
i 

m
i

vCM
 M  i dt
M
r
ur
r
ur ur
r
mi vi

  mi v i   p  p tot
p CM  M vCM  M
M
r
r
r
r
r
 1
d vCM  d  1
dvi  mi a i
m
v
i
aCM  dt dt  M  i   M  mi dt 
M
r
d r CM  d  1


dt
M
dt
ur
r dp
ur
r
 F ext  M aCM   mi ai  dttot
ur
ur
d ptot

 F ext  0 dt
ur
ptot  const
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
What about the
internal forces?
System’s momentum
is conserved.
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