PHYS 1443 – Section 001
Lecture #12
Wednesday, June 21, 2006
Dr. Jaehoon Yu
•
•
Center of Mass
CM and the Center of Gravity
Wednesday, June 21, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
1
Announcements
• Reading assignments: CH. 9 – 10.
• Quiz tomorrow
– Early in the class
– Covers Ch. 8.5 – Ch. 9
• Mid-term grade discussions today
• Exam results
– Average: 70/102
– Top score: 100/102
• Grade proportions
–
–
–
–
–
Exam constitutes 22.5% each, totaling 45%
Homework: 25%
Lab: 20%
Quizzes: 10%
Extra credit: 10%
Wednesday, June 21, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2
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
Wednesday, June 21, 2006
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
m1 x1  m2 x2
CM is closer to the
xCM 
heavier object
m1  m2
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
3
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, June 21, 2006
The motion of the center of mass
of the diver is always the same.
PHYS 1443-001, Summer 2006
4
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, June 21, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
5
Center of Mass of a Rigid Object
The formula for CM can be expanded to Rigid Object or a
system of many particles
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
rCM  xCM i  yCM j  zCM k
rCM 
i
i

m x
i
i
i  mi yi j   mi zi k
i
i
m
i
i
m r
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, June 21, 2006
i i
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 2006
Dr. Jaehoon Yu
i
i
i
rCM 
 m x
i
i
M
i

1
rdm

M
1
M
 xdm
6
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
One obtains rCM  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
i
(2,0)
m3
x=2
(1,0)
m2
x=1
i
2
3 3
3

m2  2m3
m1  m2  m3
m y

m
i
yCM
i
i
i
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
m y  m2 y2  m3 y3
2m1
 1 1

m1  m2  m3
m1  m2  m3
Wednesday, June 21, 2006
rCM
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
3i  4 j

 0.75i  j
4
7
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, June 21, 2006
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 1443-001, Summer 2006
xCM 
1
M
Dr. Jaehoon Yu
xL
1 3 
 3 ax 

 x 0
8