PHYS 1443 – Section 003
Lecture #16
Monday, Oct. 27, 2002
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
6.
Center of Mass
Motion of a group of particles
Rotational Motion
Rotational Kinematics
Rotational Energy
Moment of Inertia
Remember the 2nd term exam (ch 6 – 11), Monday, Nov. 3!
Monday, Oct. 27, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
1
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. 27, 2003
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
2
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. 27, 2003
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
3
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. 27, 2003
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
4
Example for Center of Mass
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. 27, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
r CM 
3i  4 j
 0.75i  j
4
5
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. 27, 2003
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
6
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. 27, 2003
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.
7
Fundamentals on Rotation
Linear motions can be described as the motion of the center of
mass with all the mass of the object concentrated on it.
Is this still true for
rotational motions?
P
O
r q s
No, because different parts of the object have
different linear velocities and accelerations.
Consider a motion of a rigid body – an object that
does not change its shape – rotating about the axis
protruding out of the slide.
The arc length, or sergita, is s  rq
Therefore the angle, q, is q  s . And the unit of
r
the angle is in radian.
One radian is the angle swept by an arc length equal to the radius of the arc.
Since the circumference of a circle is 2pr,
360  2pr / r  2p
The relationship between radian and degrees is 1 rad  360  / 2p  180 / p
Monday, Oct. 27, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
8
Angular Displacement, Velocity, and Acceleration
Using what we have learned in the previous slide, how
q
 qi

q

f
would you define the angular displacement?
How about the average angular speed?
And the instantaneous angular speed?

q f  qi
t f  ti
  lim
t 0
a
And the instantaneous angular
acceleration?
a  lim
t f  ti
t 0
q
t
q dq

t
dt
 f  i
By the same token, the average angular
acceleration


qf
qi

t
 d

dt
t
When rotating about a fixed axis, every particle on a rigid object rotates through
the same angle and has the same angular speed and angular acceleration.
Monday, Oct. 27, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
9
Rotational Kinematics
The first type of motion we have learned in linear kinematics was
under a constant acceleration. We will learn about the rotational
motion under constant angular acceleration, because these are the
simplest motions in both cases.
Just like the case in linear motion, one can obtain
Angular Speed under constant
angular acceleration:
Angular displacement under
constant angular acceleration:
One can also obtain
Monday, Oct. 27, 2003
 f i  at
1 2
q f  q i  i t  a t
2
    2a q f  qi 
2
f
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
i
10
Example for Rotational Kinematics
A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If
the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what
angle does the wheel rotate in 2.00s?
Using the angular displacement
formula in the previous slide, one gets
What is the angular speed at t=2.00s?
Using the angular speed and
acceleration relationship
Find the angle through which the wheel
rotates between t=2.00 s and t=3.00 s.
Monday, Oct. 27, 2003
q f qi
 t 
1 2
at
2
1
2
 2.00  2.00  3.50  2.00  11.0rad
2
11.0

rev.  1.75rev.
2p
 f   i  at
 2.00  3.50  2.00  9.00rad / s
q 2  11.0rad
q 3  2.00  3.00  3.50  3.002  21.8rad
q  q   q 2  10.8rad 
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
1
2
10.8
rev.  1.72rev.
2p
11
Relationship Between Angular and Linear Quantities
What do we know about a rigid object that rotates
about a fixed axis of rotation?
Every particle (or masslet) in the object moves in a
circle centered at the axis of rotation.
y
When a point rotates, it has both the linear and angular motion
components in its motion.
The
direction
What is the linear component of the motion you see?
P
ri
Linear velocity along the tangential direction.
q
x
of 
follows a
right-hand
rule.
How do we related this linear component of the motion
with angular component?
ds
d
dq
 rq   r
The arc-length is s  rq So the tangential speed vt is vt 
O
dt
dt
dt
 r
What does this relationship tell you about Although every particle in the object has the same
the tangential speed of the points in the angular speed, its tangential speed differs
object and their angular speed?:
proportional to its distance from the axis of rotation.
Monday, Oct. 27, 2003
The farther
away the particle is from the center of
PHYS 1443-003,
Fall 2002
Dr. rotation,
Jaehoon Yu
the higher the tangential speed.
12
How about the Accelerations?
y
How many different linear accelerations do you see
in a circular motion and what are they? Two
at
a
r
P
Tangential, at, and the radial acceleration, ar.
ar
q
Since the tangential speed vt is vt  r
x
The magnitude of tangential a  dvt  d r   r d  ra
t
acceleration at is
dt
dt
dt
What does this
relationship tell you?
Although every particle in the object has the same angular
acceleration, its tangential acceleration differs proportional to its
distance from the axis of rotation.
O
The radial or centripetal acceleration ar is
r 
v2


r
r
r
2
a
 r 2
What does The father away the particle from the rotation axis the more radial
this tell you? acceleration it receives. In other words, it receives more centripetal force.
Total linear acceleration is
Monday, Oct. 27, 2003
a
 
at2  ar2  ra 2  r 2  r a 2   4
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
13
Example for Rotational Motion
Audio information on compact discs are transmitted digitally through the readout system consisting of laser
and lenses. The digital information on the disc are stored by the pits and flat areas on the track. Since
the speed of readout system is constant, it reads out the same number of pits and flats in the same time
interval. In other words, the linear speed is the same no matter which track is played. a) Assuming the
linear speed is 1.3 m/s, find the angular speed of the disc in revolutions per minute when the inner most
(r=23mm) and outer most tracks (r=58mm) are read.
Using the relationship
between angular and
tangential speed v  r
r  23mm

v 1.3m / s
1.3


 56.5rad / s
r 23mm 23 10 3
 9.00rev / s  5.4 10 2 rev / min
r  58mm  
1.3m / s
1.3

 22.4rad / s
58mm 58 10 3

b) The maximum playing time of a standard music
CD is 74 minutes and 33 seconds. How many
revolutions does the disk make during that time?
     540  210rev / min

qf
i
f
2
2
Monday, Oct. 27, 2003
a



PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
 375rev / min
375
 q i   t  0  60 rev / s  4473s  2.8 104 rev
l  vt t
c) What is the total length of the track past through the readout mechanism?
d) What is the angular acceleration of the CD over
the 4473s time interval, assuming constant a?
 2.110 2 rev / min
f
 i 
t

 1.3m / s  4473s
 5.8  103 m
22.4  56.5rad / s
4473s
 7.6 10 3 rad / s 2
14