Physics
203
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College Physics I
Department of Physics
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The Citadel
Physics 203
College Physics I
Fall 2012
S. A. Yost
Chapter 8 Part 1
Rotational Motion
Physics
203
–
College Physics I
Department of Physics
–
The Citadel
Announcements
A problem set HW07B was due today.
Today: Rotational Motion, Ch. 10.
sec. 1
–
6
Next: sec. 7
–
8, skipping sec. 9.
Homework set HW08 on sections 1
–
6 will be due
next Thursday.
Sections 7
–
8 will be combined with chapter 9, sec.
1, 2, and 4 in a later set, HW09.
Physics
203
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Motion of the Center of Mass
If an external force
F
acts on an extended object or
collection of objects of mass
M
, the acceleration of the
CM is given by
F
=
M
a
cm
.
You can apply Newton’s 2
nd
Law as if it were a particle
located at the CM, as far as the collective motion is
concerned.
This
says nothing about the relative motion, rotation,
etc., about the CM
. That comes up in chapter 8.
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Motion of Extended Objects
The motion of extended objects or collections of
particles is such that the CM obeys Newton’s 2
nd
Law.
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203
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College Physics I
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Motion of the Center of Mass
The CM of a wrench sliding on a frictionless table will move in
a straight line because there is no external force. In this
sense, the wrench may be though of as a particle located
at the CM.
cm motion
Physics
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Motion of the Center of Mass
For example, if a hammer is thrown, its CM follows a
parabolic trajectory under the influence of gravity,
as a point object would.
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Motion of the Center of Mass
For example, if a hammer is thrown, its CM follows a
parabolic trajectory under the influence of gravity,
as a point object would.
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Rotational Motion
Everything we have done for linear motion has an
analog for rotational motion.
We will consider only fixed axis rotations.
Rotational displacements can be expressed in various
units: radians are most standard, but degrees and
revolutions are common as well.
Conversions:
1 rev = 2
p
rad
= 360
o
, 180
o
=
p
rad.
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Rotational
Kinematics
The physics of rotations of a rigid body about a fixed
axis is in many ways analogous to the one

dimensional motion of a point particle.
x
v
w
q
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Rotational
Kinematics
Linear Motion:
distance
x
velocity
v
acceleration
a
Rotational Motion:
angle
q
angular velocity
w
angular
acceleration
a
Greek letters are used for rotational quantities…
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Constant
(Angular) Acceleration
Linear Motion:
v = v
0
+ at
x = x
0
+ v
0
t +
½
at
2
v
2
= v
0
2
+
2
a
(
x
–
x
0
)
Rotational Motion:
w
=
w
0
+
a
t
q
=
q
0
+
w
0
t
+ ½
a
t
2
w
2
=
w
0
2
+
2
a
(
q
–
q
0
)
Physics
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Example
A centrifuge accelerates from rest to
20,000
rpm
in
5.0 min
.
(a)
What
is its average angular acceleration
?
(b)
How many times does it spin during these
five minutes?
Physics
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Example
(a)
a = w
/
t. t
= 300 s
.
w = 2p
f =
2p
(20,000 rev / 60 s) = 2094
rad
/s
a
= (2094
rad
/s) / 300 s = 6.98
rad
/s
2
(b)
Revolutions =
f
avg
t
= ½
f
max
t
= 10,000 rpm
×
5 min
= 50,000 rev.
Or, use
q
=
½
a
t
2
to get the same result with more effort.
0
t
5 min
f
20,000 rpm
10,000 rpm
Physics
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Rotational Energy
Consider a massless
rotating disk with small
masses
m
i
inserted in it.
The total kinetic energy is
K
=
S
½
m
i
v
i
2
.
v =
r
w
for an object a
distance
r
from the axis.
We’d like to express
K
in
terms of
w
.
w
r
4
m
2
m
3
m
4
m
5
m
1
r
1
r
2
r
3
r
5
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Rotational
Kinetic Energy
In terms of the angular velocity,
K
=
S
½
m
i
v
i
2
= ½
S
m
i
r
i
2
w
2
= ½
w
2
S
m
i
r
i
2
K =
½
I
w
2
I
is called the
moment of inertia
.
I
=
S
m
i
r
i
2
Units:
kg m
2
.
w
r
4
m
2
m
3
m
4
m
5
m
1
r
1
r
2
r
3
r
5
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Moment of Inertia
For example, the
moment of inertia of a
hoop of radius
R
and
mass
M
about an axis
through the center is
I
= MR
2
since all parts are the
same distance from
the center.
R
M
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Moments
of
Inertia
Thin Hoop Disk
Thin Solid
Rod Sphere
I = MR
2
R
R
L
I =
½
MR
2
I
=ML
2
/12
I =
2/5
MR
2
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Solid
vs
Hollow Sphere
If a solid and hollow sphere
have the
same mass and
size
, which has a bigger
moment of inertia about an
axis through the center
?
A: solid
B: hollow
C: the same
The
hollow
one, since the mass
is distributed further from the
axis.
Solid
Hollow
2/5
MR
2
2/3
MR
2
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Parallel Axis Theorem
The moment of inertia about
any axis can be found if
its
value
I
CM
is known
about
an axis through the
CM.
I
h
= I
CM
+ Mh
2
The moment of inertia is
always the smallest about
an axis through the CM.
CM
M
h
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Moment of Inertia of Rod
What is the moment
of inertia of uniform rod of mass
M
and length
L
about the
end?
I
end
=
I
CM
+
M
(
L
/2)
2
I
CM
=
ML
2
/12
I
end
=
ML
2
/12 +
ML
2
/4
=
ML
2
/3
CM
h = L
/2
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Rotational Work
An object can be given
kinetic energy by doing
work:
D
K = W
To do work, I can apply a
force
F
at a point given,
relative to the axis, by a
vector
R
.
The force does work through
a distance
d =
R
q
.
F
R
q
d
Only the component
of
F
perp. to
R
does
work:
W = F
R
q
.
┴
F
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Torque
The product
t
=
F
R
is
called
torque
.
If
f
is the angle between
R
and
F
,
t
=
RF
sin
f
Linear motion:
W = F x
Rotational motion:
W
= t q
.
R
W = F
R
q
.
F
F
f
Power:
W = Pt, P =
tw
.
_
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Torque
Equivalent expression:
If
R
is the component of
R
perpendicular to
F
, then
t
=
F
R
.
R
is the distance of the line
of the force from the axis,
and is called the
lever
arm
of the force.
R
t
= F
R
=
RF
sin
f
F
R
f
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F
Torque
R
q
t
=
R F
sin
q
.
The torque is defined to
be the perpendicular
component of the force
times the distance from
the pivot to where it acts:
t
=
R F
^
where
F
^
=
F
sin
q
.
Then
Counter

clockwise torque
is considered to be
positive, as for angles.
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F
Torque
R
q
t
=
R F
sin
q
The torque
can also be
expressed in terms of the
magnitude of the force and
the distance from the axis
to the line of the force.
The distance
R
^
is called
the
lever arm
of the torque.
t
=
R F
^
t
=
R
^
F
Physics
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Tightening a Nut with a Wrench
Which use of the
wrench is most
effective for
tightening the nut?
Which is least
effective?
Which of A and D is
more effective?
Choose E if they are
the same.
The lever arms are the same.
A B
C D
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Rotational
Dynamics
Linear Motion:
mass
m
force
F = ma
kinetic energy
K
t
=
½
mv
2
work
W
t
=
Fx
Rotational Motion:
moment of inertia
I
torque
t =
I
a
kinetic energy
K
r
=
½
I
w
2
work
W
r
=
tq
power
P
r
=
tw
power
P
t
= Fv
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Example
Mass falling on rope
wrapped around a
massive pulley.
Assume the pulley is a
uniform disk as shown.
What is the acceleration of
the hanging mass?
m
M
R
a
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Example
Isolate the hanging mass:
Newton’s Law:
M a =
F
net
= Mg
–
F
T
where
F
T
is the tension in
the rope.
M
a
m
g
F
T
Physics
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Example
Isolate the pulley:
t
= I
a
with
I =
½
m
R
2
,
t
=
RF
T
,
a
=
a/R
.
Then
RF
T
=
(½
mR
2
)(
a/R
).
Therefore,
F
T
=
½
ma
.
m
R
F
T
a
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Example
Combine results:
M a =
F
net
= Mg
–
F
T
= Mg
–
½
ma.
Then
(
M
+
m
/2)
a
=
Mg.
Result:
a =
m
R
F
T
a
1 +
m
/2
M
g
M
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Example
Note that the tension does
not need to be the same
on two sides of a
massive pulley.
Net
torque =
R
(
F
1
–
F
2
) =
Ia
.
m
R
F
1
F
2
R
a
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Rigid Body Motion
The relation
t
=
I
a
holds for rigid body rotation
in any inertial frame.
This
always
holds in the CM frame of the rigid
body, even if it is accelerating.
The energy of a rigid body can be expressed as
a sum
K
=
K
cm
+
K
rot
with
K
cm
= ½
mv
cm
2
,
K
rot
= ½
Iw
2
.
“Newton’s Law”
F
= m
a
cm
(
ch
. 9),
t
=
I
a.
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Dumbbell
A force
F
is applied for time
t
to a
dumbbell in one of two ways
shown.
Which
gives the greater speed to
the center of mass?
(a)
A
(b)
B
(c)
the same
m
m
m
m
F
F
A
B
D
p
=
F
t
→ →
→
→
→
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Dumbbell
A force
F
is applied for time
t
to a
dumbbell in one of two ways
shown.
Which
gives the greater energy to
the dumbbell?
(a)
A
(b)
B
(c)
the same
m
m
m
m
F
A
B
→
F
→
→
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Dumbbell
The total kinetic energy is
Case A:
K
=
K
trans
+
K
rot
= ½
mv
cm
2
+ ½
Iw
2
Case B:
no rotation:
K
= ½
mv
cm
2
There is more energy in
case A
.
m
m
m
m
A
B
F
→
F
→
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Dumbbell Energy
How much more
energy
does it have if I push at
the top?
The bottom mass is initially
at rest, so the speed of
any point on the
dumbbell is initially
proportional to its
distance from the bottom
mass.
m
m
2
v
cm
v
cm
v
= 0
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Dumbbell Energy
The initial kinetic energy is then
K
total
= ½
m
(
2
v
cm
)
2
= 2
m
v
cm
2
.
This is conserved as the dumbbell
rotates.
K
cm
= ½ (2
m
)
v
cm
2
=
mv
cm
2
K
rot
=
K
cm
= ½
K
total
There is twice as much total
energy.
m
m
2
v
cm
v
cm
v
= 0
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