# College Physics I

Mechanics

Nov 14, 2013 (4 years and 7 months ago)

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Physics
203

College Physics I
Department of Physics

Physics 203

College Physics I

Fall 2012

S. A. Yost

Chapter 8 Part 1

Rotational Motion

Physics
203

College Physics I
Department of Physics

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

College Physics I
Department of Physics

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,
. That comes up in chapter 8.

Physics
203

College Physics I
Department of Physics

Motion of Extended Objects

The motion of extended objects or collections of
particles is such that the CM obeys Newton’s 2
nd

Law.

Physics
203

College Physics I
Department of Physics

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
203

College Physics I
Department of Physics

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.

Physics
203

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Department of Physics

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.

Physics
203

College Physics I
Department of Physics

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

= 360
o
, 180
o

=
p

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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…

Physics
203

College Physics I
Department of Physics

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
203

College Physics I
Department of Physics

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
203

College Physics I
Department of Physics

Example

(a)
a = w

/
t. t
= 300 s
.

w = 2p

f =

2p
(20,000 rev / 60 s) = 2094
/s

a

= (2094
/s) / 300 s = 6.98
/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
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

Moment of Inertia

For example, the
moment of inertia of a
R

and
mass
M

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

Physics
203

College Physics I
Department of Physics

Solid
vs

Hollow Sphere

If a solid and hollow sphere
have the
same mass and
size
, which has a bigger
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

Physics
203

College Physics I
Department of Physics

Parallel Axis Theorem

any axis can be found if
its
value
I
CM

is known
an axis through the
CM.

I
h

= I
CM

+ Mh
2

The moment of inertia is
an axis through the CM.

CM

M

h

Physics
203

College Physics I
Department of Physics

Moment of Inertia of Rod

What is the moment
of inertia of uniform rod of mass
M

and length
L

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

Physics
203

College Physics I
Department of Physics

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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203

College Physics I
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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
.

_

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
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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.

Physics
203

College Physics I
Department of Physics

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
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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
203

College Physics I
Department of Physics

Example

Isolate the pulley:
t
= I
a
with
I =
½

m
R
2
,
t
=
RF
T

,
a

=
a/R
.

Then

RF
T

=

mR
2
)(
a/R
).

Therefore,
F
T

=
½

ma
.

m

R

F
T

a

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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.

Physics
203

College Physics I
Department of Physics

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

→ →

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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

Physics
203

College Physics I
Department of Physics

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