Lecture Notes 13



θ (t)
  
Δθ = θ 2 − θ1
(


dθ (t)
ω (t) =
dt
€
€
)
€


2

dω (t) d€
θ (t)
α (t) =
=
dt
dt 2
s = θr
  
v =ω ×r
 dθ
v=
r = ωr
dt

a tot
2
€
 2  2
= ar + at
2
= ω 2 r + αr
€
€
2

 
a tot = a t + a r

  
a r = ω × (ω × r )
  
at = α × r
€
€
Rota%onal
Kinema%cs:
(
ONLY
IF
α
=
constant)
€
ω = ω 0 + αt
θ = θ + ω 0 t + 12 αt
I = ∑ m i ri 2
If rigid body = few particles
If rigid body = too-many-to-count particles
Sum→ Integral
€
Parallel‐Axis
Theorem
I = I COM + Mh
Energy
of
rota%onal
mo%on
KE rot = 12 Iω 2
€
€
[KE
trans
2
= 12 mv 2 ]
Example
#3
A
bicycle
wheel
has
a
radius
of
0.33m
and
a
rim
of
mass
1.2
kg.
The
wheel
has
50
spokes,
each
with
a
mass
10g.
What
is
the
moment
of
iner%al
about
axis
of
rota%on?
What
is
moment
of
iner%a
about
COM?
I tot,com = I rim,center + 50I spoke
→
What
is
Ispoke
(parallel‐axis)?
2
Ispoke = Irod .com + Mh 2 = 121 ML2 + M ( 12 L ) = 13 ML2
€
2
= 13 (0.01kg)(0.33m) ≅ 3.6 ×10 −4 kg ⋅ m 2
→
PuUng
together
€
I tot,com = I rim,center + 50I spoke
= M wheel R 2 + 50I spoke
2
= (1.2kg)(0.33m) + 50(3.6 ×10 −4 kg ⋅ m 2 ) = 0.149kg ⋅ m 2
€
Example
#4
What
is
the
kine%c
energy
of
the
earth’s
rota%on
about
its
axis?
Energy
of
rota%onal
mo%on
is
found
from:
KE rot = 12 Iω 2
What
is
earth’s
moment
of
iner%a,
I?
€
Iearth = Isphere = 25 Mr 2
=
2
5
(6 ×10
24
2
kg)(6.4 ×10 6 m ) ≅ 1 ×10 38 kg ⋅ m 2
What
is
earth’s
angular
velocity,
ω?
€
From T = 2 π ω ⇒
ω = 2 π radians day = 2 π
(3600 × 24 )
= 7.3 ×10 −5 rad / s
Now
plug‐’n‐chug:
€


2
38
2
−5
29
KE rot = 12 1 ×10
€ kg ⋅ m ( 7.3 ×10 rad / s ) = 2.6 ×10 J
10‐8:
Rota;onal
dynamics

Ability
of
force
to
rotate
body
F
What
causes
something
to
rotate
What
does
it
depend
on??
€
Depends
on
distance
from
pivot
point
Depends
on
angle
between
force
and
radius
Moment
of
force
about
an
axis
=
torque
‐
“to
twist”
  
τ =r ×F
  
τ = r F sin ϑ
€
τ
depends
on:
r,
F
and
θ
Units
of
torque:
N∙m
(not
Joule)
10‐8:
Rota;onal
Dynamics
Define:
Torque
vector
=
magnitude
+
direc;on
  
τ =r ×F

vector τ is always
  
τ = r F sin ϑ
perpendicular to both


vectors r and F
Units
of
torque:
N∙m
€
€
€
  
If
Ft
is
tangen%al
component
of
force
τ = r F sin ϑ = rFt
If
r⊥
is
moment
arm


= ( r sin ϑ ) F = r⊥ F
(
)
10-9: Rotational dynamics - Newton’s 2nd Law
Force
about
a
point
causes
torque



τ net = ∑ τ i = Iα
  
τ = r×F
€

τ net =
=
∑

τi =



{Fnet = ∑ Fi = ma}
→
∑r
→
∑ (m r ) α
2
i i
→
i
× Fi =
∑
 → 
r ×  m€
i ai =
i 

→
€
∑
→ 

r ×  mi riα 
i 

→
→
= Iα
€
€
• torque
is
posi%ve
when
the
force
tends
to
produce
a
counterclockwise
rota%on
about
an
axis
is
nega%ve
when
the
force
tends
to
produce
a
clockwise
rota%on
about
an
axis
(→
VECTOR!
posi%ve
if
poin%ng
in
+z,
nega%ve
if
poin%ng
in
‐z
)
• net
torque
and
angular
accelera%on
are
parallel
A
uniform
disk,
with
mass
M
and
radius
R,
is
mounted
on
a
Example
#1
fixed
horizontal
axle.
A
block
of
mass
m
hangs
from
a
massless
cord
that
is
wrapped
around
the
rim
of
the
disk.
a)
Find
the
accelera%on
of
the
disk,
and
the
tension
in
the
cord.
b)
Assume
now
that
m
=
M.
Ini%ally
at
rest,
what
is
the
speed
of
the
block
aeer
dropping
a
distance
d?
Method
1:
Using
conserva%on
of
Energy
ΔKE tot = −ΔU
(
1
2
Mv 2 + 12 Iω 2
)
final
[
− (0 + 0) init = − (0) final − (Mgd ) init
2
1
Subs%tu%ng
v = ωr & I = 2 MR
€
2
1
2
2
Mv +
€
1
2
(
1
2
v
MR )  = Mgd
 R
2
2
v 2 + ( 12 )(v) = 2gd
v 2 (1 + 12 ) = 32 v 2 = 2gd
v=
€
€
4
3
gd
]
Method
2
Using
force
+
torque

 
ˆz : τ net = r × F
yˆ : T − Mg = −Ma
⇒ T = M (g − a )
= ( RT )(−zˆ )


τ = Iα
( RT )€(−zˆ) = ( Iα )(−zˆ)
€
a t = αr &
1
2
2
I = MR ⇒
€
a
R[ M ( g − a )] = ( 12 MR 2 ) 
 R
a ( 12 M + M ) = Mg
€
a=
Mg 2
= g
M 3
3
2
Now
use
1‐D
kinema%cs
€
v 2 = v02 + 2ad
v 2 = 0 + 2d ( 23 g)
v=
4
3
gd
Work
and
Rota;onal
Kine;c
Energy
where
τ
is
the
torque
doing
the
work
W,
and
θi
and
θf
are
the
body’s
angular
posi%ons
before
and
aeer
the
work
is
done,
respec%vely.
When
τ
is
constant,
The
rate
at
which
the
work
is
done
is
the
power
Summary:
Newton’s
2nd
law
for
rota;on



{Fnet = ∑ Fi = ma}



τ net = ∑ τ i = Iα
€Summary:
Work
and
Rota%onal
Kine%c
Energy
€
NET
Work
done
ON
system
Rota%onal
work,
fixed
axis
rota%on
€
(if
torque
is
const)
Power,
€
fixed
axis
rota%on
€
W = ΔKE trans



2
2
1
 = 2 m (v f − vi )
W = ΔKE rot
= 12 I (ω 2f − ω i2 )
θ2
W = ∫ τ net dθ
€
θ1
= τ (θ f − θ i )
dW d
P=
= (τθ ) = τω
dt dt
€
€

W =

x2
∫ Fdx
x1

1 − D motion

 

dW d  
= ( F • x ) = F • v
P =


dt dt
€
Example:
Block
and
Pulley
Accelera;ng
In the figure one block has a mass M, the other has mass m , and the pulley
(I), which is mounted in horizontal frictionless bearings, has a radius r.
When released from rest, the heavier block falls a distance d in time t.
(a) What is the magnitude of the block's acceleration?
(b) What is the tension in the part of the cord that supports the heavier block?
(c) What is the tension in the part of the cord that supports the lighter block?
(d) What is the magnitude of the pulley's angular acceleration? Force
about
a
point
causes
torque
τ = F
Torque
causes
rota%on



τ net = ∑ τ i = Iα
Trans
‐
rota%onal
rela%onship
aT = rα
T1
T2
Basic Equations
Mg-T2 = Ma
T1 − mg = ma
Ia
(T2 − T1 ) r = Iα =
r
Ia
+ ( M + m)a = ( M − m) g
2
r
M − m) g
(
a=
I
+ ( M −+
m )
2
r
Check:
I=0
Check:
M=m
Check:
m=0
Example:
Torque
or
Kine;c
Energy
α
‐
into
board
1)
What
are
forces
on
m1?
a
Massless
cord
wrapped
around
a
pulley
of
radius
R
and
mass
MW
(fric%onless
surface/bearings)
and
IW=1/2MWR2.
What
is
angular
accelera%on,
α,
of
pulley
(disc)?
2)
What
are
forces
on
m2?
xˆ : T1 − m1 g sin θ = m1 a
yˆ : N − m1 g cos θ = 0
⇒ T1 = m1 (a + g sin θ )
yˆ : T2 − m 2 g = −m 2 a
⇒ T2 = m 2 (g − a )
NOTE:
T1
&
T2
are
NOT
equal
3)
What
are
torques
about
wheel?
€
τ net
€
 
= τ1 + τ 2
= R(T1 − T2 )(+ zˆ)
NOTE:
angular
accelera%on
vector
is
in
nega%ve‐z
direc%on
4)
Solve
for
α
?
€
here a t = αR ⇒
€
€
€

τ net = RT1 sin (90  ) + RT2 sin (−90  )
IW α = R[(T2 ) − (T1 )]
IW α = ( 12 M W R 2 )α = R[( m 2 ( g − a )) − ( m1 ( a + g sin θ ))]
(
1
2
)
M W R 2 α = Rg ( m2 − m1 sin θ ) − R (α R ) ( m2 + m1 )
α=
2g  m2 − m1 sin θ 


R  M w + 2 ( m1 + m2 ) 
a)  If
Mw
=
0,
same
as
problem
5‐43
b)  If
θ
=90º,
same
as
problem
11‐55
(HW
#
9)
Problem
90
(Ch
10):
A
rigid
body
is
made
of
three
iden%cal
thin
rods,
each
with
a
length
L=0.600
m,
fastened
together
in
the
form
of
a
leoer
H.
The
body
is
free
to
rotate
about
the
axis
shown
through
on
of
the
rods.
The
body
is
allowed
to
fall
rota%ng
around
the
axis.
What
is
the
angular
velocity
when
it
is
ver%cal?
Can't do Torque--use conservation of energy!
2
Iω
L
ΔKω = ΔU g =
= mg + mgL
2
2
Iω 2 4mL2 2
3L
=
ω = mg
2
3
2
9L
ω= g
4L
I1 = 0 why?
2
mL2
mL2
 L
I2 =
+M  =
 2
12
3
I 3 = mL2
4mL2
I=
3
Chapter 11
Rolling, Torque, and Angular Momentum
In this chapter we will cover the following topics:
-Rolling of circular objects and its relationship with friction
-Redefinition of torque as a vector to describe rotational
problems that are more complicated than the rotation of a rigid
body about a fixed axis
-Angular momentum of single particles and systems of particles
-Newton’s second law for rotational motion
-
What
Is
Rolling?
‐‐
Understanding
rolling
with
wheels
‐‐
Wheel
moving
forward
with
constant
speed
vcom
displacement:
transla%on→
rota%on
s = θR
vcom =
€
These
rela%onships
€
define
“smooth
rolling
mo%on”
a com =
ds d (θR)
=
= ωR
dt
dt
dvcom d (ωR)
=
= αR
dt
dt
Only
if
NO
SLIDING
[smooth
rolling]
€
At
point
P
(point
of
contact),
wheel
does
not
move
Understanding
rolling
with
wheels
II
All
points
on
wheel
move
with
same
ω.
All
points
on
outer
rim
move
with
same
linear
speed
v
=
vcom.
All
points
on
wheel
move
Combina%on
of
“pure
to
the
right
with
same
linear
rota%on”
and
“pure
transla%on”
velocity
vcom
as
center
of
wheel
  
v =ω ×r
Note
at
point
P:
at
point
T:
€
vector
sum
of
velocity
=
0 (point
of
sta;onary
contact)
vector
sum
of
velocity
=
2vcom
(top
moves
twice
as
fast
as
com)
Kine;c
Energy
of
Rolling
1
2
2
1
2
I comω +
Mv
2
com
= KE tot
Note:
rota%on
about
COM
and
transla%on
of
COM
combine
for
total
KE
€
Remember:
vcom=ωr
€