Physics 2101
S ti 3
Section
March 5rd : Ch. 10
Announcements:
• Grades for Exam #2
posted next week.
• Today Ch. 10.7
10.7-10,
11.211.2
-3
• Next Quiz is March 12
on SHW#6
Class Website:
http://www.phys.lsu.edu/classes/spring2010/phys2101--3/
http://www.phys.lsu.edu/classes/spring2010/phys2101
http://www.phys.lsu.edu/~jzhang/teaching.html
Summary: Translational and Rotational Variables
Rotational position and distance moved
s = θr
(only radian units)
Rotational and translational speed
ddrr ds
dθ
d
v=
= =
r
dt dt dt
v= ωr
v = ω ×r
Rotational and translational acceleration
dω
at =
r = α r tangential acceleration
at = α × r
dt
v2
2
a r = = ω r radial/centripetal acceleration a r = ω × (ω × r )
r
a tot = a r + a t
2
2
2
= ω r + αr
2
2
2
a tot = a t + a r
Example: Beetle on a merrymerry-gogo-round
A beetle rides the rim of a rotating merry-go-round.
If the angular
ang lar speed of the system
s stem is constant,
constant does the beetle have
ha e
a) radial acceleration and b) tangential acceleration?
a tot = a r + a t
If ω = constant, α = 0
2
2
2
=ω r +0
2
2
2
⇒ a tot = ω 2r(−rˆ)
only radial
If the angular speed is decreasing at a constant rate, does the beetle have a)
radial acceleration and b) tangential acceleration?
If α = negative but constant
constant,
ω = ω0 + α t
a tot = a r + a t
2
2
a tot =
2
a tot = a t + a r
2
= ω r + αr
2
v
2
ar
rˆ
at
a tot
((ω 0 + αt) r) + (αr)
2
2
2
both radial and tangential
A ladybug sits at the outer edge of a merry-go-round, and a
gentleman bug sits halfway between her and the axis of
rotation.
t ti
Th merry-go-round
The
d makes
k
a complete
l t revolution
l ti
once each second. The gentleman bug’s angular speed is
ladybug
1. half the ladybug’s.
y g
2. the same as the ladybug’s.
3 twice
3.
i the
h lladybug’s.
d b ’
gentleman bug
4. impossible to determine
A ladybug sits at the outer edge of a merry-go-round, that is
turning and speeding up. At the instant shown in the figure, the
tangential component of the ladybug’s (Cartesian) acceleration
is:
1. In the +x direction
2. In the - x direction
3. In the +y direction
4. In the - y direction
5. In the +z direction
6. In the - z direction
7. zero
A ladybug sits at the outer edge of a merry-go-round that is
turning
g and is slowing
g down. The vector expressing
p
g her
angular velocity is
1. In the +x direction
2. In the - x direction
3. In the
3
e +y
yd
direction
ec o
4. In the - y direction
5. In the +z direction
6. In the - z direction
7. zero
Sec. 3.7 Multiplying vectors:
1) Dot Product or Scalar Product (Vector ⋅ Vector = scalar)
A
θ
A • B = A B cos θ
B
= Ax Bx + Ay By + Az Bz
If A⊥B then A•B=0
If A||B then A • B=max
2) Cross Product (Vector × Vector = Vector)
iˆ
C = A × B = Ax
Bx
C = A B sin θ
ˆj
Ay
By
kˆ
Az
Bz
If A⊥B then A×B=max
If A||B then A×B=0
Two vectors A and B define a plane. Vector C is perpendicular to
plane according to rightright-hand rule.
Kinetic Energy of Rotation
TRANSLATION Review:
proportionality is inertia, m ⇒ constant of an object
nd
Newton’s 2 Law ⇒ Fnet ∝ a
energy associated with state of
“motion” of particles
2
KE
∝
v
trans
translational
l i l motion
i
with
i h same v
→ mass is translational inertia ←
What about rotation?
motion?
What is energy associated with state of rotational
KEsystem = ∑ 12 mi vi2
where
= ∑ 12 mi (ωri )
vi = ωri
trans→ rot
2
= 12 [∑ (mi ri2 )]ω 2
rotational inertia (moment of inertia)
about some axis of rotation
Energy of rotational motion
≡I
KErot = 12 Iω 2
All particles have
same ω
[KE
trans
= 12 mv2 ]
Moment of Inertia
For a discrete number
of particles distributed
about
b
an axis
i off rotation
i
I≡
2
m
r
∑ ii
units of kg·m2
all
ll mass
Si l example:
Simple
l
4
I = ∑ mi ri 2 = m1r12 + m2r22 + m 2r22 + m1r12
i=1
what about
other axis?
= 2m1r12 + 2m 2r22
- Rotational inertia (moment of inertia) only valid about some axis of rotation.
- For arbitrary shape, each different axis has a different moment of inertia.
- I relates how the mass of a rotating body is distributed about a given axis.
axis
- r is perpendicular distance from mass to axis of rotation
Moment of inertia: comparison
I1 = ∑ mi ri 2
= (5 ⋅ kg)(2 ⋅ m) + (7 ⋅ kg)(2 ⋅ m)
2
= 52 ⋅ kgm
g
2
2
I 2 = (5 ⋅ kg)(0.5 ⋅ m) 2 + (7 ⋅ kg)(4.5 ⋅ m) 2
= (1.3 + 142) ⋅ kgm 2 = 144 ⋅ kgm 2
Note: 5 kg mass contributes <1% of total
- Mass close to axis of rotation contributes little to total moment of inertia.
[Fnet = ma]
- How
H ddoes a llarger moment off iinertia
i “feel”
“f l” ?
If rigid body = few particles
If rigid body = too-many-to-count particles
I = ∑ mi ri 2
Sum→ Integral
Moment of inertia: continuum mass
m
with ρ =
V
I = ∑ mi ri ⇒ ∫ r dm
d
2
2
⇒
I = ρ ∫ r 2 dV
Example:
E
l moment off inertia
i
i off thin
hi rod
d with
i h perpendicular
di l
rotation through center
I = ∫ r 2 dm = ρ ∫ r 2 dV where dV = area ⋅ dx = A ⋅ dx
I = ρ A ∫ r 2 dx
I = ρA
−L2 → x→ + L2
+L 2
∫
x dx = ρ A
2
-L 2
m ⎞⎟ ⎛ L3 ⎞ 1 ⎛ m ⋅ A ⎞ 3
I = ⎟ A⎜ ⎟ = ⎜
⎟ L
V ⎟⎠ ⎝ 12 ⎠ 12 ⎝ A ⋅ L ⎠
⎛
⎜
⎜
⎜⎝
r 2 = x 2 in this case
( x)
1
3
3
+L
-L
( ) = 121 mL2
2
(
= ρ A 13 ( 18 L3 ) − 13 (− 18 L3 )
)
2
INDEPENDENT
of cross section area
Some Rotational Inertias
Each of these rotational inertias GO THROUGH the center of mass !
Parallel--Axis Theorem
Parallel
I = I COM + Mh
2
P f
Proof:
I= ∫ r dm = ∫ {(x − a ) + (y − b ) }dm
2
2
(
)
2
(
)
I = ∫ x 2 + y2 dm − 2a ∫ x dm − 2b ∫ y dm + ∫ a 2 + b 2 dm
I = I COM + 0 + 0 + h 2 M
Moment of inertia of a Pencil
It depends on where the rotation axis is considered…
I = ML
1
3
2
I = 121 ML2
I = 12 MR2
I = 0.3 × 10 −6 kg ⋅ m 2
−6
I = 150 × 10 kg ⋅ m
2
I = 38 × 10−6 kg ⋅ m 2
Consider a 20g pencil 15cm long and 1cm wide …
… somewhat like MASS, you can feel the
diff
difference
iin th
the rotational
t ti
l iinertia
ti
Example #1
A bicycle wheel has a radius of 0.33m
0 33m and a rim of mass 1.2
1 2 kg.
kg The
wheel has 50 spokes, each with a mass 10g.
What is the moment of inertial about axis of rotation?
What is moment of inertia about COM?
I tot,com = I rim,center + 50I spoke
→ What is Ispoke (p
(parallel-axis)?
)
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
2
→ Putting together
I tot,com
= I rim,center
50I spokek
t t
i
t + 50
= M wheel R 2 + 50I spoke
= (1.2kg)(0.33m) + 50(3.6 ×10−4 kg ⋅ m 2 ) = 0.149kg ⋅ m 2
2
Example #2
How much work did Superman exert on earth in
order to stop it?
What is the kinetic energy of the earth’s rotation
about its axis?
Energy of rotational motion is found from:
KErot = 12 Iω 2
What is earth’s moment of inertia, I?
Iearth = Isphere = 25 Mr 2
= 25 (6 ×10 24 kg)(6.4 ×10 6 m) ≅ 1×10 38 kg ⋅ m 2
2
What is earth’s angular velocity, ω?
From T = 2 π
ω
⇒
di
ω = 2π radians
= 2π
day
(3600 × 24)
= 7.3 ×10−5 rad / s
Now plug-’n-chug:
1⎛
⎜
2⎝
KErot = 1×10 kg ⋅ m
38
29
7.3
×10
rad
/
s
=
2.6
×10
J
(
)
2⎞
⎟
⎠
−5
2