Example:
Assume the angle θ of rigid body is given as the following function of time
θ (t ) = θ 0 + bt + ct 2
vt
where b = 10 rad/sec, c = 5 rad/sec
2
r
1. Find ω = angular speed (differentiate) at 3 sec
2. Find α = angular acceleration (differentiate again) at 3 sec
3. Find tangential and radial accelerations, and magnitude of
acceleration at 3 sec of the point located at 2 m from the rotation axis.
Physics 106 Week 2
Rotational Dynamics I
SJ 8th Ed.: Chap 10.4 to 6
• Kinetic energy of rotation
• Moment of inertia – rotational analog of mass
• Calculating moment of inertia (rotational inertia)
– Parallel axis theorem – definition
– Examples
Today
• Torque – rotational analog of force
– Moment arm definition for 2 dimensions
– Superposition of torques
– Torque in 3D - Cross product definition
2
1
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3
Previously:
Rotational Kinematics
Dynamics Concepts
Moment of
Inertia I
same mass
small I
F
larger I
θ
F
θ
rp
measures BOTH mass and its distribution
depends on the choice of rotation axis
rp is perpendicular distance from mass point to axis
masses on axis have rp= 0, no contribution to I
I=
Torque
τ
r
r
angular variables θ, ω, α
related velocity vT and acceleration components arad, aT
but no “physical” content (dynamics)
rp
Next Time: Rotational2nd Law:
∑ m ir⊥2,i
for point masses
TWIST - measures applied force AND where it acts
MOMENT ARM (lever arm) rp = perpendicular distance
from axis to line of action of force
τ = force x lever arm = rF sin( θ ) = r × F
“moment of inertia” plays the role of mass
“torque
plays
the
“t
l
th role
l off force
f
modified statement of Second Law for rotation
Fnet = ma
τnet = Iα
rotational work:
dW = Fdx
dWrot = τdθ
power used when doing rotational work:
dW
dW
P≡
= Fo v
Prot ≡ dtrot
dt
= τω
2
Rotational Kinetic Energy
An rigid body rotating about some axis with an angular speed ω has
rotational kinetic energy...
... even though it’s translational kinetic energy may be zero
Consider the Pulley:
•
•
•
Each particle of mass has kinetic energy Ki = ½ mivi2
The tangential velocity depends on the distance, r, from the axis
of rotation, so vi = ω ri
The total rotational kinetic energy of the pulley is the sum of the
energies of all its particles
K rot =
≡
∑K
1
2
i
= ∑ 12 mi vi2 =
1
2
∑m r
ω2 =
2
i i
Iω 2 where I ≡ ∑ m r 2
ii
⎡
1 ⎢
2 ⎢
⎣
2⎤ 2
∑ m r ⎥⎥ ω
i ,i ⎦
ΔU
ω factors
I is the moment of inertia ( “rotational inertia”)
•
•
•
I
measures the resistance to torques that might be accelerating
the angular motion
Rotational kinetic energy is not a new type of energy, it looks
different because it is applied to a rotating object
Units: Joules (J), as usual
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. Ladybug and gentleman bug have the same mass.
The gentleman bug’s moment of inertia with respect to the
rotation axis is
A. half the ladybug’s.
B. the same as the ladybug’s.
C. twice the ladybug’s.
D. quarter of ladybug’s
E. four times of ladybug’s
3
Example: Find moment of inertia for a crossed dumbbell
•Four identical balls as shown:
m = 2 kg
•Connected by massless rods:
length d = 1 m.
I=
∑
m ir⊥2,i
m
2d
d
m
f point
for
i t masses
d
B
A
d
d
Rotational inertia I depends on axis chosen
C
m
m
A) Choose rotation axis perpendicular to figure through point “A”
B) Now
N
choose
h
axis
i perpendicular
di l to
t figure
fi
through
th
h point
i t “B”
C) Let rotation axis pass through points “B” and “C”
Example: Moment of inertia for 2 equal masses connected
by a massless rod
a) Rotation about axis at one end
IP = mL2
P
m
cm
m
L
b) Rotation
i
about
b
axis
i through
h
h CM
2
⎛L⎞
⎛L⎞
ICM ≡ ∑ mirp,i 2 = m⎜ ⎟ + m⎜ ⎟
2
⎝ ⎠
⎝ 2⎠
i
=
2
P
L/2
L/2
1
mL2
2
c) Rotation axis through both masses
IP = 0
m
cm
m
P
cm
since rp,i are both zero
m
L
Propose rule:
IP = Iabout CM + Itotal mass at CM
in example “a”:
IP =
2
m
Parallel Axis
Theorem
1
⎛L⎞
mL2 + 2m ⎜ ⎟ = mL2
2
⎝ 2⎠
4
Parallel Axis Theorem
The moment of inertia I
depends where the
rotation axis P is
parallel axis theorem describes variation, based on Icm
applies to point masses and to continuous bodies
M = the total mass of the body
CM = mass center
di l
di
i through
h
hP
h = perpendicular
distance ffrom CM to axis
moment of inertia about
axis through point P
parallel axes through
P and CM
h
IP = Icm + Mh
moment of inertia about
axis through CM parallel
to axis through P
CM
2
P
rotational inertia of
total mass M if at
the CM point, taken
about axis through P
Choosing the axis
Through the CM minimizes
the moment of inertia (h = 0)
“rotational inertia” and “moment of inertia” are synonyms
An example showing that the parallel axis theorem works
• Rotation axes perpendicular to plane of figure
• Masses on the corners of a rectangle, sides a & b
I=
h2 = (a/2)2 + (b/2)2
∑ m ir⊥2,i
[
]
]
b
m
• About an axis “P” through a corner:
[
m
h
⎡ a2 b2 ⎤
2
2
ICM ≡ 4 m h2 = 4m ⎢ +
⎥ = m a +b
4 ⎥⎦
⎣⎢ 4
IP = 0 + ma 2 + m b 2 + m a 2 + b 2
a
m
• About an axis through the CM:
[
X
cm
h
m
P
= 2m a2 + b 2
]
• Using the Parallel Axis Theorem directly for the same corner
axis:
⎡ a2 b2 ⎤
2
2
IP = Icm + Mtoth2 = m a 2 + b 2 + 4m ⎢ +
⎥ = 2m a + b
4
4
⎣⎢
⎦⎥
[
]
[
]
5
Moment of inertia for continuous mass distributions
∑ si2 Δmi
I ≡
∑ Δli
=
i = 1,n
i = 1,n
Continuum limit: let chunk size Æ 0
Δmi → dm, n → ∞,
ΔI i = si2 Δmi
→
2
∑
→
P
∫
Δmi
dI = r⊥2 dm
∫ r ⊥ dm = ∫ r⊥ ρ(r ) d
IP ≡
si
2
3
V
volume
density
The integral can often be evaluated, especially for constant density.
The parallel axis theorem holds, as before
moment of inertia about
axis through point P
IP = Icm + Mh 2
moment of inertia about
mass center
total
mass
perpendicular
distance
between axes
Example: Find the moment of inertia of a
uniform thin rigid rod about its center
y
2
2
∫ r dm = ∫ x dm
IP ≡
⊥
•The
The shaded area has a mass:
dm = λ dx
-L/2
L/2
λ = mass/unit length = M/L
total mass M
•Then the moment of inertia
about y-axis is:
Iy ≡
+ L/2
∫-L/2
x 2dm =
∴ Iy =
+ L/2
∫-L/2
x 2λdx =
M 1 3 +L / 2
x
L 3
−L / 2
=
M 1 ⎡ L3 (− L )3 ⎤
−
⎥
L 3 ⎢⎣ 8
8 ⎦
1
ML2
12
6
Example: Moment of inertia for a rod
rotating around one end
Use Parallel Axis Theorem
• The moment of inertia of
the rod about its center is
Icm =
1
12
ML 2
cm
h=½L
Therefore,
IP = Icm + Mh 2
IP =
1
12
⎡L ⎤
ML + M ⎢ ⎥
⎣2⎦
2
2
=
1
3
ML 2
Rotational inertias for basic shapes for use in problems
Use with parallel axis theorem to get I’s about other axes.
7
Moments of Inertia of Various Rigid Objects
Example: Application of the parallel axis theorem to a wheel and a sphere
ƒ Circular plate or cylinder: Find I about axis through P
“P” axis is normal to plane through edge of wheel
IP = Icm + Mh 2
cm
r
P
h=r
“cm”
cm axis is normal to plane through center of wheel
From Table:
IP =
Icm =
1
Mr 2
2
1
Mr 2 + Mr 2
2
∴ IP =
3
Mr 2
2
ƒ Solid sphere: Find I about a tangent line through P
“P” axis is in p
plane tangent
g
to edge
g of sphere
p
cm
r
P
IP = Icm + Mh 2
h=r
“cm” axis is in plane center of sphere
From Table:
IP =
2
5
Icm =
Mr 2 +
5
5
Mr 2
2
5
Mr 2
∴ IP =
7
Mr 2
5
8