Rotational Dynamics
8.01t
Nov 8, 2004
Fixed Axis Rotation: Angular Velocity and Angular Acceleration
Angle variable
θ
Angular velocity
dθ
ω≡
dt
Angular acceleration
d 2θ
α≡ 2
dt
Mass element
∆mi
Radius of orbit
r⊥,i
Fixed Axis Rotation: Tangential Velocity and Tangential Acceleration
• Individual mass elements
∆mi
• Tangential velocity
vtan,i = r⊥ ,iω
• Tangential acceleration
atan,i = r⊥ ,iα
• Radial Acceleration
arad ,i =
2
vtan,
i
r⊥ ,i
= r⊥,iω 2
Newton’s Second Law • Tangential force on
mass element
produces torque
• Newton’s Second
Law
Ftan,i = ∆mi atan,i
Ftan,i = ∆mi r⊥,iα
• Torque
G
G
G
τ S ,i = rS ,i × Fi
Torque Torque about is S: G
G
G
τ S ,i = rS ,i × Fi
• Counterclockwise • perpendicular to the
plane
τ S ,i = r⊥,iFtan,i = ∆mi (r⊥,i ) α
2
Moment of Inertia
• Total torque is the sum over all mass
elements
i=N
i =N i=N
i=1
i=1
i=1
τ Stotal = τ S ,1 + τ S ,2 + ... = ∑ τ S ,i = ∑ r ⊥ ,i Ftan,i = ∑ ∆mi
(r⊥ ,i
) 2 α
• Moment of Inertia about S:
i= N
I S = ∑ ∆mi
(r⊥ ,i
) 2
i=1
• Unit:
[kg − m 2
]
• Summary:
τ
total
S
= I S α
Rotational Work
G
• tangential force Ftan,i = Ftan,i θ̂
• displacement vector
• infinitesimal work
G
∆rS ,i = ( rS ,⊥ )i ∆θ θ̂
G
G
∆Wi = Ftan,i ⋅ ∆rS ,i = Ftan,i θ̂ ⋅ ( rS ,⊥ )i ∆θ θ̂ = Ftan,i ( rS ,⊥ )i ∆θ
Rotational Work
• Newton’s Second Law
Ftan,i = ∆mi atan,
i
• tangential acceleration
atan,i = ( rS ,⊥ )i α
• infinitesimal work
∆Wi = ∆mi ( rS ,⊥ )i α∆θ
2
• summation
⎛
2⎞
2⎞
⎛
∆W = ⎜ ∑ ∆mi ( rS ,⊥ )i ⎟ α∆θ = ⎜ ∫ dm ( rS ,⊥ ) ⎟ α∆θ = I Sα∆θ
⎜
⎟
⎝ i
⎠
body
⎝
⎠
Rotational Work
∆W = I Sα∆θ
• infinitesimal rotational work
• torque
τ S
= I Sα
∆W = τ S ∆θ
• infinitesimal rotational work
θ =θ f
• Integrate total work W =
∫
θ θ
=
θ =θ f
dW =
0
τ
∫
θ θ
=
0
S
dθ
Rotational Kinetic Energy
Rotational kinetic energy
K cm =
K cm ,i
2
1
1
2
= ∆mi vcm ,i = ∆mi ( rcm,⊥ )i ωcm 2
2
2
K cm = ∑ K cm ,i
i
1
I cmωcm 2
2
⎛1
2⎞
2⎞
1
⎛ 1
2
2
= ⎜ ∑ ∆mi ( rcm,⊥ )i ⎟ ωcm = ⎜ ∫ dm ( rcm,⊥ ) ⎟ ωcm = I cmωcm 2
⎜2
⎟
2
⎝ i 2
⎠
⎝ body
⎠
• Total kinetic energy
K total = K trans + K rot =
1 2 1
2
mvcm + I cmωcm
2
2
PRS - kinetic energy
A disk with mass M and radius R is spinning with
angular velocity ω about an axis that passes
through the rim of the disk perpendicular to its
plane. Its total kinetic energy is:
1. 1/4 M R2 ω2
4. 1/4M R ω2
2. 1/2M R2 ω2
5. 1/4 M R ω2
3. 3/4 M R2 ω2
6. 1/2 M R ω
Rotational Work-Kinetic Energy Theorem
• angular velocity
ω ≡ dθ dt
• angular acceleration
α ≡ dω dt
• infinitesimal rotational work
dWrot
dω
dθ
dθ = I S d ω
= I S α dθ = I S
= I S d ωω
dt
dt
• integrate rotational work
ω =ω f
Wrot =
∫
ω ω
=
0
ω =ω f
dWrot
1
1
2
= ∫ I S d ωω = I S ω f − I S ω02
2
2
ω =ω0
Rotational Work-Kinetic Energy Theorem
• Fixed axis passing through a point S in the
body
Wrot
1
1
2
2
= I cmωcm , f − I cmωcm
,0 = K rot , f − K rot ,0 ≡ ∆K rot
2
2
• Rotation and translation
Wtotal = ∆K trans + ∆K rot
1 2 ⎞ ⎛1
1
⎛1 2
2
2⎞
Wtotal = Wtrans +Wrot =⎜ mvcm
mv
I
ω
I
ω
−
+
−
,f
cm ,0 ⎟ ⎜
cm f
cm 0 ⎟
2
2
⎝2
⎠ ⎝2
⎠
Concept Question
• Two cylinders of the same size and mass
roll down an incline, starting from rest.
Cylinder A has most of its mass
concentrated at the rim, while cylinder B
has most of its mass concentrated at the
center. Which is moving faster at the
bottom?
1) A
2) B
3) Both have the same Concept Question
• Two cylinders of the same size and mass
roll down an incline, starting from rest.
Cylinder A has most of its mass
concentrated at the rim, while cylinder B
has most of its mass concentrated at the
center. Which has more total kinetic
energy at the bottom?
1) A
2) B
3) Both have the same Rotational Power
• rotational power is the time rate of doing
rotational work
dWrot
Prot ≡
dt
• product of the applied torque with the
angular velocity
dWrot
dθ
= τ S ,⊥
= τ S ,⊥ω
Prot ≡
dt
dt
Class Problem
A turntable is a uniform disc of mass m and a radius R. The turntable is
spinning initially at a constant frequency f . The motor is turned off and the
turntable slows to a stop in t seconds. Assume that the angular
acceleration is constant. The moment of inertia of the disc is I.
a) What is the initial angular velocity of the turntable? b) What is the angular acceleration of the turntable?
c) What is the magnitude of the frictional torque acting on the disc?
d) How much work is done by the frictional torque?
e) What is the change in kinetic energy of the turntable?
f) Graph the rotational power as a function of time.
Simple Pendulum
Pendulum: bob hanging from end of string
• Pivot point
• bob of negligible size • massless string
Simple Pendulum: Torque Diagram
torque about the pivot point
G
G
G
τ S = rs ,m × mg = lr̂ × mg (−
sin θ θ̂ + cos r̂) = −lmg sin θ k̂
angular acceleration
(vector quantity)
• Points along axis
• Positive or negative
G d 2θ
α = 2 k̂
dt
Simple Pendulum: Rotational Equation of Motion
• moment of inertial of a point mass about
the pivot point
2
I S = ml
• Rotational Law of Motion
G
G
τ S = I S α
• Simple pendulum oscillator equation
dθ
−lmg sin θ = ml
2
dt
2
2
Simple Pendulum: Small Angle Approximation
Angle of oscillation is small
sin θ ≅ θ
• Simple harmonic oscillator
d 2θ
g
≅− θ
2
dt
l
• Analogy to spring equation
d 2x
k
=− x
2
dt
m
• Angular frequency of oscillation
ω pendulum ≅
• Period
T0 =
2π
ωp
≅ 2π
g
l
l
g
Simple Pendulum: Mechanical Energy
• released from rest at an angle
θ0
Extra Topic: Simple Pendulum: Mechanical Energy
• Velocity
vtan = l
• Kinetic energy
• Initial energy
• Final energy
Kf =
dθ
dt
1
1 ⎛ dθ ⎞
mvtan 2 = m ⎜ l
⎟
2
2 ⎝ dt ⎠
2
E0 = K 0 + U 0 = mgl (1 −
cos θ 0 )
1 ⎛ dθ ⎞
Ef = K f +U f = m⎜l
⎟ + mgl (1 − cos θ )
2 ⎝ dt ⎠
2
• Conservation of energy
1 ⎛ dθ ⎞
m⎜l
⎟ + mgl (1 − cos θ ) = mgl (1 − cos θ 0 )
2 ⎝ dt ⎠
2
Simple Pendulum: Angular Velocity Equation of Motion
• Angular velocity
dθ
2g
(cos θ − cos θ0 )
=
dt
l
• Integral form
∫
dθ
2g
dt
=∫
l
(cos θ − cos θ 0 )
Simple Pendulum: Integral Form
b sin a = sin (θ 2 )
• Change of variables
b = sin (θ0 2 )
• Integral form
∫
(1− b
da
2
sin a )
2
12
=∫
g
dt
l
• Power series approximation
(1− b
2
• Solution
sin a )
2
−1 2
1 2 2
3 4 4
= 1+ b sin a + b sin a + ...
2
8
1
2π + π sin 2 (θ 0 2 ) + ... =
2
g
T
l
Simple Pendulum: First Order Correction
• period
T = 2π
l
g
• initial angle is small
• Approximation
⎛ 1 2
⎞
1+
sin
2
...
θ
+
(
)
0
⎜
⎟
⎝ 4
⎠
sin 2 (θ0 2 ) ≅ θ0 2 4
l
T ≅ 2π
g
• First order correction
l
T0 = 2π
g
1 2⎞
1 2⎞
⎛
⎛
⎜ 1+ θ 0 ⎟ = T0 ⎜ 1+ θ 0 ⎟
⎝ 16
⎠
⎝ 16
⎠
1 2
∆T1 ≅ θ 0 T0
16
Physical Pendulum
• pendulum pivoted
about point S
• gravitational force
acts center of mass
• center of mass
distance lcm from the
pivot point
Physical Pendulum
• torque about pivot point
G
G G
τ S = rs ,cm × mg = lcmr̂ × mg (−
sin θ θ̂ + cos r̂) = −lcm mg sin θ k̂
• moment of inertial about pivot point I S
• Example: body is a uniform rod of mass m
and length l .
1
I S = ml 2
3
Physical Pendulum
• rotational dynamical equation
G
G
τS = ISα
• small angle approximation
sin θ ≅ θ
• Equation of motion
• Angular frequency
• Period
lcm mg
d 2θ
≅
−
θ
2
dt
IS
ω pendulum
2π
lcm mg
≅
IS
IS
≅ 2π
T=
lcm mg
ωp
PRS - linear momentum
A disk with mass M and radius R is spinning with
angular velocity ω about an axis that passes
through the rim of the disk perpendicular to its
plane. The magnitude of its linear momentum
is:
1. 1/2 M R2 ω
4. M R ω2
2. M R2 ω
5. 1/2 M R ω
3. 1/2 M R ω2
6. M R ω
PRS: Rigid Body Rotation
A rigid body rotates about an axis, S.
What is the relationship between the rotation rate about S vs
about the center of mass? Also, what is the relationship
between the rotational accelerations about these points?
S
1. αcm = αS and ωcm = ωS
2. αcm = αS but ωcm ≠ ωS
3. ωcm = ωS but αcm ≠ αS
4. αcm ≠ αS and ωcm ≠ ωS
5. None of above is consistently true
Concept Question: Physical Pendulum
A physical pendulum consists of a
uniform rod of length l and
mass m pivoted at one end. A
disk of mass m1 and radius a
is fixed to the other end.
Suppose the disk is now
mounted to the rod by a
frictionless bearing so that is
perfectly free to spin. Does
the period of the pendulum
a) increase?
b) stay the same?
c) decrease?
Class Problem: Physical Pendulum
A physical pendulum consists of a
uniform rod of length l and mass
m pivoted at one end. A disk of
mass m1 and radius a is fixed to
the other end.
a)
Find the period of the pendulum.
Suppose the disk is now mounted to
the rod by a frictionless bearing
so that is perfectly free to spin.
b) Find the new period of the
pendulum.