Summary -- Translation - Rotation
translational motion Quantity
Rotational motion
x
Position
θ
∆x
Displacement
∆θ
v = dx/dt
Velocity
ω=dθ/dt
a = dv/dt
Acceleration
α=dω/dt
m
Mass Inertia
I
F = ma
Newton’s second law
W=∫
xf
xi
v v
F ⋅ dx
K = ½ mv2
v v
P = F⋅v
τ= r x F
θf
Work
W = ∫ τ dθ
Kinetic energy
K = ½ Iω2
Power (constant F or τ)
P = τω
θi
The Kinetic Energy of Rolling
View the rolling as pure rotation around P, the kinetic energy
K = ½ IP ω2
parallel axis theorem: Ip = Icom +MR2
so
K = ½ Icomω2 + ½ MR2ω2
since vcom = ωR
K = ½ Icomω2 + ½ M(vcom)2
½ Icomω2 : due to the object’s rotation about its center of mass
½ M(vcom)2 : due to the translational motion of its center of mass
Chapter 11: Rolling, Torque, and Angular
Momentum
For an object rolling smoothly, the motion of the
center of mass is pure translational.
s=θR
vcom = ds/dt = d(θ R)/dt = ωR
vcom = ωR
• Rolling viewed as a combination of pure rotation
and pure translation
• Rolling viewed as pure rotation
vtop = (ω)(2R) = 2 vcom
• Different views, same conclusion
Sample Problem: A uniform solid cylindrical
disk, of mass M = 1.4 kg and radius R = 8.5 cm,
rolls smoothly across a horizontal table at a
speed of 15 cm/s. What is its kinetic energy K?
vc.m. = 0.15m/s
Idisk =1/2MR2 = (0.5)(1.4kg)(0.085m)2 = 5.058x10−3kg m2
ω = v/R = (0.15m/s)/0.085m = 1.765 rad/s
1
1 2
2
K = K trans + K rot = Mv c.m. + Iω
2
2
1
1
2
K = K trans + K rot = (1.4)(0.15) + (5.058x10 −3 )(1.765) 2
2
2
= 15.75x10 −3 + 7.878x10 −3 = 23.63x10 −3 J
Angular momentum
Angular momentum with respect to
point O for a particle of mass m and
linear momentum p is defined as:
v v v
v v
l = r × p = m ( r × v)
Compare to the linear case p = mv
direction: right-hand rule
magnitude:
l = r p sin φ = r mv sin φ
The angular momentum of a rigid body
rotating about a fixed axis
Consider a simple case, a mass m rotating
about a fixed axis z:
l = r mv sin90o = r m r ω = mr2ω = I ω
z
v
r
y
x
In general, the angular momentum of a rigid
body rotating about a fixed axis is
L=Iω
L : angular momentum (group or body) along the rotation axis
l : angular momentum (particle) along the rotation axis
I : moment of inertia about the same axis
Question
Particles 1, 2, 3, 4, and 5 have the same mass and speed as shown in
the figure. Particles 1 & 2 move around O in opposite directions.
Particles 3, 4, and 5 move towards or away from O as shown.
2r
2r
r
r
Which of the particles has the smallest magnitude angular momentum?
1) 1
2) 2
3) 3
4) 4
5) 5
6) all have the same l
Question
Particle with smallest angular momentum (magnitude)?
1. 1
2. 2
3. 3
4. 4
5. 5
6. all have same ang. mom.
Question
Particles 1, 2, 3, 4, and 5 have the same mass and speed as shown in
the figure. Particles 1 & 2 move around O in opposite directions.
Particles 3, 4, and 5 move towards or away from O as shown.
2r
2r
r
r
l = r p sin φ = r mv sin φ
φ = 0o for 5. => l = 0
Which of the particles has the smallest magnitude angular momentum?
1) 1
2) 2
3) 3
4) 4
5) 5
6) all have the same l
Newton’s Second Law in Angular Form
v
v
v
v
v v v v v d(mv) d( r × mv) d l
τ = Iα = r × F = r ×
=
=
dt
dt
dt
r
Let τ net be the vector sum of all the torques acting on
the object.
r
τ net
r
dL total
=
dt
n r
r
L total = ∑ li
i =1
Net external torque equals to the time rate
change of the system’s total angular momentum
Torque and Angular Momentum
v
v
v
dl d ( r × mv)
v
=
τ net =
dt
dt
v
v  dv  v
v v v
= r ×  m  = r × (ma ) = r × F
 dt 
Torque is the time rate of change of
angular momentum.
Precession
mg
θ
r
Torque is the time rate of
change of angular momentum.
Falling due to torque about the pivot point.
τ = rFsinθ = rmg sinθ
Falling causes angular momentum
about the pivot point (along y-axis).
v
L
v
L
Precession
mg
θ
r
Torque is the time rate of
change of angular momentum.
Falling due to torque about the pivot point.
τ = rFsinθ = rmg sinθ
Falling causes angular momentum
about the pivot point (along y-axis).
Now set the gyroscope in motion
(along x-axis)
L=Iω
L is fixed by the spinning, so the torque
can only change the direction of L
v
L
v
L
Precession
Rate
mg
θ
r
Torque is the time rate of
change of angular momentum.
v
L
Falling due to torque about the pivot point.
Falling causes angular momentum
about the pivot point (along y-axis).
dL Mg r dt
dφ =
=
L
Iω
dφ Mg r
Ω≡
=
dt
Iω
(Precession along z-axis)
Precession
Rate
mg
θ
r
Torque is the time rate of
change of angular momentum.
Nuclei have intrinsic angular momentum.
This effect is at the core of MRI,
which is tuned to pick up the intrinsic
angular momentum of the proton in
hydrogen.
(Precession along z-axis)
v
L
Conservation of Angular Momentum
If the net external torque acting on a system is zero, the
angular momentum of the system is conserved.
If
τnet = 0
v
v
τnet
v
dL
=
dt
then
v
L = constant
v v
L i = Lf
For a rigid body rotating around a fixed axis, ( L = I ω )
the conservation of angular momentum can be written
as
Ii ω i = If ω f
Some examples involving conservation
of angular momentum
The spinning volunteer
Lf = Li => If ωf = Ii ωi
Angular momentum is
conserved
Li is in the spinning wheel
Now exert a torque to flip its rotation.
Lf, wheel = −Li.
Conservation of Angular momentum means that the person must
now acquire an angular momentum.
Lf, person = +2Li
so that Lf = Lf, person + Lf, wheel =+2Li + −Li = Li.
More examples
The springboard diver
Spacecraft orientation
Problem 11-66
Ring of R1 (=R2/2) and R2 (=0.8m),
Mass m2 = 8.00kg.
ωi = 8.00 rad/s. Cat m1 = 2kg. Find
kinetic energy change when cat walks
from outer radius to inner radius.
Problem 11-66
Ring of R1 (=R2/2) and R2 (=0.8m),
Mass m2 = 8.00kg.
ωi = 8.00 rad/s. Cat m1 = 2kg. Find
kinetic energy change when cat walks
from outer radius to inner radius.
Initial Momentum
Li = L i ,cat + Li ,ring = m1R 2 v i + Iωi
1
= m1R ωi + m 2 (R 12 + R 22 )ωi
2
2



1
m
R
2
2
1
 2 + 1 
= m1R 2 ωi 1 +

2
m
R
1 
2


2
2
Problem 11-66
Ring of R1 (=R2/2) and R2 (=0.8m),
Mass m2 = 8.00kg.
ωi = 8.00 rad/s. Cat m1 = 2kg. Find
kinetic energy change when cat walks
from outer radius to inner radius.
Final Momentum
L f = L f ,cat + L f ,ring = m1R 1v f + Iωf
1
= m1R ωf + m 2 (R 12 + R 22 )ωf
2
2



1
m
R
2
2
2
 2 + 1 
= m1R 1 ωf 1 +

2
m
R
1 
1


2
1
Problem 11-66
Problem 11-66
Ring of R1 (=R2/2) and R2 (=0.8m),
Mass m2 = 8.00kg.
ωi = 8.00 rad/s. Cat m1 = 2kg. Find
kinetic energy change when cat walks
from outer radius to inner radius.
Initial Kinetic energy Ki is: