Rigid Body: Rotational and
Translational Motion;
Rolling without Slipping
8.01
W11D1
Today’s Reading Assignment:
W11D1
Young and Freedman: 10.3-10.4
Overview: Rotation and
Translation of Rigid Body
Thrown Rigid Rod
Translational Motion: the gravitational external force
acts on center-of-mass
F ext
dV
dpsys

 m total cm  mtotal Acm
dt
dt
Rotational Motion: object rotates about center-ofmass. Note that the center-of-mass may be
accelerating
Overview: Rotation about the
Center-of-Mass of a Rigid Body
The total external torque produces an angular acceleration
about the center-of-mass

ext
cm
 I cm cm
dLcm

dt
Icm is the moment of inertial about the center-of-mass
 cm is the angular acceleration about the center-of-mass
Lcm is the angular momentum about the center-of-mass
Fixed Axis Rotation
• CD is rotating about
axis passing through
the center of the disc
and is perpendicular to
the plane of the disc.
• For straight line motion,
bicycle wheel rotates
about fixed direction
and center of mass is
translating
Review: Relatively Inertial
Reference Frames
Two reference frames.
Origins need not coincide.
One moving object has different
position vectors in different frames
r1  R  r2
Relative velocity between the two reference frames
V  dR dt
is constant since the relative acceleration is zero
A  dV dt  0
Review: Law of Addition of
Velocities
Suppose the object is moving; then, observers in
different reference frames will measure different
velocities
Velocity of the object in Frame 1:
v1  dr1 dt
Velocity of the object in Frame 2:
v 2  dr2 dt
Velocity of an object in two different reference
frames
dr1 R r2
dt

dt

dt
v1  V  v2
Center of Mass Reference Frame
Frame O: At rest with respect to
ground
Frame Ocm: Origin located at center
of mass
Position vectors in differentr frames:
r r
ri  rcm,i  R cm
Relative velocity between the two
reference framesr
r
Vcm  dR cm / dt
Law of addition of velocities:
r
r
r
v i  v cm,i  Vcm
r
r r
rcm,i  ri  R cm
r
r
r
A cm  dVcm / dt  0
r
r
r
v cm,i  v i  Vcm
Rolling Bicycle Wheel
Reference frame fixed to ground
Center of mass reference frame
Motion of point P on rim of rolling bicycle wheel
Relative velocity of point P on rim:
r
r
r
v P  v cm, P  Vcm
Rolling Bicycle Wheel
Distance traveled in center
of mass reference frame of
point P on rim in time Δt:
s  R  R cm t
Distance traveled in
ground fixed
reference frame of
point P on rim in time
Δt:
Xcm =Vcm t
Rolling Bicycle Wheel:
Constraint relations
Rolling without slipping:
s  X cm
R cm  Vcm
Rolling and Skidding
s  X cm
R cm  Vcm
Rolling and Slipping
s  X cm
R cm  Vcm
Rolling Without Slipping: velocity
of points on the rim in reference
frame fixed to ground
The velocity of the point on the rim that is in contact with
the ground is zero in the reference frame fixed to the ground.
Concept Question: Rolling
Without Slipping
When the wheel is rolling without slipping what is the
relation between the final center-of-mass velocity and
the final angular velocity?
1. v
  R f .
cm, f
2.
 f  Rvcm, f .
3. v
 R f
cm, f
4.
.
 f  Rvcm, f .
Angular Momentum for 2-Dim
Rotation and Translation
The angular momentum for a rotating and translating object
is given by (see next two slides for details of derivation)
L S  R S ,cm  p
i N
sys
  rcm,i  mi v cm,i
i 1
The first term in the expression for angular momentum about
S arises from treating the body as a point mass located at
the center-of-mass moving with a velocity equal to the
center-of-mass velocity,
L S ,cm  R S ,cm  psys
The second term is the angular momentum about the centerof mass,
i N
L cm   rcm,i  mi v cm,i
i 1
Derivation: Angular Momentum for
2-Dim Rotation and Translation
The angular momentum for a rotating and translating object is
given by
iN


L S    mi ri  mi v i 
 i 1

The position and velocity with respect to the center-of-mass
reference frame of each mass element is given by
ri  R S ,cm  rcm,i
vi  Vcm  v cm,i
So the angular momentum can be expressed as
iN
iN
 iN 
 iN

L S  R S ,cm    mi  Vcm  R S ,cm   mi v cm,i    mi rcm,i   Vcm   rcm,i  mi v cm,i
i 1
i 1
 i 1 
 i 1

Derivation: Angular Momentum for
2-Dim Rotation and Translation
iN
iN
 iN 
 iN

L S  R S ,cm    mi  Vcm  R S ,cm   mi v cm,i    mi rcm,i   Vcm   rcm,i  mi v cm,i
i 1
i 1
 i 1 
 i 1

The two middle terms in the above expression vanish because in the
center-of-mass frame, the position of the center-of-mass is at the
origin, and the total momentum in the center-of-mass frame is zero,
iN
1 iN
mi v cm,i  0

mr 0
total  i cm,i
i 1
m
i 1
Then then angular momentum about S becomes
i N
 i N 
L S  R S ,cm    mi  Vcm   rcm,i  mi v cm,i
i 1
 i 1 
The momentum of system is
p
sys
So the angular momentum about S is
 iN 
   mi  Vcm
 i 1 
L S  R S ,cm  p
i N
sys
  rcm,i  mi v cm,i
i 1
Table Problem: Angular Momentum
for Earth
What is the ratio of the spin angular momentum to the
orbital angular momentum of the Earth?
What is the vector expression for the total angular
momentum of the Earth about the center of its orbit
around the sun (you may assume the orbit is circular
and centered at the sun)?
Earth’s Motion Orbital Angular
Momentum about Sun
• Orbital angular momentum about
center of sun
• Center of mass velocity and angular
velocity
• Period and angular velocity
r orbital r
r
LS
 rS ,cm  ptotal  rs,e me vcmkφ
vcm  rs,e orbit
 orbit  (2 / Torbit )  2.0  107 rad  s1
2
r
m
r
2
• Magnitude
e s,e
orbital
2
φ
LS
 me rs,e  orbit k 
kφ
Torbit
r
40
2
1 φ
Lorbital

2.67

10
kg

m

s
k
S
Earth’s Motion
Spin Angular Momentum
• Spin angular momentum
about center of mass of earth
• Period and angular velocity
• Magnitude
r spin
r
2
L cm  I cm spin  me Re 2 spin nφ
5
 spin 
2
 7.29  10 5 rad  s1
Tspin
r
Lcmspin  7.09  1033 kg  m2  s1nφ
Earth’s Motion about Sun:
Orbital Angular Momentum
For a body undergoing orbital motion like the earth orbiting the sun, the
two terms can be thought of as an orbital angular momentum about the
center-of-mass of the earth-sun system, denoted by S,
LS ,cm  R S ,cm  psys  rs ,e me vcmkˆ
Spin angular momentum about center-of-mass of earth
Lcm spin  I cm spin 
2
me Re 2spin nˆ
5
Total angular momentum about S
r total
2
φ
L S  rs,e mevcmk  me Re2 spin nφ
5
Rotational Work-Kinetic Energy
Theorem
Change in kinetic energy of rotation about
center-of-mass
K rot  K rot, f  K rot,i
1
1
2
2
 Icm cm, f  Icm cm,i
2
2
Change in rotational and translational
kinetic energy
K  Ktrans  Krot
K  K trans  Krot
1 2
1 2  1
1
2
2 
  mvcm, f  mvcm,i    Icm cm, f  Icm cm,i 
2
2
2
 2

Rules to Live By: Kinetic Energy
of Rotation and Translation
Change in kinetic energy of rotation about
center-of-mass
K rot  K rot, f  K rot,i
1
1
2
2
 Icm cm, f  Icm cm,i
2
2
Change in rotational and translational
kinetic energy
K  Ktrans  Krot
K  K trans  Krot
1 2
1 2  1
1
2
2 
  mvcm, f  mvcm,i    Icm cm, f  Icm cm,i 
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 reaches the bottom first?
1) A
2) B
3) Both at the same time.
Problem: Cylinder on Inclined
Plane Energy Method
A hollow cylinder of outer radius R and mass m with moment of inertia I cm
about the center of mass starts from rest and moves down an incline tilted
at an angle  from the horizontal. The center of mass of the cylinder has
dropped a vertical distance h when it reaches the bottom of the incline. Let
g denote the gravitational constant. The coefficient of static friction between
the cylinder and the surface is ms. The cylinder rolls without slipping down
the incline. Using energy techniques calculate the velocity of the center of
mass of the cylinder when it reaches the bottom of the incline.
Concept Question: Angular
Collisions
A long narrow uniform stick lies motionless on ice
(assume the ice provides a frictionless surface).
The center of mass of the stick is the same as the
geometric center (at the midpoint of the stick). A
puck (with putty on one side) slides without spinning
on the ice toward the stick, hits one end of the stick,
and attaches to it.
Which quantities are constant?
1. Angular momentum of puck about center of mass of
stick.
2. Momentum of stick and ball.
3. Angular momentum of stick and ball about any
point.
4. Mechanical energy of stick and ball.
5. None of the above 1-4.
6. Three of the above 1.4
7. Two of the above 1-4.
Problem: Angular Collision
A long narrow uniform stick of length l and mass m lies
motionless on ice (assume the ice provides a
frictionless surface). The center of mass of the stick is
the same as the geometric center (at the midpoint of the
stick). The moment of inertia of the stick about its
center of mass is lcm. A puck (with putty on one side)
has the same mass m as the stick. The puck slides
without spinning on the ice with a speed of v0 toward
the stick, hits one end of the stick, and attaches to it.
You may assume that the radius of the puck is much
less than the length of the stick so that the moment of
inertia of the puck about its center of mass is negligible
compared to lcm.
What is the angular velocity of the stick plus puck after
the collision?
How far does the stick's center of mass move during
one rotation of the stick?
Next Reading Assignment:
W11D1
Young and Freedman: 10.3-10.6