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Rotational and Translational
Motion Dynamics
8.01
W11D2
Today’s Reading Assignment:
W11D2
Young and Freedman: 10.3-10.6
Review: Angular Momentum and
Torque for a System of Particles
Change in total angular momentum about
a point S equals the total torque about the
point S
iN
i  N dr
 S ,i
dLsys
dpi 
S
  L S ,i   
 pi  rS ,i 

dt
dt
dt
i 1
i 1 

iN
dLsys
dpi

S
   rS ,i 
dt
dt
i 1 
iN
 iN
total

r

F




S
  S ,i i  S ,i
 i 1
i 1

dLsys
S
  S total
dt

Torque for Rotation and
Translation
The total torque about S is given by
 S   S ,cm   cm
where the first term torque about S due to the total external
force acting at the center-of-mass and the second term is torque
about the center-of-mass is only due to the forces as seen in the
laboratory frame.
i N
 S ,cm  R S ,cm  Fext
 cm   rcm,i  Fi
i 1
The total external torque produces an angular acceleration about
the center-of-mass
 ext
cm  I cm cm 
dLcm
dt
Torque and Angular
Momentum for Rotation and
Translation
The torque about a point S is the time derivative of the angular
momentum about S,
S 
S 
dR S ,cm
dt
p
sys
dL S
dt
iN
dpsys i  N drcm,i
d

 R S ,cm 

 mi v cm,i   rcm,i   mi v cm,i 
dt
dt
 dt

i 1
i 1
Once again the first and third terms vanish because
dR S ,cm
dt
p
sys
 Vcm  m
total
Vcm  0
drcm,i
dt
 mi vcm,i  vcm,i  mi vcm,i  0
So the torque about S becomes,
dpsys i  N
d

 S  R S ,cm 
  rcm,i   mi vcm,i 
dt
 dt

i 1
Torque and Angular
Momentum for Rotation and
Translation
The torque about a point S is
dpsys i  N
d

 S  R S ,cm 
  rcm,i   mi vcm,i 
dt
 dt

i 1
Recall that the external force is the time change of the momentum of the
center-of-mass,
sys
F ext 
dp
dt
So the first term is the torque about S due to the total external force
acting at the center-of-mass
 S ,cm  R S ,cm  Fext
Torque for Rotation and
Translation
i N
d

 S   S ,cm   rcm,i   mi v cm,i 
 dt

i 1
The time derivative that appears in the second term in the above
expression, the time derivative of the momentum of a mass
element in the center-of mass-frame, is equal to the force acting
on that element which include both inertial and fictitious forces,
d
mi v cm,i  Fi
dt
The torque about the center-of-mass is then
 cm
d
 iN
 rcm,i   mi v cm,i    rcm,i  Fi
 dt
 i 1
Torque for Rotation and
Translation
When we sum the torques over all the elements in the body, the
fictitious forces act at the center-of-mass, so the torque from
these fictitious forces is zero, so, the torque about the center-ofmass is only due to the forces as seen in the laboratory frame.
i N
iN
i 1
i 1

 cm   rcm,i  Fi   rcm,i  Fi  mi A
 iN

  rcm,i  Fi    mi rcm,i   A
i 1
 i 1

iN
iN
  rcm,i  Fi
i 1

Rules to Live By: Angular
Momentum and Torque
1) About any fixed point S
LS  Labout cm  Lof cm  Labout cm  rs ,cm  mtotal vcm
r
r ext
 S    S ,i
i
r
dL S

dt
2) Independent of the CM motion, even if L about cm and 
are not parallel
 about cm 
dLabout cm
dt
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
1 2
1 2  1
1
2
2 
K  K trans  Krot   mvcm, f  mvcm,i    Icm cm, f  Icm cm,i 
2
2
2
 2

Rotational and Translational
Quantity
Momentum
Ang Momentum
Rotation
r
r
Lcm  Icm
Translation
r
r
p  mVcm
Force
Torque
cm  dLcm / dt
Fext  dpsys / dt  mtotal Acm
Kinetic Energy
K rot  (1/ 2) I cm 2
K trans  (1/ 2)mVcm2
K rot  L2cm / 2 I cm
Work
f
W    S d
0
Power
Prot   S 
K trans  p 2 / 2m
f
W   F  dr
0
r r
P  Fv
Concept Question
A cylinder is rolling without slipping down an inclined plane.
The friction at the contact point P is
1. Static and points up the inclined plane.
2. Static and points down the inclined plane.
3. Kinetic and points up the inclined plane.
4. Kinetic and points down the inclined plane.
5. Zero because it is rolling without slipping.
12
Table Problem: Cylinder on
Inclined Plane Torque About
Center of Mass
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 the torque
method about the center of mass, calculate the velocity of the center of
mass of the cylinder when it reaches the bottom of the incline.
Concept Question: Rotation
and Translation
Concept Question: Rotation
and Translation
Mini-Experiment: Pulling Spool
1. Which way does the yo-yo roll when you pull
it horizontal?
2. Is there some angle at which you can pull
the string in which the yo-yo doesn’t roll
forward or back?
Table Problem: Descending
and Ascending Yo-Yo
A Yo-Yo of mass m has an axle
of radius b and a spool of radius
R. It’s moment of inertia about
the center of mass can be taken
to be I = (1/2)mR2 and the
thickness of the string can be
neglected. The Yo-Yo is
released from rest. What is the
acceleration of the Yo-Yo as it
descends.
17
Concept Question: Constants
of the Motion
A bowling ball is initially thrown down an alley with an initial
speed v0, and it slides without rolling but due to friction it
begins to roll until it rolls without slipping. What quantities
are constant before it rolls without slipping?
1.Energy.
2.Angular momentum about the center of mass.
3.Angular momentum about a fixed point on the ground.
4.Three of the above.
5.Two of the above.
6.One of the above.
7.None of the above.
Problem: Bowling Ball
Conservation of Angular
Momentum Method
A bowling ball of mass m and radius R is initially thrown
down an alley with an initial speed v0, and it slides without
rolling but due to friction it begins to roll. The moment of
inertia of the ball about its center of mass is I = (2/5)mR2.
By cleverly choosing a point about which to calculate the
angular momentum, use conservation of angular
momentum to find is the velocity of the center-of-mass
when the wheel rolls without slipping.
Next Reading Assignment:
W11D3
Young and Freedman: 10.1-10.6