Two-Dimensional Rotational
Kinematics
8.01
W09D1
Today’s Reading Assignment:
W09D1
Young and Freedman: 1.10 (Vector
Products) 9.1-9.6, 10.5
2
Rigid Bodies
A rigid body is an extended object in which the
distance between any two points in the object is
constant in time.
Springs or human bodies are non-rigid bodies.
Demo
Center of Mass and Rotational
Motion of Baton
Overview: Rotation and Translation
of Rigid Body
Demonstration: Motion of a thrown baton
Translational motion: external force of gravity acts
on center of mass
Rotational Motion: object rotates about center of
mass
Recall: Translational Motion of
the Center of Mass
Total momentum of system of particles
r
r
psys  msys Vcm
External force and acceleration of center of
mass
r total
Fext 
r
dpsys
dt
 msys
r
dVcm
dt
r
 msys Acm
Main Idea: Rotational Motion
about Center of Mass
Torque produces angular acceleration about center of
mass

total
cm
 Icm cm
Icm is the moment of inertial about the center of mass
 cm is the angular acceleration about center of mass
Two-Dimensional Rotational
Motion
Fixed Axis Rotation:
Disc is rotating about axis passing
through the center of the disc and
is perpendicular to the plane of the
disc.
Plane of Motion is Fixed:
For straight line motion, bicycle
wheel rotates about fixed direction
and center of mass is translating
Cylindrical Coordinate
System
Coordinates
(r, , z)
Unit vectors
(rˆ , ˆ, zˆ )
Rotational Kinematics
for
Point-Like Particle
Rotational Kinematics
for Fixed Axis Rotation
A point like particle undergoing circular motion at
a non-constant speed has
(1) an angular velocity vector
(2) an angular acceleration vector
Fixed Axis Rotation: Angular
Velocity
Angle variable
SI unit:
Angular velocity
SI unit:
Vector:
Component
magnitude

[rad]
r
d φ
φ
  z k 
k
dt
 rad  s 1 
d
z 
dt
d
  z 
dt
 z  d / dt  0, direction  kφ
direction   d / dt  0, direction  kφ
z
Concept Question: Angular
Speed
Object A sits at the outer edge (rim) of a merry-go-round, and
object B sits halfway between the rim and the axis of rotation. The
merry-go-round makes a complete revolution once every thirty
seconds. The magnitude of the angular velocity of Object B is
1.
2.
3.
4.
half the magnitude of the angular velocity of Object A .
the same as the magnitude of the angular velocity of Object A .
twice the the magnitude of the angular velocity of Object A .
impossible to determine.
Example: Angular Velocity
r
v
 v φ
Consider point-like object rotating with velocity
tangent to the circle of radius r as shown in the figure
below with v  0
The angular velocity vector points in the  kφ direction,
given by
r
   z kφ  (v / r) kφ
z  0
Fixed Axis Rotation: Angular
Acceleration
2
d
 φ
φ
Angular acceleration:    z k  2 k
dt
SI unit
2
 rad  s 
r
Vector:
Component:
Magnitude:
Direction:
d 2 d z
z  2 
dt
dt
  z
d z

dt
d z
z 
 0, direction  kφ
dt
d z
z 
 0, direction  kφ
dt
Rotational Kinematics: Integral
Relations
The angular quantities  ,  z , and  z
x, vx , and ax
are exactly analogous to the quantities
for one-dimensional motion, and obey the same type of
integral relations
t
 z (t)   z,0    z (t  ) dt  ,
0
t
 (t)  0    z (t  ) dt  .
0
Example: Constant angular acceleration
 z (t)   z,0   z t
1
2
 (t)  0   z,0 t   z t 2

2
 z (t)2   z,0
 2 z ( (t)   0 )
Concept Question: Rotational
Kinematics
The figure shows a graph of  z and αz versus time for a
particular rotating body. During which time intervals is the
rotation slowing down?
1. 0 < t < 2 s
2. 2 s < t < 4 s
3. 4 s < t < 6 s
4. None of the intervals.
5. Two of the intervals.
6. Three of the intervals.
Table Problem: Rotational
Kinematics
A turntable is a uniform disc of mass m and a radius R. The
turntable is initially spinning clockwise when looked down on from
above at a constant frequency f . The motor is turned off and the
turntable slows to a stop in t seconds with constant angular
deceleration.
a) What is the direction and magnitude of the initial angular velocity of
the turntable?
b) What is the direction and magnitude of the angular acceleration of
the turntable?
c) What is the total angle in radians that the turntable spins while
slowing down?
Summary: Circular Motion for
Point-like Particle
Use plane polar coordinates: circle of radius r
Unit vectors are functions of time because
direction changes
r
r(t)  r rφ(t)
Position
Velocity
d φ
r
v(t)  r
 (t)  r  z φ(t)  vφ(t)
dt
Acceleration
r
a(t)  ar rφ(t)  at φ(t)
r
v  v(t)
r
a  a(t)  (ar2  at2 )1/ 2
at  r z , ar   v  z  r 2  (v 2 / r)
Rigid Body Kinematics
for Fixed Axis Rotation
Kinetic Energy and Moment of
Inertia
Rigid Body Kinematics
for Fixed Axis Rotation
Body rotates with angular velocity 
acceleration 
and angular
Divide Body into Small Elements
Body rotates with angular velocity,
r
   z kφ
r
   z kφ
angular acceleration
Individual elements of mass
mi
Radius of orbit
r ,i
Tangential velocity
v ,i  r,i z
Tangential acceleration
a ,i  r,i z
Radial Acceleration
ar ,i  
vi2
r ,i
 r ,i 2
Rotational Kinetic Energy and
Moment of Inertia
Rotational kinetic energy about axis
passing through S
1
1
2
 mi vi2  mi (r,i
) 2
2
2
K rot,i
Moment of Inertia about S :
SI Unit:
 kg  m 2 
Continuous body:
K cm 
1
I cmcm 2
2
i N
I S   mi (r,i )2
i1
i N
r,i  r,dm
mi  dm
Rotational Kinetic Energy:
IS 
 
i1

body
dm (r,dm )2
body
K rot   K rot,i
i
1
 2 1
2
 1
2
2
   mi r,i      dm (r,dm )    I S 2
2
 i 2

 2 body

 
Discussion: Moment of Inertia
How does moment of inertia compare to the total mass and the center
of mass?
Different measures of the distribution of the mass.
mtotal 
Total mass: scalar

dm
body
Center of Mass: vector (three components)
R cm 
1
m total

body
Moment of Inertia about axis passing through S: (nine possible
moments)
IS 

body
dm (r,dm )2
r dm
Concept Question
All of the objects below have the same mass. Which of the objects
has the largest moment of inertia about the axis shown?
(1) Hollow Cylinder
(2) Solid Cylinder
(3)Thin-walled Hollow Cylinder
Concept Question
Concept Question
Worked Example: Moment of
Inertia for Uniform Disc
Consider a thin uniform disc of radius R and mass m.
What is the moment of inertia about an axis that pass
perpendicular through the center of the disc?
Strategy: Calculating Moment of
Inertia
Step 1: Identify the axis of rotation
Step 2: Choose a coordinate system
Step 3: Identify the infinitesimal mass element dm.
Step 4: Identify the radius, r ,dm , of the circular orbit of
the infinitesimal mass element dm.
Step 5: Set up the limits for the integral over the body
in terms of the physical dimensions of the rigid body.
Step 6: Explicitly calculate the integrals.
Worked Example: Moment of
Inertia of a Disc
Consider a thin uniform disc of radius R and mass m.
What is the moment of inertia about an axis that pass
perpendicular through the center of the disc?
da  r dr d
r,dm  r
dm mtotal
M


da Area  R2
M
dm   r dr d 
r dr d
2
 2
R
r 3 d dr

M rR
Icm   (r,dm ) dm 
2 r 0  0

R
body
M r  R    2  3
M rR
2M
3

d r dr 
2 r dr  2
2 r 0   0
2 r 0

R
R
R
2
I cm
I cm
2M
 2
R

rR
r 0
4
2M r
r dr  2
R 4
rR
3
r 0

rR
r 0
2 M R4 1
 2
 MR 2
2
R 4
r 3dr
Parallel Axis Theorem
• Rigid body of mass m.
• Moment of inertia Icm about
axis through center of mass of
the body.
• Moment of inertia I S about
parallel axis through point S in
body.
• dS,cm perpendicular distance
between two parallel axes.
I S  I cm  md S2,cm
Table Problem: Moment of Inertia
of a Rod
Consider a thin uniform rod of length L and mass M.
Odd Tables : Calculate the moment of inertia about an axis
that passes perpendicular through the center of mass of
the rod.
Even Tables: Calculate the moment of inertia about an axis
that passes perpendicular through the end of the rod.
Summary: Moment of Inertia
Moment of Inertia about S:
i N
I S   mi (r,S ,i )2 
i1
body
Examples: Let S be the center of mass
• rod of length l and mass m
• disc of radius R and mass m
Parallel Axis theorem:
I cm 
Icm

1 2
ml
12
1
 mR2
2
I S  I cm  md S2,cm
r,S 2 dm
Table Problem: Kinetic
Energy of Disk
A disk with mass M and radius R is spinning
with angular speed  about an axis that
passes through the rim of the disk
perpendicular to its plane. The moment of
inertia about the cm is (1/2)M R2. What is the
kinetic energy of the disk?
Concept Question: Kinetic
Energy
A disk with mass M and radius R is spinning with
angular speed  about an axis that passes through
the rim of the disk perpendicular to its plane.
Moment of inertia about cm is (1/2)M R2. Its total
kinetic energy is:
1. (1/4)M R2  2
4. (1/4)M R 2
2. (1/2)M R2  2
5. (1/2)M R 2
3. (3/4)M R2  2
6. (1/4)M R
Summary: Fixed Axis Rotation
Kinematics

Angle variable
 z  d / dt
Angular velocity
Angular acceleration
 z  d z / dt  d 2 / dt 2
mi
Mass element
r ,i
Radius of orbit
i N
Moment of inertia
I S   mi (r,i )2 
i1
Parallel Axis Theorem

body
I S  Md 2  Icm
dm(r )2
Simple Pendulum
Simple Pendulum: bob of mass m hanging from end of massless
string of length I pivoted at S.
Angular velocity
d φ

k
dt
Angular acceleration
d 2 ˆ
 2 k
dt
Kinetic energy of rotation
K
r
1 2 2
ml 
2
Table Problem: Simple
Pendulum
A simple pendulum is released from rest at an
angle  0 . Find the angular speed at angle 
Table Problem Solution: Simple
Pendulum Energy Diagram
The energy diagrams for the pendulum are
shown in the figure below
Solution Simple Pendulum
d
v  l
dt
• Velocity
• Kinetic energy
1 2 1  d 
K f  mv  m  l 
2
2  dt 
• Initial energy
2
E0  K 0  U 0  mgl(1  cos0 )
2
1  d 
E f  K f  U f  m  l   mgl(1 cos )
2  dt 
• Final energy
• Conservation of energy
d

dt
2
1  d 
m  l   mgl(1 cos )  mgl(1 cos0 )
2  dt 

2
mg(1  cos )  g(1  cos 0 )
l

Table Problem: Pulley System
and Energy
Using energy techniques, calculate the speed of
block 2 as a function of distance that it moves down
the inclined plane using energy techniques. Let IP
denote the moment of inertia of the pulley about its
center of mass. Assume there are no energy losses
due to friction and that the rope does slip around the
pulley.
Next Reading Assignment:
W09D2
Young and Freedman: 1.10 (Vector
Product) 10.1-10.2, 10.5-10.6 ; 11.111.3