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Two-Dimensional Rotational
Dynamics
8.01
W09D2
Young and Freedman: 1.10 (Vector Product), 10.1-10.2,
10.4, 11.1-11.3;
Announcements
Next Reading Assignment
Young and Freedman: 1.10 (Vector Product) 10.110.2, 10.5-10.6 ; 11.1-11.3
Rigid Bodies
• Rigid body: An extended object in which the
distance between any two points in the object is
constant in time. Examples: sphere, disk …
• Effect of external forces (the solid arrows
represent forces):
Translate
Rotate
Both translate
and rotate
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 not slip around
the pulley.
Main Idea: Fixed Axis Rotation of
Rigid Body
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
Recall: Fixed Axis Rotation
Kinematics

Angle variable
r
   z kφ  (d / dt)kφ
Angular velocity
r
   zkφ  (d 2 / dt 2 )kφ
Angular acceleration
Mass element
mi
Radius of orbit
r ,i
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
Torque as a Vector
Force
FP
exerted at a point P on a rigid body.
Vector
rS ,P
from a point S to the point P.
Torque about point S due to the force exerted at point P:
 S  rS ,P  FP
Summary: Cross Product
Magnitude: equal to the area of the parallelogram defined by
the two vectors



r r
r r
r r
r
r
A  B  A B sin  A B sin  A sin B
Direction: determined by
the Right-Hand-Rule
(0     )
Torque: Magnitude and Direction
Magnitude of torque about a point S due to force
acting at point P
 S  rF  rF sin 
r
where F is the magnitude of the force FP .
Direction ofr torque:r Perpendicular to the plane
formed
.Determined by the
FPby rSand
,P
Right-Hand-rule.
Properties of Cross Products
A  B  B  A
c( A  B)  A  cB  cA  B
( A  B)  C  A  C  B  C
Cross Product of Unit Vectors
Unit vectors in Cartesian coordinates
ˆi  ˆj | ˆi || ˆj | sin  2   1
ˆi  ˆi | ˆi || ˆj | sin(0)  0
ˆi  ˆj  k
ˆ ˆi  ˆi  0
ˆj  k
ˆ  ˆi ˆj  ˆj  0
ˆ ˆi  ˆj k
ˆ k
ˆ 0
k
Components of Cross Product
ˆ , B  B ˆi  B ˆj  B k
ˆ
A  Axˆi  Ayˆj  Az k
x
y
z
A  B  ( Ay Bz  Az By )ˆi  ( Az Bx  Ax Bz )ˆj  ( Ax By  Ay Bx )kˆ
ˆi
 Ax
Bx
ˆj
Ay
By
kˆ
Az
Bz
Concept Question: Torque
r
r
φ
φ
Consider two vectors r  xi with x > 0 and F  F φ
i

F
k
z
r r x
with Fx > 0 and Fz > 0 . The cross product r  F
points in the
1)
2)
3)
4)
5)
6)
7)
+ x-direction
-x-direction
+y-direction
-y-direction
+z-direction
-z-direction
None of the above directions
Concept Question: Torque
Answer
Answer 4. We calculate the cross product noting that in a right handed
r
choice of unit vectors, φi  φi  0 and φi  kφ   φj
r
r
rP ,F  F  xφi  Fxφi  Fz kφ  xφi  Fxφi  xφi  Fz kφ




 xFz φj
Since x  0 and Fz  0 , the direction of the cross product is in the
negative y-direction.
Concept Question: Magnitude of
Torque
In the figure, a force of magnitude F is applied
to one end of a lever of length L. What is the
magnitude of the torque about the point S?
1.FL sin
2.FL cos
3.FL tan
4.None of the above
Concept Question: Magnitude of
Torque Answer
Answer 2
r
r
φ
 S  rS , F  F  (Lcos iφ Lsin  φ
j)  F( φ
j)  FLcos ( k)
r
r
 S  FLcos
Torque due to Uniform
Gravitational Force
The total torque on a rigid body due to the
gravitational force can be determined by placing
all the gravitational force at the center-of-mass
of the object.
N
N
N
i 1
i 1
i 1
S,grav   rS ,i  Fgrav,i   rS ,i  mi g   mi rS ,i  g
 1
  totat
m

totat
m
r

m
g

i S ,i 
i 1

N
 RS,cm  m totat g
Fixed Axis Rotational
Dynamics
Recall: Rotational Kinematics
• Individual element of mass
r ,i
• Radius of orbit
• Tangential velocity
vtan,i  r,i
• Tangential acceleration
• Radial Acceleration
mi
atan,i  r,i
arad,i 
2
vtan,i
r,i
 r,i 2
Dynamics: Newton’s Second Law
and Torque about S
Tangential force on mass element
produces torque
Newton’s Second Law
F  F ˆ  m a
tan,i
tan,i
i tan,i
ˆ
Ftan,i  mi r,i
Torque about S
 S ,i  r,i  Ftan,i
z-component of torque about S
( z,S )i  r ,i Ftan,i  mi (r,i )2 
Torque, Moment of Inertia and
Angular Acceleration
Component of the total torque about an
axis passing through S is the sum over
all elements
i N
i N
i1
i1
total
 z,S
 ( z,S )1  ( z,S )2      ( z,S )i   mi (r,i )2  z
Recall: Moment of Inertia about and
axis passing through S :
i N
I S   mi (r,i )2
i1
Summary:
total
 z,S
 I S z
Concept Question: Chrome
Inertial Wheel
A fixed torque is applied to
the shaft of the chrome
inertial wheel. If the four
weights on the arms are slid
out, the component of the
angular acceleration along
the shaft direction will
1)
increase.
2)
decrease.
3)
remain the same.
Concept Question: Chrome
Inertial Wheel Answer
Answer 2. Torque about the
central axis is proportional to
the component of the
angular acceleration along
that axis. The proportionality
constant is the moment of
inertia about that axis. By
pushing the weights out, the
moment of inertia has
increased. If the applied
torque is constant then the
component of the angular
acceleration must decrease.
Worked Example:
Moment of Inertia Wheel
An object of mass m is attached
to a string which is wound around
a disc of radius Rd. The object is
released and takes a time t to fall
a distance s. What is the moment
of inertia of the disc?
Analysis: Measuring Moment of
Inertia
Free body force diagrams and
force equations:
F  T  mp g  0
mg  T  ma
Rotational equation: RdT  I cm
Constraint: a  Rd
Solve for moment of inertia: I cm
g
 mRd (  1)
a
2
2
2s
2 gt
Time to travel distance s: a  2  I cm  mRd (  1)
2s
t
Demo: Moment of Inertia Wheel
Measuring the moment of inertia.
Radius of disc:
Rd  0.50 m
Mass of disc:
md  5.223 kg
Mass of weight holder: m  0.150 kg
Theoretical result: I cm
1
 md Rd 2
2
Problem Solving Strategy:
Two Dimensional Rotation
Step 1: Draw free body force diagrams for each object and indicate the
point of application of each force
Step 2: Select point to compute torque about (generally select center of
mass)
Step 3: Choose coordinate system. Indicate positive direction for increasing
rotational angle.
Step 4: Apply Newton’s Second Law and Torque Law to obtain equations
Step 5: Look for constraint condition between rotational acceleration and
any linear accelerations.
Step 6: Design algebraic strategy to find quantities of interest
Rotor Moment of Inertia
Table Problem: Moment of Inertia
Wheel
A steel washer is mounted on a cylindrical
rotor . The inner radius of the washer is R.
A massless string, with an object of mass m
attached to the other end, is wrapped
around the side of the rotor and passes
over a massless pulley. Assume that there
is a constant frictional torque about the axis
of the rotor. The object is released and
falls.We choose coordinates such that as
the mass falls, the rotor undergoes an
angular acceleration with a positive
component α1 > 0. After the string detaches
from the rotor, the rotor coasts to a stop
with a component of angular acceleration α2
< 0. Let g denote the gravitational constant.
What is the moment of inertia of the rotor
assembly (including the washer) about the
rotation axis?
Torque and Static Equilibrium
Conditions for Static Equilibrium
(1) Translational equilibrium: the sum of the forces
acting on the rigid body is zero.
Ftotal  F1  F2  ...  0
(2) Rotational Equilibrium: the vector sum of the
torques about any point S in a rigid body is zero.

total
S
  S ,1   S ,2  ...  0
Concept Question: Tipping
A box, with its center-ofmass off-center as
indicated by the dot, is
placed on an inclined
plane. In which of the four
orientations shown, if any,
does the box tip over?
Concept Question: Tipping
Answer
Answer 3. In order to tip
over, the box must pivot
about its bottom left corner.
Only in case 3 does the
torque about this pivot (due
to gravity) rotate the box in
such a way that it tips over.
Problem Solving Strategy: Static
Equilibrium
Force:
1.
Identify System and draw all forces and where they act on Free Body
Force Diagram
2.
Write down equations for static equilibrium of the forces: sum of forces
is zero
Torque:
1.
Choose point to analyze the torque about.
2.
Choose sign convention for torque
3.
Calculate torque about that point for each force. (Note sign of torque.)
4.
Write down equation corresponding to condition for static equilibrium:
sum of torques is zero
Table Problem: Standing on a
Hill
A person is standing on a hill that is sloped at an
angle α with respect to the horizontal. The person’s
legs are separated by a distance d, with one foot
uphill and one downhill. The center of mass of the
person is at a distance h above the ground,
perpendicular to the hillside, midway between the
person’s feet. Assume that the coefficient of static
friction between the person’s feet and the hill is
sufficiently large that the person will not slip.
a) What is the magnitude of the normal force on
each foot?
b) How far must the feet be apart so that the
normal force on the upper foot is just zero? (This is
the instant when the person starts to rotate and fall
over.)
Rotational Work
Tangential force
Ftan,i  Ftan,i ˆ
Displacement vector
rS ,i  r,i  ˆ
work for a small displacement
Wi  Ftan,i  rS ,i  Ftan,i ˆ  r,i  ˆ  r,i Ftan,i 
Rotational Work
Newton’s Second Law
Ftan,i  mi atan,i
Tangential acceleration
atan,i  r,i
Work for small displacement
Wi  mi r,i 2  
Summation becomes integration for continuous body



2
2
W    mi r,i       dm(rS , )     I S  
 i

 body

Rotational Work
Rotational work for a small displacement
W  I S  
Torque about S
 S  IS 
Infinitesimal rotational work
W   S 
  f
Integrate total work
W

  0
  f
dW 

 0
 S d
Rotational Work-Kinetic
Energy Theorem
Infinitesimal rotational work
d
d
dWrot  I S d  I S
d  I S d
 I S d 
dt
dt
Integrate rotational work
  f
Wrot 

  0
  f
dWrot 

  0
1
1
2
I S d   I S f  I S 02
2
2
Kinetic energy of rotation about S
1
1
2
Wrot  I S f  I S 02  K rot, f  K rot ,0  K rot
2
2
Rotational Power
Rotational power is the time rate of doing
rotational work
Prot 
dWrot
dt
Product of the applied torque with the
angular velocity
Prot 
dWrot
dt
d
 S
 S 
dt
Table Problem: Change in
Rotational Energy and Work
Suppose that a rotor of moment of
inertia Ir is slowing down during the
interval [t1, t2] according to
ω(t)= ω(t1)-α t,
where α = ω(t1)/(t2 ). Use work
energy techniques to find the
frictional torque acting on the rotor.
Table Problem Solution: Change
in Rotational Energy and Work
Suppose that a rotor of moment of
inertia Ir is slowing down during
the interval [t1, t2] according to
ω(t)= ω(t1)-α t,
where α = ω(t1)/(t2). Use work
energy techniques to find the
frictional torque acting on the rotor.
t2
t2
t1
t1
K  0  (1 / 2)I r 12     f  dt   f   dt
f 
(1 / 2)I r12
t2
  dt
t1

(1 / 2)I r12
t2
 (
t1
1
  t  ) dt 

(1 / 2)I r12
1 (t2  t1 )  (1 / 2) (t2  t1 )2