3-Dimensional Rotation:
Gyroscopes
8.01
W14D1
Today’s Reading Assignment
Young and Freedman: 10.7
Announcements
Problem Set 11 Due Thursday Dec 8 9 pm
Sunday Tutoring in 26-152 from 1-5 pm
W014D2 Reading Assignment Young and Freedman:
10.7
2
Demo: Gimbaled Gyroscope (B140)
Rules to Live By: Angular
Momentum and Torque
1) About any fixed point P
r
r
r
r
r
r
L P  L about cm  Lof cm  L about cm  rP,cm  mtotal v cm
r
r ext
 P    P,i 
i
r
dL P
dt
2) Independent of the CM motion, even if L about cm and 
r
are not parallel
r
 about cm 
dL about cm
dt
Mini Demo: Pivoted Falling Stick
Magnitude of the angular momentum about pivot
changes.
Direction of change of angular momentum about pivot is
the same as direction of angular momentum about pivot
Demo: Bicycle Wheel Two Cases
Case 1: Magnitude of the angular momentum about pivot
changes.
Direction of change of angular momentum about pivot is
the same as direction of angular momentum about pivot
Case 2: Direction of angular momentum about pivot
changes
Time Derivative of a Vector
with Constant Magnitude that
Changes Direction
Concept Question:
Time Derivative of Rotating Vector
r
A vector A of f ixed length A is rotating about the z-axis with the z-component
of
r
dA(t)
angular velocity  z   . At t  0 it is pointed in the positive φ
j -direction.
is
dt
given by
1)  Asin  t φ
i   Acos  t φ
j
2)  Asin  t φ
i   Acos  t φ
j
3)  Acos  t φ
i   Asin  t φ
j
4)  Acos  t φ
i   Asin  t φ
j
5)  Asin  t φ
i   Acos  t φ
j
6)  Acos  t φ
i   Asin  t φ
j
Example: Time Derivative of
Position Vector for Circular
Motion
Circular Motion: position vector
points radially outward, with constant
magnitude but changes in direction.
The velocity vector points in a
tangential direction to the circle.
r  r rˆ
dr
d ˆ
v
r

dt
dt
Generalization: Time Derivative
of a Vector
Consider a vector
r
r
where Ar  A sin   A 
A  Az kˆ  Ar rˆ
Az  A cos 
Vector can change both magnitude and
direction.
Suppose it only changes direction then
r
dA
d φ r
d φ r d φ
 Ar
  A sin 
  A

dt
dt
dt
dt
Torque and Time Derivative of
Angular Momentum
Torque about P is equal to the time derivative of the
angular momentum about P
r ext
P 
r
dL P
dt
If the magnitude of the angular momentum is constant then
the torque can cause the direction of the perpendicular
component of the angular momentum to change
r
r ext
d
 P  L P,
dt
Introduction To Gyroscopic
Motion
Gyroscopic Approximation
Flywheel is spinning with an
angular velocity
   r̂
Precessional angular velocity
   k̂
Gyroscopic approximation: the
angular velocity of precession
is much less than the component
of the spin angular velocity,
= 
Strategy
1. Calculate torque about appropriate point P
2. Calculate angular momentum about P
r
P
r
LP
3. Apply approximation that  =  to decide which
contribution to the angular momentum
about P is
r
changing in time. Calculate dL P / dt
r
4. Apply torque law  P 
r
dL P
dt
to determine direction and magnitude of angular
precessional velocity 
Table Problem: Gyroscope:
Forces and Torque
Gravitational force acts at the
center of the mass and points
downward. Pivot force acts
between the end of the axle and
the pylon. What is the torque about
the pivot point P due to
gravitational force
Table Problem Gyroscope: Time
Derivative of Angular Momentum
What is the time derivative of the angular momentum
about the pivot point for the gyroscope?
Torque and Time Derivative of
Angular Momentum
Torque about P is equal to the
time derivative of the angular
momentum about P
r ext
 SP 
r
dL P
dt
Therefore
φ I   φ
d mg 
1
Precession angular speed is
  d mg / I1
More Detailed Analysis of
Angular Momentum for
Gyroscopic Motion
Angular Momentum About Pivot Point
The total angular momentum about the pivot point P of a
horizontal gyroscope in steady state is the sum of the
rotational angular momentum and the angular momentum of
center of mass
r
r
r
r
L P  Lcm  rP,cm  MVcm
Angular Momentum about Center of Mass
.
The disk is rotating about two orthogonal axes through center of mass. It
is rotating about the axis of the shaft, with angular speed ω. The moment
of inertia of a uniform disk about this axis is I1 = (1/2) MR2. The disk is
also rotating about the z-axis with angular speed Ω. The moment of
inertia of a uniform disk about a diameter is I2 = (1/4)MR2. The angular
momentum about the center of mass is the sum of two contributions
r
Lcm  I1 rφ I2 kφ  (1/ 2)MR2 rφ (1/ 4)MR2 kφ
Angular Momentum Due to Motion of
Center of Mass
The angular momentum about the pivot point P due to the
center of mass motion is
r
r
rP,cm  MVcm  Md 2 kφ
where kφ is a unit vector in the positive z-direction and 
is the angular speed about the z-axis
Angular Momentum of Flywheel
about Pivot Point:
r
r
r
r
L P  Lcm  rP,cm  MVcm
r
2
φ
φ
L P  I1 r  I2 k  Md  kφ
r
2
φ
L P  I1 r  (I2  Md ) kφ
Gyroscope: Time Derivative of
Angular Momentum
If the angular speed (precession
angular speed) about the z-axis
is constant then only the direction
of the spin angular momentum
r
r
L P,  Lspin  I1 rφ
along the axis of the gyroscope is
changing in time hence
r
dL P
r
d φ
 L P,
  I1  φ
dt
dt
Torque and Time Derivative of
Angular Momentum
Torque about P is equal to the
time derivative of the angular
momentum about P
r ext
 SP 
r
dL P
dt
Therefore
φ I   φ
d mg 
1
Precession angular speed is
  d mg / I1
Concept Question: Gyroscope
For the simple gyroscope problem we just solved,
if the mass of the disk is doubled how will the new
precession rate  be related to the original rate 0?
1)  = 4 0
2)  = 2 0
3) = 0
4)  = (1/2) 0
5)  = (1/4) 0
Table Problem: Tilted Gyroscope
A wheel of mass M and moment of
inertia I cm about its centra l axis, through
its center of mass, is at one end of an
axle of l ength d . The a xle is pivoted at
an angle  with respect t o the vertical.
The wheel is set into motion so that it
executes uniform precession. Its spin
angular velocity has magnitude  and
is directed as sho wn in the figure below.
Find the magnitude and direction of the
precessional angular velocity  in
terms of the other parameters. Assume
r
   .
Demo: Gyroscope in a Suitcase
A gyroscope inside a suitcase is spun up via a connection to
the outside of the suitcase. The suitcase is carried across
the lecture hall. When the lecturer turns while walking, the
gyroscope causes the suitcase to rise about the handle.
Table Problem: Suspended
Gyroscope
A gyroscope wheel is at one end of
an axle of length d . The oth er end of
the axle is suspend ed from a string
of length s . The wheel is set into
motion so that it executes uniform
precession in the horizontal plane.
The string makes an angle  with
the vertical. The wheel has mass M
and moment of inertia about its
center of mass I cm . Its sp in angular
speed is  . Neglect the mass of the
shaft and the mass of the string.
Assume    . What is the
direction and magnitude of
precessional angular velocity?
the