Honors Physics 1
Class 14 Fall 2013
The rotating skew rod
Rotational Inertia Tensor
Stability of spinning objects
The spinning top in gravity
1
The angular momentum vector and angular velocity do not
necessarily point in the same direction
Consider a rigid body consisting of two particles
of equal mass on the ends of a massless rigid rod
of length 2l . The midpoint of the rod is attached
to a vertical axis which rotates at angular speed
. The rod is skewed at an angle  from the axis.
Find the angular momentum of the system.
L 
 ri  p i
i
E ach m ass m oves in a circle of radius l co s  w ith angular speed  .
p i  m  l cos 
T aking the m idpoint of the rod as origin , r  l .
2
L  2 m  l cos  .
L is perpendicular to the rod and lies in the plan e of rod and the z axis.
L turns w ith the rod and traces a circle about the z axis.
2
In the previous exam ple, w e point out th at L is not parallel to  .
T his is generally the case for non-sym m e tric bodies.
T he fact that L rotates m eans that there m ust be a torque on it.
T he com ponent o f L , L z , parallel to the z axis is constant .
T he horizontal com ponent L h = L sin  rotates w ith the rod.
C hoosing a starting phase,
L x  L h cos  t  L sin  cos  t
L y  L sin  sin  t


L  L sin  cos  tiˆ  sin  tjˆ  L cos  kˆ
and the torque is  =
dL
dt

 L  sin   sin  tiˆ  cos  tjˆ

so   L  sin 
T he larger the angular m om entum , the m or e torque to rotate.
3
Geometric interpretation of precession
of the skewed rod
Lh (t   t )
 Lh
Lh (t )
 Lh  Lh  
dL h
dt
 Lh
d
dt
 Lh
so    L sin  and points along the tangent in the xy plane.
R em em ber that   r  F , so F points at right ang les to the
change in L.
4
Moment of Inertia Tensor
B ecause w e have found a case w here L is not parallel to 
L  I  should be m odified.
T he proper expression is L  I  w here I is a 3x 3 m atrix.
 I xx

I   I yx

I
 zx
I xx 

I xy
I yy
I zy
I xz 

I yz 

I zz 

2
x  ( x , y , z ) d xdydz
body
I xy  

xy  ( x , y , z ) dxdydz  I yx   product of inertia 
body
L x  I xx  x  I xy  y  I xz  z
N ote: It is alw ays possible to find a se t of three orthogonal axes
about w hich the products of inertia are zero.
5
The spinning bicycle wheel
(Gyroscope)
T he angular m om entum is along the axis o f the spinning w heel.
(horizontal if w e did it right)
T he torque    r  m g  about the support point is d ue to the w eight
and is at right angles to L and g.
S ince  is at right angles to L, the m agnitude of L does not change.
A ssum ing that all of the m ass of the w he el is at a distance R
from the axis and a distance D from the support point.
2
L= M  R ;   D W  L  ;  
DW
MR
2
6
Moment of inertia of various objects
T he m om ent of a bicycle w heel is easy. T he m ass is
all at a distance of the radius:
I  MR
2
A disk requires a little m ore w ork:
I 
  rr
2
drd   2 
r
3
dr 
2  R
4
4

MR
2
2
A rectangular plate is solved using the para llel axis theorem
for one of the dim ensional integrals and I cm 
I p la te 
1
12
M

2
2
Lx  L y
1
12
M Lx
2
for a rod.

7
Stability of spinning objects
F
Applications: Rolling hula hoops,
flying saucers, footballs, rifle
bullets...
C onsider a cylinder m oving parallel to its axis and w hat happens if w e
exert a sm all perturbing force at right angles to the cylinder axis for a
short tim e  t.
In the case w here the cylinder is not in iti ally spinning
  F l so  L A    t  F l  t so  =
Flt
.
IA
If the cylinder is rapidly spinning w ith angular m om entum L s .
T orque causes precession w hile the torqu e is applied
=
Fl
so the axis rotates by  =   t=
Ls
Flt
.
Ls
Instead of tum bling, the cylinder change s orientation slightly and
then stops precessing.
N ote that spin has no effect on center o f m ass m otion.
8