hp1f2013_class14_rigid_body_rotation
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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
MR
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 =
Flt
.
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
Flt
.
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