lect 16 mechanics for bsc physics course.pdff

lect 16 mechanics

1. 11/3/2023
1
Sect 5.4: Eigenvalues of I & Principal
Axis Transformation
•Definition of inertia tensor (continuous body):I
jk
∫
V
ρ(r)[r
2
δ
jk
-x
j
x
k
]dV
–Clearly, I
jk
is symmetric: I
jk
= I
kj
Out of the 9 elements I
jk
only 6 are independent.
I
jk
depend on the location of the origin of the body axes set &
on the orientation of the body axes with respect to the body.
•Symmetry There exists a set of coordinates Principal Axesin which the tensor I
jk
is diagonalwith
3 Principal ValuesI
1
, I
2
, I
3
.
In this system, the angular
momentum: L =Iω becomes:
L
1
= I
1
ω
1
, L
2
= I
2
ω
2
, L
3
= I
3
ω
3
.
the KE T = (½)ωIω becomes:
T = (½)I
1
(ω
1
)
2
+ (½)I
2
(ω
2
)
2
+ (½)I
3
(ω
3
)
2
•Get principal axes set andprincipal values of tensor I by
diagonalizing I. That is, by findingeigenvalues(principal
valuesI
1
, I
2
, I
3
) & eigenvectors (definingprincipal axes).
•From Ch. 4, do this by solving determinant eigenvalue
problem or by a similarity transformation on I.
•Given inertia matrix I,principal axes & principal values can be
found byfindinga suitable rotation matrix
(findinga proper set of
Euler angles
,θ,ψ
)
Aand performing the similarity
transformation:
I
D
= AIA
-1
= AIÃ
such thatI
D
is a
diagonal matrix.That is, want I
D
to have form:
I
1
0 0 I
1
, I
2
, I
3
eigenvalues of I
I
D
= 0 I
2
0I
1
, I
2
, I
3
principal Components ofI
0 0 I
3
Directions x, y, zdefined by the eigenvectors 
principal axes ofI
•Once Iis diagonalized, Principal Components (I
1
,
I
2
, I
3
)&principal Axes (x, y, z)are known.
–Then, can get I relative to any other axes set by another
similarity transformation: I = AI(A)
-1
= AIÃ
–Also Parallel Axis Theorem might be used to shift rotation
axis.
•The matrix algebra method to diagonalize I
Solve secular eqtn: (I -I1)R = 0 (1)
The Eigenvalue Problem
Values of I which satisfy (1) Eigenvalues (I
1
, I
2
, I
3
)
Vectors R which satisfy (1) Eigenvectorsx, y, z
(I -I1)R = 0 (1)
I
xx
-II
xy
I
xz
I
xy
I
yy
-II
yz
= 0 (2)
I
xz
I
yz
I
zz
-I
–Have used I
jk
= I
kj
.
•3 solutions to (2):Eigenvalues(I
1
, I
2
, I
3
)
•Put these into (1)& get: Eigenvectors
R
i
= x, y, z= Principal Axes.
•Often, can know principal axes by the object symmetry.
•Some Properties of the Eigenvalues(I
1
, I
2
, I
3
)
1.Can’t be < 0! I
i
> 0, (i = 1,2,3)
2.If one I
i
= 0, the body is vanishingly small in the
direction given by the corresponding eigenvector.

2. 11/3/2023
2
•Principal axes from geometry:
•Moment of inertia about the rotation axis n: I nIn
•Body Cartesian axes unit vectors: i,j,kDefine:n αi +βj + γk
Work out the details of I:
I = I
xx
α
2
+I
yy
β
2
+I
zz
γ
2
+2I
xy
αβ+ 2I
yz
βγ+ 2I
xz
αγ
Definethe vector ρn/(I)
½
. |ρ| = (I)
-½
Write: ρρ
1
i+ ρ
2
j+ ρ
3
k
 I = I
xx
(ρ
1
)
2
+ I
yy
(ρ
2
)
2
+ I
zz
(ρ
3
)
2
+ 2I
xy
ρ
1
ρ
2
+ 2I
yz
ρ
2
ρ
3
+ 2I
xz
ρ
1
ρ
3
(A)
(A)  I = I(ρ
1
,ρ
2
,ρ
3
)
Equation of a surface in “ρ” space Inertial Ellipsoid
ρ= n(I)
-½
= ρ
1
i+ ρ
2
j+ ρ
3
k
I = I
xx
(ρ
1
)
2
+I
yy
(ρ
2
)
2
+I
zz
(ρ
3
)
2
+ 2I
xy
ρ
1
ρ
2
+ 2I
yz
ρ
2
ρ
3
+ 2I
xz
ρ
1
ρ
3
(A)
Equation of a surface in “ρ” space Inertial Ellipsoid
•Diagonalizing the inertia tensor I Transforms to principal
axes where the moment of inertia Ihas the form:
I = I
1
(ρ
1
)
2
+ I
2
(ρ
2
)
2
+I
3
(ρ
3
)
2
(B)
The normal form for an ellipsoid!
A geometric interpretation of the principal moments
of inertia (I
1
,I
2
,I
3
): They are exactlythe lengths of the
axes of the inertial ellipsoid. If 2 I
j
’s are equal, this
ellipsoid has 2 equal axes & this is an ellipsoid of
revolution. If all 3 are equal, this is a sphere!
•Define: The Radius of GyrationR
0
in terms of the
total mass M& the moment of inertia I: I M(R
0
)
2
Iis written as ifall mass Mwere a distance R
0
from the rotation axis.
•From the definition of the vector ρ= n(I)
-½
, we can write:
ρ= n(R
0
)
-1
(M)
-½
A radius vector to a point on the inertia ellipsoid is
inversely proportional to the radius of gyration about
the direction of that same vector.
Emphasize:
•The inertia tensor I& all associated quantities
(principal axes, principal moments, inertia ellipsoid, moment
of inertia, radius of gyration, etc.)
are defined relative to
some fixed point in the body.
•If we shift this point to somewhere else, all of thesein general
are changed. The parallel axis theorem can be used.
•The principal axis transformation which diagonalizes
Iabout an axis through CM will not necessarily
diagonalize it about another axis!
It will be diagonal with
respect to both axes only if the shift is along a vector parallel
to one of the original principal axes & only if that axis passes
through the CM.

3. 11/3/2023
3
Example 1 from Marion
•Calculate the inertia tensor of a homogeneous cube of density
ρ, mass M, and side length b. Let one corner be at the origin,
and let the 3 adjacent edges lie along the coordinate axes
(see
figure).
(For this choice of coord axes, it should be obvious that
the origin does NOTlie at the CM!)
I
jk
∫
V
ρ(r)[r
2
δ
jk
-x
j
x
k
]dV

βMb
2
I =
Symmetry:
Example 2 from Marion
Find the principal moments of inertia &
the principal axes for the same cube:
Solve secular eqtn: (I -I1)R = 0

So:
Row manipulation:
•This results in:
•Or:
•Giving:
DiagonalI =
To get the principal axes, substitute I
1
, I
2
, I
3
into the
secular equation (I -I
j
1)R
j
= 0
(j = 1,2,3)
& solve for R
j
.
Find: R
1
along cube diagonal. R
2
, R
3
each other & R
1
.
Example 3 from Marion
•Calculate the inertia tensor of the
same cube in a coord system with
origin at the CM.
(figure).
a = (½)b(1,1,1)
Again: I
jk
∫
V
ρ(r)[r
2
δ
jk
-x
j
x
k
]dV
Student exercise to show:
I
11
= I
22
= I
33
= (
1/6
)Mb
2
, I
12
= I
21
= I
13
= I
31
= I
23
= I
32
= 0
1 0 0
I
D
= =
(
1/6
)
Mb
2
0 1 0
0 0 1
Or: I
D
=
(
1/6
)
Mb
2
1

4. 11/3/2023
4
Sect 5.5: Solving Rigid Body Problems;
The Euler Equations
•We now have the tools to solve rigid body dynamics
problems. Usually assume holonomic constraints. If
not
(like rolling friction)
need to use special methods.
•Usually start by seeking a reference point in the body
such that the problem can be split into pure
translational + pure rotational parts.
•If one point in the body is fixed, then obviously all
that is needed is to treat the dynamics of the rotation
about that point.
•If one point is not fixed, it is most useful to choose
the reference point in the body to be the CM.
•We have already seen that, for case where the reference
point is the CM, the KE splits into KE of translation of
CM + KE of rotation about an axis through the CM:
•We had: T = T
trans
+ T
rot
–1
st
term = T
trans
= (½)Mv
2
Translational KE of the CM
–2
nd
term = T
rot
= (½)∑
i
m
i
(v
i
)
2
. Rotational KE about CM
–We’ve written second term as: T
rot
= (½)ωIω
–For n unit vector along the rotation axis:
T
rot
= (½)ω
2
nIn (½) Iω
2
•So:
T = (½)Mv
2
+ (½)Iω
2
•For all problems considered here, we can make a similar
division for the PE:
V = V
trans
+ V
rot
Lagrangian similarly divides: L= L
trans
+ L
rot
•More generally (Goldstein notation):L= L
trans
+ L
rot
= L
c
(q
c
,q
c
) + L
b
(q
b
,q
b
)
L
c
= CM Lagrangian, q
c
,q
c
= generalized coordinates &
velocities of the CM.
L
b
= Body Lagrangian, q
b
,q
b
= generalized coordinates &
velocities of body (rotation).
•Can use either Newtonian or Lagrangian methods, of course.
In either case, often convenient to work in the principal axes
systemso that the rotational KE takes the simple form:
T
rot
= (½)I
1
(ω
1
)
2
+ (½)I
2
(ω
2
)
2
+ (½)I
3
(ω
3
)
2
•The most convenient generalized coordinates to use are the
Euler angles: ,θ,ψ. They arecumbersome, but useable.
•If the motion is confined to 2 dimensions (the rotation axis is
fixed in direction), then only 1 angle describes motion!
•Follow Goldstein & start with the Newtonian approach to
describe the rotational motion about an axis through a fixed
point (like the CM):
•Considereitheran inertial frame with the origin at a fixed
point in bodyor a space axes system with CM as the origin.
•In this case, a Ch. 1 result is:
(dL/dt)
s
= N (1)
Newton’s 2
nd
Law (rotational motion):Time derivative of
the total angular momentum L
(taken with respect to the space
axes)
is equal to the total external torque N.
•Make use of the Ch. 4 result relating the time derivatives in the
space & body axes (angular velocity ω):
(d/dt)
s
= (d/dt)
b
+ ω (2)
 (dL/dt)
s
= (dL/dt)
b
+ ωL (3)
•Equating (1)& (3)& dropping the “body” subscript (which is
understood in what follows) gives:

5. 11/3/2023
5
(dL/dt) + ωL = N (4)
•(4):Newtonian eqtn of motion relative to body axes.
•In terms of inertia tensor, I, in general, L =Iωneeds to be
substituted into (4).
•For i
th
component, (4)becomes:
(dL
i
/dt) + 
ijk
ω
j
L
k
= N
i
(4)
–Levi-Civita density 
ijk
0, any 2 indices equal

ijk
1, if i, j, keven permutation of 1,2,3

ijk
-1, if i, j, kodd permutation of 1,2,3

122
=

313
=

211
= 0, etc.

123
=

231
=

312
=1,

132
=

213
=

321
=-1
•Take the body axes to be the principal axes ofI,
(relative to
the reference point)
L
i
= I
i
ω
i
(i = 1,2,3)
•(4)becomes (no sum on i)
I
i
(dω
i
/dt) + 
ijk
ω
j
ω
k
I
k
= N
i
(5)
•
Newtonian eqtns of motionrelative to the body axes are:
I
i
(dω
i
/dt) + 
ijk
ω
j
ω
k
I
k
= N
i
(5)
•Write component by component:
I
1
(dω
1
/dt) -ω
2
ω
3
(I
2
-I
3
) = N
1
(5a)
I
2
(dω
2
/dt) -ω
3
ω
1
(I
3
-I
1
) = N
2
(5b)
I
3
(dω
3
/dt) -ω
1
ω
2
(I
1
-I
2
) = N
3
(5c)
Euler’s Equations of Motion for a Rigid
Body with one point fixed.
•Could derive these from Lagrange’s Eqtns, treating torques as
generalized forces corresponding to Euler angles as generalized
coords.
•Special Case: I
1
= I
2
I
3
: In this case, a torque in the “1-2”
plane (N
3
= 0) causes a change in ω
1
& ω
2
while leaving ω
3
=
constant. A very important special case! (Sect. 5.7)