Paradigms in Physics: Rotational Motion
solutions to
TRANSLATING TENSORS
Worksheet 3
by Philip J. Siemens
TOPICS
page
A. Checking previous exercises
2
1. Brick
2. Cage
B. Translating the CageX inertial tensor
4
1. From corner of cage to center of mass
2. From center of cage to center of mass
C. Numerical values
© 2003 Philip J. Siemens
9
version February 11, 2003
Oregon State University
Worksheet
Solutions to Translating Tensors
page 2
A. CHECKING PREVIOUS EXERCISES
We can use formula C.7 in the chapter on Rigid Bodies to compare the tensors we
calculated in the worksheet Inertial Integrals, in the center of mass and another
coordinate system. (This equation is also 11.49 in Marion and Thornton.)
The inertial tensor in the center-of-mass coordinate system is given by
I'ij = Iij – Tij
(C.7)
where Tij is the inertia of a point mass M at the center-of-mass's position R :
Tij = M (R 2δij – R i R j ) .
A.1. Brick
X 
 
Center of mass: Y 
Z 
=
(C.7a)
z
 12 A 
 
1B
2 
 
 12 C 
A
y
C
x
The translation tensor is
M
Tij = M (R 2δij – R i R j ) = 4
B2+C2

–AB

–AC
–AB
A 2+C2
–AC
B


–BC  .


A 2+B 2
–AC
The tensors for the brick, calculated in the Worksheet Inertial Integrals , are
origin at corner, Iij =
4(B2+C2)

M  –3AB
12 
 –3AC
–3AB
4(A 2+C2)
–3BC
center of mass, I'ij =


–3BC 


4(A 2+B 2)
–3AC
M
, 12
2 2 0
 B +C
A 2+C2
 0
 0
0
These 3 matrices actually work in eq. (C.7)!
© 2003 Philip J. Siemens
Paradigms in Physics: Rotational Motion
0

0
.
A 2+B 2
√
Oregon State University
Worksheet
Solutions to Translating Tensors
A.2. Cage
X 
 
Y 
 
Z 
Center of mass:
=
 12 A 
 
1B
2 
 
 12 C 
page 3
z
A
y
C
B
x
The translation tensor is the same as for the brick. The tensors calculated in the
Worksheet on Inertial Integrals are
origin at corner, Iij =
2C+8B 3(A+C)
12AB
+12ABC2+8C3(A+B)
σ –6AB(AB+BC+AC)
12 

–6AC(AB+BC+AC)
–6AB(AB+BC+AC)
12A 2BC+8A 3(B+C)
+12ABC 2+8C3(A+B)
–6BC(AB+BC+AC)


–6BC(AB+BC+AC) 


12A 2BC+8A 3(B+C) 
+12AB 2C+8B 3(A+C)
–6AC(AB+BC+AC)
center of mass, I'ij =
σ
6
2C
B+C3(A+C)+3AB
0 3(A+B)+3ABC2

0

The difference is
Iij – I'ij
M
= 4
© 2003 Philip J. Siemens
0


0

A 3(B+C)+3A 2BC 
+B 3(A+C)+3AB 2C
0
A 3(B+C)+3A 2BC
+C3(A+B)+3ABC 2
0
B2+C2

–AB

–AC
–AB
A 2+C2
–AC


–BC  .


A 2+B 2
–AC
Paradigms in Physics: Rotational Motion
check
comparison:
the difference is
equal to T i j
Oregon State University
Worksheet
Solutions to Translating Tensors
page 4
B. TRANSLATING THE CAGEX INERTIAL TENSOR
Use formula (C.7) in the chapter on Rigid Bodies to compare the tensors you
calculated in the worksheet Inertial Integrals, in the center of mass and another
coordinate system.
The inertial tensor in the center-of-mass coordinate system is given by
I'ij = Iij – Tij
(C.7)
where Tij is the inertia of a point mass M at the center-of-mass's position R :
Tij = M (R 2δij – R i R j ) .
(C.7a)
B.1. From corner of cage to center of mass
The CageX apparatus consists of a hollow rectangular
cage with two square faces and four narrow
rectangular faces.
z
L
Its length in the x and y directions is L, and the
length in the z direction is h .
The walls of the cage have a uniform surface mass
density, and its total mass is M .
In addition there is a clay sphere of mass m fastened
to the corner of the cage at the origin. The radius of
this sphere is r .
L
h
x
y
mass m
radius r
The expression for the center of mass of the cage-ball system in terms of M , m , L ,
h , and r is:
M
L L h
R = M+m  2 , 2 , 2 
The numerical values from
the Inertial Integrals Worksheet are
© 2003 Philip J. Siemens
X 
 
Y 
 
Z 
 ________ 


=  ________  .


________
Paradigms in Physics: Rotational Motion
Oregon State University
Worksheet
Solutions to Translating Tensors
page 5
B.1. From corner of cage to center of mass (continued)
The translation tensor for the cage-ball apparatus expressed in terms of M , m , L ,
and h is
M2 1
MT (R 2δij – R i R j ) = Tij = M+m 4
 L2+h2

 –L2

 –Lh
–L2
L2+h2
–Lh


–Lh 


2L2 
–Lh
The inertial tensor for the cage-ball apparatus in the original coordinate system from
section C of the Worksheet on Inertial Integrals expressed in terms of M , m , L ,
and h is
4+20L3h
–6L3(L+2h)
–6L2h(L+2h)
8L
2
2
3
+12L h +16Lh

M
8L4+20L3h
–6L2h(L+2h)
Iij = 24(L2+2Lh)–6L3(L+2h)


+12L2h2+16Lh3
 2

–6L h(L+2h) –6L2h(L+2h) 16L4+40L3h 
_________________________________
The center-of-mass inertial tensor calculated from these results is I'ij =
Mm 1
M+m 4
 L2+h2

 –L2

 –Lh
–L2
L2+h2
–Lh


–Lh 


2L2 
–Lh
+
M
12(L2+2Lh)
© 2003 Philip J. Siemens
L4+L33h
 +8Lh
0

0
Paradigms in Physics: Rotational Motion
0



0


2L4+8L3h 
0
L4+L3h
+8Lh3
0
Oregon State University
Worksheet
Solutions to Translating Tensors
page 6
B.2. From center of cage to center of mass
This time we work in a coordinate system with origin
at the center of the cage.
z
L
The expression for the center of mass of the cage-ball
system in terms of M , m , L , and h is:
L
y
h
m
L L h
R = – M+m  2 , 2 , 2  .
The numerical values from
the Inertial Integrals Worksheet are
x
X 
 
Y 
 
Z 
 ________ 


=  ________  .


________
The translation tensor
for the cage-ball apparatus
expressed in terms of
M, m , L , and h is
MT (R 2δij – R i R j ) =
© 2003 Philip J. Siemens
mass m
radius r
m2 1
Tij = M+m 4
 L2+h2

 –L2

 –Lh
Paradigms in Physics: Rotational Motion
–L2
L2+h2
–Lh


–Lh 


2L2 
–Lh
Oregon State University
Worksheet
Solutions to Translating Tensors
page 7
B.2. From center of cage to center of mass (continued)
The inertial tensor for the ball only in the cage-centered coordinate system of
section B of the Worksheet on Inertial Integrals may be approximately expressed in
terms of m , L , and h as
Iij =
 L2+h2

m  –L2
4 
 –Lh
–L2
L2+h2
–Lh


–Lh 


2L2 
–Lh
The inertial tensor ,for the cage only in the cage-centered coordinate system found
in section B of the Worksheet on Inertial Integrals, expressed in terms of M , m , L
, h , and r is
M
Iij = 12(L2+2Lh)
3h
L4+4L
+3L2h2+2Lh3
0

0



4
3
L +4L h
0

+3L2h2+2Lh3

0
2L4+8L3h
0
0
–––––––––––––––––––––––––––––––––
The inertial tensor for the cage plus ball in the cage-centered coordinate system of
section B of the Worksheet on Inertial Integrals, expressed in terms of m , L , and
h is therefor
 L2+h2 –L2 –Lh 


m  –L2
2
2
L +h –Lh  +
Itot = = Iball + Icage = 4 

 –Lh –Lh 2L2 
M
12(L2+2Lh)
© 2003 Philip J. Siemens
3h
L4+4L
2
2
+3L h +2Lh3
0

0


.
4
3
L +4L h
0

2
2
3
+3L h +2Lh

0
2L4+8L3h
0
Paradigms in Physics: Rotational Motion
0
Oregon State University
Worksheet
Solutions to Translating Tensors
page 8
B.2. From center of cage to center of mass (concluded)
The translation tensor for the cage-ball apparatus, expressed in terms of M , m , L ,
and h , found on page 6, is
m2 1
Tij = M+m 4
 L2+h2

 –L2

 –Lh
–L2
L2+h2
–Lh


–Lh 


2L2 
–Lh
The inertial tensor for the cage plus ball in the cage-centered coordinate system,
calculated on the previous page, expressed in terms of m , L , and h is
m
It = 4
 L2+h2

 –L2

 –Lh
–L2


–Lh 


2L2 
–Lh
L2+h2
–Lh
+
M
12(L2+2Lh)
3h
L4+4L
+3L2h2+2Lh3
0

0


.
4
3
L +4L h
0

+3L2h2+2Lh3

0
2L4+8L3h
0
0
–––––––––––––––––––––––––––––––––
The center-of-mass inertial tensor calculated from these results is I'ij =
Mm 1
M+m 4
 L2+h2

 –L2

 –Lh
–L2
L2+h2
–Lh


–Lh 


2L2 
–Lh
+
M
12(L2+2Lh)
L4+L33h
 +8Lh
0

0
Comparing this result to that of part B.1, we observe that
© 2003 Philip J. Siemens
0



0


2L4+8L3h 
0
L4+L3h
+8Lh3
0
they are the same .
Paradigms in Physics: Rotational Motion
Oregon State University
Worksheet
Solutions to Translating Tensors
page 9
C. Numerical values
The inertial tensor for the cage-ball
apparatus in the original coordinate
system, found in section C.3 of
Worksheet 3 Inertial Integrals, is Iij =
numerically expressed in units of____ .
The measured numerical values of the
center-of-mass coordinates, from page 5
of the CageX1 Workbook, are
X 
 
Y 
 
Z 
 ________ 


=  ________  .


________
The corresponding translation tensor for
the cage-ball apparatus, evaluated
numerically from the expression
Tij = M (R 2δij – R i R j )
isTij =
The center-of-mass inertial tensor
calculated from these results, is I'ij =
numerically expressed in units of ____.
© 2003 Philip J. Siemens
Paradigms in Physics: Rotational Motion
Oregon State University
Worksheet
Solutions to Translating Tensors
page 10
B.3. Numerical values continued
We can compare some of the elements of the inertial tensor to the measured results
from the CageX1 Workbook, page 10:
Cases
Observed
moment I'obs
(units
© 2003 Philip J. Siemens
Corresponding Calculated
element I'??
value above
)
(units
Paradigms in Physics: Rotational Motion
Comparison
)
Oregon State University
InertiaCalcs_cage.nb
1
CageX Inertia Tensor in Center-of-Mass frame (correction to Translating Tensors worksheet
solution, p. 5 and p. 8, to Rotating Tensors worksheet solution, p. 4, and to Principal Axes
worksheet solution, p. 6)
Written as one piece:
M H3 h L H2 m+ML+2 h H4 m+ML+L H4 m+ML+2 h L H5 m+2 MLL
i
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
j
12 H2 h+LL Hm+ML
j
j
j
j
2
j
L mM
j
- ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
j
4 Hm+ML
j
j
j
j
j
hLmM
- ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
4 Hm+ML
k
2
3
3
L mM
- ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
4 Hm+ML
2
2
M H3 h2 L H2 m+ML+2 h3 H4 m+ML+L3 H4 m+ML+2 h L2 H5 m+2 MLL
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
12 H2 h+LL Hm+ML ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
hLmM
- ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
4 Hm+ML
y
z
z
z
z
z
hLmM
z
z
- ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
z
4 Hm+ML
z
z
z
z
L2 M H10 h m+4 L m+4 h M+L ML z
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
Å
ÅÅÅÅÅ
6 H2 h+LL Hm+ML
{
hLmM
- ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
4 Hm+ML
Written as two pieces:
h2 + L2
-L2
-h L y
i
j
z
j
z
mM
j
z
2
2
2
j
z
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ j
+
-L
h
+
L
-h
L
z
j
z
z
4 Hm + ML j
j
z
-h L
2 L2 {
k -h L
2 h3 L + 3 h2 L2 + 4 h L3 + L4
i
j
j
M
j
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ j
j
0
2
j
12 H2 h L + L L j
j
0
k
0
3
2
2
3
2h L+3h L +4hL +L
0
y
z
z
z
z
0
z
z
z
3
4 z
8hL +2L {
0
4