Physics 105B Problem Set 4
April 29, 2008
Jef Schonert:
schonert (at) physics.ucsb.edu
Taylor 10.19
a)
We will work out the
x
component oF the cross product; the other components Follow
identically. Given two vectors
±ω
and
±
r
, their cross product in component notation is
×
±
r
=(
ω
y
z

ω
x
y, ω
z
x

ω
x
z, ω
x
y

ω
y
x
)
(1)
Now consider the
x
component oF the cross product
±
r
×
(
×
±
r
):
(
±
r
×
(
×
±
r
))
x
=
y
(
ω
x
y

ω
y
x
)

z
(
ω
z
x

ω
x
z
)=(
y
2
+
z
2
)
ω
x

xyω
y

xzω
z
(2)
This is the
x
component oF equation 10.35; the calculation oF the
y
and
z
components is now
straightForward.
b)
The
BAC

CAB
rule For the cross product in this problem is
±
r
×
(
×
±
r
)=
(
±
r
·
±
r
)

±
r
(
·
±
r
).
Then the
x
component is
(
±
r
×
(
×
±
r
))
x
=
ω
x
(
x
2
+
y
2
+
z
2
)

x
(
ω
x
x
+
ω
y
y
+
ω
z
z
y
2
+
z
2
)
ω
x

xyω
y

xzω
z
(3)
which is the same as in a). Again, the calculation oF the
y
and
z
components is identical.
Taylor 10.24
a)
Let the center oF mass be located at
±
r
0
x
0
,y
0
,z
0
). We are told that the point
P
is
displaced by
Δ
ξ, η, ζ
) From this point. Now consider the displacement oF the mass From
both oF these points. IF
q
q
1
,q
2
3
) is the displacement oF the mass
M
From
P
, and
q
±
q
±
1
±
2
±
3
) is the displacement oF
M
From the center oF mass, then the two displacement
vectors are related by
q
±

q
=
Δ
. This allows us to work out how the components oF
q
and
q
±
are related to each other:
q
1
=
q
±
1

ξ
q
2
=
q
±
2

η
q
3
=
q
±
3

ζ
So now we have written the displacement oF the mass From
P
in terms oF the known quantities
(since we are assuming we already know the moment oF inertia about the center oF mass).
Now plug these into the de±nitions oF the inertia tensor components:
I
xx
=
±
i
m
i
(
q
±
2
2

η
)
2
+(
q
±
3
2

ζ
)
2
=
±
i
m
i
(
q
±
2
2
+
q
±
2
3
+
η
2
+
ζ
2

2
ηq
±
2

2
ζq
±
3
)
=
±
i
m
i
(
q
±
2
2
+
q
±
2
3
)+
±
i
m
i
(
η
2
+
ζ
2
)
(4)
1
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View Full DocumentThe sums involving
ηq
±
2
and
ζq
±
3
were dropped because they are zero according to equation
(10.7). Therefore,
I
xx
is related to the moment of inertia about the center of mass
I
cm
xx
by
I
xx
=
I
cm
xx
+
M
(
η
2
+
ζ
2
)
(5)
Similarly,
I
yy
=
±
i
m
i
(
(
q
±
1

ξ
)
2
+(
q
±
3

ζ
)
2
)
(6)
=
±
i
m
i
(
q
±
2
1

2
q
±
1
ξ
+
ξ
2
+
q
±
2
3

2
q
±
3
ζ
+
ζ
2
)
(7)
=
±
i
m
i
(
(
q
±
2
1
+
q
±
2
3
)+
ξ
2
+
ζ
2
)
(8)
=
I
cm
+
M
(
ξ
2
+
ζ
2
)
(9)
The calculation of the other diagonal component follows identically:
I
zz
=
I
cm
+
M
(
η
2
+
ξ
2
)
(10)
Now turn to the products of inertia, which are calculated in the same way. For example,
I
xz
=

±
i
m
i
q
1
q
3
(11)
=

±
i
m
i
(
q
±
1
q
±
3

q
±
1
ζ

q
±
3
ξ
+
ξζ
)
(12)
=
I
xz

Mξζ,
(13)
where again we used (10.7) to drop the
q
±
1
ζ
and
q
±
3
ξ
terms. Since the inertia tensor is
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 Spring '08
 MARTINIS
 mechanics, Angular Momentum, Work, Moment Of Inertia, Ixx, inertia tensor

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