PHYS 230
Fall 2007
Assignment 9
1. French 1.24
A uniform rod of mass
M
and length 2
L
spins with angular velocity
ω
c/L
about
its centre. In Special Relativity its angular momentum and kinetic energy are given by
=
ML
2
ω
3
ˆ
1 +
A
ω
2
L
2
c
2
+
· · ·
!
K
=
ML
2
ω
2
6
ˆ
1 +
B
ω
2
L
2
c
2
+
· · ·
!
What are the values of
A
and
B
?
You will need: (1

α
)

1
2
1 +
1
2
α
+
3
8
α
2
+
· · ·
, where
α
1.
We need the SR definitions of momentum and energy, namely
p
=
γm
0
v
;
and
E
=
γm
0
c
2
as well as the preEinstein definition of angular momentum, namely
=
mrv
=
rp
(where we have assumed that r and v are mutually perpendicular).
Since the answers are expressed as series expansions in powers of
v
2
/c
2
, we will expand
γ
immediately to give
γ
1 +
1
2
v
2
/c
2
+
3
8
v
4
/c
4
+
· · ·
Then, the angular momentum of the stick is
total
Z
L

L
[1 +
1
2
v
2
/c
2
]
rvμdr
where
μ
is the restmass per unit length,
μ
=
M
0
/
2
L
. The velocity
v
is simply
v
=
ωr
.
Thus
total
[
ωM
0
/
2
L
]
Z
L

L
r
2
dr
+ [
ω
3
M
0
/
4
Lc
2
]
Z
L

L
r
4
dr
or
total
1
3
M
0
L
2
ω
+
1
10
M
0
L
4
ω
3
/c
2
so that
A
= 3
/
10
.
Notice that the first term is just the “preEinstein”, nonrelativistic value.
Next: the energy. Here the first term will be the restmass energy
M
0
c
2
. So, to get the
two required terms, we will need to keep the next term in the expansion of
γ
.
1
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Thus:
E
Z
L

L
[1 +
1
2
v
2
/c
2
+
3
8
v
4
/c
4
]
c
2
μdr
E
M
0
c
2
+
1
2
(
M
0
/L
)
ω
2
Z
L

L
r
2
dr
+
3
8
(
M
0
/L
)
ω
4
Z
L

L
r
4
dr/c
2
E
M
0
c
2
+
1
6
M
0
L
2
ω
2
+
3
40
M
0
L
4
ω
4
/c
2
so that
B
= 9
/
20
.
This time, notice that the
second
term is the nonrelativistic one.
2. Do all of these calculations using
only
the parallel axis theorem, and no other mathe
matics. Especially not any calculus.
The parallel axis theorem applies to the moment of inertia of a twodimensional object
rotating about an axis perpendicular to its plane. It states that: the moment of inertia
for rotation about a given axis is
I
=
I
C
+
MD
2
where
M
is the mass,
I
C
is the moment of inertia for rotation about the centre of mass,
and
D
is the distance which separates the given axis from the axis through the centre
of mass. (By construction, these two axes are parallel to each other.)
Figure out the moments of inertia, for axes perpendicular to their planes and through
their respective centres of mass, of the following “simple” objects.
(a) A uniform stick of mass
M
, length 2
L
.
Hint: a stick of length
2
L
is two sticks, each of length
L
, placed endtoend. What
is the ratio of the moments of inertia for two sticks, one twice the length of the
other?
Think of this problem as follows: there are two unknowns – the moment of inertia
of the stick itself, about its own CofM, and the moments of inertia – around their
own CofM’s – of the two halfsize sticks from which it is constructed. Call these
I
big
and
I
small
.
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 Fall '07
 Harris
 Angular Momentum, Energy, Kinetic Energy, Mass, Momentum, Special Relativity, Moment Of Inertia, Centre of mass

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