Physics 318: Solutions to Homework #2
February 11, 2008
1
M
=
N
s
n
=1
m
n
R
(
t
) =
1
M
N
s
n
=1
m
n
r
n
(
t
)
r
′
n
(
t
) =
r
n
(
t
)
−
R
(
t
)
(1)
a.)
Simplifying:
T
=
1
2
M
˙
R
2
+
1
2
N
s
i
=1
m
i
˙r
′
2
i
(2)
=
1
2
M
˙
R
2
+
1
2
N
s
i
=1
m
i
(
˙r
2
i
−
2
˙r
i
·
˙
R
+
˙
R
2
)
(3)
=
1
2
M
˙
R
2
+
N
s
i
=1
1
2
m
i
˙r
2
i
−
1
M
N
s
i,j
=1
m
i
m
j
˙r
i
·
˙r
j
+
1
2
M
˙
R
2
(4)
=
1
2
M
˙
R
2
+
N
s
i
=1
1
2
m
i
˙r
2
i
−
M
˙
R
2
+
1
2
M
˙
R
2
(5)
=
N
s
i
=1
1
2
m
i
˙r
2
i
(6)
b.)
Simplifying:
L
=
M
R
×
˙
R
+
N
s
i
=1
m
i
r
′
i
×
˙r
′
i
(7)
=
M
R
×
˙
R
+
N
s
i
=1
m
i
(
r
i
×
˙r
i
−
R
×
˙r
i
−
r
i
×
˙
R
+
R
×
˙
R
)
(8)
=
M
R
×
˙
R
+
N
s
i
=1
m
i
r
i
×
˙r
i
−
1
M
N
s
i,j
=1
(
m
i
m
j
r
j
×
˙r
i
+
m
i
m
j
r
i
×
˙r
j
) +
M
R
×
˙
R
(9)
=
M
R
×
˙
R
+
N
s
i
=1
m
i
r
i
×
˙r
i
−
M
R
×
˙
R
−
M
R
×
˙
R
+
M
R
×
˙
R
(10)
=
N
s
i
=1
m
i
r
i
×
˙r
i
(11)
1
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View Full Documentc.)
The strong form of Newton’s third law can be written
F
nm
=
−
F
mn
=
f
nm
(

r
n
−
r
m

)(
r
n
−
r
m
)
(12)
Let’s see if the Frst term of Eqn.(2) is conserved (time independent).
d
dt
(
1
2
M
˙
R
2
) =
M
˙
R
·
¨
R
(13)
With
¨
R
=
1
M
N
s
i
=1
m
i
¨
r
i
(14)
=
1
M
N
s
i
=1
N
s
j
=1
F
ij
=
−
1
M
N
s
i
=1
N
s
j
=1
F
ji
=
−
1
M
N
s
j
=1
N
s
i
=1
F
ji
(15)
⇒
¨
R
= 0
(16)
⇒
d
dt
(
1
2
M
˙
R
2
) = 0
(17)
We have used the fact that a (symmetric) sum over antisymmetric indices must be zero.
Now let’s check the Frst term in Eqn.(7).
d
dt
(
M
R
×
˙
R
) =
M
˙
R
×
˙
R
+
M
R
×
¨
R
(18)
=
M
R
×
(
1
M
N
s
n
=1
m
n
¨
r
n
)
(19)
=
R
×
N
s
n
=1
N
s
m
=1
F
nm
(20)
= 0
(21)
(Again by symmetry).
We will now show that the second term is conserved by showing that
L
is conserved and using
the previous result.
d
dt
L
=
d
dt
(
N
s
n
=1
m
n
r
n
×
˙r
n
)
(22)
=
N
s
n
=1
m
n
˙r
n
×
˙r
n
+
N
s
n
=1
m
n
r
n
×
¨
r
n
(23)
=
N
s
n
=1
r
n
×
N
s
m
=1
F
nm
(24)
=
N
s
n
=1
r
n
×
N
s
m
=1
f
nm
(
r
n
−
r
m
)
(25)
=
−
N
s
n
=1
N
s
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 Spring '08
 FLANAGAN
 mechanics, Force, Momentum, Work, Trigraph, fnm

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