example:
A chain of uniform mass density
ρ
, length
b
, and mass
M
(where
ρ
=
M/b
)
hangs from both ends.
At time
t
= 0, the ends are adjacent, but one is released.
Find
the tension in the chain at the fixed point, after the other has fallen a distance
x
. Assume
freefall.
solution
: Assume freefall: that is, the only forces acting on the system at time
t
are
the tension (vertically upward at the fixed end) and the gravitational force
Mg
pulling the
chain down. The center of mass momentum reacts to these forces such that
˙
P
=
Mg

T
(20)
The free side of the chain, with mass
ρ
(
b

x
)
/
2, moves at the speed ˙
x
, and the other side
is not moving. The total momentum of the system is therefore
P
=
ρ
b

x
2
˙
x
(21)
so
˙
P
=
ρ
2

˙
x
2
+ ¨
x
(
b

x
)
(22)
Now, the kinematic equations for freefall are
x
=
gt
2
/
2, so
˙
x
=
gt
=
2
gx,
¨
x
=
g
(23)
so
˙
P
=
ρ
2
(
gb

3
gx
) =
Mg

T
(24)
finally
T
=
Mg
2
3
x
b
+ 1
(25)
1.2.1
Rocket Motion
An important application of linear momentum conservation is rocket propulsion. Consider
the rocket as a rigid body of mass
M
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 Spring '10
 RogerMelko
 Force, Mass, Momentum, General Relativity

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