Physics 325
Lecture 6
Finally, the most general case is:
f)
(
)
, ,
F
F r v t
=
G
G
G G
In general, we can’t solve this analytically.
But there are some special cases which can
be.
If
G
can be factorized such that the variables separate then things simplify
considerably.
We work in one dimension for simplicity. Suppose
F
(
)
( )
(
,
F v t
f
v g t
=
)
.
Then
( )
( )
( )
( )
dv
dv
f
v g t
m
g t dt
m
dt
f
v
=
→
=
∫
∫
And with a little luck, these integrals can be done.
Similarly, suppose
(
)
(
)
( )
,
F x v
h x
f
v
=
.
Then
(
)
( )
dv
h x
f
v
m
dt
=
Invoking again the chain rule
dv
dx dv
dv
v
dt
dt dx
dx
=
=
Gives
(
)
( )
(
)
( )
dv
vdv
h x
f
v
mv
h x dx
m
dx
f
v
=
→
=
∫
∫
The integration yields
(
)
v x
, and
x(t)
is then found from
dx dt
=
(
)
dx
dt
v x
=
∫
∫
Finally, the last variation on this,
(
)
(
)
( )
,
F x t
h x g t
=
, is not solvable analytically unless
either
h
or
g
is a constant.
Instead, numerical techniques must be applied.
Next I want to illustrate another technique for solving equation of motion problems other
than the integration or energy techniques we’ve just used a few times.
This problem will
also allow us to exercise our tools in a non-cartesian coordinate system.
Bead on a Rod Example (courtesy Jim Wiss)
Consider a bead on a thin rod that is rotating around a frictionless pivot such that it has a
constant angular velocity:
t
φ
ω
=
r

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