Saturday, December 12, 2009
Professor Philip S. Marcus
ME 106
Solution to Final Examination
1a) The relevant “input” parameters are:
ρ
,
R
outer
, Ω
inner
,
H
,
R
inner
, Ω
outer
,
and
ν
. Specifcally, we do
not
include
g
, and the air pressure. We choose units
oF
{
Length
} ≡
R
outer
;
{
Time
} ≡
1
/
Ω
inner
; and
{
Mass
} ≡
ρ
(
R
outer
)
3
.
These
choices are arbitrary, and you are free to choose any other units that make
sense
. With these units, an expression For the period
τ
oF the ±ow is
τ
=
τ
*
{
Time
}
=1
/
Ω
inner
ˆ
f
±
ρ
*
,R
*
outer
,
Ω
*
inner
,
Ω
*
outer
,H
*
,R
*
inner
,ν
*
²
(1)
Expressing the dimensionless Form (with asterisks) oF the input parameters in
the units chosen above, eq. (1) becomes
τ
=1
/
Ω
inner
ˆ
f
±
1
,
1
,
1
,
Ω
outer
/
Ω
inner
,H/R
outer
,R
inner
/R
outer
,ν/
[(
R
2
outer
Ω
inner
]
²
(2)
and dropping the “slots” flled with numerical values rather than dimensionless
groupings oF parameters, we obtain
τ
=1
/
Ω
inner
f
±
Ω
outer
/
Ω
inner
,H/R
outer
,R
inner
/R
outer
,ν/
[(
R
2
outer
Ω
inner
]
²
(3)
where
f
is a dimensionless Function oF dimensionless “slots”. Note that the 4
dimensionless grouping in the 4 slots oF
f
are independent oF each other, as
required.
b) You must make all Four oF the expressions in the slots oF
f
the same in the
small model and in the Etcheverry design; otherwise, the physics will di²er,
and the ±ows will di²er and not be similar. In Fact, unless all Four oF the slots
have the same numerical values For the small apparatus and For the Etcheverry
apparatus, it is possible that the ±ow in Etcheverry won’t be periodic in time.
By your design requirement that
all
lengths are 10 times bigger in the Etchev
erry version,
H/R
outer
and
R
inner
/R
outer
are same in the Etcheverry and in the
small apparati. In addition, you must keep Ω
outer
/
Ω
inner
≡
0 in both apparati.
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View Full DocumentThus, keeping the values in the slots the same requires that Ω
outer
= 0 in the
Etcheverry apparatus and
ν/
[(
R
2
outer
Ω
inner
] be the same in both apparati. Or,
ν
Etch
(
R
Etch
outer
)
2
Ω
Etch
inner
=
ν
small
(
R
small
outer
)
2
Ω
small
inner
.
(4)
Or
ν
Etch
ν
small
=
±
R
Etch
outer
R
small
outer
²
2
Ω
Etch
inner
Ω
small
inner
= 100
Ω
Etch
inner
Ω
small
inner
.
(5)
Note that to enforce the correct period of 25 seconds, we use eq. (3), remember
ing that we will have the same numerical value of
f
in both apparati (because
the numerical values in the 4 slots of
f
are the same in both apparati)
τ
=1
/
Ω
inner
f,
(6)
so
Ω
small
inner
Ω
Etch
inner
=
τ
Etch
τ
small
≡
25
/
10 = 2
.
5
.
(7)
Using this ratio of the inner cylinder speeds in eq. (5), we obtain
ν
Etch
ν
small
= 40
.
(8)
Equations (5) and (8) along with the parameter values in the small apparatus
give Ω
Etch
inner
=0
.
4 radians/second and
ν
Etch
=0
.
4 cm
2
/sec.
No experiments
are needed.
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 Fall '08
 Morris
 numerical values, inertial frame

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