SME
Classical Mechanics
Noninertial frames
Use this for “At latitude 30°, find the centrifugal acceleration, the effective g and how far a plumb line tips from the true vertical.”
Diagram
g = 9.81 m/s² at 45° from the horizontal: (−6.937, −6.937) m/s².
ac = 0.02395 m/s² at 0° from the horizontal: (0.02395, 0) m/s².
ΣFₓ
=
−6.913
m/
s
2
and
ΣFᵧ
ΣFₓ
=
−6.937
m/
s
2
geff
=
√
−
6.913
2
+ −
6.937
2
=
9.793
m/
s
2
at
225.1°
from
+x.
This is an example. Type your own values to start a new problem.
a
c
=
Ω²
R (km)
cos
φ
°
=
ac (m/s²)
g
True gravity
example · m/s²
Ω²R
Centrifugal acceleration at the equator
worked out (not typed) · m/s²
geff
Effective gravity
worked out (not typed) · m/s²
α
Plumb line’s tilt
worked out (not typed) · °
Formulas
Ω
2
R
=
(7.292 ×
10
-5
)
2
×
R
×
1,000
0.03388
=
(7.292 ×
10
-5
)
2
×
6,371
×
1,000
ac
=
Ω
2
R
×
cos(
φ
)
0.02395 = 0.03388 × cos(45°)
geff
=
√((
g
−
ac
×
cos(
φ
))²
+
(ac
×
sin(
φ
))²)
9.7931 = √((9.81 − 0.02395 × cos(45°))² + (0.02395 × sin(45°))²)
α
=
tan⁻¹(ac
×
sin(
φ
)
g
− ac × cos(
φ
)
)
0.0991
=
tan⁻¹(0.02395
×
sin(45°)
9.81 − 0.02395 × cos(45°)
)
Assumptions
•
In the turning frame a centrifugal term −Ω × (Ω × r) acts on every body, straight out from the axis: Ω²R cos φ.
•
Earth is taken as a sphere of radius R turning at Ω = 7.292 × 10⁻⁵ rad/s; true gravity g points to its center. R is 6,357 km (polar), 6,371 km (mean) or 6,378 km (equatorial).
•
Latitude is entered as a size, 0° to 90°: the south is the same, the plumb line tipping toward the equator.
Step-by-step
WE KNOW
φ = 45°, g = 9.81 m/s², R = 6,371 km
FIND
centrifugal acceleration at the equator (Ω²R), centrifugal acceleration (ac), effective gravity (geff), plumb line’s tilt (α)
Step 1 · Find centrifugal acceleration at the equator (Ω²R)
Use Ω²R = (7.292 × 10⁻⁵)² × R × 1,000
The
centrifugal
term
on
the
equator,
Ω
2
R
:
the
radius
in
meters,
so
km
times
1000.
Ω
2
R
=
(7.292 ×
10
-5
)
2
×
6,371
×
1,000
Ω
2
R
=
(5.317
×
10
-9
)
×
6,371
×
1,000
Ω
2
R
=
(3.387
×
10
-5
)
×
1,000
Ω
2
R
=
0.033877
m/
s
2
Step 2 · Find centrifugal acceleration (ac)
Use ac = Ω²R × cos(φ)
The
distance
to
the
axis
is
R
cos
φ
,
so
the
centrifugal
term
is
Ω
2
times
it:
largest
on
the
equator,
0
at
the
poles.
ac = 0.033877 × cos(45°)
ac = 0.03388 × 0.7071
ac
=
0.02395
m/
s
2
Step 3 · Find effective gravity (geff)
Use geff = √((g − ac × cos(φ))² + (ac × sin(φ))²)
Split
the
centrifugal
term:
ac
cos
φ
straight
up,
against
g
,
and
ac
sin
φ
along
the
ground
toward
the
equator;
then
add
the
parts
by
Pythagoras.
geff = √((9.81 − 0.02395 × cos(45°))² + (0.02395 × sin(45°))²)
geff
=
√(
(9.81 − 0.02395 × 0.7071)
2
+
(0.02395 × 0.7071)
2
)
geff
=
√(
(9.81 − 0.01694)
2
+
0.01694
2
)
geff
=
√
9.793
2
+
0.01694
2
geff
=
√
95.9 + 0.000287
geff
=
√
95.9
geff
=
9.7931
m/
s
2
Step 4 · Find plumb line’s tilt (α)
Use α = tan⁻¹(ac × sin(φ) ÷ (g − ac × cos(φ)))
The plumb line hangs along geff: its sideways part over its downward part is the tangent of the tilt from the line to the center.
α
=
tan⁻¹(0.02395
×
sin(45°)
9.81 − 0.02395 × cos(45°)
)
α
=
0.0991°
Check
Put the numbers back in. Both sides should match.
0.033877
=
(7.292 ×
10
-5
)
2
×
6,371
×
1,000
✓
0.02395 = 0.033877 × cos(45°) ✓
9.7931 = √((9.81 − 0.02395 × cos(45°))² + (0.02395 × sin(45°))²) ✓
0.0991
=
tan⁻¹(0.02395
×
sin(45°)
9.81 − 0.02395 × cos(45°)
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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