SME
Classical Mechanics · Problem type
The Coriolis acceleration and the Foucault pendulum
Use this for “A train runs at 80 m/s at latitude 52° N. How big is its sideways Coriolis acceleration, and how long does a Foucault pendulum there take to turn once?”
Table
Latitude (λ)
(°)
Coriolis acceleration, sideways (ac)
(m/s²)
This is an example. Type your own values to start a new problem.
a
c
=
2Ω
v (
)
sin
λ
°
=
ac (m/s²)
T
F
=
23.93
h
sin
°
=
TF (
)
Formulas
ac
=
2
×
7.292 ×
10
-5
×
v
×
sin(
λ
)
0.05156
=
2
×
7.292 ×
10
-5
×
500
×
sin(45°)
TF
=
23.93
sin(
λ
)
33.8421
=
23.93
sin(45°)
Assumptions
•
Earth turns once a sidereal day, 23.93 h: Ω = 7.292 × 10⁻⁵ rad/s.
•
Only the horizontal part of the Coriolis acceleration −2Ω × v is found, for motion along the ground: 2Ωv sin λ, whichever way the body heads.
•
North of the equator it deflects motion to the right; south of it, by the same amount, to the left. Latitude is entered as a size, 0° to 90°.
•
The Foucault pendulum’s swing plane turns at Ω sin λ, so one full turn takes 23.93 h ÷ sin λ.
Step-by-step
WE KNOW
λ = 45°, v = 500 m/s
FIND
Coriolis acceleration, sideways (ac), turn time of a Foucault pendulum (TF)
Step 1 · Find Coriolis acceleration, sideways (ac)
Use ac = 2 × 7.292 × 10⁻⁵ × v × sin(λ)
Only
the
part
of
Earth’s
spin
square
to
the
ground,
Ω
sin
λ
,
turns
motion
sideways:
twice
that
times
the
speed.
ac
=
2
×
7.292 ×
10
-5
×
500
×
sin(45°)
ac
=
2
×
(7.292
×
10
-5
)
×
500
×
0.70711
ac = 0.00014584 × 500 × 0.70711
ac = 0.07292 × 0.70711
ac
=
0.05156
m/
s
2
Step 2 · Find turn time of a Foucault pendulum (TF)
Use TF = 23.93 ÷ sin(λ)
The
swing
plane
turns
at
Ω
sin
λ
,
so
one
turn
takes
the
day
23.93
h
over
sin
λ
:
once
a
day
at
a
pole,
never
at
the
equator.
TF
=
23.93
sin(45°)
TF
=
23.93
0.7071
TF = 33.8421 h
Check
Put the numbers back in. Both sides should match.
0.05156
=
2
×
7.292 ×
10
-5
×
500
×
sin(45°)
✓
33.8421
=
23.93
sin(45°)
✓
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