SME
Classical Mechanics · Problem type
The surface of a spinning liquid
Use this for “Water turns at 8 rad/s in a bucket. How much higher is it 15 cm from the axis than at the center?”
Chart
z(0.1) ≈ 0.05097
This is an example. Type your own values to start a new problem.
z
=
(
ω (rad/s)
)
2
(
r (m)
)
2
2g
=
z (m)
n
Turns each minute
worked out (not typed) · rpm
Formulas
z
=
ω
2
×
r
2
2 × 9.81
0.05097
=
10
2
×
0.1
2
2 × 9.81
n
=
60
×
ω
2 × π
95.493
=
60
×
10
2 × π
Assumptions
•
The liquid turns as one body with its bucket at a steady ω, so it is at rest in the turning frame; g = 9.81 m/s².
•
There each bit feels g down and the centrifugal ω²r outward; the surface lies square to their sum.
•
Its slope dz ÷ dr = ω²r ÷ g, so the surface is the paraboloid z = ω²r² ÷ (2g), measured from its lowest point.
Step-by-step
WE KNOW
ω = 10 rad/s, r = 0.1 m
FIND
height above the center (z), turns each minute (n)
Step 1 · Find height above the center (z)
Use z = ω² × r² ÷ (2 × 9.81)
The
surface
rises
with
slope
ω
2
r
g
;
adding
up
that
slope
from
the
axis
out
to
r
gives
ω
2
r
2
2g
.
z
=
10
2
×
0.1
2
2 × 9.81
z
=
10
2
×
0.1
2
19.62
z
=
100
×
0.01
19.62
z
=
1
19.62
z
=
0.05097
m
Step 2 · Find turns each minute (n)
Use n = 60 × ω ÷ (2 × π)
One turn is 2π radians and one minute 60 s.
n
=
60
×
10
2 × π
n
=
600
2π
n
=
95.493
rpm
Check
Put the numbers back in. Both sides should match.
0.05097
=
10
2
×
0.1
2
2 × 9.81
✓
95.493
=
60
×
10
2 × π
✓
Topic check
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