SME
Classical Mechanics
Rigid-body motion
Use this for “A 0.8 kg disk 5 cm in radius spins at 250 rad/s on a level axle 6 cm from its pivot. How fast does it precess, and how long is one turn round the post?”
Diagram
I = ½mR² = 4 × 10⁻⁴ kg·m², so L = Iω = 0.12 kg·m²/s along the axle.
τ = mgr = 0.5 × 9.81 × 0.05 = 0.245 N·m, sideways: it turns L, not down.
Ω
=
τ
L
=
2.04
rad/s,
one
turn
round
the
post
every
3.07
s.
This is an example. Type your own values to start a new problem.
Ω
=
m (kg)
×
g
×
r (
)
I (kg·m²)
ω (
)
=
Ω (rad/s)
R
Rotor radius
example
L
Spin angular momentum
calculated · kg·m²/s
τ
Torque of the weight
calculated · N·m
Tₚ
Precession period
calculated
Formulas
I
=
0.5
×
m
×
R
2
4 ×
10
-4
=
0.5
×
0.5
×
0.04
2
L
=
I
×
ω
0.12
=
4 ×
10
-4
×
300
τ
=
m
×
9.81
×
r
0.2453 = 0.5 × 9.81 × 0.05
Ω
=
τ
L
2.0438
=
0.2453
0.12
Tₚ
=
2
×
π
Ω
3.0743
=
2
×
π
2.0438
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
The rotor is a uniform disk, so I = ½mR², and g = 9.81 m/s².
•
Fast spin: L = Iω is far larger than the angular momentum of the slow swing round the post.
•
The axle stays level: the torque τ = mgr is sideways, so it turns L round the post instead of tipping it down (τ = dL/dt).
Step-by-step
WE KNOW
m = 0.5 kg, R = 4 cm, ω = 300 rad/s, r = 5 cm
FIND
moment of inertia (I), spin angular momentum (L), torque of the weight (τ), precession rate (Ω), precession period (Tₚ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
R = 4 cm = 0.04 m (1 m = 100 cm)
r = 5 cm = 0.05 m (1 m = 100 cm)
Step 2 · Find moment of inertia (I)
Use I = 0.5 × m × R²
A
uniform
disk
about
its
own
axle:
half
of
m
times
R
squared.
I
=
0.5
×
0.5
×
0.04
2
I
=
0.5
×
0.5
×
0.0016
I
=
0.25
×
0.0016
I
=
4 ×
10
-4
kg·
m
2
Step 3 · Find spin angular momentum (L)
Use L = I × ω
The
spin
angular
momentum
points
along
the
axle:
I
times
the
spin
rate.
L
=
4 ×
10
-4
×
300
L
=
0.12
kg·
m
2
/s
Step 4 · Find torque of the weight (τ)
Use τ = m × 9.81 × r
The
weight
mg
pulls
straight
down
at
the
rotor,
r
out
from
the
pivot:
its
torque
is
mgr.
τ
=
0.5
×
9.81
×
0.05
τ
=
4.905
×
0.05
τ
=
0.24525
N·
m
Step 5 · Find precession rate (Ω)
Use Ω = τ ÷ L
Each
second
the
torque
adds
τ
to
L
at
right
angles,
turning
it
by
the
angle
τ
L
.
Ω
=
0.24525
0.12
Ω
=
2.0438
rad/s
Step 6 · Find precession period (Tₚ)
Use Tₚ = 2 × π ÷ Ω
One
full
turn
round
the
post
is
2π
radians,
at
Ω
radians
each
second.
Tₚ
=
2
×
π
2.0438
Tₚ = 3.0743 s
Check
Put the numbers back in. Both sides should match.
4 ×
10
-4
=
0.5
×
0.5
×
0.04
2
✓
0.12
=
4 ×
10
-4
×
300
✓
0.24525 = 0.5 × 9.81 × 0.05 ✓
2.0438
=
0.24525
0.12
✓
3.0743
=
2
×
π
2.0438
✓
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