SME
Mechanical Vibrations · Problem type
A compound pendulum: a rod swinging from its end
Use this for “A uniform 2 kg rod 1 m long swings from a pin at one end. Find IO, ωₙ, fₙ and the period of small swings.”
Diagram
Pinned
at
its
end:
d
=
L
2
=
0.5
m
I
=
mL²
12
+
md²
=
2
×
1
2
12
+
2
×
0.5
2
I
=
0.6667
kg·
m
2
T
=
2π√(
I
mgd
)
=
2π√(
0.6667
2 × 9.81 × 0.5
)
T = 1.638 s
A
simple
pendulum
of
length
I
md
=
0.6667
m
(dashed)
swings
in
step.
Drawn at 12°; T holds for small swings.
This is an example. Type your own values to start a new problem.
d (m)
=
L (m)
÷
2
IO (kg·m²)
=
m (kg)
×
(
L (m)
)
2
÷
12
+
m (kg)
×
(
d (m)
)
2
ωₙ (rad/s)
=
m (kg)
×
g
×
d (m)
÷
IO (kg·m²)
fₙ
Natural frequency in hertz
calculated · Hz
T
Period
calculated
Formulas
d
=
L
2
0.5
=
1
2
IO
=
m
×
L
2
12
+
m
×
d
2
0.6667
=
2
×
1
2
12
+
2
×
0.5
2
ωₙ
=
√(
m
×
9.81
×
d
IO
)
3.836
=
√(2
×
9.81
×
0.5
0.6667
)
fₙ
=
ωₙ
2π
0.6105
=
3.836
2π
T
=
1
fₙ
1.6379
=
1
0.6105
Assumptions
•
Small swings, so sin θ ≈ θ and IO θ̈ + mgdθ = 0, with g = 9.81 m/s² and no friction.
•
A uniform rod pinned at its end: its center of mass is d = L ÷ 2 below the pin.
•
Parallel axes: IO = mL² ÷ 12 + md², which is mL² ÷ 3 for this rod.
Step-by-step
WE KNOW
m = 2 kg, L = 1 m
FIND
pin to center of mass (d), moment of inertia about the pin (IO), natural frequency (ωₙ), natural frequency in hertz (fₙ), period (T)
Step 1 · Find pin to center of mass (d)
Use d = L ÷ 2
A uniform rod’s center of mass is at its middle, half its length below the end pin.
d
=
1
2
d
=
0.5
m
Step 2 · Find moment of inertia about the pin (IO)
Use IO = m × L² ÷ 12 + m × d²
Parallel
axes:
the
rod’s
own
mL
²
12
about
its
middle,
plus
md²
for
the
pin
d
away.
IO
=
2
×
1
2
12
+
2
×
0.5
2
IO
=
2
12
+
2
×
0.25
IO = 0.1667 + 0.5
IO
=
0.6667
kg·
m
2
Step 3 · Find natural frequency (ωₙ)
Use ωₙ = √(m × 9.81 × d ÷ IO)
Gravity’s moment mgd sin θ ≈ mgdθ pulls the rod back, so IO θ̈ = −mgdθ.
ωₙ
=
√(2
×
9.81
×
0.5
0.6667
)
ωₙ
=
√(19.62
×
0.5
0.6667
)
ωₙ
=
√(
9.81
0.6667
)
ωₙ
=
√
14.714
ωₙ = 3.836 rad/s
Step 4 · Find natural frequency in hertz (fₙ)
Use fₙ = ωₙ ÷ 2π
One cycle is 2π radians, so divide the radians each second by 2π.
fₙ
=
3.836
2π
fₙ = 0.61052 Hz
Step 5 · Find period (T)
Use T = 1 ÷ fₙ
The time for one cycle is one over the cycles each second.
T
=
1
0.61052
T
=
1.6379
s
Check
Put the numbers back in. Both sides should match.
0.5
=
1
2
✓
0.6667
=
2
×
1
2
12
+
2
×
0.5
2
✓
3.836
=
√(2
×
9.81
×
0.5
0.6667
)
✓
0.61052
=
3.836
2π
✓
1.6379
=
1
0.61052
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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