SME
University Physics I: Mechanics · Problem type
A physical pendulum: a rod swinging from its end
Use this for “A 0.3 kg rod 1.2 m long swings from a pin at one end. Find its period and the length of a simple pendulum that keeps time with it.”
Diagram
Pinned
at
its
end:
d
=
L
2
=
0.5
m
I
=
mL²
12
+
md²
I
=
0.5
×
1
2
12
+
0.5
×
0.5
2
I
=
0.1667
kg·
m
2
T
=
2π√(
I
mgd
)
T
=
2π√(
0.1667
0.5 × 9.81 × 0.5
)
=
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.
T (s)
=
2π
I (kg·m²)
m (kg)
×
9.81
×
d (m)
L
Length of the rod
example
Leq
Equivalent simple length
worked out (not typed)
Formulas
d
=
L
2
0.5
=
1
2
I
=
m
×
L
2
12
+
m
×
d
2
0.16667
=
0.5
×
1
2
12
+
0.5
×
0.5
2
T
=
2π
×
√(
I
m
× 9.81 ×
d
)
1.6379
=
2π
×
√(
0.16667
0.5 × 9.81 × 0.5
)
Leq
=
I
m
×
d
0.6667
=
0.16667
0.5 × 0.5
Assumptions
•
A uniform rod swings through small angles about a fixed pin at its end, with no friction and g = 9.81 m/s².
•
I about the pin by parallel axes: I = mL² ÷ 12 + md², the rod’s own I about its middle plus md².
•
A simple pendulum of length I ÷ (md) keeps the same time; for this rod it is 2L ÷ 3.
Step-by-step
WE KNOW
m = 0.5 kg, L = 1 m
FIND
pin to center of mass (d), moment of inertia about the pin (I), period (T), equivalent simple length (Leq)
Step 1 · Find pin to center of mass (d)
Use d = L ÷ 2
A
uniform
rod’s
center
of
mass
is
at
its
middle,
L
2
below
the
pin
at
its
end.
d
=
1
2
d
=
0.5
m
Step 2 · Find moment of inertia about the pin (I)
Use I = m × L² ÷ 12 + m × d²
Parallel
axes:
I
about
the
middle,
mL
²
12,
plus
md²
for
the
pin
d
away.
I
=
0.5
×
1
2
12
+
0.5
×
0.5
2
I
=
0.5
12
+
0.5
×
0.25
I
=
0.04167
+
0.125
I
=
0.166667
kg·
m
2
Step 3 · Find period (T)
Use T = 2π × √(I ÷ (m × 9.81 × d))
Gravity’s
torque
mgd
sin
θ
≈
mgdθ
pulls
it
back,
so
ω
2
=
mgd
I
and
T
=
2π
ω
.
T
=
2π
×
√(
0.166667
0.5 × 9.81 × 0.5
)
T
=
1.6379
s
Step 4 · Find equivalent simple length (Leq)
Use Leq = I ÷ (m × d)
A
simple
pendulum
has
T
=
2π√(
L
g
):
set
L
=
I
md
to
keep
the
same
time.
Leq
=
0.166667
0.5 × 0.5
Leq
=
0.1667
0.25
Leq = 0.6667 m
Check
Put the numbers back in. Both sides should match.
0.5
=
1
2
✓
0.166667
=
0.5
×
1
2
12
+
0.5
×
0.5
2
✓
1.6379
=
2π
×
√(
0.166667
0.5 × 9.81 × 0.5
)
✓
0.6667
=
0.166667
0.5 × 0.5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.