SME
University Physics I: Mechanics · Problem type
The parallel-axis theorem for a rod
Use this for “A 2 kg rod 1.5 m long turns about an axis 0.5 m from its center. Find its moment of inertia about that axis.”
Diagram
About
the
center:
ML²
12
=
1.2
×
0.9
2
12
=
0.081
kg·
m
2
.
Moved d = 0.45 m: I = 0.081 + 1.2 × 0.45² = 0.081 + 0.243 = 0.324 kg·m².
At
the
end,
I
=
ML²
3
:
four
times
the
center’s.
This is an example. Type your own values to start a new problem.
I (kg·m²)
=
Icm (kg·m²)
+
M (kg)
×
(
d (m)
)
2
L
Rod length
example
Formulas
Icm
=
M
×
L
2
12
0.081
=
1.2
×
0.9
2
12
I
=
Icm
+
M
×
d
2
0.324
=
0.081
+
1.2
×
0.45
2
Assumptions
•
The rod is thin and uniform, so about its center Icm = ML² ÷ 12.
•
The new axis is parallel to the one through the center, a distance d from it: I = Icm + Md².
•
d = L ÷ 2 puts the axis at one end, where I = ML² ÷ 3; the axis may also lie past the rod.
Step-by-step
WE KNOW
M = 1.2 kg, L = 0.9 m, d = 0.45 m
FIND
moment of inertia about the center (Icm), moment of inertia about the new axis (I)
Step 1 · Find moment of inertia about the center (Icm)
Use Icm = M × L² ÷ 12
A uniform rod about its center: the mass times the length squared, over 12.
Icm
=
1.2
×
0.9
2
12
Icm
=
1.2
×
0.81
12
Icm
=
0.972
12
Icm
=
0.081
kg·
m
2
Step 2 · Find moment of inertia about the new axis (I)
Use I = Icm + M × d²
The
parallel-axis
theorem:
add
the
mass
times
the
shift
squared
to
the
center’s
I
.
I
=
0.081
+
1.2
×
0.45
2
I
=
0.081
+
1.2
×
0.2025
I
=
0.081
+
0.243
I
=
0.324
kg·
m
2
Check
Put the numbers back in. Both sides should match.
0.081
=
1.2
×
0.9
2
12
✓
0.324
=
0.081
+
1.2
×
0.45
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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