Move the axis: the parallel-axis theorem
You will be able to: Use I=I_CM+Md² for axes parallel to the center-of-mass axis.
Why is the same rod harder to rotate about its end?
Hold a uniform rod at its center and imagine turning it in its plane. Now pivot it at one end. More of its mass is far from the end axis, so its rotational inertia is greater even though the rod itself is unchanged.
A useful starting point: Rotational inertia: where the mass matters →
Words and symbols before equations
- I_CM
- Inertia about an axis through the center of mass.
- M
- Total mass of the rigid system, in kg.
- d
- Perpendicular separation between the two parallel axes, in m.
- Parallel axes
- Axes with the same orientation, separated by a sideways shift.
What this picture assumes
Uniform 2 kg rod, length 2 m. Given I_CM=ML²/12=2/3 kg·m². All compared axes are perpendicular to the drawing, hence parallel. Offset d is measured from the center of mass.
Connect the picture to the physics
For parallel axes, I=I_CM+Md². Start with a center-of-mass-axis value and add the positive shift term. Moving the axis does not change the mass or the body’s physical shape.
For a given axis orientation, I is smallest at the center-of-mass axis because d=0 there. Equal offsets to opposite sides have equal d². This comparison does not say all differently oriented axes have the same minimum.
Do not apply the formula with an arbitrary starting axis labeled I_CM. If a noncentral value is given, first subtract its Md² to recover the central value, then add the new offset term. Extended-body formulas used here are provided.
A worked example, step by step
A uniform 2 kg rod is 2 m long. Given I_CM=ML²/12, find the inertia about a parallel axis through one end.
- I_CM=2(2²)/12=2/3 kg·m².
- Center-to-end distance d=L/2=1 m, not the full 2 m.
- I_end=I_CM+Md²=2/3+2(1²)=8/3≈2.67 kg·m².
- That is four times the central-axis inertia; the two axes are both perpendicular to the rod’s plane of motion.
The offset d is measured between parallel axes, one through the center of mass. It is not automatically the object’s full length.
If the axis moves from d to −d on the other side of the center, does I change?
Compare with an explanation
No. The squared separation d² is unchanged, provided the axes remain parallel.
Predict. Change one thing. Explain.
Shift the axis of the same 2 kg, 2 m rod from its center to its end. Compare the central term and the added Md² term. Predict the change when d doubles.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
I_CM=2/3 kg·m²; added Md²=0.5 kg·m²; I=1.17 kg·m². All compared axes are parallel.
Uniform 2 kg rod, length 2 m. Given I_CM=ML²/12=2/3 kg·m². All compared axes are perpendicular to the drawing, hence parallel. Offset d is measured from the center of mass.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use an angular-motion or torque relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA rigid body has M=3 kg and I_CM=0.6 kg·m². (a) Find I at d=0.2 m. (b) Find I at d=0.4 m. (c) Compare the added terms. (d) State the required axis relationship.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: I=0.6+3(0.2²)=0.72 kg·m².
- 1 point: I=0.6+3(0.4²)=1.08 kg·m².
- 1 point: Added terms 0.12 and 0.48 kg·m² differ by a factor of four.
- 1 point: The axes are parallel; I_CM is about the center-of-mass axis.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Parallel-axis formula?
I=I_CM+Md².
RECALL 2Which inertia is smallest for parallel axes?
The center-of-mass-axis value.
RECALL 3Double d does what?
Quadruples the added Md² term, not generally the entire I.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Move the axis: the parallel-axis theorem
- I=I_CM+Md², for parallel axes.
- At fixed axis orientation, the center-of-mass axis has minimum I.
Remember: The offset d is measured between parallel axes, one through the center of mass. It is not automatically the object’s full length.
Conditions: Uniform 2 kg rod, length 2 m. Given I_CM=ML²/12=2/3 kg·m². All compared axes are perpendicular to the drawing, hence parallel. Offset d is measured from the center of mass.
Refresh Kid · Unit 5 · Objectives 5.4.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.4, objectives 5.4.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
Want to work through this with a tutor?
Bring your question about move the axis: the parallel-axis theorem. Your explanation and answers remain free to access.