A solid disk of mass M and radius R has I_CM = ½MR² about its central axis. What is its moment of inertia about an axis tangent to its rim, parallel to the central axis?
A½MR² — the same as the center-of-mass value, since the mass distribution doesn't change
BMR² — just the Md² term, since d = R
C(3/2)MR² — applying I = I_CM + Md² with d = R gives ½MR² + MR²
D2MR² — you double I_CM when moving to the rim
The parallel axis theorem gives I = I_CM + Md². The axis at the rim is parallel to the central axis and displaced by d = R. So I = ½MR² + M(R²) = (3/2)MR². Option A forgets the Md² term — a common error from not realizing that displacing the axis always adds inertia. Option B forgets I_CM. Option D has no derivation; doubling is incorrect. The extra MR² represents the additional inertia from the fact that all the mass is now at least distance R from the new axis.
Question 2 Multiple Choice
Why does the parallel axis theorem include a Md² term?
AIt corrects for rotational friction at the new axis location
BEven if all the mass were concentrated at the center of mass, it would now be at distance d from the new axis — this additional contribution is exactly Md², and the theorem's proof shows the cross-term vanishes
CIt reduces I_CM to account for mass that is now closer to the new axis than to the old one
DIt converts from the center-of-mass reference frame to an inertial lab frame
The proof of the theorem expands I = Σmᵢ(rᵢ − d)² = Σmᵢrᵢ² − 2d·Σmᵢrᵢ + Md². The first term is I_CM, the middle term vanishes because the center of mass is at the origin by definition (Σmᵢrᵢ = 0 in the CM frame), and the last term Md² remains. Physically: when you shift the axis by d, even a point mass at the center of mass is now at distance d from the axis, contributing Md² to the total moment of inertia. The spread of mass around the CM (I_CM) is unchanged; you've simply added the inertia of the CM's own displacement.
Question 3 True / False
Among all axes parallel to a given direction, the axis through the center of mass gives the smallest moment of inertia.
TTrue
FFalse
Answer: True
This follows directly from I = I_CM + Md². Since M > 0 and d² ≥ 0, we have I ≥ I_CM for every parallel axis, with equality only when d = 0 (i.e., the axis passes through the center of mass). Any displacement strictly increases the moment of inertia. This makes the center-of-mass axis special: it is the axis of minimum rotational inertia among all parallel axes, which has physical consequences for natural rotation and the design of rotating machinery.
Question 4 True / False
The parallel axis theorem can be applied to find the moment of inertia about any new axis, whether or not it is parallel to the original axis through the center of mass.
TTrue
FFalse
Answer: False
The parallel axis theorem applies only to parallel axes — axes that have the same orientation in space, differing only in their location (perpendicular displacement d). The formula I = I_CM + Md² does not apply when the new axis is tilted relative to the center-of-mass axis, because the derivation assumes the coordinate shift is purely perpendicular. For axes in different orientations, you need the full inertia tensor and the general transformation rules.
Question 5 Short Answer
Why does the parallel axis theorem make physical sense? What does the Md² term represent, and why does the I_CM term remain unchanged?
Think about your answer, then reveal below.
Model answer: When the rotation axis is displaced from the center of mass by distance d, two contributions add independently. First, I_CM captures how the mass is internally distributed around the center of mass — this doesn't change when you shift the axis, because the relative positions of all the mass elements to each other are unchanged. Second, Md² captures what would happen even if all the mass were concentrated at the center of mass: a point mass M at distance d from the axis contributes Md² to rotational inertia. The theorem says the total effect is the sum of these two independent contributions, because the cross-term in the algebraic expansion vanishes when you work in the center-of-mass frame.
The physical intuition is cleanest this way: shifting the axis has two effects that can be computed separately — the internal distribution (unchanged, so I_CM) and the displacement of the whole mass distribution's center (adding Md²). The vanishing cross-term is the mathematical expression of the fact that the center of mass is exactly the 'balance point' — there is no net first moment around it.