Parallel-Axis Theorem - Statics and Dynamics

1. Rod mass $m$, length $L$; $I$ about one end is $mL^2/3$. Find $I$ about a parallel axis $L/3$ from that end.
   Answer: $mL^2/9$ The new axis is L/3 from the end, so it is L/6 from the rod's center. Use the parallel-axis theorem from the center: I = $mL^2$/12 + $m(L/6)^2$ = $mL^2$/12 + $mL^2$/36 = $mL^2$/9. The tempting error is adding $m(L/3)^2$ to the given $mL^2$/3, but that shifts from the end axis instead of the center, giving $4mL^2$/9.
2. Disk mass $m$, radius $R$; $I$ about a diameter is $mR^2/4$. Find $I$ about a parallel tangent in the disk plane.
   Answer: $5mR^2/4$ The tangent line in the disk plane is a distance R from the center, so you add $mR^2$ to the given diameter moment. Thus I = $mR^2$/4 + $mR^2$ = $5mR^2$/4. A tempting wrong answer is $3mR^2$/2, which comes from using a tangent perpendicular to the plane instead of one in the disk plane.
3. A student computes the moment of inertia of the T-section about the centroidal axis of the composite section (34.6 mm from the bottom). She correctly identifies the parallel-axis distances as $d_F = 110 - 34.6 = 75.4$ mm for the flange and $d_W = 50 - 34.6 = 15.4$ mm for the web. However, she then writes: $I_x = I_{\bar{x},F} + A_F(75.4)^2 + I_{\bar{x},W} + A_W(15.4)^2$ where $A_F = 200(20) = 4000 \text{ mm}^2$ and $A_W = 20(100) = 2000 \text{ mm}^2$. Is this expression correct, and what result does it yield (approximately)?
   Answer: The expression is correct. It yields approximately $I_x \approx 2.41 \times 10^7 \text{ mm}^4$, obtained by adding each part's own centroidal inertia to the parallel-axis term $A d^2$, with distances measured from each part's centroid to the composite centroid. When computing the moment of inertia of a composite section about its centroidal axis, you apply the parallel-axis theorem to each part independently: $I_x = \sum\left(I_{\bar{x},i} + A_i d_i^2\right)$, where $d_i$ is the distance from each part's own centroid to the composite centroid. The key insight is that each part gets its own $I_{\bar{x}}$ plus its own $Ad^2$ term — they're simply summed together. The student's expression is exactly this. For the flange: $I_{\bar{x},F} + A_F d_F^2 = \frac{200(20)^3}{12} + 4000(75.4)^2 \approx 133{,}333 + 22{,}740{,}640 \approx 22.87 \times 10^6 \text{ mm}^4$. For the web: $I_{\bar{x},W} + A_W d_W^2 = \frac{20(100)^3}{12} + 2000(15.4)^2 \approx 1{,}666{,}667 + 474{,}320 \approx 2.14 \times 10^6 \text{ mm}^4$. Summing gives approximately $2.41 \times 10^7 \text{ mm}^4$, confirming A is correct. Choice B mistakes the reference point — distances in the parallel-axis theorem must be measured to the composite centroid, not to an arbitrary datum like the bottom. Using raw datum distances would double-count offsets already embedded in the centroid calculation. Choice C commits a classic error of combining the two offsets into a single squared term, as if both parts share one displacement — they don't; each part has its own independent $d$. Choice D wrongly treats a negative $d$ as a special case requiring different handling. Since $d$ is always squared, its sign is irrelevant — $(-15.4)^2 = (15.4)^2$. Your study tip: always square each part's distance independently, and remember that sign never matters in $Ad^2$ — negative distances contribute positively to inertia just like positive ones.
4. 3 kg rod (center $I=4$ kg·m², length 2 m) with a 2 kg particle at one end; find $I$ for parallel axis at other end.
   Answer: 15 kg·m² Move the rod from its center to the end axis using parallel axis: 4 + 3$(1)^2$ = 7 kg $m^2$. The 2 kg particle is 2 m from that end axis, so it adds 2$(2)^2$ = 8. Total 15. A common mistake is stopping at 7 and forgetting the particle.
5. For a uniform slender rod of mass $m$ and length $L$, the moment of inertia about its centroidal axis (perpendicular to the rod, through its midpoint) is $I_{\bar{x}} = \frac{1}{12}mL^2$. An engineer uses the parallel-axis theorem to find the moment of inertia about an axis located at distance $d$ from the centroid and obtains $I = \frac{1}{12}mL^2 + md^2$. She then applies the theorem a second time from this new (non-centroidal) axis to another axis a distance $s$ further away, writing $I' = I + ms^2$. Under what condition, if any, is this second step valid?
   Answer: The second step is never valid in this form, because the parallel-axis theorem requires the starting axis to be centroidal. To find $I'$, one must return to the centroid: $I' = \frac{1}{12}mL^2 + m(d+s)^2$, using the total distance from the centroid to the final axis. Whenever you encounter the parallel-axis theorem, the single most important rule to memorize is this: the starting axis must always be centroidal. The theorem states $I = I_{\bar{x}} + md^2$, where $I_{\bar{x}}$ is specifically the moment of inertia about the center of mass. It is not a general "shift by any distance" formula between arbitrary axes. This is exactly why D is correct. Once the engineer has moved to a non-centroidal axis and obtained $I = \frac{1}{12}mL^2 + md^2$, she cannot treat that axis as a new starting point for a second application. The theorem's derivation relies on the cross-product term vanishing — which only happens when the reference axis passes through the centroid. To find $I'$, she must return to the centroid and apply the theorem once using the total distance: $I' = \frac{1}{12}mL^2 + m(d+s)^2$. You can verify this is different from $I + ms^2 = \frac{1}{12}mL^2 + md^2 + ms^2$, which is missing the cross term $2mds$. A is wrong because the parallel-axis theorem has no requirement about the destination axis being farther from the centroid than the starting axis — that condition is invented and meaningless. B is the classic trap: the theorem is not a general shift theorem between any two parallel axes. It only works when one axis is centroidal. C is partially insightful — if $d = 0$ the intermediate axis is centroidal and the second step would be valid — but this is a special case, not the condition that makes the procedure generally valid. Your study tip: always ask yourself, "Is my starting axis through the centroid?" before applying the parallel-axis theorem. If not, go back to the centroid first.
6. An engineer needs the moment of inertia of a thin rectangular plate (width $b$, height $h$) about an axis parallel to its width and located a distance $h/2$ below the bottom edge. The centroidal moment of inertia about the horizontal centroidal axis is $I_{\bar{x}} = \frac{bh^3}{12}$. She first applies the parallel-axis theorem with $d = h/2$ to shift to the bottom edge, obtaining $I_{\text{edge}} = \frac{bh^3}{12} + bh\left(\frac{h}{2}\right)^2 = \frac{bh^3}{3}$. She then applies the theorem a second time from the bottom edge to the target axis (another $h/2$ further away), writing $I_{\text{target}} = \frac{bh^3}{3} + bh\left(\frac{h}{2}\right)^2$. Which statement best describes the validity of her second application?
   Answer: The second application is invalid because the parallel-axis theorem must start from a centroidal axis. Since $I_{\text{edge}}$ is not a centroidal value, using it as the base for another shift double-counts part of the centroid-to-edge distance. The correct result is found in one step: $I_{\text{target}} = \frac{bh^3}{12} + bh(h)^2 = \frac{13bh^3}{12}$, using the total distance $d = h$ from the centroid to the target axis. Whenever you see the parallel-axis theorem applied in sequence, your first instinct should be to check whether each application starts from a centroidal axis — because that is a strict requirement of the theorem. The parallel-axis theorem states $I = I_{\bar{x}} + Ad^2$, where $I_{\bar{x}}$ must be the moment of inertia about the axis passing through the centroid. The distance $d$ is always measured from the centroid to the new axis. This means you cannot "chain" shifts by using a non-centroidal intermediate result as your new base — doing so double-counts the centroid-to-intermediate distance. The correct approach is always a single shift from the centroid to the final target, using the total distance. Here, the centroid is at the plate's midpoint, and the target axis is $h/2$ below the bottom edge, so the centroid-to-target distance is $d = h/2 + h/2 = h$. The correct result is: $I_{\text{target}} = \frac{bh^3}{12} + bh(h)^2 = \frac{13bh^3}{12}$ This confirms A is correct. The engineer's second application is invalid because $I_{\text{edge}}$ is not a centroidal moment of inertia. B is the key distractor — it sounds reasonable but is flatly wrong. The theorem is not valid between arbitrary parallel axes; the reference must always be centroidal. C is incorrect because the theorem imposes no restriction on whether the axis lies inside or outside the body. D is wrong on two counts: the method is invalid and the arithmetic produces the wrong answer ($\frac{bh^3}{3} + \frac{bh^3}{4} = \frac{7bh^3}{12}$, not $\frac{5bh^3}{12}$). Study tip: Always return to the centroid. No matter how many axes you need, apply the parallel-axis theorem exactly once per target, using the direct centroid-to-target distance.
7. What is the moment of inertia of the composite section about the centroidal horizontal axis of the full rectangle (i.e., the x-axis through the rectangle's centroid, before the hole is introduced)? Note that removing material subtracts its contribution, and the parallel-axis theorem must be applied to the circle's centroid, which is offset from this reference axis.
   Answer: $I_x = \frac{1}{12}(120)(80)^3 - \left[\frac{\pi}{4}(20)^4 + \pi(20)^2(20)^2\right] \text{ mm}^4$, because the circular hole's centroid is 20 mm above the rectangle's centroid, so the parallel-axis shift uses $d = 20$ mm. Whenever you encounter a composite section with removed material, your go-to framework is: subtract the hole's contribution, applying the parallel-axis theorem correctly. The key insight is that the parallel-axis theorem for a horizontal (x-axis) moment of inertia only cares about the vertical distance between the hole's centroid and the reference axis — horizontal offsets are irrelevant. For the composite section, you start with the full rectangle's moment of inertia and subtract the circle's contribution about the same reference axis. The circle's centroid sits 20 mm above the rectangle's centroidal x-axis (the vertical component of its offset). Applying the parallel-axis theorem to the circle gives $I_{\bar{x},\text{circle}} + A_{\text{circle}} \cdot d^2 = \frac{\pi}{4}(20)^4 + \pi(20)^2(20)^2$, where $d = 20$ mm is that vertical offset. Subtracting yields answer A, which is correct. B is wrong because it uses $d = 30$ mm — the horizontal offset — in the parallel-axis theorem. The horizontal distance between centroids has zero effect on $I_x$; only vertical separation matters. C is wrong for a similar reason: it uses the total radial distance $\sqrt{30^2 + 20^2} = \sqrt{1300}$ mm. The parallel-axis theorem is not about radial distance — it's axis-specific. For a horizontal axis, only the perpendicular (vertical) distance applies. D is wrong because it ignores the parallel-axis shift entirely. Just because the circle has its own centroidal axis doesn't mean $d = 0$; the circle's centroid is offset 20 mm from the reference axis, so the shift term $Ad^2$ is nonzero and must be included. Study tip: Always ask yourself, "What is the perpendicular distance from this sub-shape's centroid to my reference axis?" For $I_x$, that's the vertical gap — ignore horizontal offsets completely.
8. A student needs the moment of inertia of the semicircular area about its own centroidal axis (parallel to the diameter, passing through $\bar{y} = \frac{4R}{3\pi}$). She correctly applies the parallel-axis theorem in reverse: $I_{\bar{x}} = I_x - A d^2$, where $A = \frac{\pi R^2}{2}$ and $d = \frac{4R}{3\pi}$. Which of the following correctly evaluates this expression?
   Answer: $I_{\bar{x}} = \frac{\pi R^4}{8} - \frac{\pi R^2}{2}\left(\frac{4R}{3\pi}\right)^2 = \frac{\pi R^4}{8} - \frac{8R^4}{9\pi}$, which simplifies to $I_{\bar{x}} = R^4\left(\frac{\pi}{8} - \frac{8}{9\pi}\right)$. Whenever you see a moment of inertia problem involving a non-standard axis, your first instinct should be the parallel-axis theorem: $I = I_{\bar{x}} + Ad^2$, where $I_{\bar{x}}$ is the centroidal moment of inertia, $A$ is the area, and $d$ is the perpendicular distance between the two axes. Here, you already know $I_x$ about the diameter (a non-centroidal axis), so you rearrange to find the centroidal value: $I_{\bar{x}} = I_x - Ad^2$. Plugging in, the $Ad^2$ correction term becomes $\frac{\pi R^2}{2} \cdot \left(\frac{4R}{3\pi}\right)^2 = \frac{\pi R^2}{2} \cdot \frac{16R^2}{9\pi^2} = \frac{16R^4}{18\pi} = \frac{8R^4}{9\pi}$. Therefore $I_{\bar{x}} = \frac{\pi R^4}{8} - \frac{8R^4}{9\pi} = R^4\!\left(\frac{\pi}{8} - \frac{8}{9\pi}\right)$, confirming D is correct. Choice A makes an arithmetic error when squaring: it writes $\frac{16R^2}{9\pi^2}$ but then cancels only one factor of $\pi$, leaving $\frac{4R^4}{9\pi}$ instead of the correct $\frac{8R^4}{9\pi}$. The student forgot the factor of 2 from the area term. Choice B reverses the sign, adding $Ad^2$ instead of subtracting it — this would be correct if you were shifting away from the centroid, but here you're shifting toward it, so you subtract. Choice C misapplies the theorem entirely by using $d$ to the first power instead of $d^2$; the parallel-axis theorem always involves the square of the distance. As a study tip, always track the direction of your axis shift: moving away from the centroid means you add $Ad^2$; moving toward it means you subtract. Writing the standard form $I_{\text{far}} = I_{\bar{x}} + Ad^2$ and solving for what you need prevents sign errors every time.
9. Using the parallel-axis theorem, what is the total moment of inertia of the three-disk assembly about the central axis of the assembly (the axis equidistant from all three disk centers, perpendicular to all disk faces)?
   Answer: $I_{\text{total}} = 3\left(\frac{1}{2}mr^2 + mR^2\right) = \frac{3}{2}mr^2 + 3mR^2$, because each disk contributes its own centroidal polar moment $\frac{1}{2}mr^2$ plus the parallel-axis shift $mR^2$, and the three shifts add rather than cancel due to the symmetric but non-collinear arrangement. When a problem asks for the moment of inertia of an assembly of objects about a common axis, your go-to tool is the parallel-axis theorem: $I = I_{cm} + md^2$, where $I_{cm}$ is the object's moment of inertia about its own center of mass, and $d$ is the distance from that center to the new axis. For a system of objects, you simply sum each contribution. For a thin disk spinning about its own central axis (perpendicular to the face), the centroidal moment of inertia is $\frac{1}{2}mr^2$. Here, the assembly's central axis is parallel to each disk's own axis, so this formula applies directly. Each disk center sits a distance $R$ from the central axis, contributing a parallel-axis term of $mR^2$. One disk contributes $\frac{1}{2}mr^2 + mR^2$, and since all three disks are identical with the same $R$, the total is $3\left(\frac{1}{2}mr^2 + mR^2\right) = \frac{3}{2}mr^2 + 3mR^2$, confirming C. A is a common and tempting trap: the displacement vectors of the three disk centers do sum to zero by symmetry, but moment of inertia uses $d^2$, a scalar distance — not a vector. Squared distances are always positive and never cancel. B uses $\frac{1}{4}mr^2$ as the centroidal term, which is correct only for rotation about a diameter of the disk, not about the axis perpendicular to its face — a completely different geometry. D invents a fictional "equivalent single disk" with no physical basis; you cannot combine radius and mass this way. Your key takeaway: the parallel-axis shift $mR^2$ always adds — it can never be negative or cancel, because $d^2 \geq 0$ always. If symmetry is tempting you to cancel these terms, that's your signal to slow down and check whether you're working with vectors or scalars.
10. A 4 kg body has $I=3.0$ kg·m² about a parallel axis 0.5 m from its center. Find $I$ about its center.
    Answer: 2.0 kg·m² The parallel-axis theorem says I about a shifted axis equals I about the center plus m $d^2$. Here m $d^2$ = 4$(0.5)^2$ = 1.0 kg $m^2$, so the center value is 3.0 - 1.0 = 2.0 kg $m^2$. The tempting wrong answer is 4.0 kg $m^2$, from adding 1.0 instead of subtracting; the shifted axis always has the larger moment of inertia.
11. A hollow square tube has outer side length $a$ and inner side length $b$ ($b < a$), with the centroid at its geometric center. Its area moment of inertia about the horizontal centroidal axis is $I_{\bar{x}} = \frac{a^4 - b^4}{12}$. An engineer wants the moment of inertia about a horizontal axis located at the outer bottom edge of the tube (distance $a/2$ below the centroid). She writes: $I_{\text{bottom edge}} = \frac{a^4 - b^4}{12} + (a^2 - b^2)\left(\frac{a}{2}\right)^2$ A colleague claims this is wrong because the parallel-axis distance should be $\frac{a-b}{2}$ (half the wall thickness), not $\frac{a}{2}$. Who is correct and why?
    Answer: The engineer is correct. The parallel-axis distance is measured from the composite section's centroid to the target axis. The centroid of the hollow tube lies at its geometric center, which is $a/2$ from the outer bottom edge, regardless of the wall thickness or inner dimension $b$. Whenever you apply the parallel-axis theorem, the single most important thing to nail down is what distance you're actually measuring. The theorem states $I = I_{\bar{x}} + Ad^2$, where $d$ is the perpendicular distance from the section's centroid to the target axis — full stop. It has nothing to do with wall thickness, material distribution, or any average dimension. For a hollow square tube with outer side $a$, the centroid sits at the geometric center by symmetry. The outer bottom edge is exactly $a/2$ below that centroid — the inner dimension $b$ is irrelevant to this distance. The area of the hollow section is correctly $a^2 - b^2$. So the engineer's formula, $\frac{a^4-b^4}{12} + (a^2-b^2)\left(\frac{a}{2}\right)^2$, applies the theorem perfectly. C is correct. A is wrong because $\frac{a+b}{4}$ has no geometric meaning here — it isn't the centroid location of any relevant axis or area. It's a fabricated distractor that sounds plausible by mixing inner and outer dimensions. B is wrong because the colleague confuses the centroid of the wall material alone with the centroid of the entire cross-section. The parallel-axis distance must reference the composite section's centroid, not the mid-thickness of the wall. $\frac{a-b}{2}$ is the wall thickness, not a centroidal distance. D is wrong because the engineer's area $a^2 - b^2$ is exactly right for the hollow section. You don't need to split the calculation into outer minus inner for the $Ad^2$ term — that's only necessary when computing $I_{\bar{x}}$ using the subtraction method. Study tip: Always ask yourself "where is the centroid of the whole section?" before writing down $d$. The parallel-axis distance is purely geometric — centroid to target axis.
12. The moment of inertia of a solid circular disk of radius $R$ and mass $m$ about its central axis (perpendicular to the disk face) is $I_c = \frac{1}{2}mR^2$. A second, identical disk is placed such that its center is a distance $2R$ from the first disk's center, and the two disks are rigidly connected. Using the parallel-axis theorem, an analyst computes the combined system's moment of inertia about an axis through the midpoint between the two centers (perpendicular to both disks). Which expression is correct?
    Answer: $I_{\text{total}} = 2\left(\frac{1}{2}mR^2\right) + 2m(R)^2 = mR^2 + 2mR^2 = 3mR^2$, because each disk's center is $R$ from the midpoint axis, so each parallel-axis term contributes $m(R)^2$. Whenever you encounter a moment of inertia problem involving multiple bodies, the parallel-axis theorem is your framework: $I = I_c + md^2$, where $d$ is the distance from each body's own center of mass to the new axis of rotation. The key is identifying that distance correctly. Here, the two disk centers are separated by $2R$, so the midpoint axis sits exactly $R$ from each disk's center. Applying the parallel-axis theorem to each disk: $I_{\text{each}} = \frac{1}{2}mR^2 + m(R)^2$. Summing both disks gives $I_{\text{total}} = 2\left(\frac{1}{2}mR^2\right) + 2m(R)^2 = mR^2 + 2mR^2 = 3mR^2$, confirming that A is correct. B is tempting because $2R$ appears prominently in the problem, but that's the center-to-center distance between the disks — not the distance from either disk's center to the midpoint axis. Each disk's center is only $R$ from the midpoint, so using $2R$ in the parallel-axis term double-counts the separation. C reflects a fundamental misunderstanding: the parallel-axis corrections don't "cancel" because both terms are positive ($md^2$ is always additive). Symmetry means the two disks contribute equally, not that their corrections eliminate one another. D incorrectly treats the two-disk system as a single solid disk with radius $2R$. That formula only applies to a uniform disk's own central axis — you cannot simply substitute an arbitrary effective radius for a composite system. Study tip: Always sketch the geometry and label the distance $d$ explicitly before plugging into $I = I_c + md^2$. The most common mistake on parallel-axis problems is confusing total separation with the distance to the specific reference axis.
13. The moment of inertia of a solid sphere of mass $m$ and radius $R$ about any axis through its center is $I_c = \frac{2}{5}mR^2$. A solid hemisphere (half the sphere, mass $m_{h} = m/2$) has its flat face in the xz-plane, with its centroid located at $\bar{y} = \frac{3R}{8}$ above the flat face. An engineer needs the moment of inertia of the hemisphere about the flat-face diameter axis (the x-axis at $y = 0$, lying in the flat face). She reasons: (1) the full sphere's $I$ about this diametral axis is $\frac{2}{5}mR^2$; (2) by symmetry, each hemisphere contributes half: $I_{\text{hemi, flat face}} = \frac{1}{5}mR^2$; (3) applying the reverse parallel-axis theorem with the hemisphere's own mass and centroid offset gives $I_{\bar{x},\text{hemi}} = \frac{1}{5}mR^2 - \frac{m}{2}\left(\frac{3R}{8}\right)^2$. Is step (3) valid, and what is $I_{\bar{x},\text{hemi}}$?
    Answer: Step (3) is valid. $I_{\bar{x},\text{hemi}} = \frac{mR^2}{5} - \frac{9mR^2}{128} = mR^2\left(\frac{128 - 45}{640}\right) = \frac{83mR^2}{640}$, obtained by correctly subtracting $A_{\text{hemi}}\,d^2$ from the flat-face inertia. Whenever you encounter moment-of-inertia problems involving composite or partial bodies, your go-to tool is the parallel-axis theorem: $I_{\text{axis}} = I_{\bar{x}} + md^2$, where $d$ is the perpendicular distance between the centroidal axis and the parallel axis of interest. Running it in reverse — subtracting $md^2$ to find the centroidal inertia — is perfectly valid, as long as you correctly identify $d$. Here, the flat-face x-axis passes directly through the sphere's center (the flat face lies in the xz-plane, which contains the sphere's center). The hemisphere's centroid sits at $\bar{y} = \frac{3R}{8}$ above that flat face. So the distance from the flat-face axis to the hemisphere's centroidal axis is exactly $d = \frac{3R}{8}$ — no adjustment needed. The hemisphere's moment of inertia about the flat-face axis is $\frac{1}{2} \cdot \frac{2}{5}mR^2 = \frac{mR^2}{5}$ (half the full sphere's value, by symmetry). Applying the reverse parallel-axis theorem with the hemisphere's mass $m_h = m/2$: $I_{\bar{x}} = \frac{mR^2}{5} - \frac{m}{2}\left(\frac{3R}{8}\right)^2 = \frac{mR^2}{5} - \frac{9mR^2}{128} = \frac{128mR^2 - 45mR^2}{640} = \frac{83mR^2}{640}$ This confirms B is correct. A is wrong because it incorrectly subtracts $md$ (linear) instead of $md^2$ — the parallel-axis theorem always involves distance squared. C is wrong because it misidentifies $d$: the flat-face axis already passes through the sphere's center, so $d = 3R/8$ directly, not $R - 3R/8$. D suggests a mass error exists, but steps (2) and (3) already correctly use $m/2$; it merely relabels B's valid reasoning without identifying a real flaw. Study tip: Always sketch the geometry and explicitly label which axis is the centroidal axis versus the reference axis — confusing these two is the most common source of parallel-axis errors on dynamics exams.
14. An L-shaped (angle) section is formed by two rectangular legs: a horizontal leg (width 100 mm, thickness 10 mm) and a vertical leg (width 10 mm, height 90 mm), joined at their inner corner so that the section's outer corner is at the origin. The centroid of the composite section is located at $\bar{x} = 17.4$ mm and $\bar{y} = 22.4$ mm from the outer corner (origin). A student needs the moment of inertia about the vertical centroidal axis (the axis parallel to y, passing through $\bar{x} = 17.4$ mm). She computes $I_{\bar{y}}$ for each leg about its own centroidal vertical axis and then applies the parallel-axis theorem. The horizontal leg's own centroid is at $x = 50$ mm from the origin; the vertical leg's own centroid is at $x = 5$ mm from the origin. Which parallel-axis distances $d$ are correct for shifting each leg's inertia to the composite vertical centroidal axis at $\bar{x} = 17.4$ mm?
    Answer: $d_{\text{horiz leg}} = |50 - 17.4| = 32.6$ mm and $d_{\text{vert leg}} = |5 - 17.4| = 12.4$ mm, since the parallel-axis distance for a vertical axis is the horizontal offset between each part's centroid and the composite centroidal vertical axis. Whenever you encounter a moment of inertia problem involving composite sections, the key question to ask is: which direction does the axis run, and which coordinate offset matters? For a vertical centroidal axis (parallel to y), the parallel-axis theorem states $I_{\bar{y}} = I_{y,\text{own}} + A d^2$, where $d$ is the horizontal (x-direction) distance between the part's own centroidal vertical axis and the composite's vertical centroidal axis. This is because shifting a vertical axis left or right requires measuring that horizontal gap — not any vertical distance. For the horizontal leg, its centroid sits at $x = 50$ mm, so its distance from the composite centroidal axis at $\bar{x} = 17.4$ mm is $d = |50 - 17.4| = 32.6$ mm. For the vertical leg, its centroid sits at $x = 5$ mm, giving $d = |5 - 17.4| = 12.4$ mm. That confirms B as correct. Choice A swaps the logic entirely — it uses the *y*-coordinate offsets (17.4 mm and 27.6 mm) to shift a vertical axis, which is backward. Those y-distances would be relevant for shifting a horizontal axis, not a vertical one. Choice C uses full Euclidean (diagonal) distances, which have no place in the parallel-axis theorem; the theorem always requires the perpendicular distance between two parallel axes, not a 2D diagonal. Choice D measures from the origin rather than from the composite centroidal axis — this conflates the centroid-location calculation with the parallel-axis shift, producing the wrong reference point. Study tip: Before applying $Ad^2$, always identify the axis direction first. Vertical axis → use horizontal (x) offsets. Horizontal axis → use vertical (y) offsets. Writing this as a two-column rule on your formula sheet can prevent the axis-confusion trap that makes A so tempting.
15. A 2 kg body has $I=5$ kg·m² about its center and $8$ kg·m² about parallel axis B. Find distance from B to center.
    Answer: 1.22 m Use the parallel-axis theorem: I about B equals I about center plus m $d^2$. Subtract the center value: 8 - 5 = 3 kg $m^2$, then divide by mass 2 kg: 3 / 2 = 1.5 $m^2$. Taking the square root gives d = 1.22 m. The tempting error is 1.50 m, which is $d^2$, not the distance d.

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