Using Symmetry - Statics and Dynamics

1. The engineer invokes the vertical axis of symmetry at $x = 0$ to conclude $\bar{x} = 0$, and then computes $\bar{y}$ using the composite-area method. Which of the following gives the correct $\bar{y}$?
   Answer: $\bar{y} = \dfrac{(6b)(b)(4.5b) + (b)(4b)(2b)}{(6b)(b) + (b)(4b)} = \dfrac{27b^3 + 8b^3}{10b^2} = 3.5b$, where the top bar (area $6b^2$, centroid at $y = 4.5b$) and stem (area $4b^2$, centroid at $y = 2b$) are treated as two non-overlapping rectangles. When a composite shape can be divided into simple sub-regions with no overlap, the composite-area centroid formula is straightforward: multiply each sub-area by its centroid location, sum those products, then divide by the total area. The T-shape here divides cleanly into two non-overlapping rectangles — the horizontal bar occupying $y = 4b$ to $y = 5b$, and the vertical stem occupying $y = 0$ to $y = 4b$ — so long as you define the division correctly. Answer A does exactly this. The top bar has area $6b \times b = 6b^2$ with its centroid at $y = 4.5b$ (midpoint of the bar). The stem has area $b \times 4b = 4b^2$ with its centroid at $y = 2b$ (midpoint of the stem). Plugging into the formula: $\bar{y} = \frac{(6b^2)(4.5b) + (4b^2)(2b)}{6b^2 + 4b^2} = \frac{27b^3 + 8b^3}{10b^2} = 3.5b$. This is correct. Answer B raises a false concern. If you define the bar as only the portion above $y = 4b$ (the full $6b \times b$ strip) and the stem as the full $b \times 4b$ column below, there is no geometric overlap — so no subtraction is needed. Overlap only arises if you define sub-regions carelessly. Answer C correctly computes $3.5b$ but then discards it in favor of a simple average of centroid heights, $3.25b$. Simple averaging ignores the areas as weights — that's the entire point of the weighted formula. Answer D invokes a false "vertical symmetry implies midpoint" argument. Vertical symmetry only guarantees $\bar{x} = 0$; it says nothing about $\bar{y}$. Study tip: Always check whether your sub-regions actually overlap before applying corrections — and never confuse symmetry about a vertical axis (which fixes $\bar{x}$) with symmetry about a horizontal axis (which would fix $\bar{y}$).
2. A student places a thin uniform disk of radius $R$, surface density $\sigma$, and negligible thickness on the flat face of the hemisphere (coincident with the $xz$-plane, centered at the origin), creating a composite body. The student claims: 'The hemisphere is axially symmetric about the y-axis, so $\bar{x} = \bar{z} = 0$ for the hemisphere alone, and the disk is centered at the origin so $\bar{x} = \bar{z} = 0$ for the disk alone. Therefore the composite has $\bar{x} = \bar{z} = 0$, and $\bar{y}$ for the composite is found by weighting each component centroid by its mass.' What is $\bar{y}$ for the composite?
   Answer: $\bar{y} = \dfrac{\rho \cdot \tfrac{2}{3}\pi R^3 \cdot \tfrac{3R}{8} + \sigma \pi R^2 \cdot 0}{\rho \cdot \tfrac{2}{3}\pi R^3 + \sigma \pi R^2} = \dfrac{\tfrac{\rho \pi R^4}{4}}{\pi R^2\!\left(\tfrac{2\rho R}{3} + \sigma\right)} = \dfrac{\rho R^2}{4\!\left(\tfrac{2\rho R}{3} + \sigma\right)}$, since the disk lies in the plane $y = 0$ and its centroid contributes zero to the first moment. When combining bodies into a composite, the centroid of each component is its own geometric center of mass — determined entirely by that component's shape and position in space, independent of what it's attached to. The thin disk lies flat in the $xz$-plane, meaning every point on the disk has $y = 0$. By symmetry, the disk's centroid sits at the origin: $\bar{y}_{\text{disk}} = 0$. The hemisphere's centroid is the given $\bar{y}_{\text{hemi}} = 3R/8$. Applying the composite centroid formula by weighting each centroid by its mass: $\bar{y} = \frac{\rho \cdot \frac{2}{3}\pi R^3 \cdot \frac{3R}{8} + \sigma \pi R^2 \cdot 0}{\rho \cdot \frac{2}{3}\pi R^3 + \sigma \pi R^2} = \frac{\frac{\rho\pi R^4}{4}}{\pi R^2\!\left(\frac{2\rho R}{3}+\sigma\right)} = \frac{\rho R^2}{4\!\left(\frac{2\rho R}{3}+\sigma\right)}$ This is answer A, and the student's reasoning is sound. Choice B is wrong because it assigns the disk a centroid height of $3R/8$. The disk does not inherit the hemisphere's centroid — it has its own, located at $y = 0$ because that's where the disk actually lives in space. Choice C arrives at the same correct numerical answer as A, but claims the student's symmetry argument is incomplete. In fact, both the hemisphere and disk are individually axially symmetric about the $y$-axis, so their composite is too — the argument requires no additional verification. This makes C's criticism incorrect. Choice D confuses the disk's radius $R$ (a length within the $xz$-plane) with a displacement along $y$. The disk has zero thickness and zero $y$-position, so $\bar{y}_{\text{disk}} = 0$, not $R$. Study tip: Always ask "where is this component located in space?" before assigning its centroid. A component's centroid is its own center — it never borrows another body's centroid just because they're in contact.
3. A student argues: 'Because the generating region is bounded on the right by $x \geq 0$ only, and the rotation is about the y-axis, the resulting 3-D solid is axially symmetric about the y-axis. Therefore, by symmetry, both $\bar{x} = 0$ and $\bar{z} = 0$, and only $\bar{y}$ requires integration.' Is this reasoning valid, and what is $\bar{y}$?
   Answer: The reasoning is valid. The solid has full rotational symmetry about the y-axis, guaranteeing $\bar{x} = \bar{z} = 0$. Computing $\bar{y}$ by the disk/shell method gives $\bar{y} = \dfrac{8}{3}$, placing the centroid closer to the top face ($y=4$) than to the vertex ($y=0$) due to the increasing cross-sectional area with height. When a solid is formed by rotating a 2D region a full 360° about an axis, the result has complete rotational symmetry about that axis. This means every point in the solid has a mirror image across any plane containing the y-axis, which forces the centroid to lie exactly on that axis — so $\bar{x} = 0$ and $\bar{z} = 0$ by symmetry alone. The student's reasoning here is sound. For $\bar{y}$, you must integrate because there is no up-down symmetry — the cross-sectional area grows with height (wider disks near $y = 4$, narrower near $y = 0$), pulling the centroid upward. Using the disk method, each horizontal slice at height $y$ has radius $x = \sqrt{y}$, so: $V = \int_0^4 \pi y \, dy = \pi \left[\frac{y^2}{2}\right]_0^4 = 8\pi$ $\bar{y} = \frac{1}{V}\int_0^4 \pi y \cdot y \, dy = \frac{1}{8\pi} \cdot \pi \left[\frac{y^3}{3}\right]_0^4 = \frac{64/3}{8} = \frac{8}{3}$ Since $\tfrac{8}{3} \approx 2.67 > 2$, the centroid sits above the midpoint — exactly what the expanding cross-section predicts. This confirms A is correct. B is wrong because $\bar{y} = 2$ assumes uniform cross-sections, ignoring that the disk area increases with $y$. C compounds that error by claiming axial symmetry forces $\bar{y}$ to the midpoint — axial symmetry says nothing about the vertical location. D is wrong because rotating even a one-sided region a full 360° absolutely produces rotational symmetry; the generating curve's shape doesn't break that. Study tip: Distinguish the two symmetry questions separately — rotational symmetry about the axis kills $\bar{x}$ and $\bar{z}$, but $\bar{y}$ always requires checking whether mass is evenly distributed vertically.
4. Which of the following correctly assesses the student's symmetry claim and its implications for the centroid?
   Answer: The student is correct that the triangle has a line of symmetry along $y = x$, which forces the centroid to lie on that line, meaning $\bar{x} = \bar{y}$. Symmetry alone does not determine the actual value; the standard centroid formula (average of vertex coordinates) gives $\bar{x} = \bar{y} = \frac{0 + 2a + 0}{3} = \frac{2a}{3}$, confirming the centroid at $\left(\frac{2a}{3},\, \frac{2a}{3}\right)$. When a shape has a line of symmetry, the centroid must lie on that line — this is the key principle being tested. Your first task is always to verify whether the symmetry claim is valid, then determine what it actually tells you (and doesn't tell you) about the centroid's location. The line $y = x$ genuinely is an axis of symmetry for this triangle. Reflecting across $y = x$ swaps x- and y-coordinates, sending $(2a, 0) \to (0, 2a)$ and $(0, 2a) \to (2a, 0)$, while fixing the origin. The triangle maps onto itself, so the centroid must satisfy $\bar{x} = \bar{y}$. To find the actual value, use the standard result: the centroid of any triangle is the average of its three vertex coordinates. This gives $\bar{x} = \bar{y} = \frac{0 + 2a + 0}{3} = \frac{2a}{3}$, placing the centroid at $\left(\frac{2a}{3}, \frac{2a}{3}\right)$. This is answer B, and it's correct precisely because it separates what symmetry proves from what still requires calculation. Answer A conflates two separate ideas: the symmetry argument correctly gives $\bar{x} = \bar{y}$, but the "1/3 rule" is what produces the numerical value — A wrongly implies symmetry alone yields the number without further work. Answer C correctly identifies the symmetry but then misapplies the 1/3 rule. One-third of the leg length from the origin gives $a/3$, not $2a/3$ — this confuses measuring from a vertex versus measuring from the opposite side. Answer D is simply wrong: $y = x$ does function as a line of symmetry here, and independent integration is unnecessary once symmetry reduces the problem to a single unknown. Study tip: Always treat symmetry as a two-step tool — it reduces unknowns, but you still need a formula or integration to pin down the actual coordinate value.
5. Which of the following statements correctly identifies what symmetry can and cannot determine about the centroid of this wire, and gives the correct centroid location?
   Answer: The wire has a vertical axis of symmetry at $x = 0$, so $\bar{x} = 0$ by symmetry. Integration gives $\bar{y} = \dfrac{2R}{\pi} \approx 0.637R$, which is above the midpoint $R/2$ because the upper portion of the arc (near the apex) has larger y-coordinates, weighting the arc-length integral toward higher values. When finding the centroid of a wire (not an area), you must integrate arc-length elements — not area elements. This distinction drives nearly every error on this type of problem. The semicircular wire has a clear vertical axis of symmetry at $x = 0$: the left half mirrors the right half perfectly, so $\bar{x} = 0$ by inspection. No integration needed there. For $\bar{y}$, you parameterize using angle $\theta$ from $0$ to $\pi$, where each arc-length element is $dL = R\,d\theta$ and the height of each element is $y = R\sin\theta$. The centroid formula gives: $\bar{y} = \frac{\int_0^{\pi} R\sin\theta \cdot R\,d\theta}{\int_0^{\pi} R\,d\theta} = \frac{R^2[-\cos\theta]_0^{\pi}}{\pi R} = \frac{2R^2}{\pi R} = \frac{2R}{\pi} \approx 0.637R$ This confirms C is correct. The centroid sits above the midpoint $R/2 \approx 0.5R$ because more arc-length near the apex ($\theta \approx 90°$) carries large $y$-values, pulling the centroid upward. A is wrong in its interpretation: it correctly computes $\bar{y} = 2R/\pi$ but falsely claims this is less than $R/2$. In fact $2/\pi \approx 0.637 > 0.5$, so the centroid is above the midpoint, not closer to the diameter. B incorrectly gives $\bar{y} = R/2$, as if the wire were uniformly spread along the y-axis rather than curved along an arc — a geometry confusion. D gives $4R/3\pi$, which is the centroid of a solid semicircular area, not a wire. Sharing a boundary does not mean sharing a centroid. Study tip: Always ask yourself — is this a wire (integrate arc length) or an area (integrate area elements)? These yield different centroids even for identical shapes.
6. A student claims: 'Rod 1 is symmetric about the y-axis, so it contributes $\bar{x} = 0$. Rod 2 lies entirely on the y-axis, so it also contributes $\bar{x} = 0$. Therefore by superposition, $\bar{x} = 0$ for the composite system, and I can find $\bar{y}$ using only the y-coordinates of each rod's centroid.' Which of the following is true?
   Answer: The student's argument for $\bar{x} = 0$ is valid, and $\bar{y}$ is correctly computed as $\bar{y} = \dfrac{(3\lambda L)(0) + (4\lambda L)(L)}{3\lambda L + 4\lambda L} = \dfrac{4L}{7}$, where the mass of Rod 2 is $m_2 = 2\lambda \cdot 2L = 4\lambda L$ and its centroid is at $y = L$. When finding the centroid of a composite body, you need two things for each component: its total mass and the location of its own centroid. The superposition formula is $\bar{y} = \frac{\sum m_i \bar{y}_i}{\sum m_i}$, where each $m_i$ is the full mass of that piece. For this system, the argument that $\bar{x} = 0$ is sound: Rod 1 is symmetric about the y-axis (centroid at $x = 0$), and Rod 2 lies on the y-axis (every point has $x = 0$), so both contributions vanish and $\bar{x} = 0$ by superposition. For $\bar{y}$, calculate each mass carefully. Rod 1 has linear density $\lambda$ and length $3L$, so $m_1 = 3\lambda L$, with centroid at $y = 0$ (it lies on the x-axis). Rod 2 has linear density $2\lambda$ and length $2L$, so $m_2 = 2\lambda \cdot 2L = 4\lambda L$. Its centroid sits at the midpoint of $y = 0$ to $y = 2L$, which is $\bar{y}_2 = L$. Applying superposition: $\bar{y} = \frac{(3\lambda L)(0) + (4\lambda L)(L)}{3\lambda L + 4\lambda L} = \frac{4L}{7}$. This confirms B is correct. Choice A incorrectly uses $m_2 = 2\lambda L$, as if Rod 2's length were only $L$ — confusing the distance from centroid to origin with the rod's actual length. Choice C uses $\bar{y}_2 = 2L$, mistaking the rod's endpoint for its centroid. Choice D ignores Rod 2's actual density, using $\lambda$ instead of $2\lambda$, which halves the mass. A reliable habit: always write out $m_i = \rho_i \times L_i$ explicitly before plugging into the centroid formula — density and length are both easy to misread under pressure.
7. Which part(s) of the student's symmetry argument are correct?
   Answer: Only the first part is correct: $\bar{y} = 0$ follows validly from x-axis symmetry. The second part is wrong because 'fore-aft symmetry' would require the region to map onto itself under $x \to 9 - x$, which it does not since the parabola becomes $9 - x = y^2$, a different curve. The correct $\bar{x}$, found by integration, is $\bar{x} = \dfrac{27}{5}$. Centroid problems test two distinct skills: recognizing valid symmetry arguments and performing integration correctly. When evaluating a symmetry claim, always ask: does the region map exactly onto itself under the proposed transformation? The first part of the student's argument is sound. The region bounded by $x = y^2$ and $x = 9$ is symmetric about the x-axis because for every point $(x, y)$ in the region, $(x, -y)$ is also in the region — the parabola and the vertical line both satisfy this. Mass distributed symmetrically about the x-axis produces $\bar{y} = 0$ with no integration required. The second part fails. "Fore-aft symmetry" would require the region to map onto itself under the substitution $x \to 9 - x$. But this transforms the boundary $x = y^2$ into $9 - x = y^2$, or $x = 9 - y^2$ — a leftward-opening parabola, which is a completely different curve. The region is narrow near $x = 0$ and wide near $x = 9$, so more area (and mass) lies toward the right, pulling $\bar{x}$ past the midpoint 4.5. Integration gives $\bar{x} = \dfrac{27}{5} = 5.4$, confirming this. Answer C captures both points correctly. Answer A fails because it accepts the false midpoint argument. Answer B correctly rejects the midpoint claim but gives no value for $\bar{x}$ and doesn't explain why the symmetry fails structurally. Answer D wrongly rejects the x-axis symmetry — a rightward-opening parabola is absolutely symmetric about the x-axis. Strategy tip: Always verify a symmetry claim by substituting the transformation into every boundary equation. If even one boundary changes form, the symmetry — and any centroid shortcut derived from it — is invalid.
8. The engineer claims that by symmetry, the centroid of the resulting plate lies at the origin, and that no integration is required to determine the centroid. Which of the following best evaluates this claim?
   Answer: The claim is correct: the plate has both $x = 0$ and $y = 0$ as axes of reflective symmetry. Under reflection about $x = 0$, each hole maps to another hole (e.g., $(c,c) \leftrightarrow (-c,c)$), so the composite plate is symmetric about $x=0$, forcing $\bar{x} = 0$. Under reflection about $y = 0$, similarly $(c,c) \leftrightarrow (c,-c)$, so the plate is also symmetric about $y = 0$, forcing $\bar{y} = 0$. Two independent symmetry axes together locate the centroid at the origin with no integration required. Centroid problems are really symmetry problems in disguise. The key principle: if a shape (including any cutouts) is symmetric about a line, the centroid must lie on that line. If it lies on two independent symmetry axes, it sits at their intersection — no integration needed. Here, the square is clearly symmetric about both $x = 0$ and $y = 0$. The critical question is whether removing the four circular holes preserves those symmetries. Check each reflection: when you reflect about $x = 0$, the hole at $(c, c)$ maps to $(-c, c)$ — which is exactly another hole — and $(c, -c)$ maps to $(-c, -c)$, again another hole. The composite plate looks identical after this reflection, so $\bar{x} = 0$. Repeating the argument for reflection about $y = 0$: $(c, c) \leftrightarrow (c, -c)$ and $(-c, c) \leftrightarrow (-c, -c)$, both swapping existing holes. The plate is also symmetric about $y = 0$, forcing $\bar{y} = 0$. The centroid is the origin. This is exactly what D captures. A is partially right in reasoning but incomplete — it only invokes one symmetry axis and treats the second conclusion as a "similar argument" without spelling out that both axes independently apply. The centroid logic requires both axes explicitly confirmed. B introduces a false rule: rotational symmetry is sufficient but not necessary for a centroid to lie at a geometric center. Reflective symmetry axes work perfectly well, as this problem demonstrates. C is simply wrong about the geometry. The holes come in symmetric pairs above and below the x-axis, so the plate absolutely is symmetric about $y = 0$. Your study tip: always ask whether cutouts or additions preserve the original symmetry lines. If a removed region has a mirror image also removed, the symmetry — and the centroid argument — survives intact.
9. A student notes that the equilateral triangle has a vertical axis of symmetry along $x = 0$. Which of the following correctly describes whether and how this symmetry can be used to find the centroid of the composite body?
   Answer: The axis of symmetry $x = 0$ of the triangle is broken by the off-center hole, so the composite body has no line of symmetry. Both $\bar{x}$ and $\bar{y}$ must be computed via the subtraction method: $\bar{x} = \dfrac{A_{\triangle}\cdot 0 - A_{\circ}\cdot(a/4)}{A_{\triangle} - A_{\circ}}$, which is nonzero, and $\bar{y}$ likewise requires explicit calculation. When finding the centroid of a composite body formed by subtraction, symmetry arguments only hold if the entire composite — not just one component — possesses that symmetry. This question tests whether you can distinguish between a component's symmetry and the composite's symmetry. Since the hole is centered at $\left(\dfrac{a}{4},\, \dfrac{a}{\sqrt{3}}\right)$ rather than on the triangle's axis of symmetry $x = 0$, the composite body is no longer symmetric about any vertical line. This means $\bar{x} \neq 0$ and must be computed directly using the subtraction method: $\bar{x} = \frac{A_{\triangle}\cdot(0) - A_{\circ}\cdot(a/4)}{A_{\triangle} - A_{\circ}}$ Because the hole pulls mass to the right of $x = 0$, removing it shifts the centroid leftward — but away from zero, making $\bar{x}$ nonzero. Similarly, $\bar{y}$ requires explicit calculation. This is exactly what B describes, making it correct. A is wrong because "dominates the area" is never a valid symmetry argument. Symmetry is a geometric property — not a weighted vote. Even a tiny off-center hole breaks the axis of symmetry completely. C is wrong for the same reason: the hole's centroid does not lie on $x = 0$, so horizontal balance is not preserved. The claim that any subtracted region through the triangle's centroid preserves symmetry is fabricated reasoning. D is wrong because a shared $y$-coordinate between two points does not create a horizontal axis of symmetry for the composite. A true axis of symmetry requires the entire shape to be a mirror image across that line, which it is not here. Your key takeaway: symmetry belongs to the composite, not its parts. Always check whether every feature of the final body respects the proposed axis before invoking symmetry to skip a calculation.

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