Statics and Dynamics Quiz: Using Symmetry

What this quiz covers

This quiz focuses on Using Symmetry, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 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. (correct answer)
2. The reasoning is valid. The solid has full rotational symmetry about the y-axis, guaranteeing $\bar{x} = \bar{z} = 0$. Computing $\bar{y}$ gives $\bar{y} = 2$, which is the midpoint of the y-extent $[0,4]$, consistent with the parabolic solid's fore-aft symmetry.
3. The reasoning is partially valid: axial symmetry about the y-axis does guarantee $\bar{x} = \bar{z} = 0$, but it also forces $\bar{y}$ to equal the midpoint of the y-extent by the same symmetry argument, giving $\bar{y} = 2$ without any integration needed.
4. The reasoning is invalid: rotating only the $x \geq 0$ half-region about the y-axis does not produce a body with true rotational symmetry, because the generating curve $y = x^2$ is asymmetric. Therefore $\bar{x}$ and $\bar{z}$ cannot be determined by symmetry and all three coordinates require integration.

Question 2

1. $\bar{y} = \dfrac{(6b)(b)(4.5b) + (b)(4b)(2b)}{(6b)(b) + (b)(4b)} = \dfrac{27b^3 + 8b^3}{10b^2} = 3.5b$, 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. (correct answer)
2. $\bar{y} = \dfrac{(6b)(b)(4.5b) + (b)(4b)(2b)}{(6b)(b) + (b)(4b)} = 3.5b$, but this result is invalid because the two rectangles overlap at a $b \times b$ square, so the overlapping area must be subtracted once and its centroid accounted for separately.
3. $\bar{y} = \dfrac{(6b^2)(4.5b) + (4b^2)(2b)}{6b^2 + 4b^2} = 3.5b$, but since both sub-centroids lie on the vertical axis of symmetry, the correct $\bar{y}$ must equal the simple average of the two centroid heights: $\bar{y} = \dfrac{4.5b + 2b}{2} = 3.25b$.
4. $\bar{y} = \dfrac{(6b)(b)(4.5b) + (b)(4b)(2b)}{(6b)(b) + (b)(4b)} = \dfrac{27b^3 + 8b^3}{10b^2} = 3.5b$, but the correct centroid height is actually $\bar{y} = 2.5b$ because the overall shape has a vertical extent of $5b$, and by the vertical symmetry of the T-shape the centroid must lie at the midpoint $5b/2$.

Question 3

1. Both parts are correct: the x-axis symmetry argument validly gives $\bar{y} = 0$, and the midpoint argument validly gives $\bar{x} = 4.5$, so neither coordinate requires direct integration.
2. Only the first part is correct: the region is indeed symmetric about the x-axis, so $\bar{y} = 0$ by valid symmetry argument. The second part is incorrect because the region is not symmetric about any vertical line $x = c$; the horizontal extent of the region is non-uniform in $y$, so $\bar{x}$ must be found by integration and will not equal 4.5.
3. 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 $\bar{x} = \dfrac{27}{5}$. (correct answer)
4. Both parts are incorrect: the region bounded by $x = y^2$ and $x = 9$ is not symmetric about the x-axis because the parabola opens rightward and is therefore asymmetric. The correct centroid must be found by full double integration with no symmetry simplifications.

Question 4

1. The wire has a vertical axis of symmetry at $x = 0$, so $\bar{x} = 0$ by symmetry. The value $\bar{y}$ cannot be found by symmetry and requires integration, yielding $\bar{y} = \dfrac{2R}{\pi}$, which is less than $R/2$ — closer to the diameter than to the top of the arc.
2. The wire has a vertical axis of symmetry at $x = 0$, so $\bar{x} = 0$ by symmetry. Integration gives $\bar{y} = \dfrac{R}{2}$, the midpoint of the y-extent $[0, R]$, reflecting the uniform distribution of the wire along the arc.
3. 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. (correct answer)
4. The wire has a vertical axis of symmetry at $x = 0$, so $\bar{x} = 0$ by symmetry. Integration gives $\bar{y} = \dfrac{4R}{3\pi}$, which equals the centroid of a solid semicircular area — since both shapes share the same outer boundary and symmetry axis, their centroids must coincide.

Question 5

1. The axis of symmetry $x = 0$ still applies to the composite body because the triangle dominates the area; therefore $\bar{x} = 0$ for the composite, and only $\bar{y}$ requires direct calculation using the subtraction method.
2. 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. (correct answer)
3. The axis of symmetry $x = 0$ of the triangle guarantees $\bar{x} = 0$ for the composite regardless of hole placement, because the centroid of any subtracted region lies on a line through the triangle's centroid, preserving the horizontal balance of the body.
4. The axis of symmetry $x = 0$ is broken by the hole, so $\bar{x} \neq 0$; however, because the hole center and the triangle centroid share the same $y$-coordinate, there remains a horizontal axis of symmetry at $y = a/\sqrt{3}$ for the composite body, meaning $\bar{y} = a/\sqrt{3}$ can be determined by symmetry alone without computation.

Question 6

1. The claim is correct: the plate has an axis of symmetry at $x = 0$, which forces $\bar{x} = 0$. Since all four holes lie at equal distances from this axis, the left-right balance is preserved. By this single symmetry argument alone the centroid is pinned to the y-axis, and a similar argument about the upper and lower holes then places it at the origin.
2. The claim is correct for $\bar{x} = 0$ and $\bar{y} = 0$, but only because the four holes are arranged with 4-fold rotational symmetry ($90^\circ$ rotation maps the plate onto itself). Reflective symmetry axes alone are insufficient to locate the centroid — rotational symmetry is the necessary condition for a centroid to lie at the geometric center of a body.
3. The claim is partially correct: the plate has symmetry about $x = 0$, forcing $\bar{x} = 0$, but the four holes break the symmetry about $y = 0$ because two holes lie in the upper half-plane and two in the lower, making the mass distribution unequal above and below the x-axis. Therefore $\bar{y}$ requires integration.
4. 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. (correct answer)

Question 7

1. $\bar{y} = \dfrac{\rho \cdot \tfrac{2}{3}\pi R^3 \cdot \tfrac{3R}{8} + \sigma \pi R^2 \cdot 0}{\rho \cdot \tfrac{2}{3}\pi R^3 + \sigma \pi R^2} = \dfrac{\tfrac{\rho \pi R^4}{4}}{\pi R^2\!\left(\tfrac{2\rho R}{3} + \sigma\right)} = \dfrac{\rho R^2}{4\!\left(\tfrac{2\rho R}{3} + \sigma\right)}$, since the disk lies in the plane $y = 0$ and its centroid contributes zero to the first moment. (correct answer)
2. $\bar{y} = \dfrac{\rho \cdot \tfrac{2}{3}\pi R^3 \cdot \tfrac{3R}{8} + \sigma \pi R^2 \cdot \tfrac{3R}{8}}{\rho \cdot \tfrac{2}{3}\pi R^3 + \sigma \pi R^2} = \dfrac{3R}{8}$, because the disk lies on the flat face of the hemisphere and therefore shares the hemisphere's centroid height $3R/8$, not $y=0$.
3. $\bar{y} = \dfrac{\rho \cdot \tfrac{2}{3}\pi R^3 \cdot \tfrac{3R}{8} + \sigma \pi R^2 \cdot 0}{\rho \cdot \tfrac{2}{3}\pi R^3 + \sigma \pi R^2}$, which simplifies to $\dfrac{\rho R^2}{4\left(\tfrac{2\rho R}{3} + \sigma\right)}$; however, the student's symmetry argument is incomplete because it neglects to verify that the composite body (hemisphere plus disk) retains axial symmetry about the y-axis, which must be confirmed separately before concluding $\bar{x} = \bar{z} = 0$.
4. $\bar{y} = \dfrac{\rho \cdot \tfrac{2}{3}\pi R^3 \cdot \tfrac{3R}{8} + \sigma \pi R^2 \cdot R}{\rho \cdot \tfrac{2}{3}\pi R^3 + \sigma \pi R^2}$, because the disk's centroid is located at $y = R$ — the radius of the disk measured perpendicular to its face — rather than at the origin.

Question 8

1. The student is correct that the triangle has a line of symmetry along $y = x$: reflecting the triangle across $y = x$ swaps the vertices $(2a, 0)$ and $(0, 2a)$ while fixing the origin, mapping the triangle onto itself. By the '1/3 rule' for triangles, this symmetry directly gives $\bar{x} = \bar{y} = 2a/3$ with no further calculation needed.
2. 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 $\left(\frac{2a}{3},\, \frac{2a}{3}\right)$. (correct answer)
3. The student is correct that reflecting across $y = x$ maps the triangle onto itself, so $\bar{x} = \bar{y}$ and the centroid lies on $y = x$. Because the right-angle vertex is at the origin and each leg has length $2a$, the centroid is at $\bar{x} = \bar{y} = a/3$, located one-third of the leg length from the right-angle vertex.
4. The student is incorrect: the line $y = x$ is not an axis of symmetry for this right triangle because a true axis of symmetry must pass through a vertex and bisect the opposite side perpendicularly, which $y = x$ does not do here. Therefore both $\bar{x}$ and $\bar{y}$ must be found independently by integration.

Question 9

1. The student's argument for $\bar{x} = 0$ is valid, but $\bar{y}$ is incorrectly computed as $\bar{y} = \dfrac{(3\lambda L)(0) + (2\lambda L)(L)}{3\lambda L + 2\lambda L} = \dfrac{2L}{5}$, because Rod 2's linear density is $2\lambda$ but its length is only $L$ when measuring from the centroid, so its effective mass is $2\lambda L$.
2. 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 $y = L$. (correct answer)
3. The student's argument for $\bar{x} = 0$ is valid, and $\bar{y}$ is correctly computed as $\bar{y} = \dfrac{(3\lambda L)(0) + (4\lambda L)(2L)}{3\lambda L + 4\lambda L} = \dfrac{8L}{7}$, where the centroid of Rod 2 is taken as its top endpoint $y = 2L$ rather than its midpoint.
4. The student's argument for $\bar{x} = 0$ is valid, and $\bar{y} = \dfrac{(3\lambda L)(0) + (2\lambda L)(L)}{3\lambda L + 2\lambda L} = \dfrac{2L}{5}$, where the mass of Rod 2 is $2\lambda L$ using its length $2L$ and the base linear density $\lambda$ rather than its actual density $2\lambda$.