Statics and Dynamics Quiz: Center Of Mass

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Center Of MassQuestion 1 of 4

A planar composite plate is made by starting with a uniform solid square plate of side $a = 0.6\text{ m}$ and areal mass density $\sigma = 15\text{ kg/m}^2$ , with its center at the origin. A uniform solid circular disk of radius $r = 0.1\text{ m}$ and the same areal density $\sigma$ is then attached concentrically at each of the four corners of the square. The four corner circles partially overlap the square plate (each circle's center is exactly at a corner of the square at coordinates $\pm 0.3\text{ m}, \pm 0.3\text{ m}$ ).

Given the four-fold symmetry of the assembly, the center of mass must lie at the origin. A student instead claims the center of mass shifts toward the corner at $(+0.3, +0.3)\text{ m}$ because the added disks have non-zero moments. What is the most precise rebuttal?

The student is correct: the four added disks each have centroids offset from the origin, creating non-zero individual moments, so the center of mass must shift toward the centroid of the added disks, which lies at the average of the four corner coordinates.

The student is wrong because the four added disks all have equal mass, and equal masses at symmetric locations produce moment contributions that cancel pairwise, leaving the total moment unchanged and the center of mass at the origin regardless of the square's own contribution.

The student is wrong because each added disk's centroid is at a corner, and the centroid of the set of four corner positions is

[(+0.3,+0.3)+(-0.3,+0.3)+(-0.3,-0.3)+(+0.3,-0.3)]/4 = (0,0)

, so the net moment added by the four disks is zero, keeping the overall center of mass at the origin — which already coincided with the square's centroid.

The student is wrong because the added disks overlap the square plate, and the overlapping area reduces the effective mass of each disk, making their net contribution to the total moment smaller than the student assumed, thereby keeping the center of mass near but not exactly at the origin.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Center Of Mass

Practice Center Of Mass in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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

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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.

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Question 1

The student is correct: the four added disks each have centroids offset from the origin, creating non-zero individual moments, so the center of mass must shift toward the centroid of the added disks, which lies at the average of the four corner coordinates.
The student is wrong because the four added disks all have equal mass, and equal masses at symmetric locations produce moment contributions that cancel pairwise, leaving the total moment unchanged and the center of mass at the origin regardless of the square's own contribution.
The student is wrong because each added disk's centroid is at a corner, and the centroid of the set of four corner positions is $[(+0.3,+0.3)+(-0.3,+0.3)+(-0.3,-0.3)+(+0.3,-0.3)]/4 = (0,0)$ , so the net moment added by the four disks is zero, keeping the overall center of mass at the origin — which already coincided with the square's centroid. (correct answer)
The student is wrong because the added disks overlap the square plate, and the overlapping area reduces the effective mass of each disk, making their net contribution to the total moment smaller than the student assumed, thereby keeping the center of mass near but not exactly at the origin.

Explanation: Whenever you see a center-of-mass problem involving a symmetric arrangement of added components, your first instinct should be to check the geometry's symmetry before doing any arithmetic. The center of mass of a composite body is found by computing the weighted sum of each part's centroid position. Here, the square plate already has its centroid at the origin. The four circular disks are centered at

(+0.3,+0.3)

(-0.3,+0.3)

(-0.3,-0.3)

, and

(+0.3,-0.3)

. Because all four disks share the same radius and the same areal density, they have equal masses. The net moment they contribute is proportional to the vector sum of their centroid positions:

\frac{(+0.3,+0.3)+(-0.3,+0.3)+(-0.3,-0.3)+(+0.3,-0.3)}{4} = (0,0)

The four positions cancel exactly, adding zero net moment. Combined with the square's centroid already at the origin, the overall center of mass remains at the origin. That makes C the correct and most precise answer. A is wrong because the student's reasoning only looks at individual disk moments in isolation — it ignores that the four moments are vectors that must be summed, and their vector sum is zero. B is partially correct in spirit but imprecise: it says contributions "cancel pairwise," which is true, but it skips explicitly computing the centroid of the four corner positions, making it a less rigorous rebuttal than C. D introduces the overlap as a relevant factor, but overlap doesn't break the four-fold symmetry — whatever mass reduction occurs happens equally at all four corners, so the center of mass still stays at the origin. The key study tip: symmetry arguments are your fastest tool. If a configuration has four-fold (or any rotational) symmetry about a point, the center of mass must lie on that point — no calculation required.

Question 2

A composite beam cross-section consists of three rectangular sections arranged along the y-axis. Section 1 has mass $m_1 = 3\text{ kg}$ centered at $y_1 = 0\text{ cm}$ . Section 2 has mass $m_2 = 5\text{ kg}$ centered at $y_2 = 8\text{ cm}$ . Section 3 has mass $m_3 = 2\text{ kg}$ centered at $y_3 = 14\text{ cm}$ . An engineer then removes a small plug of mass $m_4 = 1\text{ kg}$ from Section 2, where the plug's centroid was located at $y_4 = 6\text{ cm}$ .

What is the y-coordinate of the center of mass of the final composite body after the plug is removed?

$\bar{y} = \dfrac{3(0) + 5(8) + 2(14)}{3+5+2} = 6.8\text{ cm}$ , since removing the plug does not shift the centroid of the remaining body once total mass is adjusted.
$\bar{y} = \dfrac{3(0) + 5(8) + 2(14) - 1(6)}{3+5+2-1} = \dfrac{62}{9} \approx 6.89\text{ cm}$ , treating the removed plug as a negative mass contribution subtracted from both numerator and denominator. (correct answer)
$\bar{y} = \dfrac{3(0) + 4(8) + 2(14)}{3+4+2} = \dfrac{60}{9} \approx 6.67\text{ cm}$ , by reducing the mass of Section 2 from 5 kg to 4 kg while keeping its centroid at $y_2 = 8\text{ cm}$ .
$\bar{y} = \dfrac{3(0) + 5(8) + 2(14) - 1(8)}{3+5+2-1} = \dfrac{60}{9} \approx 6.67\text{ cm}$ , subtracting the plug's mass at Section 2's centroid location rather than the plug's actual centroid.

Explanation: When a composite body has material removed, the most reliable approach is the negative mass method: treat the removed piece as a negative contribution in both the numerator (moment sum) and denominator (total mass). This keeps your centroid formula consistent and avoids guessing how removal shifts individual section properties. For this problem, start with the full composite moment sum and total mass, then subtract the plug's contribution at its actual centroid location (

y_4 = 6\text{ cm}

, not Section 2's centroid):

\bar{y} = \frac{3(0) + 5(8) + 2(14) - 1(6)}{3 + 5 + 2 - 1} = \frac{0 + 40 + 28 - 6}{9} = \frac{62}{9} \approx 6.89\text{ cm}

This confirms B is correct. A is wrong because simply adjusting the total mass without subtracting the plug's moment ignores the spatial contribution of the removed material — the centroid absolutely does shift. C reduces Section 2's mass by 1 kg but keeps its centroid at

y = 8\text{ cm}

, implicitly assuming the plug was centered at 8 cm. Since the plug's centroid is actually at 6 cm, this misrepresents which moment is being removed. D makes exactly that same location error — it subtracts the plug's mass but applies it at

y = 8\text{ cm}

(Section 2's centroid) instead of the plug's true centroid at

y = 6\text{ cm}

. Always use the plug's own centroid, not the host section's. Study tip: In composite body problems involving removal, always ask: "Where exactly was the removed piece centered?" That location — not the parent section's centroid — is what enters your moment calculation.

Question 3

A spacecraft is modeled as two modules connected by a rigid massless truss. Module A has mass $M_A = 500\text{ kg}$ and its center of mass is at position $\vec{r}_A = (1, 2, 0)\text{ m}$ . Module B has mass $M_B = 300\text{ kg}$ and its center of mass is at $\vec{r}_B = (7, -1, 4)\text{ m}$ . Fuel is stored in a spherical tank (uniform density) of mass $m_f = 200\text{ kg}$ centered at $\vec{r}_f = (4, 1, 2)\text{ m}$ . During a burn, fuel is consumed uniformly, reducing the tank mass to $m_f' = 50\text{ kg}$ while the tank's centroid location does not change (the tank remains centered at the same point).

What is the displacement vector $\Delta\vec{r}_{cm}$ (final minus initial center of mass position) of the spacecraft system's center of mass due to fuel consumption?

$\Delta\vec{r}_{cm} = \vec{0}$ , since fuel consumption is an internal process and the center of mass of a system cannot move due to internal mass redistribution alone, per Newton's laws.
$\Delta\vec{r}_{cm} = 150\vec{r}_f\!\left(\dfrac{1}{950}-\dfrac{1}{1000}\right)$ , pointing in the direction of $\vec{r}_f$ , obtained by differencing only the fuel term's contribution in numerator while keeping denominators fixed.
$\Delta\vec{r}_{cm} = \dfrac{(m_f - m_f')}{M_A+M_B+m_f'}\bigl(\vec{r}_{cm,i} - \vec{r}_f\bigr)$ , representing the shift of the CM toward the structural modules as fuel mass is removed, with magnitude proportional to the fuel lost.
$\Delta\vec{r}_{cm} = \left(\dfrac{M_A\vec{r}_A + M_B\vec{r}_B + m_f'\vec{r}_f}{M_A+M_B+m_f'}\right) - \left(\dfrac{M_A\vec{r}_A + M_B\vec{r}_B + m_f\vec{r}_f}{M_A+M_B+m_f}\right)$ , evaluating each state's CM separately and subtracting, which yields a shift directed away from $\vec{r}_f$ as the fuel-tank contribution diminishes. (correct answer)

Explanation: When a system loses mass (like a spacecraft burning fuel), the center of mass shifts because you're recalculating a weighted average with different weights. The key framework here is the center of mass formula:

\vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}

. To find how it changes, compute it at each state independently and subtract — there's no shortcut that avoids evaluating both denominators. Answer D applies this correctly. The initial CM uses total mass

M_A + M_B + m_f = 1000\text{ kg}

, and the final CM uses

M_A + M_B + m_f' = 850\text{ kg}

. Because the fuel tank's centroid stays fixed at

\vec{r}_f

while its mass drops from 200 kg to 50 kg, the fuel term contributes less to the numerator and the denominator shrinks, so the structural modules (A and B) pull the CM toward them. The displacement points away from

\vec{r}_f

— exactly what D states. Answer A is tempting but wrong. Newton's law about internal processes conserving CM momentum applies when no mass leaves the system. Here, fuel mass is removed from the system entirely, changing the system's total mass and therefore its CM location. Answer B is sloppy algebra — it treats only the numerator's fuel term as changing while leaving both denominators unchanged, which is physically and mathematically inconsistent. Answer C has intuitive appeal but uses an incorrect formula. The actual displacement isn't simply proportional to

({\vec{r}_{cm,i} - \vec{r}_f})

— the correct expression requires evaluating both CM states fully, as D does. Study tip: Whenever system mass changes (fuel burn, jettisoned module), always recompute CM from scratch at each state. Never assume the denominator stays constant.

Question 4

An engineer models a machine component as three uniform solid cylinders sharing a common axis (the x-axis). Cylinder 1: radius $r_1 = 0.10\text{ m}$ , length $L_1 = 0.20\text{ m}$ , density $\rho_1 = 7800\text{ kg/m}^3$ , centered at $x_1 = 0.10\text{ m}$ . Cylinder 2: radius $r_2 = 0.05\text{ m}$ , length $L_2 = 0.40\text{ m}$ , density $\rho_2 = 7800\text{ kg/m}^3$ , centered at $x_2 = 0.40\text{ m}$ . Cylinder 3: radius $r_3 = 0.08\text{ m}$ , length $L_3 = 0.10\text{ m}$ , density $\rho_3 = 2700\text{ kg/m}^3$ , centered at $x_3 = 0.65\text{ m}$ . All cylinders are solid with no voids.

A technician argues that because Cylinders 1 and 2 have the same density, he can replace them with a single equivalent mass at their combined centroid (simple average of $x_1$ and $x_2$ ) before computing the system's center of mass. Which statement best describes the technician's approach?

The approach is valid because equal densities imply equal mass, so the simple average of $x_1 = 0.10\text{ m}$ and $x_2 = 0.40\text{ m}$ gives the correct combined centroid at $x = 0.25\text{ m}$ , regardless of cylinder geometry.
The approach is invalid because the centroid of two sub-bodies is always the geometric midpoint of their centroids only when they have identical shapes; since the cylinders have different radii and lengths, a mass-weighted average must be used even for equal densities.
The approach is invalid because combining sub-bodies before computing the system centroid changes the total moment arm and introduces an error proportional to the density ratio $\rho_1/\rho_3$ , making the final answer density-dependent in a way the simple average does not capture.
The approach is valid provided the two cylinders are combined using a mass-weighted average of their centroids (not a simple average), since equal density means mass scales with volume, so $x_{12} = (m_1 x_1 + m_2 x_2)/(m_1+m_2)$ is still required and differs from $(x_1+x_2)/2$ . (correct answer)

Explanation: When finding the center of mass of a composite body, your fundamental tool is the mass-weighted average of centroids:

\bar{x} = \frac{\sum m_i x_i}{\sum m_i}

. The key insight is that this weighting is always required — equal density does not exempt you from it. Here's why D is correct: Even though Cylinders 1 and 2 share the same density

\rho

, their masses still differ because their volumes differ. With

m = \rho V = \rho \pi r^2 L

, you get

m_1 = \rho \pi (0.10)^2(0.20)

and

m_2 = \rho \pi (0.05)^2(0.40)

, which are not equal. So the combined centroid requires

x_{12} = (m_1 x_1 + m_2 x_2)/(m_1 + m_2)

, not simply

(x_1 + x_2)/2 = 0.25\text{ m}

. The technician's instinct to combine first is actually fine — it's the method of combining that fails. A is wrong because it assumes equal density implies equal mass, which is false when geometry differs. Equal density and equal volume would imply equal mass, but these cylinders have different radii and lengths. B is wrong in its reasoning. The flaw isn't that the cylinders have different shapes per se — it's specifically that their masses are unequal. A mass-weighted average is required whenever masses differ, full stop. C is a distractor that invents a spurious error proportional to

\rho_1/\rho_3

. Combining sub-bodies before the final calculation introduces no such density-ratio error; the issue is purely about using unweighted vs. weighted averaging. Study tip: On composite-body problems, always ask "are the masses equal?" not "are the densities equal?" before choosing a simple average.