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📝 Center of Gravity and Centroid of a Solid (15 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 15 questions available

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What is Center of Gravity and Centroid of a Solid?

Definition:
The centroid is the center of gravity for a uniform density solid. For variable density, use weighted averages: $\bar{x} = \frac{\iiint_E x\rho \, dV}{\iiint_E \rho \, dV}$ .

Example:
The centroid of a uniform tetrahedron is the average of its vertices' coordinates.

Reason:
Centroids represent geometric centers, useful in statics and dynamics for simplifying force and motion analysis.

Easy

Medium

Hard

📝 All Center of Gravity and Centroid of a Solid MCQs

Q1. A homogeneous solid occupies a region $E$ . Which statement best explains why its center of gravity and centroid coincide?

A.Both are determined only by the surface area of the solid.

B.Uniform density makes the mass distribution proportional to volume, so the mass-weighted position equals the geometric average position. ✅

C.The centroid is always located at the geometric center of the bounding box.

D.Gravity has no effect on the location of the centroid.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a homogeneous solid, density is constant throughout the region. Therefore, each small volume element contributes mass in direct proportion to its volume. In the center-of-gravity integrals, the constant density cancels from the numerator and denominator, leaving the volume-weighted coordinates that define the centroid.

Q2. For a solid with density $\rho(x,y,z)$ , which expression correctly represents its $x$ -coordinate of the center of gravity?

x_G=\frac{\iiint_E x\,dV}{\iiint_E dV}

x_G=\frac{\iiint_E \rho\,dV}{\iiint_E x\,dV}

x_G=\frac{\iiint_E x\rho(x,y,z)\,dV}{\iiint_E \rho(x,y,z)\,dV}

✅

x_G=\iiint_E x\rho(x,y,z)\,dV

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The center of gravity is a mass-weighted average position. A small element has mass $dm=\rho\,dV$ , so its contribution to the first moment about the relevant plane is $x\rho\,dV$ . Dividing the total first moment by total mass gives $x_G=\frac{\iiint_E x\rho\,dV}{\iiint_E\rho\,dV}$ .

Q3. A homogeneous solid is symmetric about the $yz$ -plane, but its shape is not symmetric about either the $xz$ - or $xy$ -plane. What can be concluded about its centroid?

A.Only its

x

-coordinate is zero. ✅

B.Only its

y

-coordinate is zero.

C.Only its

z

-coordinate is zero.

D.All three coordinates must be zero.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Reflection across the $yz$ -plane changes $x$ to $-x$ while leaving $y$ and $z$ unchanged. Because corresponding volume elements occur in symmetric pairs, their $x$ -moments cancel. Thus $x_G=0$ . No corresponding symmetry is given for the other coordinates, so they cannot automatically be assumed to vanish.

Q4. A homogeneous solid has a centroid at $(2,-1,4)$ . The solid is translated by the vector $(-3,5,2)$ without changing its shape or density. Where is the new centroid?

(5,-6,2)

(-1,4,6)

✅

(-1,-5,8)

(2,4,6)

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Translation moves every point of a solid by the same displacement, so the centroid moves by exactly that displacement. Adding $(-3,5,2)$ to the original centroid $(2,-1,4)$ gives $(2-3,-1+5,4+2)=(-1,4,6)$ . No recalculation of triple integrals is necessary.

Q5. A manufacturer models a solid component by $E$ and assigns density $\rho(x,y,z)=kz$ , where $k>0$ . Which modeling consequence is most important when locating the center of gravity?

A.The value of

k

changes the center of gravity substantially.

B.The center of gravity must lie on the

xy

-plane because density depends on

z

C.Regions at larger positive

z

contribute more strongly to the mass-weighted position. ✅

D.The centroid and center of gravity are necessarily identical.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since $\rho=kz$ , mass density increases as $z$ increases within the region where the model is physically meaningful. Consequently, elements farther in the positive $z$ -direction receive greater mass weighting. The constant $k$ cancels when coordinates are divided by total mass, while the spatial variation in $z$ remains important.

Q6. A solid occupies $0\le x\le2$ , $0\le y\le1$ , and $0\le z\le3$ , with density increasing linearly with $x$ . Without evaluating every integral, which prediction is most reasonable for the $x$ -coordinate of its center of gravity?

A.It must be less than

1

B.It must equal

1

because the geometric midpoint is

1

C.It should be greater than

1

, because larger

x

-values carry greater density and therefore greater mass. ✅

D.It must equal

2

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For uniform density, the rectangular region would have its $x$ -coordinate of centroid at $1$ . When density increases with $x$ , mass is redistributed toward the larger- $x$ side. Therefore the center of gravity shifts rightward from $1$ , but it cannot exceed the boundary $x=2$ .

Q7. An engineer computes $M=\iiint_E\rho\,dV$ and $M_{yz}=\iiint_E x\rho\,dV$ , then reports $x_G=M_{yz}M$ . What is the error?

A.The first moment should be divided by total mass, not multiplied by it. ✅

B.The density should be removed from both integrals.

C.The first moment should be integrated only over the boundary.

D.The

x

-coordinate must always be zero for a solid.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The total mass is $M=\iiint_E\rho\,dV$ , while the first moment about the $yz$ -plane is $M_{yz}=\iiint_E x\rho\,dV$ . The coordinate of the center of gravity is the ratio $x_G=M_{yz}/M$ . Multiplying by $M$ produces incorrect dimensions and does not represent a weighted average.

Q8. A student claims: 'If a solid is symmetric about the $xy$ -plane, its center of gravity must have $z_G=0$ , even if density is greater above the plane.' Which evaluation is correct?

A.The claim is correct because geometric symmetry always determines the center of gravity.

B.The claim is incorrect because density symmetry is also required; asymmetric density can shift

z_G

away from zero. ✅

C.The claim is correct only when the solid is a cube.

D.The claim is incorrect because symmetry has no role in center-of-gravity calculations.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Geometric symmetry alone is sufficient for a centroid of a homogeneous solid, but a center of gravity depends on mass distribution. If density is greater above the $xy$ -plane than below it, corresponding volume elements do not have equal masses. Their $z$ -moment therefore fails to cancel, shifting the center of gravity upward.

Q9. A graph of a solid's cross-sectional density shows density increasing steadily as $z$ increases, while the solid extends equally from $z=-2$ to $z=2$ . Which qualitative graph or conclusion best represents $z_G$ ?

A.The center of gravity remains exactly at

z=0

B.The center of gravity lies below

z=0

C.The center of gravity lies above

z=0

, because the upper portions have greater density. ✅

D.The center of gravity must be at

z=2

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Although the geometry is vertically symmetric, the density graph is not. The upper half contains more mass per unit volume than the lower half. Therefore the positive $z$ -contributions to the first moment receive greater weights, producing a positive $z_G$ . The center of gravity remains inside the solid rather than reaching the boundary.

Q10. Two methods are proposed for a homogeneous solid: Method I evaluates three triple integrals directly for the centroid coordinates; Method II finds the centroid of each horizontal slice and then averages the slice centroids without weighting by slice volume. Which assessment is best?

A.Both methods are always equivalent.

B.Method II is valid only if every horizontal slice has the same volume. ✅

C.Method I is invalid because centroid calculations require surface integrals.

D.Method II is always better because it uses fewer integrals.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A centroid is a volume-weighted average. If horizontal slices have different volumes, simply averaging their individual centroids gives every slice equal influence, which is generally incorrect. Method II becomes valid when the slices being averaged contribute equally to volume or when the appropriate volume weighting is explicitly included.

Q11. A solid has uniform density and is divided into two disjoint parts with masses $m_1,m_2$ and centroids $G_1=(x_1,y_1,z_1)$ , $G_2=(x_2,y_2,z_2)$ . Which expression correctly determines the combined center of gravity?

G=\frac{G_1+G_2}{2}

G=\frac{m_1G_1+m_2G_2}{m_1+m_2}

✅

G=m_1G_1+m_2G_2

G=\frac{G_1}{m_1}+\frac{G_2}{m_2}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The combined center of gravity is a mass-weighted average of the centers of gravity of the two parts. Each component's location must be multiplied by its mass before the contributions are added. Dividing by $m_1+m_2$ normalizes the result, producing the correct weighted coordinate in all three directions.

Q12. A designer removes a small spherical cavity from one side of a homogeneous solid. The cavity was originally part of the solid and has centroid $C$ and volume $V_c$ . Which strategy correctly accounts for the removed material when finding the new centroid?

A.Treat the cavity as having negative mass located at

C

. ✅

B.Ignore the cavity because it has zero density after removal.

C.Move the original centroid toward

C

D.Treat the cavity as positive mass located at the new centroid.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A removed region can be modeled using negative mass or negative volume in the moment calculation. If the original solid has mass $M$ and centroid $G$ , the new first moment is obtained by subtracting the cavity's contribution $m_cC$ . This avoids recomputing the entire remaining solid from scratch.

Q13. Consider a homogeneous solid bounded by $z=0$ and $z=4-x^2-y^2$ , where the upper surface is a downward-opening paraboloid. From the geometry alone, which conclusion about the centroid is justified?

A.The centroid must be at

z=4

B.The centroid must lie below

z=2

because the solid contains more volume near

z=0

than near the top.

C.The centroid must have

x_G=y_G=0

, while its

z

-coordinate lies between

0

and

4

. ✅

D.All three centroid coordinates must equal

2

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The solid is rotationally symmetric about the $z$ -axis, so reflection and rotational symmetry force $x_G=y_G=0$ . The centroid must lie inside the solid, so $0<z_G<4$ . Its exact vertical position requires volume and moment analysis; the geometry alone does not justify assigning a specific value such as $2$ .

Q14. A solid has density $\rho(x,y,z)=1+x^2$ and occupies a region symmetric about the $yz$ -plane. A student argues that $x_G$ must be positive because density increases as $|x|$ increases. What is the best response?

A.The student is correct because greater density always shifts the center toward positive

x

B.The student is incorrect because the density is symmetric in

x

, so positive and negative

x

-moments cancel. ✅

C.The student is correct because

x^2

is always positive.

D.The center of gravity must be at

x=1

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The density $1+x^2$ is an even function of $x$ , so points at $x$ and $-x$ have equal density. Because the solid is also symmetric about the $yz$ -plane, their weighted $x$ -moments are equal in magnitude and opposite in sign. Thus they cancel and $x_G=0$ , despite higher density farther from the plane.

Q15. A homogeneous solid is formed by joining two regions whose masses are in the ratio $1:3$ . Their centroids lie at $x=5$ and $x=1$ , respectively. Without calculating the other coordinates, where must the combined $x$ -coordinate lie?

A.Exactly at

3

B.Exactly at

2

C.Between

1

and

2

, closer to

1

✅

D.Between

4

and

5

, closer to

5

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Let the masses be $m$ and $3m$ . The combined coordinate is $x_G=(m(5)+3m(1))/(4m)=8/4=2$ . Thus it lies between $1$ and $5$ , and because the second part has three times the mass, the result is pulled toward $1$ . This illustrates mass-weighted positioning rather than an ordinary average.

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←Previous Topic:Density and Mass of an Inhomogeneous Lamina📝 Center of Gravity and Centroid of a SolidCenter of Gravity of an Inhomogeneous Lamina:Next Topic→