📝 Moments and centers of gravity calculus (37 MCQs)

📖 Explanation: The combined center of gravity is the weighted average of the individual centers. For the x-coordinate: 0 = (12*2 + 3*x)/15 -> 0 = 24 + 3x -> x = -8. For the y-coordinate: 0 = (12*5 + 3*y)/15 -> 0 = 60 + 3y -> y = -20. Therefore, the mass must be placed at (-8, -20).

📖 Explanation: The formula for the y-coordinate of the centroid of a region under a curve is $\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2} [f(x)]^2 dx$. The student omitted the 1/2 factor, which is essential for calculating the first moment about the x-axis. The given formula would calculate twice the correct value of $\bar{y}$.

📖 Explanation: The moment about the x-axis is calculated by summing the product of each mass and its y-coordinate. For the 4 kg mass at y=2, the moment is 4*2=8. For the 6 kg mass at y=4, the moment is 6*4=24. The total moment about the x-axis is 8+24=32. Wait, the correct calculation is 8 + 24 = 32. Let's re-evaluate: 4*2 = 8 and 6*4 = 24. The sum is 32, not 22. The correct answer should be 32. Let's correct the options. The correct option is 32. However, I need to adjust the options. Let's set the correct answer to 32.

📖 Explanation: The centroid of a symmetric figure like a circle or a uniformly dense triangle is a point that balances the shape. When a hole is cut, the remaining shape's center of gravity shifts away from the removed mass. Since the hole is centered exactly at the centroid, the removed mass is symmetrically distributed around that point, and the remaining shape will still be balanced at the same point. The centroid remains at (2/3, 4/3).

📖 Explanation: The region is symmetric about the y-axis. For any point (x, y) in the region, the point (-x, y) is also in the region. This symmetry implies that the first moment about the y-axis is zero. Therefore, the x-coordinate of the centroid, which is the moment about the y-axis divided by the area, is 0.

📖 Explanation: The center of gravity of a rigid body moves with the body under translation. When the entire lamina is shifted by a vector (4,3), every point, including the center of gravity, shifts by the same amount. Therefore, the new center of gravity is (3+4, 2+3) = (7,5).

📖 Explanation: The first moment of a lamina about the x-axis is defined as $M_x = \iint_R y \, dA$. For a lamina with total mass M and centroid $\bar{y}$, we have $M_x = M \bar{y}$. A lamina with uniform density has its center of gravity at its centroid, and the mass is proportional to area. If the density is 1 (or we are considering area moments), $M_x = A \bar{y} = 10 \times 2 = 20$.

📖 Explanation: The center of mass for a continuous object is found using integrals. For a rod from 0 to 10, the total mass M = ∫ ρ(x) dx = ∫₀¹⁰ x dx = 50. The first moment about the origin is M₀ = ∫ x ρ(x) dx = ∫₀¹⁰ x² dx = 1000/3. The center of mass is $\bar{x} = M_0 / M = (1000/3) / 50 = 20/3 \approx 6.67$ m.

📖 Explanation: The student's setup and calculation are correct for finding the centroid of the region under the curve y = f(x). The formula for $\bar{x}$ is $\frac{1}{A} \int_a^b x f(x) dx$, and for $\bar{y}$ it is $\frac{1}{A} \int_a^b \frac{1}{2} [f(x)]^2 dx$. The student used the correct formula for $\bar{x}$, integrated correctly, and arrived at the correct value of 6/5.

📖 Explanation: This is a problem of finding the centroid of a composite body with a hole. The area of the large disk is πR². The area of the hole is π(R/2)² = πR²/4. The centroid of the large disk is at (0,0). The centroid of the hole is at (R/2, 0). The remaining area is 3πR²/4. The new x-coordinate is $\bar{x} = \frac{0 - (πR²/4)(R/2)}{3πR²/4} = \frac{-πR³/8}{3πR²/4} = -R/6$.

📖 Explanation: The centroid is the geometric center of the square. The center of gravity depends on the mass distribution. Since the density increases linearly from bottom to top, the mass is concentrated more towards the top. This will shift the center of gravity upwards, above the geometric centroid. The x-coordinate remains centered due to symmetry, but the y-coordinate increases.

📖 Explanation: The moment of inertia, or second moment of mass, is a measure of an object's resistance to rotational acceleration about a specific axis. For a point mass m at a perpendicular distance r from the axis, it is defined as the product of the mass and the square of the distance, $I = mr^2$.

📖 Explanation: The mass of a lamina with variable density $\rho(x,y)$ over a region R is given by the double integral $M = \iint_R \rho(x,y) \, dA$. The region is bounded by $y = x^2$ (bottom) and $y = x$ (top) for $0 \le x \le 1$. Therefore, the correct integral is $\int_{0}^{1} \int_{x^2}^{x} xy \, dy \, dx$.

📖 Explanation: The center of gravity is the weighted average of the positions of the masses, where the weights are the masses themselves. For two masses, $\bar{x} = (m_1 x_1 + m_2 x_2)/(m_1 + m_2)$. This point is indeed closer to the mass with the larger weight. If $m_2 > m_1$, then $\bar{x}$ will be closer to $x_2$ than to $x_1$.

📖 Explanation: The region bounded by $y = 1 - x^2$ and the x-axis is symmetric about the y-axis. The curve is even ($f(-x) = f(x)$), so for every point (x, y) in the region, (-x, y) is also in the region. This symmetry means the first moment about the y-axis is zero, and the x-coordinate of the centroid is 0.

📖 Explanation: The y-coordinate of the centroid for a region under $y = f(x)$ is $\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2} [f(x)]^2 dx$, where $A = \int_a^b f(x) dx$. For f(x) = sin x from 0 to π, this becomes $\bar{y} = \frac{\int_0^\pi \frac{1}{2} \sin^2 x \, dx}{\int_0^\pi \sin x \, dx}$.

📖 Explanation: For a system to be in equilibrium about the origin, the sum of the moments about the origin must be zero. For three masses, the condition is $m_1 (2,1) + m_2 (-1,3) + m_3 (x,y) = (0,0)$. Without knowing the values of $m_1$ and $m_2$, the location of the third mass (x, y) cannot be uniquely determined. It depends on the masses and the positions of the first two.

📖 Explanation: When a rod is bent into a semicircle, the center of gravity moves. It remains on the axis of symmetry (the line through the center of the circle and the midpoint of the arc). It is not at the center of the circle because the mass is distributed along the arc, not the diameter. For a uniform semicircular wire, the center of gravity is located at a distance $2R/\pi$ from the center along the axis of symmetry, which is inside the semicircle.

📖 Explanation: For a solid hemisphere of radius R with its flat face on the xy-plane and its axis of symmetry along the z-axis, the center of mass is located at a distance $3R/8$ from the base along the z-axis. The x and y coordinates are zero by symmetry.

📖 Explanation: Scaling a 2D region by a factor of 2 about the origin multiplies all coordinates by 2, so the centroid will move to (2*2, 2*3) = (4,6). The area of the region scales by the square of the scaling factor, so the new area is $10 \times 2^2 = 40$.

📖 Explanation: While it's often said that weight acts at the center of gravity, the precise definition is that the center of gravity is the point about which the sum of the moments of the weights of all the individual particles is zero. This ensures that the gravitational forces can be considered to act as a single resultant force at that point. The geometric center (centroid) is only the center of gravity if the density is uniform.

📖 Explanation: The centroid's x-coordinate is the average of the x-distances of the area. For a curve that is higher (like y=2x compared to y=x² for x in [0,2]), more area is concentrated towards the right, pulling the x-coordinate of the centroid to the right. Since y=2x is above y=x² in the interval (0,2), the x-coordinate of the centroid for y=2x is larger than that for y=x². By symmetry, we can reason that the x-coordinate for y=x² is 1.5, and for y=2x it's 4/3, so the statement is correct.

📖 Explanation: For a rectangle with uniform density, the center of gravity coincides with its geometric center, which is the point of intersection of its diagonals. The coordinates of this point are the averages of the x and y coordinates of the vertices. In this case, the center of gravity is at $(a/2, b/2)$.

📖 Explanation: The center of gravity is a weighted average. If all masses are multiplied by the same factor (2 in this case), the factor cancels out in both the numerator and denominator of the weighted average formula. The ratio remains the same, so the center of gravity does not change.

📖 Explanation: The area of the region is $A = \int_{-2}^{2} (4 - x^2) dx = 32/3$. The moment about the x-axis is $M_x = \int_{-2}^{2} \frac{1}{2} (4 - x^2)^2 dx = \frac{1}{2} \int_{-2}^{2} (16 - 8x^2 + x^4) dx = \frac{1}{2} [16x - \frac{8x^3}{3} + \frac{x^5}{5}]_{-2}^{2} = \frac{1}{2} [\frac{256}{15}] = \frac{128}{15}$. The y-coordinate of the centroid is $\bar{y} = M_x / A = (128/15) / (32/3) = 8/5$.

📖 Explanation: The centroid (center of gravity for uniform density) of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is the average of the coordinates of the vertices: $((x1+x2+x3)/3, (y1+y2+y3)/3)$. For (0,0), (2,0), and (0,4), the centroid is $((0+2+0)/3, (0+0+4)/3) = (2/3, 4/3)$.

📖 Explanation: For a composite lamina, the centroid is the area-weighted average of the centroids of its parts. The x-coordinate is $\bar{x} = (2*1 + 4*3)/(2+4) = (2+12)/6 = 14/6 = 7/3 \approx 2.33$. The y-coordinate is $\bar{y} = (2*1 + 4*2)/6 = (2+8)/6 = 10/6 = 5/3 \approx 1.67$. The centroid is (7/3, 5/3).

📖 Explanation: The moment of a point mass about the origin is defined as $M = m \times r$, where $r$ is the distance from the origin. If the moment is 0, then either the mass is 0, or the distance from the origin is 0. Since the coordinates are (4,6), the distance is not 0. Therefore, the mass must be 0 to have a moment of 0.

📖 Explanation: The area is $A = \int_0^4 \sqrt{x} dx = 16/3$. The moment about the x-axis is $M_x = \frac{1}{2} \int_0^4 (\sqrt{x})^2 dx = \frac{1}{2} \int_0^4 x dx = \frac{1}{2} * 8 = 4$. The y-coordinate is $\bar{y} = M_x / A = 4 / (16/3) = 12/16 = 3/4 = 0.75$. The correct answer is 0.75, which is not listed. Let's check the options. None match. I need to adjust the options. The correct answer should be 0.75. Let's set the correct answer to 0.75 and adjust options. The correct option should be 0.75.

📖 Explanation: The x-coordinate of the centroid is given by $\bar{x} = \frac{1}{A} \int_0^1 x \cdot x^3 dx = \frac{1}{A} \int_0^1 x^4 dx = \frac{1}{A} \cdot \frac{1}{5}$. We are given that $\bar{x} = 4/5$. Therefore, $4/5 = (1/5)/A$, which implies $A = 1/4$.

📖 Explanation: The original rectangle has area 2 and centroid (0.5,1). The cut-off triangle has area 1 and centroid (1/3, 2/3). The remaining area is 1. The new centroid is the area-weighted average of the original rectangle and the negative area of the triangle. $\bar{x} = (2*0.5 - 1*(1/3))/(2-1) = (1 - 1/3) = 2/3$. $\bar{y} = (2*1 - 1*(2/3))/(2-1) = (2 - 2/3) = 4/3$. The centroid is (2/3, 4/3). Let's re-evaluate the problem. The remaining shape is a triangle with vertices at (0,0), (1,0), and (0,2). The centroid of this triangle is (1/3, 2/3). The correct answer is (1/3, 2/3).

📖 Explanation: The center of mass of a solid cone with uniform density lies on its axis of symmetry. It is located at a distance $h/4$ from the base (or $3h/4$ from the apex). This is a standard result for the centroid of a cone.

📖 Explanation: The first moment of a lamina about the x-axis is defined as $M_x = \iint_R y \rho(x,y) \, dA$. Since $\rho(x,y) = y$, the integral becomes $\iint_R y^2 \, dA$, which is the second moment about the x-axis. The integral $\iint_R y \rho(x,y) \, dA$ is the moment about the x-axis. If $\rho(x,y) = y$, then the integral is $\iint_R y^2 \, dA$, which is the second moment, not the first moment. The correct interpretation is the second moment.

📖 Explanation: For a uniform semicircular wire of radius R, the center of gravity lies on the axis of symmetry at a distance $2R/\pi$ from the straight edge (diameter). This is found by integrating the position of each infinitesimal length element along the wire and dividing by the total length.

📖 Explanation: For two identical masses, the center of gravity is the midpoint of the line segment joining them. The midpoint of (0,0) and (2,0) is (1,0).

📖 Explanation: The center of gravity is defined as the point about which the sum of the moments of the weights of all particles is zero. If the origin is chosen at this point, the condition for equilibrium is satisfied, and the sum of the moments about the origin is zero.

📖 Explanation: The area A is $\int_0^1 e^x dx = e - 1$. The first moment about the y-axis is $M_y = \int_0^1 x e^x dx$. Using integration by parts, $M_y = [x e^x]_0^1 - \int_0^1 e^x dx = e - (e-1) = 1$. The x-coordinate of the centroid is $\bar{x} = M_y / A = 1 / (e-1)$.