Center of Mass and Centroids – Study Notes

Distribution of Force

• Line distribution: when a force is distributed over a line. Intensity expressed as force per unit length, denote as
  $f<em>L(s) \text{ (N/m)}$. Total force along the line: $F = \int f</em>L(s)\, ds.$
• Area distribution: when a force is distributed over an area. Intensity expressed as force per unit area, denote as
  $p(s) \text{ or } \sigma(s) \text{ (Pa)}$. For example, hydrostatic pressure acts as a force per unit area on surfaces:
  $F = \iint_A p\, dA.$
• Volume distribution: force distributed over a body (body force). Intensity expressed as force per unit volume, denote as
  $b(s) \text{ (N/m}^3)$. For gravity, the elemental weight is
  $dW = \rho g \, dV,$
  so the total weight is
  $W = \int dW = \int \rho g \, dV.$
• Special case for gravity: if the gravitational field is uniform (constant g) and density ρ is uniform, the weight density is constant; many expressions reduce to mass-based moments with dm = ρ dV.
• Summary of force distributions:
  • Line: line density or line force per length, use dx, dy, or ds as the differential element.
  • Area: surface pressure or surface stress, use dA.
  • Volume: body force density, use dV.

Center of Mass and Centroids

• Center of mass (CM) is the mass-weighted average position of all particles in the body:
  • For two discrete masses m₁ and m₂ at positions x₁ and x₂ along a chosen axis:
    $X<em>{cm} = \frac{m</em>1 x<em>1 + m</em>2 x<em>2}{m</em>1 + m_2}.$
  • In 3D, similarly for y and z coordinates, using all masses.
• Center of gravity (CG) is the point where the resultant gravitational torque vanishes; in a uniform and parallel gravity field CG ≡ CM.
• Principle of Moments: the moment of the resultant about an axis equals the sum of the moments of the individual forces about that axis. For a single axis (e.g., the x-axis):
  $X<em>{cm} = \frac{\sum</em>i x<em>i dW</em>i}{\sum<em>i dW</em>i} = \frac{\int x \, dW}{\int dW}.$
• When weight density is uniform and gravity g is constant, w ∝ dm and
  $X_{cm} = \frac{1}{W} \int x \, dW = \frac{1}{M} \int x \, dm.$
• If dm = ρ dV, then the coordinates of the center of mass are
  $X<em>{cm} = \frac{1}{M} \int x \rho \, dV , \quad Y</em>{cm} = \frac{1}{M} \int y \rho \, dV , \quad Z_{cm} = \frac{1}{M} \int z \rho \, dV.$
• In a uniform, parallel gravitational field and with uniform density ρ throughout the body, Center of Mass ⇔ Center of Gravity ⇔ Centroid.
• Three important notes:
  • CM/CG can lie outside the physical body for irregular distributions.
  • Often symmetry and a convenient choice of axes simplify calculations.
  • CM is a point in space determined by mass distribution alone (independent of gravity) when formulated without gravity terms.

CM, CG and Centroid: Conceptual distinctions

• Center of Gravity (CG): point where the weight vector can be considered to act; depends on gravity field and density distribution.
• Center of Mass (CM): purely a mass distribution property; in uniform gravity CG = CM.
• Centroid: geometric center for uniform density shapes; for 2D areas and 3D volumes, it is the geometric mean position (density cancels when ρ is uniform).
• Summary statement:
  • Under uniform and parallel gravitational field and uniform density, CM ≡ CG ≡ Centroid.
  • In general, CM/CG/centroid definitions align when these conditions hold; otherwise they may differ.

Centroids of Lines, Areas, and Volumes

• Centroid of a line (1D): for a line of length L with constant density, the centroid coordinates are
  $\text{Centroid of line } C:\ (x<em>c, y</em>c, z_c) = \left( \frac{1}{L} \int x \, dL, \frac{1}{L} \int y \, dL, \frac{1}{L} \int z \, dL \right).$
  If the line has cross-sectional area A and density ρ, dm = ρ A dL, which cancels in the ratio, leaving the same geometric result.

• Centroid of an area (2D): for a flat region A with constant density across thickness, the centroid is
  $x<em>c = \frac{1}{A} \iint</em>A x \, dA, \quad y<em>c = \frac{1}{A} \iint</em>A y \, dA,$
  with the density factor canceling out.

• Centroid of a volume (3D): for a solid region V with constant density, the centroid is
  $x<em>c = \frac{1}{V} \iiint</em>V x \, dV, \quad y<em>c = \frac{1}{V} \iiint</em>V y \, dV, \quad z<em>c = \frac{1}{V} \iiint</em>V z \, dV.$

• Practical note: when density ρ is not uniform, CM/CG are found with dm = ρ dV and weighted integrals:
  x{cm} = \frac{1}{M} \iiintV x \rho \, dV,
  \quad M = \iiint_V \rho \, dV.

Choice of Element for Integration (five criteria)

• 1. Order of Element
  • Prefer first-order differential elements to reduce the number of integrals.
  • Examples:
  • For area: dA = length l × dy (first-order strip) vs higher-order elements.
  • For volume: a first-order element dV = A(y) dy or dV = (cross-sectional area) dy uses fewer integrals than higher-order forms.
• 2. Continuity
  • Choose elements that can be integrated in one continuous operation over the entire figure.
  • Avoid split integrals caused by discontinuities in the height or thickness functions.
• 3. Discarding Higher-Order Terms
  • Accept the lower-order terms and neglect higher-order contributions when appropriate (e.g., triangular area approximations like ½ dx dy).
• 4. Choice of Coordinates
  • Choose coordinates that best match the boundaries of the region (rectangular, polar, cylindrical, etc.).
• 5. Centroidal Coordinate of Element
  • Use the centroid coordinates of the elemental region for the moment arm in the differential moment: for an element with centroid at (xc, yc, zc), the moment about an axis uses xc dA (or y_c dA, etc.).
• Practical takeaway: a well-chosen element reduces computational effort and improves accuracy.

Composite Bodies and Figures

• Useful for irregular objects;
• The principle of moments applies to the composite: the total centroid is the weighted average of the centroids of the parts, weighted by their masses.
• Cavities or holes act as negative mass in the sum: treat the filled part as positive mass and subtract the mass corresponding to cavities.
• Diagrammatic idea: many shapes (arc segments, circular areas, rectangular areas, triangular areas, sectors, parabolic and other shapes) have known centroid formulas that can be combined to obtain the centroid of a composite body.

Key (typical) results from sample problems

• Centroid of a circular arc (arc of radius R, subtending angle 2α):
  • Using a symmetry axis along the x-axis (y = 0), the x-coordinate of the centroid is
    $\bar{x} = \frac{\int x \, dL}{\int dL} = \frac{r\sin \alpha}{\alpha}.$
  • For a semicircular arc (2α = π, α = π/2):
    $\bar{x} = \frac{2R}{\pi}.$
  • Note: polar coordinates simplify the arc length calculation.
• Centroid of a right triangle-like area with a boundary x(y) = b(h - y)/h (derived from similar triangles):
  • If the area is A = b h and dA = x(y) dy, the x-coordinate of the centroid can be obtained by evaluating
    $\bar{x} = \frac{1}{A} \int x \, dA,\quad x(y) = \frac{b}{h}(h - y).$
  • The corresponding y-coordinate can be found similarly by choosing horizontal strips.
• Centroid of a circular sector (radius R, angle 2α):
  • The centroid lies along the symmetry axis at a distance from the center given by a standard sector formula (requires deriving from dA = R dθ, etc.); commonly used results yield the centroid position along the radius depending on α. (Refer to problem-style derivations in the notes.)
• Centroid of a hemisphere (solid) of radius R (distance from the flat face):
  $\bar{y} = \frac{3R}{8},$
  where y is measured along the axis from the base (flat face) toward the dome.
• Composite centroid example: bracket-and-shaft or bracket-with-holes problems illustrate combining multiple materials with different densities; the steps are:
  • Compute each part’s mass: mi = densityi × volumei (or densityi × area_i × thickness).
  • Determine each part’s centroid coordinates (xi, yi, z_i).
  • Compute total mass M = Σ mi and the centroid: $x</em>{cm} = \frac{\sum m<em>i x</em>i}{M}, \quad y<em>{cm} = \frac{\sum m</em>i y<em>i}{M}, \quad z</em>{cm} = \frac{\sum m<em>i z</em>i}{M}.$
  • Cavities are treated as negative mass (subtract their contributions).
• Relation among CM, CG, and centroid for common engineering problems:
  • If gravity is uniform and density is uniform throughout, CM ≡ CG ≡ centroid.
  • In general, CM/CG can be located by integrating over dm with dm = ρ dV when ρ is not uniform.

Illustrative sample interpretations (from the provided problems)

• Problem 5/1: Centroid of a circular arc
  • For a semicircular arc of radius R, the centroid distance along the symmetry axis from the center is $\bar{x} = \frac{2R}{\pi}.$
  • General result for an arc subtending angle 2α: $\bar{x} = \frac{R \sin \alpha}{\alpha}.$
• Problem 5/5 (solid hemisphere centroid):
  • The vertical coordinate of the centroid from the base is $\bar{y} = \frac{3R}{8}.$
• Problem 5/8 (bracket-and-shaft):
  • The center of mass is found by summing the moments of each component about a chosen origin and dividing by the total mass; negative contributions are used for voids or cavities.
  • Example quantities shown include masses per area, densities, and resulting coordinates (x, y, z). The computed totals yield the centroid coordinates (e.g., Y, Z values). The method demonstrates assembling multiple materials into a single composite with the overall centroid determined by weighted averages.

Practical notes and reminders

• Key definition: Center of Gravity (CG) is the point where the gravitational moment vanishes; for uniform gravity, CG = CM.
• In many engineering problems, choosing axes along symmetry lines simplifies integration and often places the centroid on a symmetry line.
• For irregular objects, the centroid can lie outside the physical object; in such cases the geometry and symmetry help guide the calculation.
• When dealing with density variations, always use dm = ρ dV and compute
  $x_{cm} = \frac{\int x \rho \, dV}{\int \rho \, dV},$
  and analogous expressions for y and z.

Quick reference formulas

• Center of mass in 3D with density ρ:
  $x<em>{cm} = \frac{1}{M} \iiint</em>V x \rho \ dV, \; y<em>{cm} = \frac{1}{M} \iiint</em>V y \rho \ dV, \; z<em>{cm} = \frac{1}{M} \iiint</em>V z \rho \ dV, \quad M = \iiint_V \rho \, dV.$
• For uniform density ρ over a volume, this reduces to
  $x<em>{cm} = \frac{1}{V} \iiint</em>V x \, dV, \; y<em>{cm} = \frac{1}{V} \iiint</em>V y \, dV, \; z<em>{cm} = \frac{1}{V} \iiint</em>V z \, dV.$
• Centroid of a line (length L) with constant density:
  $x<em>c = \frac{1}{L} \int x \, dL, \; y</em>c = \frac{1}{L} \int y \, dL, \; z_c = \frac{1}{L} \int z \, dL.$
• Centroid of an area A:
  $x<em>c = \frac{1}{A} \iint</em>A x \, dA, \; y<em>c = \frac{1}{A} \iint</em>A y \, dA, \; z<em>c = \frac{1}{A} \iint</em>A z \, dA.$
• Centroid of a volume V:
  $x<em>c = \frac{1}{V} \iiint</em>V x \, dV, \; y<em>c = \frac{1}{V} \iiint</em>V y \, dV, \; z<em>c = \frac{1}{V} \iiint</em>V z \, dV.$