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IAL M2-Center of Mass
Centre of Mass (COM) - Mechanics Notes
The Centre of Mass is the point where the entire mass of a system can be considered to be concentrated for analyzing translational motion.
1. COM for Discrete Particles
In One Dimension (1D):
xcom = Σ mixiΣ mi
In Two Dimensions (2D):
rcom = Σ miriΣ mi
xcom = Σ mixiΣ mi
ycom = Σ miyiΣ mi
ycom = Σ miyiΣ mi
Example: Particles: A=2kg (at x=2), B=3kg (at x=4), C=5kg (at x=9)
xcom = (2)(2) + (3)(4) + (5)(9)2 + 3 + 5 = 4 + 12 + 4510 = 6110 = 6.1
2. COM for Uniform Plane Laminas
For a continuous, uniform object (lamina), the COM is found by integration. For standard shapes, it lies at the geometric centroid.
| Shape | COM Position |
|---|---|
| Rectangle/Square | Intersection of diagonals |
| Circle | At its center |
| Triangle | Intersection of medians |
| Circular Sector | On the axis of symmetry |
For triangle vertices A(x1,y1), B(x2,y2), C(x3,y3):
G = x1+x2+x33 , y1+y2+y33
G = x1+x2+x33 , y1+y2+y33
Key Property: The COM (G) of a triangle divides each median in a 23 ratio: AG = 23 AA'
3. COM for Circular Sector & Arc
For sector with radius r and half-angle θ:
OG = 2r sin θ3θ
OG = 2r sin θ3θ
Special Cases:
- Semi-Circle (θ = π2): OG = 4r3π
- Quarter-Circle (θ = π4): OG = 4r√23π
For circular arc with radius r and half-angle θ:
OG = r sin θθ
OG = r sin θθ
Special Case:
- Semi-Circular Arc (θ = π2): OG = 2rπ
Note: θ is HALF of the total angle (in radians)
4. COM of Composite Bodies
To find the COM of a complex shape made from simpler parts (or with parts removed):
- Divide the body into standard parts
- Find the area (Ai) and individual COM for each part
- Treat each part as a point mass at its own COM
- Use the standard COM formula
xcom = Σ AixiΣ Ai
ycom = Σ AiyiΣ Ai
ycom = Σ AiyiΣ Ai
Example with removed section: For a rectangular plate with a circular hole, treat the hole as a negative area.
Important: For removed sections (holes), treat area as negative
5. COM for Uniform Frameworks (Rods/Wires)
For a uniform rod/wire, mass is proportional to length. The COM of each straight rod is at its midpoint.
rcom = Σ Li riΣ Li
Where Li is the length of each rod, and ri is the position vector of its COM.
Note: The COM of each straight rod is at its midpoint. For curved rods, use the appropriate formula.
6. Equilibrium of Laminas
Conditions for Equilibrium:
Σ F = 0 (Translational Equilibrium)
Σ M = 0 (Rotational Equilibrium)
Σ M = 0 (Rotational Equilibrium)
Applications:
- Suspension: COM hangs directly below pivot
- Toppling: COM vertically above lowest contact point
Adding mass m at (xm, ym) to lamina of mass M:
xcom_new = M xcom + m xmM + m
ycom_new = M ycom + m ymM + m
xcom_new = M xcom + m xmM + m
ycom_new = M ycom + m ymM + m
Additional Mass Attachment: If a mass m is attached at point B to a lamina with mass M and COM at point C, the new COM will lie along the line connecting C and B, closer to the heavier mass.
7. Additional Mass Attachments
When an additional mass is attached to a lamina, the new COM can be calculated using the standard COM formula, treating the original lamina and the additional mass as two separate entities.
xcom_new = M xcom + m xmM + m
ycom_new = M ycom + m ymM + m
ycom_new = M ycom + m ymM + m
Where M is the mass of the lamina, (xcom, ycom) is its COM, m is the additional mass, and (xm, ym) is its position.
Example: A square lamina of mass 5kg has its COM at (2,3). An additional mass of 2kg is attached at (5,7). The new COM is:
xcom_new = 5×2 + 2×55+2 = 10+107 = 207 ≈ 2.86
ycom_new = 5×3 + 2×75+2 = 15+147 = 297 ≈ 4.14
ycom_new = 5×3 + 2×75+2 = 15+147 = 297 ≈ 4.14
Centre of Mass Summary
| Concept | Formula/Principle |
|---|---|
| Discrete Particles (1D) | xcom = Σ mixiΣ mi |
| Discrete Particles (2D) | rcom = Σ miriΣ mi |
| Triangle COM | G = x1+x2+x33, y1+y2+y33 |
| Circular Sector | OG = 2r sin θ3θ |
| Circular Arc | OG = r sin θθ |
| Composite Bodies | xcom = Σ AixiΣ Ai |
| Equilibrium | Σ F = 0, Σ M = 0 |
Remember: For sectors and arcs, always use the half-angle (θ) in formulas, where θ is half of the total angle in radians.
Rember too to multiply my mass/ total mass to get the exact center of mass when dealing with combined systems.
For a single system,this product reduces to just the cordinates of the COM, this includes arcs and sectors too.
Abel Masitsa (MAP)