# Centres of Mass

> Mathematics · Edexcel IAL Math M2
> Source: https://www.owlsprep.com/study/edexcel-ial-math-m2-centres-of-mass/

This guide covers all Edexcel IAL M2 centre of mass content, including discrete mass distributions, uniform laminas, composite figures, and lamina equilibrium problems, with exam-aligned worked examples.

**Prerequisites:** [Ability to calculate moments (Edexcel IAL M1)](https://www.owlsprep.com/study/edexcel-ial-math-m1-moments/); [Understanding of static equilibrium (Edexcel IAL M1)](https://www.owlsprep.com/study/edexcel-ial-math-m1-static-equilibrium/)

## Learning objectives

- Calculate centre of mass of 1D and 2D discrete mass distributions
- Find CoM of uniform plane laminas and composite figures using symmetry and formula book results
- Solve equilibrium problems for laminas suspended from a point, rotating on a fixed axis, or placed on an inclined plane

## Centre of Mass of Discrete Mass Distributions

**Centre of Mass (CoM)** — The single point at which the entire mass of a system can be considered to act for moment and gravitational calculations.

*Notation:* (\bar{x}, \bar{y})

*Example:* For 2 point masses of 2 kg at (0,0) and 3 kg at (5,0), CoM is at (3, 0).

For discrete masses distributed in 1D or 2D, the CoM coordinates are calculated as the weighted average of the coordinates of each mass, weighted by the mass of each component. For uniform laminas, mass is proportional to area, so you can use area instead of mass in calculations to simplify working.

$$\bar{x} = \frac{\sum m_i x_i}{\sum m_i}, \quad \bar{y} = \frac{\sum m_i y_i}{\sum m_i}$$

**Worked example:** Find the CoM of the following 2D discrete mass system: 1 kg at (1, 2), 2 kg at (3, 4), 3 kg at (5, 1). Give your answers as exact fractions.

1. Step 1: Calculate total mass of the system: $M = 1 + 2 + 3 = 6$ kg
2. Step 2: Calculate weighted sum of x coordinates: $\sum m_i x_i = (1 \times 1) + (2 \times 3) + (3 \times 5) = 1 + 6 + 15 = 22$
3. $$\bar{x} = \frac{22}{6} = \frac{11}{3}$$
4. Step 3: Calculate weighted sum of y coordinates: $\sum m_i y_i = (1 \times 2) + (2 \times 4) + (3 \times 1) = 2 + 8 + 3 = 13$
5. $$\bar{y} = \frac{13}{6}$$
6. Step 4: Final CoM is at $\left(\frac{11}{3}, \frac{13}{6}\right)$

> **Exam tip:** Always show full working for sum of masses and weighted position sums; method marks are awarded even if arithmetic errors occur.

*Calculator:* allowed

## CoM of Uniform Laminas and Composite Figures

**Uniform Plane Lamina** — A thin 2D object with constant mass per unit area, so mass is directly proportional to its surface area.

*Example:* A thin sheet of card of uniform thickness is a uniform lamina.

> **tip**
>
> All standard lamina CoM results (triangle, circular arc, sector) are provided in the Edexcel M2 formula booklet; you do not need to memorize or derive them.

For composite laminas made of multiple standard shapes, treat each shape as a discrete mass with mass proportional to its area, and apply the discrete CoM formula. If a section is removed from a lamina, treat the removed section as a negative mass in your calculations.

**Worked example:** A composite lamina is made of a square of side length 4 cm (mass per unit area ρ) attached to an isosceles triangle of base 4 cm and height 3 cm along the top edge of the square. Find the CoM of the composite lamina, taking the bottom left corner of the square as the origin (0,0), x-axis along the base of the square, y-axis vertically upwards. Give your answers as exact fractions.

1. Step 1: Calculate area and CoM of each component:
2. Square: Area $A_1 = 4 \times 4 = 16$ cm², CoM at (2, 2)
3. Triangle: Area $A_2 = 0.5 \times 4 \times 3 = 6$ cm². CoM is 1/3 of the height from the base (a triangle's centroid lies 1/3 of the way from the base to the apex), so y-coordinate = $4 + (\frac{1}{3} \times 3) = 5$, x-coordinate = 2, so CoM at (2, 5)
4. Step 2: Total area = $16 + 6 = 22$ cm²
5. $$\bar{x} = \frac{(16 \times 2) + (6 \times 2)}{22} = \frac{44}{22} = 2$$
6. $$\bar{y} = \frac{(16 \times 2) + (6 \times 5)}{22} = \frac{62}{22} = \frac{31}{11}$$
7. Step 3: Final CoM is at $\left(2, \frac{31}{11}\right)$ cm

> **Exam tip:** If a lamina has an axis of symmetry, your final CoM must lie on that axis; use this to check your answer for errors quickly.

*Calculator:* allowed

## Equilibrium of Plane Laminas

There are three types of equilibrium problems for laminas in M2: suspension from a fixed point, rotation about a fixed horizontal axis, and placement on an inclined plane. For all cases, the lamina is in equilibrium if the line of action of its weight (vertical line through the CoM) passes through the pivot point or contact surface.

- **Suspended from a fixed point**: CoM lies directly vertically below the suspension point in equilibrium. Angles are calculated using trigonometry with coordinates of the pivot and CoM.
- **Rotating about fixed horizontal axis**: Same as suspension: CoM lies vertically below the axis for equilibrium.
- **Placed on inclined plane**: Equilibrium exists if the line of action of the weight falls within the contact area of the lamina with the plane. If it falls outside, the lamina topples.

**Worked example:** The composite lamina from the previous example is suspended freely from the origin (0,0). Find the angle that the bottom edge of the square (along the x-axis) makes with the vertical in equilibrium. Give your answer to 3 significant figures.

1. Step 1: We know CoM is at $(2, \frac{31}{11})$
2. Step 2: When suspended from (0,0), the line connecting (0,0) to the CoM is vertical in equilibrium.
3. Step 3: The angle $\theta$ between the x-axis and the vertical line has $\tan\theta = \frac{\text{y-coordinate of CoM}}{\text{x-coordinate of CoM}} = \frac{31/11}{2} = \frac{31}{22}$
4. $$\theta = \arctan\left(\frac{31}{22}\right) = 54.6^\circ \text{ (3 s.f.)}$$
5. Step 4: This is the required angle between the bottom edge of the square and the vertical.

> **Exam tip:** Always draw a clear sketch marking the pivot point, CoM, and vertical line of action; this will make angle calculations much simpler and avoid sign errors.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting that removed sections of a lamina are treated as negative mass.
  - Why it fails: A cut-out section no longer contributes mass to the lamina, so adding its weighted position instead of subtracting leads to incorrect CoM coordinates.
  - Correct: Subtract the product of the removed area and its CoM position from your numerator, and subtract its area from the total area denominator.
- **Wrong:** Using 1/3 of the median from the vertex for a triangle's CoM instead of 2/3.
  - Why it fails: The formula book states the CoM of a triangle is 2/3 of the median length from the vertex; mixing this up is one of the most frequent arithmetic errors in CoM questions.
  - Correct: Confirm the reference point (base vs vertex) for your coordinate system before applying the triangle CoM result from the formula book.
- **Wrong:** Placing the CoM above the suspension point in equilibrium problems.
  - Why it fails: For stable equilibrium of a freely suspended lamina, the CoM must lie below the pivot point to minimize gravitational potential energy; an upper CoM would lead to unstable equilibrium, which is not tested in M2.
  - Correct: Always check that your CoM coordinate is vertically below the suspension/axis point in your equilibrium sketch.
- **Wrong:** Calculating the angle to the horizontal instead of the vertical when explicitly asked for the angle to the vertical.
  - Why it fails: Exam questions frequently specify the reference line for angle calculations, and misreading this leads to lost marks even if your trigonometry is correct.
  - Correct: Highlight the reference line (vertical/horizontal/edge) in the question, and match your trigonometric ratio to that line.
- **Wrong:** Assuming all composite laminas have a horizontal or vertical axis of symmetry.
  - Why it fails: If components are attached asymmetrically, there is no axis of symmetry, so you must calculate both x and y coordinates of the CoM fully.
  - Correct: Only use symmetry to skip coordinate calculations if you can explicitly identify an axis of symmetry shared by all components of the composite lamina.

## Cheatsheet

| Concept | Formula/Rule | Exam Tip |
| --- | --- | --- |
| Discrete 2D CoM | $\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$, $\bar{y} = \frac{\sum m_i y_i}{\sum m_i}$ | Use area instead of mass for uniform laminas to simplify calculations |
| Standard Lamina Results | Triangle: 2/3 of median from base; Sector: $\frac{2r \sin \alpha}{3\alpha}$ from centre | All results provided in formula booklet, no memorization required |
| Removed Lamina Sections | Treat removed area as negative mass in CoM calculation | Subtract removed area from both numerator and denominator |
| Suspended Lamina Equilibrium | CoM lies vertically below suspension point | Angle between edge and vertical = arctan of (horizontal separation / vertical separation) of CoM and pivot |
| Inclined Plane Lamina Equilibrium | Line of action of weight must lie within contact area | Toppling occurs when the line of action passes outside the contact edge |

## What's next

Now that you have mastered centres of mass for Edexcel IAL M2, you are ready to move on to more advanced mechanics topics that build on this foundational knowledge. Centres of mass are frequently combined with moment calculations and energy principles in 4-6 mark exam questions, so it is critical that you practice past paper questions on this topic to consolidate your understanding before progressing. Remember that all CoM problems in M2 can be solved without integration, so stick to the discrete mass and formula book results covered in this guide to maximize your marks. If you struggled with the equilibrium section, revise moment calculations from M1 first, as these are core to solving CoM equilibrium problems.

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