Welcome to Centre of Mass (Mechanics 1)
Welcome to one of the most practical and visual chapters in CCEA A2 Mechanics 1! Have you ever tried balancing a ruler or a book on the tip of your finger? If you found the exact sweet spot where it didn't tip over, you found its centre of mass.
In this chapter, we will learn how to calculate the exact position of this point for:
• Individual particles arranged on a grid
• Wire frameworks made of thin rods
• 2D flat shapes known as laminae
• Complex composite shapes with holes or pieces cut out
Don't worry if mechanics sometimes feels abstract—centre of mass problems follow a very logical, step-by-step method that you can master with practice!
1. What is the Centre of Mass?
Definition: The centre of mass of a body or system of particles is the unique point at which the entire mass of the system can be considered to be concentrated for translational motion.
In coordinates, we represent this point as \((\bar{x}, \bar{y})\) in 2D space, or with a position vector \(\bar{\mathbf{r}}\).
The Fundamental Law: Principle of Moments
To find the centre of mass, we use the Principle of Moments. This states that the total moment of the whole system about any axis is equal to the sum of the individual moments of its parts about that same axis.
Mathematically, for a set of masses \(m_1, m_2, \dots, m_n\) at coordinates \((x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)\):
Taking moments about the \(y\)-axis (to find \(\bar{x}\)):
\(\left(\sum m_i\right) \bar{x} = \sum (m_i x_i) = m_1 x_1 + m_2 x_2 + \dots + m_n x_n\)
Taking moments about the \(x\)-axis (to find \(\bar{y}\)):
\(\left(\sum m_i\right) \bar{y} = \sum (m_i y_i) = m_1 y_1 + m_2 y_2 + \dots + m_n y_n\)
Key Takeaway: Centre of mass is simply a weighted average of position, where each point's position is weighted by its mass: \(\bar{x} = \frac{\sum m_i x_i}{\sum m_i}\) and \(\bar{y} = \frac{\sum m_i y_i}{\sum m_i}\).
2. Discrete Systems of Particles
A discrete system is simply a collection of separate, individual point masses located at specific coordinates.
Vector Formulation
If particles with masses \(m_1, m_2, \dots, m_n\) have position vectors \(\mathbf{r}_1, \mathbf{r}_2, \dots, \mathbf{r}_n\), then the position vector of the centre of mass \(\bar{\mathbf{r}}\) is given by:
\(\bar{\mathbf{r}} = \frac{\sum_{i=1}^n m_i \mathbf{r}_i}{\sum_{i=1}^n m_i} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 + \dots + m_n \mathbf{r}_n}{m_1 + m_2 + \dots + m_n}\)
Step-by-Step Example
Example: Three particles of masses \(2\text{ kg}\), \(3\text{ kg}\), and \(5\text{ kg}\) are placed at coordinates \((1, 4)\), \((4, 2)\), and \((2, -1)\) respectively. Find the coordinates of their centre of mass \((\bar{x}, \bar{y})\).
Step 1: Calculate the total mass \(M\):
\(M = \sum m_i = 2 + 3 + 5 = 10\text{ kg}\)
Step 2: Apply moments in the \(x\)-direction:
\(M\bar{x} = (2 \times 1) + (3 \times 4) + (5 \times 2)\)
\(10\bar{x} = 2 + 12 + 10 = 24\)
\(\bar{x} = \frac{24}{10} = 2.4\)
Step 3: Apply moments in the \(y\)-direction:
\(M\bar{y} = (2 \times 4) + (3 \times 2) + (5 \times (-1))\)
\(10\bar{y} = 8 + 6 - 5 = 9\)
\(\bar{y} = \frac{9}{10} = 0.9\)
Result: The centre of mass is at \((2.4, 0.9)\).
3. Frameworks of Uniform Thin Rods or Wires
When thin rods are welded or bent together to form a shape (like a wire triangle or an L-shaped bracket), we treat each straight segment as a separate component.
Key Rules for Uniform Rods:
• Mass is proportional to length: Because the rod is uniform, its mass is directly proportional to its length (\(m_i \propto L_i\)). We can use the length \(L_i\) in place of mass in our formulas!
• Individual Centre of Mass: For any uniform straight rod, its individual centre of mass is located at its geometric midpoint.
Formula for Wire Frameworks:
\(\bar{x} = \frac{\sum L_i x_i}{\sum L_i}\) and \(\bar{y} = \frac{\sum L_i y_i}{\sum L_i}\)
where \(L_i\) is the length of rod \(i\), and \((x_i, y_i)\) is the midpoint of rod \(i\).
Quick Tip: Always write down the midpoint coordinates of each rod before plugging values into the equation. It prevents silly arithmetic slips!
4. Standard Uniform Planar Laminae
A lamina is an idealized 2D flat sheet of material. For a uniform lamina, the mass is evenly distributed across its surface area. This means mass is directly proportional to surface area (\(m_i \propto A_i\)).
You need to know the standard centres of mass for three fundamental geometric shapes:
1. Uniform Rectangle or Square
• The centre of mass lies at the geometric centre (the intersection of the diagonals or lines of symmetry).
• For a rectangle of width \(b\) and height \(h\) aligned with axes from the origin: \((\bar{x}, \bar{y}) = \left(\frac{b}{2}, \frac{h}{2}\right)\).
2. Uniform Circular Lamina / Disc
• The centre of mass lies at the geometric centre of the circle.
3. Uniform Triangular Lamina
• The centre of mass lies at the centroid of the triangle (where the three medians intersect).
• Distance Rule: It lies \(\frac{1}{3}\) of the perpendicular distance from any base to the opposite vertex along the median.
• Coordinate Centroid Formula: If a triangle has vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\), its centre of mass is simply the average of the coordinates:
\(\bar{x} = \frac{x_1 + x_2 + x_3}{3}, \quad \bar{y} = \frac{y_1 + y_2 + y_3}{3}\)
Memory Trick for Triangles: Remember "one-third from the base, two-thirds from the point!"
5. Composite and Truncated Laminae (The Tabular Method)
Most exam questions in CCEA Mechanics 1 involve shapes formed by joining simpler shapes together (composite laminae) or cutting pieces out of a larger shape (truncated laminae).
The Examiner-Recommended Tabular Method
Examiners highly recommend setting up a clean table to organize your working. For each component, record its relative mass (or area), individual centre of mass coordinates, and mass moments:
Standard Table Layout:
• Component / Part: (e.g., Rectangle \(A\), Triangle \(B\), Cut-out Hole \(C\))
• Mass / Relative Area (\(m_i\) or \(A_i\)): (Use positive for added shapes, negative for removed parts)
• Distance from \(y\)-axis (\(x_i\)): individual centre of mass \(x\)-coordinate
• Distance from \(x\)-axis (\(y_i\)): individual centre of mass \(y\)-coordinate
• Moment about \(y\)-axis (\(m_i x_i\)): product of mass and \(x_i\)
• Moment about \(x\)-axis (\(m_i y_i\)): product of mass and \(y_i\)
Handling Cut-Outs (The Subtraction Rule)
When a piece is removed from a lamina, treat the removed portion as a negative mass (or negative area):
\(M_{\text{remaining}} \bar{x} = M_{\text{original}} x_{\text{original}} - M_{\text{removed}} x_{\text{removed}}\)
\(\bar{x} = \frac{M_{\text{original}} x_{\text{original}} - M_{\text{removed}} x_{\text{removed}}}{M_{\text{original}} - M_{\text{removed}}}\)
Worked Example: Truncated Lamina
Problem: A uniform rectangular lamina \(ABCD\) has length \(AB = 12\text{ cm}\) along the \(x\)-axis and height \(AD = 8\text{ cm}\) along the \(y\)-axis, with vertex \(A\) at the origin \((0,0)\). A square of side length \(4\text{ cm}\) is cut out of the top-right corner, with its edges parallel to the axes (spanning from \(x = 8\) to \(12\) and \(y = 4\) to \(8\)). Find the coordinates of the centre of mass of the remaining lamina.
Step 1: Identify components and their areas:
• Original Rectangle: Area \(A_1 = 12 \times 8 = 96\text{ cm}^2\)
• Cut-out Square: Area \(A_2 = 4 \times 4 = 16\text{ cm}^2\) (to be subtracted)
• Total Remaining Area \(A_{\text{total}} = 96 - 16 = 80\text{ cm}^2\)
Step 2: Find individual centres of mass:
• Original Rectangle: \((x_1, y_1) = \left(\frac{12}{2}, \frac{8}{2}\right) = (6, 4)\)
• Cut-out Square: \((x_2, y_2) = \left(8 + \frac{4}{2}, 4 + \frac{4}{2}\right) = (10, 6)\)
Step 3: Calculate moments about the axes:
• For \(\bar{x}\):
\(80\bar{x} = (96 \times 6) - (16 \times 10)\)
\(80\bar{x} = 576 - 160 = 416\)
\(\bar{x} = \frac{416}{80} = 5.2\text{ cm}\)
• For \(\bar{y}\):
\(80\bar{y} = (96 \times 4) - (16 \times 6)\)
\(80\bar{y} = 384 - 96 = 288\)
\(\bar{y} = \frac{288}{80} = 3.6\text{ cm}\)
Result: The centre of mass of the remaining shape is \((5.2\text{ cm}, 3.6\text{ cm})\).
6. Common Pitfalls & Examiner Warnings
Here are the most frequent mistakes flagged in CCEA examiner reports—make sure you watch out for them!
1. Adding Instead of Subtracting Cut-Outs:
When a section is removed, you must subtract both the area and the moment. Adding them is the single most common error on truncated laminae questions.
2. Inconsistent Origins / Reference Axes:
Always state your reference origin clearly (e.g., "Taking moments about point \(O\)"). Every distance \(x_i\) and \(y_i\) must be measured from this exact same origin.
3. Measuring Triangle Centroids from the Wrong End:
Remember: the centroid is \(\frac{1}{3}h\) from the flat base and \(\frac{2}{3}h\) from the opposite vertex. Confusing these two distances will throw off your entire calculation.
4. Conflating Mass and Area when Densities Differ:
If a question combines two different materials (e.g., a steel plate joined to a wooden plate), area alone is not proportional to mass. You must multiply each area by its surface density (\(\sigma\)) to obtain the relative mass: \(m = \sigma A\).
5. Forgetting Negative Signs in Other Quadrants:
If your coordinate origin is placed in the middle of a framework or lamina, parts lying to the left of the \(y\)-axis or below the \(x\)-axis must have negative coordinate values (e.g., \(x = -3\)).
Quick Chapter Summary
• Principle of Moments: \((\sum m_i)\bar{x} = \sum (m_i x_i)\) and \((\sum m_i)\bar{y} = \sum (m_i y_i)\)
• Particles: Use masses \(m_i\) and given coordinates \((x_i, y_i)\).
• Uniform Rods / Wires: Use lengths \(L_i\) as mass, with centres at the rod midpoints.
• Uniform Laminae: Use areas \(A_i\) as mass. Rectangle centre is at \(\left(\frac{b}{2}, \frac{h}{2}\right)\); triangle centroid is at \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\) or \(\frac{1}{3}\) height from base.
• Cut-outs / Holes: Treat removed areas as negative in your table and formulas!