Further Mathematics | Centres of mass

Hello and Welcome to Centres of Mass!

Hey there! In this chapter of Mechanics 2, we are going to dive into the concept of the Centre of Mass (CoM). Don't worry if the name sounds intimidating—it's essentially about finding the special balancing point of an object or a system of particles.

Think of it this way: if you could support an object at just one single point, and it remained perfectly balanced, that point would be the Centre of Mass. This concept is fundamental to understanding stability, equilibrium, and rotation in physics and engineering.

What is the Centre of Mass (\( \text{CoM} \))?

The Centre of Mass of a body (or system of particles) is the single point where the entire mass of the body can be considered to be concentrated.

• Notation: We usually denote the position vector of the Centre of Mass by \( \bar{r} \), or its coordinates as \( (\bar{x}, \bar{y}) \).
• Centre of Gravity: For all problems encountered in M2, the Centre of Mass is in the same location as the Centre of Gravity (CoG). The CoG is the point through which the resultant force of gravity acts. Since we assume gravity acts uniformly (constant \(g\)), the two points coincide.

Section 1: Centre of Mass for Discrete Particles

This is the simplest starting point. We have several individual particles, each with its own mass and position. Our goal is to find the average position weighted by mass.

1. The 1-Dimensional Case (Along a Line)

Suppose we have \(n\) particles, where the \(i\)-th particle has mass \(m_i\) and position \(x_i\). The total mass is \( M = \sum m_i \).

The position of the Centre of Mass, \( \bar{x} \), is calculated using the formula:

$$ \bar{x} = \frac{\sum m_i x_i}{\sum m_i} = \frac{m_1 x_1 + m_2 x_2 + \dots + m_n x_n}{M} $$

Memory Aid: "Mass Moment divided by Total Mass." The product \(m_i x_i\) is sometimes called the mass moment about the origin.

2. The 2-Dimensional Case (In a Plane)

If the particles are scattered across a plane (with coordinates \((x_i, y_i)\)), we just apply the 1D formula separately to the \(x\) and \(y\) coordinates.

$$ \bar{x} = \frac{\sum m_i x_i}{M} \quad \text{and} \quad \bar{y} = \frac{\sum m_i y_i}{M} $$

Step-by-Step Process for Discrete Particles

1. Choose an Origin: Always define a clear reference point \((0, 0)\). This is crucial!
2. List Data: Create a table listing \(m_i\), \(x_i\), and \(y_i\) for every particle.
3. Calculate Moments: Calculate the products \(m_i x_i\) and \(m_i y_i\) for each particle.
4. Sum Totals: Find the total mass \( M = \sum m_i \), the total x-moment \( \sum m_i x_i \), and the total y-moment \( \sum m_i y_i \).
5. Divide: Calculate \( \bar{x} \) and \( \bar{y} \).

Section 2: Centre of Mass for Uniform Rigid Bodies (Laminae)

When dealing with large objects, like a flat plate (called a lamina), we can't treat it as discrete particles. However, if the lamina is uniform, the mass is distributed evenly.

What does 'uniform' mean here?
It means the density (\(\rho\)) is the same everywhere. Since mass is proportional to area (for 2D shapes) or length (for 1D rods), we can simply use the Area (or Length) of the shape instead of the actual mass in our calculations. This saves us from needing to know the density!

Standard Shapes and Symmetry

For simple, uniform shapes, the Centre of Mass can be found by symmetry alone.

1. Uniform Rod (1D)

If a rod of length \(L\) is uniform, its CoM is simply at its geometric center, \( \frac{L}{2} \) from either end.

2. Uniform Rectangle or Square (2D)

The CoM lies at the intersection of its lines of symmetry—i.e., the center point where the diagonals cross.

3. Uniform Triangle (2D)

This is the most important standard shape to remember and is often tested!

• The CoM is located at the intersection of the medians (lines joining a vertex to the midpoint of the opposite side).
• The location is one-third (\(\frac{1}{3}\)) of the way from the base to the opposite vertex.

Example: If a triangle has its vertices at \(A, B, C\). If you choose the base to be the side \(BC\), the CoM is \( \frac{1}{3} \) of the way along the median line starting at \(A\).

Section 3: Centre of Mass for Compound Bodies

Most challenging exam questions involve compound bodies—objects made by combining several standard shapes (like rectangles and triangles).

Don't panic! The beautiful thing about CoM is that the compound body can be treated exactly like a system of discrete particles, where each standard shape acts as a particle located at its own individual CoM.

The Compound Body Method (A 5-Step Strategy)

Let's find the CoM of a complex shape \(S\) made up of components \(S_1, S_2, S_3, \dots\).

Step 1: Define the Coordinate System

• Always draw a diagram.
• Choose a convenient origin \((0, 0)\). Often, this is a corner of the object, as it minimizes negative coordinates.
• Establish the \(x\) and \(y\) axes.

Step 2: Calculate Area (or Mass) for Each Component

• Determine the Area \(A_i\) for each standard shape \(S_i\). Since the lamina is uniform, we use Area instead of mass.
• Find the Total Area \( A = \sum A_i \).

Step 3: Find the Individual Centre of Mass for Each Component

• For each shape \(S_i\), find its individual coordinates \((x_i, y_i)\) relative to the origin chosen in Step 1.
• Remember the rules for rectangles (center) and triangles (one-third rule or vertex average).

Step 4: Set up the Calculation Table and Sum Moments

Organizing your work in a table prevents errors!

Example Table Structure:

Component	Area \(A_i\)	\(x_i\)	\(y_i\)	Moment \(A_i x_i\)	Moment \(A_i y_i\)
Shape 1	...	...	...	...	...
Shape 2	...	...	...	...	...
TOTALS	\( A = \sum A_i \)	N/A	N/A	\( \sum A_i x_i \)	\( \sum A_i y_i \)

Step 5: Calculate the Final CoM Coordinates

Use the generalized CoM formula, replacing \(m\) with \(A\):

$$ \bar{x} = \frac{\sum A_i x_i}{A} \quad \text{and} \quad \bar{y} = \frac{\sum A_i y_i}{A} $$

Dealing with Holes or Removed Sections

What if a standard shape (e.g., a circle) is cut out of a larger shape (e.g., a rectangle)?

We use the principle of negative mass (or negative area).

• Treat the original large shape as Component 1 (positive area).
• Treat the hole or removed section as Component 2, giving it a negative area (\(-A_2\)).
• The Total Area \(A\) is \(A_1 - A_2\).
• The standard CoM formula still works, resulting in a subtraction in the numerator: $$ \bar{x} = \frac{(A_1 x_1) + ((-A_2) x_2)}{A_1 - A_2} $$

Section 4: Applications and Stability (Linking CoM to Equilibrium)

Finding the CoM is often the first step in solving equilibrium problems.

Equilibrium and Tilting

If a rigid body is resting on a support (or hinged), it will be in equilibrium only if the vertical line passing through the Centre of Mass passes through the base of support.

• If the CoM falls outside the base of support, the body will tilt or topple.
• When a body is suspended from a single point, the CoM will always settle directly vertically below the point of suspension.

This connection is important: once you find \((\bar{x}, \bar{y})\), you can use moments and forces from M1 to determine when the object might slip or tilt.