Mathematics | Centres of mass

Welcome to Centres of Mass: Finding the Balance Point!

Hello future engineers and physicists! This chapter, Centres of Mass (CM), is fundamental to understanding how objects behave under gravity and how stable they are. In Mechanics 2 (M2), we move beyond point masses and start analyzing the structure and geometry of real-world objects.

Don't worry if this seems abstract at first. The concept is simple: The Centre of Mass is the single point where you can perfectly balance an object, regardless of how complicated its shape is. Once you find the CM, you can treat the entire mass of the object as concentrated at that point for most mechanics calculations.

Let’s dive in and master the art of balancing!

1. Defining the Centre of Mass (CM)

What is the Centre of Mass?

The Centre of Mass (often denoted by \((\bar{x}, \bar{y})\) or just \(G\)) is the point at which the entire mass of a system or body can be considered to act.

Think of it as the balancing point. If you can balance a complicated object (like a hammer or a cricket bat) on your finger, the point resting on your finger is its Centre of Mass.

The Role of Moments

In Mechanics, we calculate the position of the CM using the principle of Moments.
A Moment is defined as Force multiplied by perpendicular distance. However, when calculating CM, we use Mass Moments:
\(\text{Mass Moment} = \text{Mass} \times \text{Distance from a chosen reference point (axis)}\)

The key principle is:
The moment of the total mass (acting at the CM) is equal to the sum of the moments of all the individual parts.

Did you know? We often calculate the Centre of Mass, but technically, gravity acts upon the Centre of Gravity. For objects on Earth, these two points are almost always the same, so we treat them synonymously in M2.

2. Centre of Mass of Discrete Particles

This is the simplest case. If we have several separate particles (point masses), we find their CM using a weighted average based on their mass and position.

Step-by-Step Calculation

Step 1: Define the Coordinate System

Choose a clear reference point (the origin, \((0, 0)\)) and defined axes (\(x\) and \(y\)). This is usually an end point or a corner of the system.

Step 2: Calculate Total Mass (\(M\))

The total mass \(M\) is the sum of all individual masses: \[M = m_1 + m_2 + m_3 + \dots \]

Step 3: Calculate Total Moments

We calculate the moments separately for the \(x\) direction and the \(y\) direction.
For particles \(m_i\) located at \((x_i, y_i)\):
\(\text{Total Moment about the y-axis (for } x \text{ coordinate)} = \sum m_i x_i\)
\(\text{Total Moment about the x-axis (for } y \text{ coordinate)} = \sum m_i y_i\)

Step 4: Find the Coordinates of the CM \((\bar{x}, \bar{y})\)

Using the principle of moments:
\[M \bar{x} = \sum m_i x_i\] \[\bar{x} = \frac{\sum m_i x_i}{M}\]
\[M \bar{y} = \sum m_i y_i\] \[\bar{y} = \frac{\sum m_i y_i}{M}\]

Analogy: Imagine two children on a seesaw (a massless rod). A heavier child needs to sit closer to the pivot point (the CM) to balance the lighter child who sits further out. The mass-distance product (the moment) must be equal on both sides.

3. Centre of Mass of Standard Uniform Rigid Bodies

For uniform rigid bodies, the CM is purely dependent on the body's geometry (shape), since the mass is evenly distributed. You must know the CM positions for these standard shapes.

Uniform means Mass \(\propto\) Length/Area/Volume

A body is uniform if its density is constant throughout.

• Uniform Rod (1D): Mass is proportional to length (\(\mathrm{d}m = \lambda \, \mathrm{d}x\), where \(\lambda\) is the linear mass density).
• Uniform Lamina (2D): Mass is proportional to area (\(\mathrm{d}m = \sigma \, \mathrm{d}A\), where \(\sigma\) is the area mass density).
• Uniform Solid (3D): Mass is proportional to volume (\(\mathrm{d}m = \rho \, \mathrm{d}V\), where \(\rho\) is the volume mass density).

Key Standard Results (to Memorise!)

The CM location is usually relative to a defined vertex, base, or centre.

1. Uniform Rod:
The CM is at its geometric midpoint.

2. Uniform Rectangular Lamina (Thin Plate):
The CM is at the intersection of its diagonals (the centre).

3. Uniform Triangular Lamina:
The CM is at the intersection of its medians (lines joining a vertex to the midpoint of the opposite side). The CM is located \(\frac{1}{3}\) of the way along any median, starting from the base.

4. Uniform Solid Cone (or Pyramid) of height \(h\):
The CM lies on the axis of symmetry, at a distance of \(\frac{1}{4} h\) from the base.

5. Uniform Hollow Cone (Conical Shell) of height \(h\):
The CM lies on the axis of symmetry, at a distance of \(\frac{1}{3} h\) from the base.

6. Uniform Solid Hemisphere of radius \(r\):
The CM lies on the axis of symmetry, at a distance of \(\frac{3}{8} r\) from the centre of the base.

7. Uniform Hemispherical Shell (Hollow) of radius \(r\):
The CM lies on the axis of symmetry, at a distance of \(\frac{1}{2} r\) from the centre of the base.

Memory Aid: Solid shapes are 'heavier' at the base, meaning their CM is closer to the base compared to hollow shells of the same shape. E.g., Cone Solid (\(h/4\)) vs. Cone Shell (\(h/3\)).

4. Finding CM Using Integration (Calculus in M2)

The most powerful technique in M2 for finding the CM of continuous bodies (rods, laminae defined by curves, non-uniform bodies) is integration. We essentially treat the body as a collection of infinitesimally small particles, \(\delta m\).

The Fundamental Formula

Instead of summing discrete particles (\(\sum m_i x_i\)), we integrate the mass moments: \[M \bar{x} = \int x \, \mathrm{d}m\] Therefore: \[\bar{x} = \frac{\int x \, \mathrm{d}m}{\int \mathrm{d}m}\]

Application 1: Uniform Rod (1D Integration)

Consider a uniform rod of length \(L\). Let \(\lambda\) be the constant linear mass density (mass per unit length).
1. The mass of a small element \(\mathrm{d}x\) is \(\mathrm{d}m = \lambda \, \mathrm{d}x\). 2. The total mass \(M = \int_0^L \lambda \, \mathrm{d}x = \lambda L\). 3. The moment is: \(M \bar{x} = \int_0^L x (\lambda \, \mathrm{d}x)\)
Since \(\lambda\) is constant, it cancels out when finding \(\bar{x}\): \[\bar{x} = \frac{\int_0^L x \, \mathrm{d}x}{\int_0^L \mathrm{d}x} = \frac{\left[ \frac{x^2}{2} \right]_0^L}{[x]_0^L} = \frac{L^2/2}{L} = \frac{L}{2}\]
This confirms the CM of a uniform rod is at its midpoint, \(L/2\).

Application 2: Uniform Lamina (2D Integration)

For a thin uniform lamina (plate) defined by a curve \(y = f(x)\) between \(x=a\) and \(x=b\). Let \(\sigma\) be the constant area mass density.
1. We usually slice the lamina into thin vertical strips of width \(\mathrm{d}x\). 2. The area of the strip is \(\mathrm{d}A = y \, \mathrm{d}x\). 3. The mass of the strip is \(\mathrm{d}m = \sigma \, \mathrm{d}A = \sigma y \, \mathrm{d}x\). 4. The CM of this small strip is located at \((x, y/2)\).

Finding \(\bar{x}\): The element's x-distance is \(x\). \[\bar{x} = \frac{\int x \, \mathrm{d}m}{\int \mathrm{d}m} = \frac{\int_a^b x (\sigma y \, \mathrm{d}x)}{\int_a^b \sigma y \, \mathrm{d}x} = \frac{\int_a^b x y \, \mathrm{d}x}{\int_a^b y \, \mathrm{d}x}\] (Remember, \(\int y \, \mathrm{d}x\) is the total area, \(A\)).

Finding \(\bar{y}\): The element's y-distance is \(y/2\). \[\bar{y} = \frac{\int \text{(element distance)} \, \mathrm{d}m}{\int \mathrm{d}m} = \frac{\int_a^b \left( \frac{y}{2} \right) (\sigma y \, \mathrm{d}x)}{\int_a^b \sigma y \, \mathrm{d}x} = \frac{\int_a^b \frac{1}{2} y^2 \, \mathrm{d}x}{\int_a^b y \, \mathrm{d}x}\]

Encouraging Note: Integration looks scary, but for uniform bodies, the constant density (\(\lambda\) or \(\sigma\)) always cancels out! You are essentially just calculating the "geometric centre" (centroid). Practice setting up the integrals correctly – that's the key skill here!

5. Centre of Mass of Composite Bodies

Most exam questions involve composite bodies—objects made up of two or more standard shapes joined together, or standard shapes with holes cut out.

The Principle of Addition and Subtraction

We use the formula for discrete particles, but we treat the CM of each standard shape as if its total mass were concentrated at that single point.

Step-by-Step for Composite Bodies

Step 1: Divide the Body
Break the complex body down into its component standard shapes (e.g., a rectangle and a triangle).

Step 2: Assign Mass/Ratio
Since the body is usually uniform, the mass of each part is proportional to its area (for laminae) or length (for rods). Use the area (or length) ratios instead of calculating the exact mass.
Example: If Component A has Area 10 and Component B has Area 5, use mass ratio 10:5 (or 2:1).

Step 3: Define Coordinates of Each CM
Establish a clear origin. Find the coordinates \((x_i, y_i)\) for the known CM of each component using the geometry results from Section 3.

Step 4: Create a Table (Highly Recommended!)
Organizing your data prevents errors:

Step 5: Calculate Final CM
Use the discrete particle formulas: \[\bar{x} = \frac{\sum m_i x_i}{M} \quad ; \quad \bar{y} = \frac{\sum m_i y_i}{M}\]

Handling Holes or Removed Sections

If a section is removed (e.g., a circle cut out of a square), treat the removed section as having negative mass.
1. Calculate the CM of the whole original body (\(m_{\text{total}}\)). 2. Calculate the CM of the removed section (\(m_{\text{hole}}\)). 3. The total mass of the final body is \(M = m_{\text{total}} - m_{\text{hole}}\). 4. Calculate the total moment: \(\sum m_i x_i = (m_{\text{total}} x_{\text{total}}) - (m_{\text{hole}} x_{\text{hole}})\).

The method of negative mass is incredibly useful and often required in M2 composite body questions.

6. Consolidation and Key Takeaways

A Note on Symmetry

Always check for symmetry first! If a body is symmetric about an axis (e.g., the y-axis), its CM must lie on that axis. This immediately tells you one coordinate (e.g., \(\bar{x}=0\)) and halves the calculation work!

Checklist for Success

• Units and Axes: Always establish a clear origin \((0, 0)\) and measure distances consistently.
• Standard Shapes: Know the CM formulae for the standard M2 shapes (especially the \(1/3\) for the triangle and the \(3r/8\) for the solid hemisphere).
• Integration Setup: When integrating, ensure you are using the correct expression for \(\mathrm{d}m\) (linear, area, or volume density) and the correct distance for the moment.
• Subtraction: Use negative mass/area ratio for holes or removed sections.