Welcome to Further Centre of Mass
Welcome to one of the most practical and satisfying topics in CCEA A2 Mechanics 2! You have already met the basics of finding centres of mass for simple particles and standard shapes. In this chapter, we level up by exploring composite 3D solids, using calculus and single integration to find centroids of curved laminae and solids of revolution, and examining real-world applications like freely suspended bodies and sliding versus toppling.
Don't worry if calculus-based mechanics sounds intimidating at first. We will break every single derivation and method down into simple, repeatable steps. Let's dive in!
---1. Core Concepts, Definitions, and Symmetry
What is the Centre of Mass?
The centre of mass (CoM), often denoted by \(G\), is the unique point through which the resultant gravitational force acts on a body, regardless of its orientation in space. In a uniform gravitational field, this is identical to the centre of gravity.
Key Assumptions in CCEA Mechanics 2:
• Uniform Density: Unless stated otherwise, bodies are assumed to have uniform mass distribution. That means mass per unit length (\(\lambda\)), mass per unit area (\(\sigma\)), or mass per unit volume (\(\rho\)) is constant throughout.
• The Principle of Symmetry: If an object has a line, plane, or axis of symmetry, its centre of mass must lie on that line, plane, or axis. This saves huge amounts of calculation!
Did you know? If an object has two lines of symmetry (like a uniform rectangular lamina or a solid cylinder), the centre of mass is simply located at their exact intersection!
Key Takeaway: Always look for symmetry first before doing any math. If a 3D solid is symmetrical about the \(x\)-axis, immediately write down \(\bar{y} = 0\) and \(\bar{z} = 0\).
---2. Standard Geometric Results You Must Know
For CCEA A2 2 Section B, you need to know and use standard centroid results for regular shapes. Keep these memorised:
1. Uniform Solid Cone / Pyramid of height \(h\):
• Distance from vertex along axis: \(\bar{x} = \frac{3}{4}h\)
• Distance from base along axis: \(\bar{x} = \frac{1}{4}h\)
2. Uniform Solid Hemisphere of radius \(r\):
• Distance from the flat circular base along axis of symmetry: \(\bar{x} = \frac{3}{8}r\)
3. Uniform Hemispherical Shell (Hollow Bowl) of radius \(r\):
• Distance from the base rim along axis of symmetry: \(\bar{x} = \frac{1}{2}r\)
4. Uniform Semicircular Lamina of radius \(r\):
• Distance from the straight diameter edge along axis of symmetry: \(\bar{y} = \frac{4r}{3\pi}\)
Memory Trick: For solid cone vs. solid hemisphere distances from the base, notice the denominators: Cone is \(\frac{1}{4}h = \frac{2}{8}h\), and Hemisphere is \(\frac{3}{8}r\). The hemisphere's mass is pulled slightly further away from the base (\(\frac{3}{8}\)) than the cone's mass (\(\frac{2}{8}\))!
---3. Composite Bodies & Bodies with Cut-Outs
A composite body is made of two or more standard parts joined together (or hollowed out). We find the overall centre of mass \(\mathbf{\bar{r}} = (\bar{x}, \bar{y}, \bar{z})\) by taking moments of mass about a chosen reference origin:
\(\left(\sum m_i\right)\mathbf{\bar{r}} = \sum (m_i \mathbf{r}_i)\)
A. Composite 2D Laminae (Uniform Area Density \(\sigma\))
Because density \(\sigma\) cancels out everywhere, mass is proportional to area \(A\):
\(\bar{x} = \frac{\sum A_i \bar{x}_i}{\sum A_i}, \quad \bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i}\)
B. Composite 3D Solids (Uniform Volume Density \(\rho\))
Because density \(\rho\) cancels out everywhere, mass is proportional to volume \(V\):
\(\bar{x} = \frac{\sum V_i \bar{x}_i}{\sum V_i}, \quad \bar{y} = \frac{\sum V_i \bar{y}_i}{\sum V_i}, \quad \bar{z} = \frac{\sum V_i \bar{z}_i}{\sum V_i}\)
C. Bodies with Removed / Hollow Sections (Negative Mass Method)
When a hole is cut out or a section is removed from a larger object, treat the removed piece as a negative mass (or negative area/volume):
\(M_{\text{remaining}} \bar{x}_{\text{remaining}} = M_{\text{total}} \bar{x}_{\text{total}} - M_{\text{removed}} \bar{x}_{\text{removed}}\)
Step-by-Step Table Method:
To avoid mistakes in exams, always construct a clear table:
1. Label each component (e.g., Cylinder, Cone, Removed Hemisphere).
2. Choose a clear reference origin (e.g., \(O\) at the base) and stick to it for every component!
3. Record the volume/area of each part (use a minus sign for cut-outs).
4. Record the distance of each part's CoM from your origin.
5. Compute the moment (\(\text{Volume} \times \text{Distance}\)) for each row.
6. Sum the total volume column and the total moment column, then divide!
Key Takeaway: Never switch your reference point mid-question! If your origin is at the base, measure every single distance from that same base.
---4. Calculus: Centre of Mass by Integration
When a body has curved edges defined by an equation \(y = f(x)\), we use calculus to integrate across thin elemental strips or slices.
A. Uniform 2D Lamina Bounded by \(y = f(x)\), \(y = 0\), \(x = a\), and \(x = b\)
Imagine dividing the area under the curve into vertical strips of width \(\mathrm{d}x\) and height \(y\):
• The area of a thin strip is \(\mathrm{d}A = y\,\mathrm{d}x\).
• The centroid of this vertical strip is at \((x, \frac{1}{2}y)\).
Using the principle of moments:
Total Area:
\(A = \int_a^b y \, \mathrm{d}x\)
\(\bar{x}\)-coordinate:
\(\bar{x} = \frac{1}{A} \int_a^b x y \, \mathrm{d}x = \frac{\int_a^b x y \, \mathrm{d}x}{\int_a^b y \, \mathrm{d}x}\)
\(\bar{y}\)-coordinate:
\(\bar{y} = \frac{1}{A} \int_a^b \left(\frac{1}{2}y\right) y \, \mathrm{d}x = \frac{\frac{1}{2} \int_a^b y^2 \, \mathrm{d}x}{\int_a^b y \, \mathrm{d}x} = \frac{\int_a^b y^2 \, \mathrm{d}x}{2 \int_a^b y \, \mathrm{d}x}\)
Common Examiner Trap: Students constantly forget the factor of \(\frac{1}{2}\) when computing \(\bar{y}\)! Always remember that a vertical strip's mass acts halfway up its height, at \(\frac{y}{2}\).
---B. Uniform Solid of Revolution (Rotated \(2\pi\) about the \(x\)-axis)
When the region bounded by \(y = f(x)\), the \(x\)-axis, \(x = a\), and \(x = b\) is rotated completely around the \(x\)-axis:
• By rotational symmetry about the \(x\)-axis: \(\bar{y} = 0\) and \(\bar{z} = 0\).
• We divide the solid into thin circular discs of thickness \(\mathrm{d}x\) and radius \(y\).
• The volume of each elemental disc is \(\mathrm{d}V = \pi y^2 \, \mathrm{d}x\), and its centre of mass is at \(x\).
Total Volume:
\(V = \pi \int_a^b y^2 \, \mathrm{d}x\)
\(\bar{x}\)-coordinate:
\(\bar{x} = \frac{\pi \int_a^b x y^2 \, \mathrm{d}x}{\pi \int_a^b y^2 \, \mathrm{d}x} = \frac{\int_a^b x y^2 \, \mathrm{d}x}{\int_a^b y^2 \, \mathrm{d}x}\)
Note: The \(\pi\) factor cancels out in the numerator and denominator, leaving a neat and clean formula!
Key Takeaway:
• For a 2D Lamina: \(\bar{x}\) uses \(\int x y \, \mathrm{d}x\), and \(\bar{y}\) uses \(\frac{1}{2} \int y^2 \, \mathrm{d}x\).
• For a 3D Solid of Revolution: \(\bar{x}\) uses \(\int x y^2 \, \mathrm{d}x\) divided by \(\int y^2 \, \mathrm{d}x\).
5. Applications to Statics and Equilibrium
A. Freely Suspended Bodies
When a lamina or solid is suspended freely from a pivot point \(A\), it will rotate until it reaches equilibrium. In equilibrium:
The centre of mass \(G\) must lie vertically below the point of suspension \(A\).
How to calculate the angle of tilt \(\theta\):
1. Draw a clear diagram showing the suspension point \(A\) and the centre of mass \(G\).
2. Draw a vertical line connecting \(A\) and \(G\).
3. Form a right-angled triangle between \(A\), \(G\), and a reference edge or corner.
4. Use basic trigonometry:
\(\tan \theta = \frac{\text{horizontal distance from } A \text{ to } G}{\text{vertical distance from } A \text{ to } G}\)
---B. Sliding vs. Toppling on an Inclined Plane
Suppose a uniform body rests on a rough plane inclined at an angle \(\alpha\) to the horizontal. As \(\alpha\) is gradually increased, the body will eventually either slide down or topple over. Which one happens first?
1. Limiting Equilibrium for Sliding:
The body is on the verge of sliding when the friction force reaches its maximum (\(F_{\text{max}} = \mu R\)):
\(\tan \alpha = \mu\)
2. Limiting Equilibrium for Toppling (Tipping):
A body topples when the vertical line of action through its centre of mass \(G\) passes outside the lower edge of its base of support.
If the base has width \(2b\) (so distance from centre of base to edge is \(b\)) and the centre of mass is at height \(h\) above the base:
\(\tan \alpha = \frac{b}{h}\)
3. Determining What Happens First:
We compare the critical sliding angle (\(\tan^{-1}\mu\)) with the critical toppling angle (\(\tan^{-1}(b/h)\)):
• Condition for Sliding before Toppling: \(\mu < \frac{b}{h}\) (low friction, wide/short body).
• Condition for Toppling before Sliding: \(\mu > \frac{b}{h}\) (high friction, tall/narrow body).
Real-World Analogy: Think of an empty tall bookcase vs. an ice puck on a ramp. The tall bookcase has a small base and high \(G\) (\(\mu > b/h\)), so it topples. The flat ice puck has almost no friction (\(\mu < b/h\)), so it slides effortlessly without tipping.
---6. Common Exam Pitfalls & How to Avoid Them
Keep these five classic examiner traps in mind during your CCEA A2 2 exam:
• Trap 1: Adding instead of subtracting cut-out shapes. Always double-check your signs in tabular moment calculations when a hollow hemisphere or hole is removed.
• Trap 2: Forgetting the \(\frac{1}{2}\) in \(\bar{y}\) for laminae. Remember that the strip centroid is at \(\frac{y}{2}\), which gives \(\frac{1}{2}y^2\) in the integral.
• Trap 3: Origin inconsistency. A cone's CoM is \(\frac{1}{4}h\) from the base, but \(\frac{3}{4}h\) from the vertex. Make sure your distance matches the origin you set in your table!
• Trap 4: Drawing the vertical line incorrectly in suspension problems. The vertical line must pass through both the suspension pivot \(A\) and \(G\). Always mark the right angle carefully when solving for \(\tan \theta\).
• Trap 5: Assuming toppling happens without checking friction. Always compare \(\mu\) with \(\frac{b}{h}\) before making a conclusion about sliding versus toppling.
7. Chapter Quick Review
• Composite Centroid Formula: \(\bar{x} = \frac{\sum V_i \bar{x}_i}{\sum V_i}\) (use \(-V\) for hollow parts).
• Lamina Integrals: \(\bar{x} = \frac{\int x y\,\mathrm{d}x}{\int y\,\mathrm{d}x}\), \quad \(\bar{y} = \frac{\int \frac{1}{2}y^2\,\mathrm{d}x}{\int y\,\mathrm{d}x}\).
• Solid of Revolution Integral: \(\bar{x} = \frac{\int x y^2\,\mathrm{d}x}{\int y^2\,\mathrm{d}x}\) with \(\bar{y} = 0\).
• Freely Suspended: \(G\) hangs vertically below suspension point \(A\).
• Sliding vs. Toppling: Slides first if \(\mu < \frac{b}{h}\); Topples first if \(\mu > \frac{b}{h}\).