Centres of Mass – Position, Uniform & Composite Bodies (Cambridge 9709 – Mechanics, Paper 4)
1. Why the Centre of Mass (COM) is Central to Mechanics
- The resultant external force on any system of particles acts through its COM. This links the COM directly to equilibrium and Newton’s 2nd law.
- The motion of the COM follows the same kinematic equations as a single particle of total mass \(M\):
\[
M\mathbf{a}_{\text{COM}}=\sum\mathbf{F}_{\text{ext}} .
\]
- For an isolated system linear momentum is conserved:
\[
M\mathbf{v}_{\text{COM}}=\text{constant}.
\]
- The total external work equals the change in kinetic energy of the COM:
\[
W_{\text{ext}}=\Delta\!\left(\tfrac12 M v_{\text{COM}}^{2}\right).
\]
2. Definition of the Centre of Mass
2.1 Discrete system of particles
For \(n\) particles with masses \(m_i\) and position vectors \(\mathbf{r}_i\) (measured from a common origin), the COM position vector \(\mathbf{R}\) is
\[
\boxed{\displaystyle \mathbf{R}= \frac{\sum_{i=1}^{n} m_i\,\mathbf{r}_i}{\sum_{i=1}^{n} m_i}}
\]
2.2 Continuous body (general vector form)
For a three‑dimensional body of total mass \(M\) and density \(\rho(\mathbf{r})\)
\[
\boxed{\displaystyle \mathbf{R}= \frac{1}{M}\iiint\limits_{V}\mathbf{r}\,\rho(\mathbf{r})\,\mathrm{d}V},
\qquad
M=\iiint\limits_{V}\rho(\mathbf{r})\,\mathrm{d}V .
\]
If the body is uniform (\(\rho=\text{constant}\)), the density cancels and the COM depends only on geometry:
\[
\mathbf{R}= \frac{1}{V}\iiint\limits_{V}\mathbf{r}\,\mathrm{d}V .
\]
3. Uniform Rigid Bodies – Using Symmetry
When the density is constant, symmetry often gives the COM instantly.
| Shape | COM location (relative to geometric centre) |
|---|
| Thin uniform rod (length \(L\)) | Mid‑point: \((L/2,0,0)\) |
| Rectangle (width \(w\), height \(h\)) | Intersection of diagonals: \((w/2,\;h/2,0)\) |
| Uniform semicircle (radius \(R\)) | On the symmetry axis at \(\displaystyle\frac{4R}{3\pi}\) from the flat side |
| Uniform thin equilateral triangle (side \(a\)) | At a distance \(\displaystyle\frac{h}{3}\) from the base, where \(h=\frac{\sqrt3}{2}a\) |
4. Forces & Equilibrium (Paper 4 4.1)
4.1 Free‑Body Diagrams (FBD)
- Identify every external force acting on the body (weight, normal reaction, tension, friction, applied forces, etc.).
- Represent each force as a vector arrow starting from the point of application.
- Label magnitude, direction and line of action. Remember: the weight \(\mathbf{W}=M\mathbf{g}\) always acts through the COM.
4.2 Component resolution
Resolve each force into orthogonal components (usually \(x\)‑ and \(y\)‑axes). For a force \(\mathbf{F}\) making an angle \(\theta\) with the positive \(x\)‑axis:
\[
F_x = F\cos\theta,\qquad F_y = F\sin\theta .
\]
4.3 Equilibrium condition
A body is in static equilibrium when the vector sum of all external forces is zero:
\[
\sum\mathbf{F}= \mathbf{0}\quad\Longleftrightarrow\quad
\begin{cases}
\displaystyle\sum F_x =0,\\[4pt]
\displaystyle\sum F_y =0 .
\end{cases}
\]
For a particle or rigid body the resultant force must pass through the COM; this is often used to locate the COM of an irregular shape.
4.4 Friction
When a surface is in contact, the limiting frictional force is
\[
F_{\text{fr}}^{\max}= \mu N,
\]
where \(\mu\) is the coefficient of friction (static \(\mu_s\) or kinetic \(\mu_k\)) and \(N\) is the normal reaction.
In equilibrium, the actual frictional force satisfies \(|F_{\text{fr}}|\le \mu_s N\).
4.5 Worked Example – Particle on an Inclined Plane (with & without friction)
Mass \(m\) on a plane inclined at \(\theta\) to the horizontal.
- Draw the FBD – weight \(\mathbf{W}=mg\) vertically down, normal \(N\) perpendicular to the plane, friction \(F_{\text{fr}}\) parallel to the plane (up the plane if the particle tends to slide down).
- Resolve \(\mathbf{W}\) into components:
\[
W_{\parallel}=mg\sin\theta,\qquad
W_{\perp}=mg\cos\theta .
\]
- Equilibrium (no motion):
\[
\sum F_{\parallel}=0\;\Rightarrow\;F_{\text{fr}}=mg\sin\theta,
\qquad
\sum F_{\perp}=0\;\Rightarrow\;N=mg\cos\theta .
\]
The condition \(|F_{\text{fr}}|\le\mu_s N\) gives the limiting angle \(\theta_{\max}\) for which the particle just begins to move:
\[
\tan\theta_{\max}= \mu_s .
\]
- When the plane is smooth (\(\mu=0\)) the particle slides. The net force down the plane is \(mg\sin\theta\) and Newton’s 2nd law gives
\[
ma = mg\sin\theta \;\Longrightarrow\; a = g\sin\theta .
\]
5. Kinematics of the COM (Paper 4 4.2)
5.1 Scalar–vector distinction
- Displacement \(\mathbf{s}\) and velocity \(\mathbf{v}\) are vectors; speed \(|\mathbf{v}|\) and distance travelled are scalars.
- Acceleration \(\mathbf{a}\) is a vector; its magnitude is the scalar acceleration.
5.2 Constant‑acceleration formulae (for the COM)
\[
\begin{aligned}
\mathbf{v}_{\text{COM}} &= \mathbf{u}_{\text{COM}} + \mathbf{a}_{\text{COM}}t,\\[4pt]
\mathbf{s}_{\text{COM}} &= \mathbf{u}_{\text{COM}}t + \tfrac12 \mathbf{a}_{\text{COM}}t^{2},\\[4pt]
v_{\text{COM}}^{2} &= u_{\text{COM}}^{2}+2a_{\text{COM}}s_{\text{COM}} .
\end{aligned}
\]
5.3 Graph interpretation
- Velocity‑time (v‑t) graph: the gradient gives the acceleration; the area under the curve gives the displacement.
- Displacement‑time (s‑t) graph: the gradient gives the instantaneous velocity; the area under a velocity‑time graph between two times equals the change in displacement.
5.4 Differentiation / Integration
If the displacement of the COM is given as a function of time, \( \mathbf{s}(t) \), then
\[
\mathbf{v}(t)=\frac{d\mathbf{s}}{dt},\qquad
\mathbf{a}(t)=\frac{d\mathbf{v}}{dt}.
\]
Conversely, if the acceleration is known, integrate twice to obtain \(\mathbf{s}(t)\).
5.5 Worked Example – COM of a Sliding Block
A block of mass \(m\) slides down a smooth plane of angle \(\theta=30^{\circ}\) from rest. Find its speed after travelling \(4\;\text{m}\) along the plane.
- From Section 4.5, the acceleration of the COM is \(a=g\sin\theta = 9.8\sin30^{\circ}=4.9\;\text{m s}^{-2}\).
- Using \(v^{2}=u^{2}+2as\) with \(u=0\) and \(s=4\) m:
\[
v = \sqrt{2(4.9)(4)} = \sqrt{39.2}\approx 6.26\;\text{m s}^{-1}.
\]
- The corresponding v‑t graph is a straight line through the origin with gradient \(4.9\;\text{m s}^{-2}\); the area under the line up to the time when \(s=4\) m gives the displacement, confirming the calculation.
6. Momentum & Work‑Energy (Paper 4 4.3 & 4.5)
- Linear momentum of the whole system is \( \mathbf{P}=M\mathbf{v}_{\text{COM}} \). In the absence of external forces, \(\mathbf{P}\) is constant.
- For collisions, treat each body as a particle located at its COM and apply \(\sum \mathbf{p}_{\text{initial}} = \sum \mathbf{p}_{\text{final}}\).
- The work‑energy theorem for a system of particles can be written in terms of the COM:
\[
W_{\text{ext}} = \Delta\!\left(\tfrac12 M v_{\text{COM}}^{2}\right).
\]
7. Non‑Uniform Bodies (Extension)
If the density varies, \(\rho(\mathbf{r})\) must be retained inside the integral. Example – linear density along a rod:
\[
\rho(x)=\rho_0\!\left(1+\frac{x}{L}\right)
\quad\Longrightarrow\quad
x_{\text{COM}}=\frac{1}{M}\int_{0}^{L}x\,\rho(x)\,\mathrm{d}x .
\]
8. Composite Bodies – Treating Sub‑bodies as Point Masses (Paper 4 4.4)
When a body can be divided into \(N\) simpler parts whose individual COMs \(\mathbf{r}_k\) and masses \(M_k\) are known, the overall COM is
\[
\boxed{\displaystyle \mathbf{R}= \frac{\sum_{k=1}^{N} M_k\,\mathbf{r}_k}{\sum_{k=1}^{N} M_k}} .
\]
All position vectors must be measured from the **same origin**.
9. Worked Examples (All Topics Integrated)
Example 1 – Uniform Thin Rod (Discrete Integration)
Rod of length \(L\), mass \(M\), lying on the \(x\)-axis from \(x=0\) to \(x=L\).
\[
x_{\text{COM}}=\frac{1}{M}\int_{0}^{L}x\left(\frac{M}{L}\right)\!\mathrm{d}x
=\frac{L}{2}.
\]
Example 2 – L‑Shaped Composite Plate (Composite Method)
Two uniform rectangles joined at a right‑angle corner (origin at the inner corner).
| Component | Dimensions | Mass | COM position |
|---|
| Plate A | \(a\times b\) | \(M_A=\rho a b\) | \(\mathbf{r}_A=(a/2,\;b/2,0)\) |
| Plate B | \(c\times d\) | \(M_B=\rho c d\) | \(\mathbf{r}_B=(a+c/2,\;d/2,0)\) |
\[
\mathbf{R}= \frac{M_A\mathbf{r}_A+M_B\mathbf{r}_B}{M_A+M_B}
=\left(\frac{ab\,\tfrac{a}{2}+cd\!\left(a+\tfrac{c}{2}\right)}{ab+cd},
\frac{ab\,\tfrac{b}{2}+cd\,\tfrac{d}{2}}{ab+cd},0\right).
\]
Example 3 – Conservation of Momentum Using the COM
Two particles, \(m_1=2\;\text{kg}\) moving east at \(3\;\text{m s}^{-1}\) and \(m_2=3\;\text{kg}\) moving north at \(2\;\text{m s}^{-1}\) stick together.
\[
\mathbf{v}_{\text{COM}}=
\frac{2(3\hat{i})+3(2\hat{j})}{5}
=\frac{6}{5}\hat{i}+\frac{6}{5}\hat{j}\;\text{m s}^{-1}.
\]
The combined \(5\;\text{kg}\) mass moves with this velocity – a direct application of linear momentum conservation.
Example 4 – Work‑Energy for a Falling Composite Body
An L‑shaped plate (as in Example 2) is released from rest with its corner at height \(h\). The external work is the loss of gravitational potential energy of the COM:
\[
W_{\text{ext}} = M g \Delta y_{\text{COM}} .
\]
Since the plate starts from rest,
\[
\tfrac12 M v_{\text{COM}}^{2}=M g \Delta y_{\text{COM}} .
\]
Example 5 – Non‑Uniform Rod (Linear Density)
Rod of length \(L\) with density \(\rho(x)=\rho_0(1+x/L)\).
\[
M=\int_{0}^{L}\rho(x)\,\mathrm{d}x=\rho_0\!\left(L+\frac{L}{2}\right)=\tfrac32\rho_0L,
\]
\[
x_{\text{COM}}=\frac{1}{M}\int_{0}^{L}x\rho(x)\,\mathrm{d}x
=\frac{1}{\tfrac32\rho_0L}\,\rho_0\!\left[\tfrac{x^{2}}{2}+\tfrac{x^{3}}{3L}\right]_{0}^{L}
=\frac{5}{8}L .
\]
10. Summary of Key Formulas
| Situation | COM Formula | Comments |
|---|
| Discrete particles |
\(\displaystyle \mathbf{R}= \frac{\sum m_i\mathbf{r}_i}{\sum m_i}\) |
Valid for any system, uniform or not. |
| Continuous uniform body (1‑D) |
\(\displaystyle x_{\text{COM}}=\frac{1}{L}\int_{0}^{L}x\,\mathrm{d}x\) |
Linear density cancels. |
| Continuous uniform body (2‑D) |
\(\displaystyle \mathbf{R}= \frac{1}{A}\iint\mathbf{r}\,\mathrm{d}A\) |
Area density cancels. |
| Continuous uniform body (3‑D) |
\(\displaystyle \mathbf{R}= \frac{1}{V}\iiint\mathbf{r}\,\mathrm{d}V\) |
Volume density cancels. |
| Non‑uniform body |
\(\displaystyle \mathbf{R}= \frac{1}{M}\iiint\mathbf{r}\,\rho(\mathbf{r})\,\mathrm{d}V\) |
Keep \(\rho(\mathbf{r})\) inside the integral. |
| Composite body |
\(\displaystyle \mathbf{R}= \frac{\sum M_k\mathbf{r}_k}{\sum M_k}\) |
Each sub‑body treated as a point mass at its own COM. |
| Resultant force on a system |
\(\displaystyle \sum\mathbf{F}_{\text{ext}} = M\mathbf{a}_{\text{COM}}\) |
Newton’s 2nd law for the COM. |
| Momentum conservation |
\(\displaystyle M\mathbf{v}_{\text{COM}}=\text{constant}\) |
Isolated system. |
| Work‑energy (COM) |
\(\displaystyle W_{\text{ext}} = \Delta\!\left(\tfrac12 M v_{\text{COM}}^{2}\right)\) |
External work equals change in kinetic energy of the COM. |
11. Common Pitfalls
- Mixing origins: All position vectors must be measured from the same reference point.
- Forgetting density in non‑uniform problems: Keep \(\rho(\mathbf{r})\) inside the integral.
- Assuming symmetry without proof: Verify that an axis of symmetry really passes through the COM; otherwise perform the integral.
- Units: Keep mass (kg) and length (m) consistent throughout.
- Force line of action: Remember that the weight always acts through the COM; this is crucial for equilibrium diagrams.
- Sign conventions in graphs: Positive direction must be chosen consistently when reading or drawing \(v\)-\(t\) and \(s\)-\(t\) graphs.
12. Practice Questions (Cambridge 9709 Paper 4 style)
- Find the centre of mass of a uniform thin triangular plate of base \(b\) and height \(h\), with the base on the \(x\)-axis.
- A solid cylinder of mass \(M\) and radius \(R\) is cut into two equal halves along a plane through its axis. Determine the COM of one half (give coordinates relative to the flat face).
- Three uniform rods of lengths \(2\;\text{m},\;3\;\text{m}\) and \(5\;\text{m}\) are joined end‑to‑end along a straight line. Find the COM of the whole assembly measured from the left‑most end.
- A particle of mass \(m\) slides down a smooth incline of angle \(30^{\circ}\) from rest. Using the COM concept, write the equation of motion and calculate its speed after travelling \(4\;\text{m}\) along the plane.
- A composite body consisting of a uniform square plate (side \(a\)) attached to a uniform right‑angled triangular plate (legs \(a\) and \(a\)) is released from rest at height \(h\). Show that the work done by gravity equals the increase in kinetic energy of the COM.
- For a non‑uniform rod of length \(L\) with linear density \(\lambda(x)=\lambda_0(1+x/L)\), calculate (a) its total mass, (b) the position of its COM, and (c) the acceleration of the COM when the rod is released from rest and falls vertically under gravity (ignore air resistance).
- Two particles, \(m_1=1.5\;\text{kg}\) moving east at \(4\;\text{m s}^{-1}\) and \(m_2=2.0\;\text{kg}\) moving west at \(2\;\text{m s}^{-1}\), collide elastically. Use the COM frame to find the velocities after the collision.
13. Further Reading
- Cambridge International AS & A Level Mathematics (9709) – Mechanics, Chapter 6 (Centres of Mass & Equilibrium).
- “Mathematical Methods for Physics and Engineering” – sections on mass moments, centroids and applications to non‑uniform bodies.
- IB Physics & Cambridge A‑Level resources on rotational dynamics for an extension to moments of inertia about the COM.