| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina in equilibrium with applied force |
| Difficulty | Challenging +1.2 This is a standard M3 centre of mass question with three parts: (a) proving a standard result by integration (bookwork), (b) applying symmetry and composite body techniques to find centre of mass, and (c) taking moments about a hinge for equilibrium. While it requires multiple techniques and careful coordinate work, all methods are standard M3 procedures with no novel insight required. The multi-part structure and integration push it slightly above average difficulty. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mass of quadrant \(= \rho\dfrac{\pi a^2}{4}\) | B1 | |
| \(\int_0^a \rho x\sqrt{a^2-x^2}\,dx = \rho\left[-\dfrac{1}{3}(a^2-x^2)^{\frac{3}{2}}\right]_0^a\) | M1A1 | |
| \(= \rho\left[0 + \dfrac{1}{3}a^3\right]\) | A1 | |
| \(\rho\dfrac{\pi a^2}{4}\bar{x} = \rho\dfrac{a^3}{3}\) | M1 | |
| \(\bar{x} = \dfrac{4a}{3\pi}\), \(\quad \bar{y} = \dfrac{4a}{3\pi}\) by symmetry | A1, A1 | *AG* (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mass of quadrant \(= \rho\dfrac{\pi a^2}{4}\) | B1 | |
| \(\int_0^{\frac{\pi}{2}} \rho\cdot\tfrac{1}{2}a^2\cdot\tfrac{2}{3}a\cos\theta\,d\theta = \left[\dfrac{a^3}{3}\sin\theta\right]_0^{\frac{\pi}{2}} = \rho\dfrac{a^3}{3}\) | M1A1, =A1 | |
| \(\rho\dfrac{\pi a^2}{4}\bar{x} = \rho\dfrac{a^3}{3}\) | M1 | |
| \(\bar{x} = \dfrac{4a}{3\pi}\), \(\quad \bar{y} = \dfrac{4a}{3\pi}\) by symmetry | A1A1 | *AG* (7) [15] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Table: Area \(2a^2\), \(\dfrac{\pi a^2}{4}\), \(-\dfrac{\pi a^2}{4}\); Distance to \(AE\): \(\dfrac{a}{2}\), \(a+\dfrac{4a}{3\pi}\), \(a-\dfrac{4a}{3\pi}\) | B1 | |
| Moments about \(AE\): \(2a^2\bar{x} = 2a^2\cdot\dfrac{a}{2} + \dfrac{\pi a^2}{4}\left(a+\dfrac{4a}{3\pi}\right) - \dfrac{\pi a^2}{4}\left(a-\dfrac{4a}{3\pi}\right)\) | M1A2 | |
| \(= a^3 + \dfrac{2a^3}{3} = \dfrac{5a^3}{3}\) | A1 | |
| \(\bar{x} = \dfrac{5a^3}{3}\times\dfrac{1}{2a^2} = \dfrac{5a}{6}\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Taking moments about \(E\): \(2aX = \dfrac{5a}{6}W\) | M1A1ft | their \(\bar{x}\) |
| \(X = \dfrac{5}{12}W\) | A1 | (3) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mass of quadrant $= \rho\dfrac{\pi a^2}{4}$ | B1 | |
| $\int_0^a \rho x\sqrt{a^2-x^2}\,dx = \rho\left[-\dfrac{1}{3}(a^2-x^2)^{\frac{3}{2}}\right]_0^a$ | M1A1 | |
| $= \rho\left[0 + \dfrac{1}{3}a^3\right]$ | A1 | |
| $\rho\dfrac{\pi a^2}{4}\bar{x} = \rho\dfrac{a^3}{3}$ | M1 | |
| $\bar{x} = \dfrac{4a}{3\pi}$, $\quad \bar{y} = \dfrac{4a}{3\pi}$ by symmetry | A1, A1 | *AG* **(7)** |
**v2 alternative:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mass of quadrant $= \rho\dfrac{\pi a^2}{4}$ | B1 | |
| $\int_0^{\frac{\pi}{2}} \rho\cdot\tfrac{1}{2}a^2\cdot\tfrac{2}{3}a\cos\theta\,d\theta = \left[\dfrac{a^3}{3}\sin\theta\right]_0^{\frac{\pi}{2}} = \rho\dfrac{a^3}{3}$ | M1A1, =A1 | |
| $\rho\dfrac{\pi a^2}{4}\bar{x} = \rho\dfrac{a^3}{3}$ | M1 | |
| $\bar{x} = \dfrac{4a}{3\pi}$, $\quad \bar{y} = \dfrac{4a}{3\pi}$ by symmetry | A1A1 | *AG* **(7) [15]** |
---
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Table: Area $2a^2$, $\dfrac{\pi a^2}{4}$, $-\dfrac{\pi a^2}{4}$; Distance to $AE$: $\dfrac{a}{2}$, $a+\dfrac{4a}{3\pi}$, $a-\dfrac{4a}{3\pi}$ | B1 | |
| Moments about $AE$: $2a^2\bar{x} = 2a^2\cdot\dfrac{a}{2} + \dfrac{\pi a^2}{4}\left(a+\dfrac{4a}{3\pi}\right) - \dfrac{\pi a^2}{4}\left(a-\dfrac{4a}{3\pi}\right)$ | M1A2 | |
| $= a^3 + \dfrac{2a^3}{3} = \dfrac{5a^3}{3}$ | A1 | |
| $\bar{x} = \dfrac{5a^3}{3}\times\dfrac{1}{2a^2} = \dfrac{5a}{6}$ | A1 | **(5)** |
---
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Taking moments about $E$: $2aX = \dfrac{5a}{6}W$ | M1A1ft | their $\bar{x}$ |
| $X = \dfrac{5}{12}W$ | A1 | **(3)** |
---6. (a) A uniform lamina is in the shape of a quadrant of a circle of radius $a$. Show, by integration, that the centre of mass of the lamina is at a distance of $\frac { 4 a } { 3 \pi }$ from each of its straight edges.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{daa795f0-2c5e-4617-a295-fbe74c22be4a-10_809_802_484_571}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A second uniform lamina $A B C D E F A$ is shown shaded in Figure 3. The straight sides $A C$ and $A E$ are perpendicular and $A C = A E = 2 a$. In the figure, the midpoint of $A C$ is $B$, the midpoint of $A E$ is $F$, and $A B D F$ and $D G E F$ are squares of side $a$. $B C D$ is a quadrant of a circle with centre $B$. $D G E$ is a quadrant of a circle with centre $G$.\\
(b) Find the distance of the centre of mass of the lamina from the side $A E$.
The lamina is smoothly hinged to a horizontal axis which passes through $E$ and is perpendicular to the plane of the lamina. The lamina has weight $W$ newtons. The lamina is held in equilibrium in a vertical plane, with $A$ vertically above $E$, by a horizontal force of magnitude $X$ newtons applied at $C$.\\
(c) Find $X$ in terms of $W$.\\
\hfill \mbox{\textit{Edexcel M3 2013 Q6 [15]}}