| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina with particle attached |
| Difficulty | Standard +0.8 This is a multi-step centre of mass problem requiring composite body techniques with negative masses (removed discs), coordinate geometry to locate centres, and then equilibrium analysis with an attached particle. Part (a) involves careful bookkeeping of three circular regions with different radii and positions, requiring algebraic manipulation to reach a specific numerical result. Part (b) adds equilibrium conditions with moments about a suspension point. While the techniques are standard M2 content, the complexity of the geometry, the precision required to show the exact fractional answer, and the two-part structure with the particle attachment make this significantly harder than routine centre of mass questions. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Mass ratios: Disc = 64, \(C_1\) = 1, \(C_2\) = 4, \(L\) = 59 | B1 | Mass ratio |
| Distances from \(R\): Disc = \(8a\), \(C_1\) = \(12a\), \(C_2\) = \(4a\), \(L\) = \(d\) | B1 | Distances |
| Moments about tangent through \(C\) | M1 | Or a parallel axis. Needs all terms. Must be dimensionally correct. Condone sign errors |
| \(64 \times 8a - 1 \times 12a - 4 \times 4a = 59d\) | A1 | Correct unsimplified equation. Accept a vector equation. \(\frac{460}{59}a\) from \(P\), \(\frac{12}{59}a\) from \(O\) |
| \(d = \frac{484}{59}a\) Given Answer | A1 | Need to see supporting working, e.g. a scalar equation in \(d\) |
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Distance of combined c.o.m. from \(R\) | M1 | |
| Horizontal: \(\bar{x} = \dfrac{\frac{484}{59}aM + 8akM}{(1+k)M}\) | A1 | |
| Vertical: \(\bar{y} = \dfrac{8akM}{(1+k)M}\) | A1 | |
| Use of \(\tan\alpha = \frac{1}{4} = \dfrac{8k}{\frac{484}{59} + 8k}\) | M1 | Condone use of tan the wrong way up |
| \(32k = \frac{484}{59} + 8k\), \(k = \frac{121}{354}\) | A1 | 0.342 |
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(C\) lies at intersection of (line through pt from (a) and \(S\)) and (line through \(R\) with gradient \(\frac{1}{4}\)) | M1 | |
| \(y = -\frac{118}{3}x - 8a\) and \(y = \frac{1}{4}x - 2a\) | A1 | |
| \(\Rightarrow x = -\frac{72}{475}a\) | A1 | |
| Moments about \(O\): \(-\frac{12}{50}a = -(k+1)\frac{72}{475}a\) | M1 | |
| \(k = \frac{121}{354}\) | A1 | |
| (5 marks) See over page for more alternative methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Take moments about axis through \(R\): \(M \times \frac{484}{59}a\sin\alpha = kM \times 6a\cos\alpha\) | M1A2 | Condone sin/cos confusion |
| Solve for \(k\): \(6k = \frac{484}{59}\tan\alpha = \frac{121}{59}\), \(k = \frac{121}{354}(= 0.342)\) | M1A1 | |
| Total: (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Moments: \(Mg \times \frac{484}{59}a\sin\alpha = kMg \times 8a\sqrt{2}\sin(45-\alpha)\) | M1A2 | |
| \(\frac{484}{59}\sin(14.036...) = k \times 8\sin(30.96...)\) | M1 | |
| \(k = 0.342\) | A1 | |
| Total: (5) [10] |
# Question 6:
## Part 6a:
| Working | Mark | Notes |
|---------|------|-------|
| Mass ratios: Disc = 64, $C_1$ = 1, $C_2$ = 4, $L$ = 59 | B1 | Mass ratio |
| Distances from $R$: Disc = $8a$, $C_1$ = $12a$, $C_2$ = $4a$, $L$ = $d$ | B1 | Distances |
| Moments about tangent through $C$ | M1 | Or a parallel axis. Needs all terms. Must be dimensionally correct. Condone sign errors |
| $64 \times 8a - 1 \times 12a - 4 \times 4a = 59d$ | A1 | Correct unsimplified equation. Accept a vector equation. $\frac{460}{59}a$ from $P$, $\frac{12}{59}a$ from $O$ |
| $d = \frac{484}{59}a$ **Given Answer** | A1 | Need to see supporting working, e.g. a scalar equation in $d$ |
| **(5 marks)** | | |
## Part 6b:
| Working | Mark | Notes |
|---------|------|-------|
| Distance of combined c.o.m. from $R$ | M1 | |
| Horizontal: $\bar{x} = \dfrac{\frac{484}{59}aM + 8akM}{(1+k)M}$ | A1 | |
| Vertical: $\bar{y} = \dfrac{8akM}{(1+k)M}$ | A1 | |
| Use of $\tan\alpha = \frac{1}{4} = \dfrac{8k}{\frac{484}{59} + 8k}$ | M1 | Condone use of tan the wrong way up |
| $32k = \frac{484}{59} + 8k$, $k = \frac{121}{354}$ | A1 | 0.342 |
| **(5 marks)** | | |
## Part 6b (alt):
| Working | Mark | Notes |
|---------|------|-------|
| $C$ lies at intersection of (line through pt from (a) and $S$) and (line through $R$ with gradient $\frac{1}{4}$) | M1 | |
| $y = -\frac{118}{3}x - 8a$ and $y = \frac{1}{4}x - 2a$ | A1 | |
| $\Rightarrow x = -\frac{72}{475}a$ | A1 | |
| Moments about $O$: $-\frac{12}{50}a = -(k+1)\frac{72}{475}a$ | M1 | |
| $k = \frac{121}{354}$ | A1 | |
| **(5 marks)** See over page for more alternative methods | | |
## Question 6b alt:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Take moments about axis through $R$: $M \times \frac{484}{59}a\sin\alpha = kM \times 6a\cos\alpha$ | M1A2 | Condone sin/cos confusion |
| Solve for $k$: $6k = \frac{484}{59}\tan\alpha = \frac{121}{59}$, $k = \frac{121}{354}(= 0.342)$ | M1A1 | |
| **Total: (5)** | | |
## Question 6balt (second method):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Moments: $Mg \times \frac{484}{59}a\sin\alpha = kMg \times 8a\sqrt{2}\sin(45-\alpha)$ | M1A2 | |
| $\frac{484}{59}\sin(14.036...) = k \times 8\sin(30.96...)$ | M1 | |
| $k = 0.342$ | A1 | |
| **Total: (5) [10]** | | |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be16c17a-c4db-4f0c-9f32-8d5614b4f2f3-16_1031_915_116_513}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The uniform lamina $L$ shown shaded in Figure 2 is formed by removing two circular discs, $C _ { 1 }$ and $C _ { 2 }$, from a circular disc with centre $O$ and radius $8 a$. Disc $C _ { 1 }$ has centre $A$ and radius $a$. Disc $C _ { 2 }$ has centre $B$ and radius $2 a$. The diameters $P R$ and $Q S$ are perpendicular. The midpoint of $P O$ is $A$ and the midpoint of $O R$ is $B$.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of $L$ is $\frac { 484 } { 59 } a$ from $R$.
The mass of $L$ is $M$. A particle of mass $k M$ is attached to $L$ at $S$. The lamina with the attached particle is suspended from $R$ and hangs freely in equilibrium with the diameter $P R$ at an angle of arctan $\left( \frac { 1 } { 4 } \right)$ to the downward vertical through $R$.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q6 [10]}}