| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Deriving standard centre of mass formulae by integration |
| Difficulty | Challenging +1.2 Part (a) is a standard M3 centre of mass derivation requiring disc integration and algebraic manipulation (5 marks). Parts (b) and (c) apply the result using composite body formulae and equilibrium conditions. While requiring multiple techniques, this follows a predictable M3 template with no novel insights needed—moderately above average difficulty for A-level. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits6.04b Find centre of mass: using symmetry6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((\pi\rho)\int_0^r xy^2 \, dx\) | ||
| \(= (\pi\rho)\int_0^r x(r^2 - x^2) \, dx\) | M1 | |
| \(= (\pi\rho)\left[\frac{1}{2}x^2r^2 - \frac{x^4}{4}\right]_0^r\) | A1 | |
| \(= (\pi\rho)\frac{r^4}{4}\) | A1 | |
| \(M\bar{x} = \pi\rho\int xy^2 \, dx\) | M1 | |
| \(\bar{x} = \frac{\pi\rho r^4}{4} \div \frac{2\pi\rho r^3}{3} = \frac{3}{8}r\) * | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Distances from \(O\): mass \(m\) at \(\frac{3}{8}r + 4r\); mass \(M\) at \(\frac{3}{4} \times 4r\) | B1 | |
| \(\frac{35}{8}rm + 3rM = (m+M)\bar{x}\) | M1A1ft | |
| \(\bar{x} = \frac{(35m + 24M)r}{8(m+M)}\) | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\bar{x}\cos\theta \leqslant OA\) | M1 | |
| \(\cos\theta = \frac{4r}{OA}\) | B1 | |
| \(\bar{x} \leqslant \frac{OA^2}{4r}\) | A1 | |
| \(\bar{x} = \frac{(35m+24M)r}{8(m+M)} \leqslant \frac{17r^2}{4r}\) | M1 | |
| \(35m + 24M \leqslant 34(m+M)\) | ||
| \(M \geqslant \frac{m}{10}\) * | A1 (5) [14] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Using \(\int xy^2\,dx\) | M1 | \(\pi\) and/or \(\rho\) may be missing; limits need not be shown |
| Correct integration | A1 | \(\pi\), \(\rho\) and limits need not be shown |
| Use limits to obtain \(\dfrac{r^4}{4}\) | A1 | |
| Use \(M\bar{x} = \pi\rho\int xy^2\,dx\) with their previous result | M1 | \(\pi\) and \(\rho\) must be seen in both sides or neither |
| Obtain \(\dfrac{3}{8}r\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Correct distances from \(O\) or centre of plane face | B1 | Can both be positive |
| Construct a moments equation with their distances using their masses (which may be volumes) | M1 | |
| Correct moments equation, follow through their distances (but not their masses) | A1ft | If working from centre of plane face one term must be negative |
| Correct result for \(\bar{x}\) (any equivalent) | A1 | Including fractions within fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempting an inequality with \(\bar{x}\) and \(OA\) | M1 | Sign either way round or \(=\) |
| A correct trig function connecting the angle used and \(OA\) | B1 | |
| Obtain \(\bar{x} \leqslant \dfrac{OA^2}{4r}\) | A1 | |
| Use their expression for \(\bar{x}\) in their inequality/equality | M1 | |
| Obtain the given result | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use distance from centre of common face and tangents; \(M\) is min when \(\dfrac{\bar{x}-4r}{r} = \dfrac{r}{4r}\ (=\tan\theta)\) |
## Question 6:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $(\pi\rho)\int_0^r xy^2 \, dx$ | | |
| $= (\pi\rho)\int_0^r x(r^2 - x^2) \, dx$ | M1 | |
| $= (\pi\rho)\left[\frac{1}{2}x^2r^2 - \frac{x^4}{4}\right]_0^r$ | A1 | |
| $= (\pi\rho)\frac{r^4}{4}$ | A1 | |
| $M\bar{x} = \pi\rho\int xy^2 \, dx$ | M1 | |
| $\bar{x} = \frac{\pi\rho r^4}{4} \div \frac{2\pi\rho r^3}{3} = \frac{3}{8}r$ * | A1 (5) | |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Distances from $O$: mass $m$ at $\frac{3}{8}r + 4r$; mass $M$ at $\frac{3}{4} \times 4r$ | B1 | |
| $\frac{35}{8}rm + 3rM = (m+M)\bar{x}$ | M1A1ft | |
| $\bar{x} = \frac{(35m + 24M)r}{8(m+M)}$ | A1 (4) | |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\bar{x}\cos\theta \leqslant OA$ | M1 | |
| $\cos\theta = \frac{4r}{OA}$ | B1 | |
| $\bar{x} \leqslant \frac{OA^2}{4r}$ | A1 | |
| $\bar{x} = \frac{(35m+24M)r}{8(m+M)} \leqslant \frac{17r^2}{4r}$ | M1 | |
| $35m + 24M \leqslant 34(m+M)$ | | |
| $M \geqslant \frac{m}{10}$ * | A1 (5) [14] | |
# Question (previous question - Centre of Mass):
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using $\int xy^2\,dx$ | M1 | $\pi$ and/or $\rho$ may be missing; limits need not be shown |
| Correct integration | A1 | $\pi$, $\rho$ and limits need not be shown |
| Use limits to obtain $\dfrac{r^4}{4}$ | A1 | |
| Use $M\bar{x} = \pi\rho\int xy^2\,dx$ with their previous result | M1 | $\pi$ and $\rho$ must be seen in both sides or neither |
| Obtain $\dfrac{3}{8}r$ | A1cso | |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct distances from $O$ or centre of plane face | B1 | Can both be positive |
| Construct a moments equation with their distances using their masses (which may be volumes) | M1 | |
| Correct moments equation, follow through their distances (but not their masses) | A1ft | If working from centre of plane face one term must be negative |
| Correct result for $\bar{x}$ (any equivalent) | A1 | Including fractions within fractions |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempting an inequality with $\bar{x}$ and $OA$ | M1 | Sign either way round or $=$ |
| A correct trig function connecting the angle used and $OA$ | B1 | |
| Obtain $\bar{x} \leqslant \dfrac{OA^2}{4r}$ | A1 | |
| Use their expression for $\bar{x}$ in their inequality/equality | M1 | |
| Obtain the given result | A1cso | |
**ALT for (c):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use distance from centre of common face and tangents; $M$ is min when $\dfrac{\bar{x}-4r}{r} = \dfrac{r}{4r}\ (=\tan\theta)$ | | |
---6. (a) Use algebraic integration to show that the centre of mass of a uniform solid hemisphere of radius $r$ is at a distance $\frac { 3 } { 8 } r$ from the centre of its plane face.\\[0pt]
[You may assume that the volume of a sphere of radius $r$ is $\frac { 4 } { 3 } \pi r ^ { 3 }$ ]\\
(5)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ffe0bc72-3136-48d9-9d5b-4a364d134070-09_351_597_598_678}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A uniform solid hemisphere of mass $m$ and radius $r$ is joined to a uniform solid right circular cone to form a solid $S$. The cone has mass $M$, base radius $r$ and height $4 r$. The vertex of the cone is $O$. The plane face of the cone coincides with the plane face of the hemisphere, as shown in Figure 3.\\
(b) Find the distance of the centre of mass of $S$ from $O$.
The point $A$ lies on the circumference of the base of the cone. The solid is placed on a horizontal table with $O A$ in contact with the table. The solid remains in equilibrium in this position.\\
(c) Show that $M \geqslant \frac { 1 } { 10 } m$\\
\hfill \mbox{\textit{Edexcel M3 2016 Q6 [14]}}