| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cylinder or hemisphere from solid |
| Difficulty | Standard +0.8 This M3 centre of mass question requires multiple steps: calculating volumes, applying the removal method with careful coordinate geometry, then using equilibrium conditions with moments. The specific geometry (off-axis cylindrical hole) and the suspended equilibrium part requiring trigonometry make it more challenging than standard CoM problems, but it follows established M3 techniques without requiring novel insight. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| \(\pi r^2 h\) | B2 | |
| \(\pi(\frac{1}{3}r)^2(\frac{1}{4}h)\) | B2 | |
| \(\bar{y} = \frac{\pi r^2 h \cdot \frac{1}{2}h - \pi(\frac{1}{3}r)^2(\frac{1}{4}h) \cdot h}{[\pi r^2 - \pi(\frac{1}{3}r)^2(\frac{1}{4}h)]h}\) | M1 A1 ft | |
| \(\bar{y} = \frac{85h}{168}\) ** | A1 | Correct result with no errors in the working |
| (7) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 - \pi(\frac{1}{3}r)^2(\frac{1}{4}h) \cdot r = [\pi r^2 - \pi(\frac{1}{3}r)^2(\frac{1}{4}h)]\bar{x}\) | M1 A1 | |
| \(\bar{x} = -\frac{r}{252}\) | A1 | |
| \(\tan \alpha = \frac{\frac{85h}{168}}{\frac{r}{252}} = 17\) | DM1 A1 ft | Ratio is correct (inc correct way up) with their distances |
| \(r = 7.5h\) | A1 | Correct answer |
| (6) | ||
| 13 |
## 5(a)
$\pi r^2 h$ | B2 |
$\pi(\frac{1}{3}r)^2(\frac{1}{4}h)$ | B2 |
$\bar{y} = \frac{\pi r^2 h \cdot \frac{1}{2}h - \pi(\frac{1}{3}r)^2(\frac{1}{4}h) \cdot h}{[\pi r^2 - \pi(\frac{1}{3}r)^2(\frac{1}{4}h)]h}$ | M1 A1 ft |
$\bar{y} = \frac{85h}{168}$ ** | A1 | Correct result with no errors in the working
| (7) |
## 5(b)
$0 - \pi(\frac{1}{3}r)^2(\frac{1}{4}h) \cdot r = [\pi r^2 - \pi(\frac{1}{3}r)^2(\frac{1}{4}h)]\bar{x}$ | M1 A1 |
$\bar{x} = -\frac{r}{252}$ | A1 |
$\tan \alpha = \frac{\frac{85h}{168}}{\frac{r}{252}} = 17$ | DM1 A1 ft | Ratio is correct (inc correct way up) with their distances
$r = 7.5h$ | A1 | Correct answer
| (6) |
| 13 |
---\includegraphics{figure_3}
A uniform solid right circular cylinder has height $h$ and radius $r$. The centre of one plane face is $O$ and the centre of the other plane face is $Y$. A cylindrical hole is made by removing a solid cylinder of radius $\frac{1}{4}r$ and height $\frac{1}{4}h$ from the end with centre $O$. The axis of the cylinder removed is parallel to $OY$ and meets the end with centre $O$ at $X$, where $OX = \frac{1}{4}r$. One plane face of the cylinder removed coincides with the plane face through $O$ of the original cylinder. The resulting solid $S$ is shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of $S$ is at a distance $\frac{85h}{168}$ from the plane face containing $O$. [7]
\end{enumerate}
The solid $S$ is freely suspended from $O$. In equilibrium the line $OY$ is inclined at an angle arctan(17) to the horizontal.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $r$ in terms of $h$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q5 [13]}}