| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with cone and cylinder |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass question requiring application of the composite body formula (using known results for cone and cylinder centres of mass) and a toppling condition. Part (a) is algebraic manipulation with given formulae, part (b) applies the standard toppling criterion (vertical through COM passes through edge). Slightly easier than average due to being a textbook application of standard techniques with no novel insight required. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Volumes: hemisphere \(\pi r^3\), cone \(\dfrac{1}{3}\pi r^2 h\), combined \(\pi r^3 + \dfrac{1}{3}\pi r^2 h\) | M1 A1 | |
| Distances of CM: hemisphere \(\dfrac{r}{2}\), cone \(r+\dfrac{h}{4}\), combined \(\bar{x}\) | B1 B1 | |
| \(\dfrac{\pi r^4}{2} + \dfrac{1}{3}\pi r^2 h\left(r+\dfrac{h}{4}\right) = \left(\pi r^3 + \dfrac{1}{3}\pi r^2 h\right)\bar{x}\) | M1 A1 A1ft | |
| \(\bar{x} = \dfrac{6r^2 + 4hr + h^2}{4(3r+h)}\) | A1 | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(h = 2r \Rightarrow \bar{x} = \dfrac{18r}{20} = \dfrac{9r}{10}\) | M1 A1 | |
| \(\tan\alpha = \dfrac{r}{9r/10} = \dfrac{10}{9}\) | M1 A1 ft | |
| \(\alpha \approx 48°\) | A1 | (5) |
# Question 5:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| Volumes: hemisphere $\pi r^3$, cone $\dfrac{1}{3}\pi r^2 h$, combined $\pi r^3 + \dfrac{1}{3}\pi r^2 h$ | M1 A1 | |
| Distances of CM: hemisphere $\dfrac{r}{2}$, cone $r+\dfrac{h}{4}$, combined $\bar{x}$ | B1 B1 | |
| $\dfrac{\pi r^4}{2} + \dfrac{1}{3}\pi r^2 h\left(r+\dfrac{h}{4}\right) = \left(\pi r^3 + \dfrac{1}{3}\pi r^2 h\right)\bar{x}$ | M1 A1 A1ft | |
| $\bar{x} = \dfrac{6r^2 + 4hr + h^2}{4(3r+h)}$ | A1 | (8) |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $h = 2r \Rightarrow \bar{x} = \dfrac{18r}{20} = \dfrac{9r}{10}$ | M1 A1 | |
| $\tan\alpha = \dfrac{r}{9r/10} = \dfrac{10}{9}$ | M1 A1 ft | |
| $\alpha \approx 48°$ | A1 | (5) |
---5.\\
\includegraphics[max width=\textwidth, alt={}, center]{e256678d-89e8-48eb-aa8a-b8e027b62ef1-3_423_357_918_847}
A uniform solid, $S$, is placed with its plane face on horizontal ground. The solid consists of a right circular cylinder, of radius $r$ and height $r$, joined to a right circular cone, of radius $r$ and height $h$. The plane face of the cone coincides with one of the plane faces of the cylinder, as shown in Fig. 3.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of $S$ from the ground is
$$\frac { 6 r ^ { 2 } + 4 r h + h ^ { 2 } } { 4 ( 3 r + h ) }$$
(8)
The solid is now placed with its plane face on a rough plane which is inclined at an angle $\alpha$ to the horizontal. The plane is rough enough to prevent $S$ from sliding. Given that $h = 2 r$, and that $S$ is on the point of toppling,
\item find, to the nearest degree, the value of $\alpha$.\\
(5)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5 [13]}}