| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with hemisphere and cylinder/cone |
| Difficulty | Standard +0.8 This is a multi-part M3 centre of mass question requiring: (1) calculation of composite centre of mass with different densities (standard but algebraically involved), (2) equilibrium analysis with friction at limiting equilibrium involving moments and force resolution with trigonometry. The algebra is substantial and the second part requires careful setup of multiple equations, making it moderately challenging but still within standard M3 scope. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Masses: \(3k\frac{2}{3}\pi r^3\), \(k\pi r^2 \cdot 3r\), \(3k\frac{2}{3}\pi r^3 + k\pi r^2 \cdot 3r\) i.e. \((2), (3), (5)\) | B1 | correct ratio of masses |
| Distances from \(O\): \(\left(\frac{3r}{8}+3r\right)\), \(\frac{3r}{2}\), \(\bar{x}\) | B1 | correct distances of centres of mass of two components from \(O\) |
| \(\left(\frac{3r}{8}+3r\right)\cdot 2 + \frac{3r}{2}\cdot 3 = 5\bar{x}\) | M1 A1ft | M1: moments equation about \(O\), must have three terms, dimensionally correct. A1ft: correct equation, follow through ratio of masses and distances but not 1:3:4 |
| \(\frac{9r}{4} = \bar{x}\) PRINTED ANSWER | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(R = W\); \(F = P\) | B1 | for the two shown equations |
| \(P\cdot 2r\sin\alpha = W\left(\frac{9r}{4}\sin\alpha - r\cos\alpha\right)\) | M1 A1 A1 | M1: moments equation about point of contact. A1A1: award A2 if fully correct; A1A0 if one error |
| \(P = W\left(\frac{9}{8} - \frac{1}{2}\cot\alpha\right)\) | A1 | re-arranging to obtain this form |
| \(F = \mu R\) | ||
| \(\frac{1}{8}(9 - 4\cot\alpha) = \mu\) PRINTED ANSWER | M1dep A1cso | M1dep: using \(F=\mu R\) with expression for \(P\) and first two equations. A1cso: must be in this form |
## Question 5(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Masses: $3k\frac{2}{3}\pi r^3$, $k\pi r^2 \cdot 3r$, $3k\frac{2}{3}\pi r^3 + k\pi r^2 \cdot 3r$ i.e. $(2), (3), (5)$ | B1 | correct ratio of masses |
| Distances from $O$: $\left(\frac{3r}{8}+3r\right)$, $\frac{3r}{2}$, $\bar{x}$ | B1 | correct distances of centres of mass of two components from $O$ |
| $\left(\frac{3r}{8}+3r\right)\cdot 2 + \frac{3r}{2}\cdot 3 = 5\bar{x}$ | M1 A1ft | M1: moments equation about $O$, must have three terms, dimensionally correct. A1ft: correct equation, follow through ratio of masses and distances but not 1:3:4 |
| $\frac{9r}{4} = \bar{x}$ **PRINTED ANSWER** | A1cso | |
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## Question 5(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $R = W$; $F = P$ | B1 | for the two shown equations |
| $P\cdot 2r\sin\alpha = W\left(\frac{9r}{4}\sin\alpha - r\cos\alpha\right)$ | M1 A1 A1 | M1: moments equation about point of contact. A1A1: award A2 if fully correct; A1A0 if one error |
| $P = W\left(\frac{9}{8} - \frac{1}{2}\cot\alpha\right)$ | A1 | re-arranging to obtain this form |
| $F = \mu R$ | | |
| $\frac{1}{8}(9 - 4\cot\alpha) = \mu$ **PRINTED ANSWER** | M1dep A1cso | M1dep: using $F=\mu R$ with expression for $P$ and first two equations. A1cso: must be in this form |5. A solid $S$ consists of a uniform solid hemisphere of radius $r$ and a uniform solid circular cylinder of radius $r$ and height $3 r$. The circular face of the hemisphere is joined to one of the circular faces of the cylinder, so that the centres of the two faces coincide. The other circular face of the cylinder has centre $O$. The mass per unit volume of the hemisphere is $3 k$ and the mass per unit volume of the cylinder is $k$.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of $S$ from $O$ is $\frac { 9 r } { 4 }$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2c0bb9ea-31a6-42f1-9e2e-d792eee8fd10-07_501_1082_653_422}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The solid $S$ is held in equilibrium by a horizontal force of magnitude $P$. The circular face of $S$ has one point in contact with a fixed rough horizontal plane and is inclined at an angle $\alpha$ to the horizontal. The force acts through the highest point of the circular face of $S$ and in the vertical plane through the axis of the cylinder, as shown in Figure 2. The coefficient of friction between $S$ and the plane is $\mu$. Given that $S$ is on the point of slipping along the plane in the same direction as $P$,
\item show that $\mu = \frac { 1 } { 8 } ( 9 - 4 \cot \alpha )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q5 [12]}}