| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Conical or hemispherical shell composite |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question using given formulae for composite bodies. Part (i) requires straightforward application of the centre of mass formula with two components (hemisphere and cylinder), involving basic algebraic manipulation. Part (ii) is simple substitution and arithmetic. The question provides all necessary formulae and follows a typical textbook pattern, making it slightly easier than average for A-level. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Taking \(y\)-axis vert downwards from O | Allow areas used as masses | |
| \(2\pi\sigma \times 8^2 \times 4 + 2\pi\sigma \times 8 \times k \times \frac{k}{2}\) | M1 | Method for c.m. |
| B1 | '4' used | |
| B1 | \(16\pi k\) | |
| B1 | \(k/2\) used | |
| \(= (2\pi\sigma \times 8^2 + 2\pi\sigma \times 8k)\,\bar{y}\) | B1 | Masses correct |
| so \(\bar{y} = \dfrac{64 + k^2}{16 + 2k}\) | E1 | Must see some evidence of simplification. Need no reference to axis of symmetry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k = 12\) gives OG as \(5.2\) and mass as \(320\pi\sigma\) | B1 | Allow for either. Allow \(\sigma = 1\) |
| \(320\pi\sigma \times 5.2 + \pi\sigma \times 8^2 \times 12\) | M1 | Method for c.m. combining with (i) or starting again |
| B1 | One term correct | |
| B1 | Second term correct | |
| \(= (320\pi\sigma + 64\pi\sigma)\,\bar{y}\) | ||
| \(\bar{y} = 6\dfrac{1}{3}\) | E1 | Some simplification shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| G above edge of base | B1 | |
| \(12 - 6\frac{1}{3} = 5\frac{2}{3}\) seen | B1 | |
| \(8\) seen | B1 | |
| \(\tan\theta = \dfrac{8}{5\frac{2}{3}}\) | M1 | Accept \(\frac{5\frac{2}{3}}{8}\) or attempts based on \(6\frac{1}{3}\) and \(8\) |
| \(\theta = 54.6887\ldots\) so \(54.7°\) (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Slips when \(\mu = \tan\theta\) | M1 | Or …. |
| \(\dfrac{8}{5\frac{2}{3}} = 1.4117\ldots\) | B1 | |
| \(< 1.5\) so does not slip | A1 | There must be a reason |
# Question 4:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Taking $y$-axis vert downwards from O | | Allow areas used as masses |
| $2\pi\sigma \times 8^2 \times 4 + 2\pi\sigma \times 8 \times k \times \frac{k}{2}$ | M1 | Method for c.m. |
| | B1 | '4' used |
| | B1 | $16\pi k$ |
| | B1 | $k/2$ used |
| $= (2\pi\sigma \times 8^2 + 2\pi\sigma \times 8k)\,\bar{y}$ | B1 | Masses correct |
| so $\bar{y} = \dfrac{64 + k^2}{16 + 2k}$ | E1 | Must see some evidence of simplification. Need no reference to axis of symmetry |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $k = 12$ gives OG as $5.2$ and mass as $320\pi\sigma$ | B1 | Allow for either. Allow $\sigma = 1$ |
| $320\pi\sigma \times 5.2 + \pi\sigma \times 8^2 \times 12$ | M1 | Method for c.m. combining with (i) or starting again |
| | B1 | One term correct |
| | B1 | Second term correct |
| $= (320\pi\sigma + 64\pi\sigma)\,\bar{y}$ | | |
| $\bar{y} = 6\dfrac{1}{3}$ | E1 | Some simplification shown |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| G above edge of base | B1 | |
| $12 - 6\frac{1}{3} = 5\frac{2}{3}$ seen | B1 | |
| $8$ seen | B1 | |
| $\tan\theta = \dfrac{8}{5\frac{2}{3}}$ | M1 | Accept $\frac{5\frac{2}{3}}{8}$ or attempts based on $6\frac{1}{3}$ and $8$ |
| $\theta = 54.6887\ldots$ so $54.7°$ (3 s.f.) | A1 | cao |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Slips when $\mu = \tan\theta$ | M1 | Or …. |
| $\dfrac{8}{5\frac{2}{3}} = 1.4117\ldots$ | B1 | |
| $< 1.5$ so does not slip | A1 | There must be a reason |4 In this question you may use the following facts: as illustrated in Fig. 4.1, the centre of mass, G, of a uniform thin open hemispherical shell is at the mid-point of OA on its axis of symmetry; the surface area of this shell is $2 \pi r ^ { 2 }$, where $r$ is the distance OA.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{81efb50d-c89d-4ce1-94d7-592c946f6176-5_344_542_445_804}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
A perspective view and a cross-section of a dog bowl are shown in Fig. 4.2. The bowl is made throughout from thin uniform material. An open hemispherical shell of radius 8 cm is fitted inside an open circular cylinder of radius 8 cm so that they have a common axis of symmetry and the rim of the hemisphere is at one end of the cylinder. The height of the cylinder is $k \mathrm {~cm}$. The point O is on the axis of symmetry and at the end of the cylinder.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{81efb50d-c89d-4ce1-94d7-592c946f6176-5_494_947_1238_267}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{81efb50d-c89d-4ce1-94d7-592c946f6176-5_478_492_1238_1370}
\captionsetup{labelformat=empty}
\caption{Fig. 4.3}
\end{center}
\end{figure}
(i) Show that the centre of mass of the bowl is a distance $\frac { 64 + k ^ { 2 } } { 16 + 2 k } \mathrm {~cm}$ from O .
A version of the bowl for the 'senior dog' has $k = 12$ and an end to the cylinder, as shown in Fig. 4.3. The end is made from the same material as the original bowl.\\
(ii) Show that the centre of mass of this bowl is a distance $6 \frac { 1 } { 3 } \mathrm {~cm}$ from O .
This bowl is placed on a rough slope inclined at $\theta$ to the horizontal.\\
(iii) Assume that the bowl is prevented from sliding and is on the point of toppling.
Draw a diagram indicating the position of the centre of mass of the bowl with relevant lengths marked.
Calculate the value of $\theta$.\\
(iv) If the bowl is not prevented from sliding, determine whether it will slide when placed on the slope when there is a coefficient of friction between the bowl and the slope of 1.5.
\hfill \mbox{\textit{OCR MEI M2 2009 Q4 [19]}}