| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| 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 | Challenging +1.2 This is a standard M2 centre of mass question requiring calculation of composite solid COM (part i is a 'show that'), followed by equilibrium with friction. The calculations are methodical but involve multiple steps: volumes, moments, resolving forces, taking moments, and friction inequality. More demanding than basic COM questions due to the equilibrium analysis with tilted axis and friction bounds, but follows standard M2 templates without requiring novel insight. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x_H = 3x0.6/8\) | B1 | CoM hemisphere (\(x_H = 0.225\)), may be implied. Use of table of moments idea. SC Volume of sphere used, max B1M1A1, moment equation fully correct for A1 (3/5). Accept -0.09 |
| \(π(0.6^2x0.6)(0.6/2) - (0.6^2π/3)(0.225)\) | M1 | |
| \(= πx0.6^2(1+2/3)x_G\) | A1 | |
| \(x_G = 0.09\) m | A1 | |
| A1 | ||
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg(0.09\cos45°) = 2(0.6+0.6\cos45°+0.6\sin45°)\) | M1 | Attempt at moments (must resolve), allow without \(g\). \(2(0.6+\sqrt[]{(0.6^2+0.6^2)})\) \((4.6451...)\) |
| \(m = 4.65kg\) | A1 | |
| A1 | ||
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2/4.6451g\) | M1 | Ratio force/weight. \(cv(4.65)\). Correct inequality sign, accept 0.044 |
| \(μ ≥ 0.0439\) | A1 | |
| A1 | ||
| [3] |
**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_H = 3x0.6/8$ | B1 | CoM hemisphere ($x_H = 0.225$), may be implied. Use of table of moments idea. SC Volume of sphere used, max B1M1A1, moment equation fully correct for A1 (3/5). Accept -0.09 |
| $π(0.6^2x0.6)(0.6/2) - (0.6^2π/3)(0.225)$ | M1 | |
| $= πx0.6^2(1+2/3)x_G$ | A1 | |
| $x_G = 0.09$ m | A1 | |
| | A1 | |
| | **[5]** | |
**Part (ii)(a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg(0.09\cos45°) = 2(0.6+0.6\cos45°+0.6\sin45°)$ | M1 | Attempt at moments (must resolve), allow without $g$. $2(0.6+\sqrt[]{(0.6^2+0.6^2)})$ $(4.6451...)$ |
| $m = 4.65kg$ | A1 | |
| | A1 | |
| | **[4]** | |
**Part (ii)(b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2/4.6451g$ | M1 | Ratio force/weight. $cv(4.65)$. Correct inequality sign, accept 0.044 |
| $μ ≥ 0.0439$ | A1 | |
| | A1 | |
| | **[3]** | |5 A uniform solid is made of a hemisphere with centre $O$ and radius 0.6 m , and a cylinder of radius 0.6 m and height 0.6 m . The plane face of the hemisphere and a plane face of the cylinder coincide. (The formula for the volume of a sphere is $\frac { 4 } { 3 } \pi r ^ { 3 }$.)\\
(i) Show that the distance of the centre of mass of the solid from $O$ is 0.09 m .\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{941c0c81-a74f-49c0-acb7-1c23266fc2c8-03_636_1036_982_593}
The solid is placed with the curved surface of the hemisphere on a rough horizontal surface and the axis inclined at $45 ^ { \circ }$ to the horizontal. The equilibrium of the solid is maintained by a horizontal force of 2 N applied to the highest point on the circumference of its plane face (see diagram). Calculate
\begin{enumerate}[label=(\alph*)]
\item the mass of the solid,
\item the set of possible values of the coefficient of friction between the surface and the solid.
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2011 Q5 [12]}}