| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with hemisphere and cylinder/cone |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question requiring formula recall (hemisphere COM at 3r/8 from base) and straightforward moment calculations. Part (i) is a 'show that' using moments about O with given COM position. Parts (ii)-(iii) involve taking moments and resolving forces in equilibrium—routine techniques for M2. The multi-part structure and need for careful bookkeeping elevate it slightly above average, but no novel insight is required. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| com of hemisphere \(0.3\) from \(O\) | B1 | or \(0.5\) from base |
| com of cylinder \(h/2\) from \(O\) | B1 | |
| \(0.6\times45 = 40\times0.5 + (0.8+h/2)\times5\) or | M1 | or \(40\times0.3 - 5\times h/2 = 45\times0.2\) |
| \(45(h+0.2) = 5h/2 + 40(h+0.3)\) | A1 | or \(5(0.2+h/2) = 40\times0.1\) |
| \(27 = 20 + (0.8+h/2)\times5\) | M1 | solving |
| \(h = 1.2\) | A1 6 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1.2\ T\) | B1 | |
| \(0.8\ F\) | B1 | |
| \(0.8F = 1.2T\) | M1 | |
| \(F = 3T/2\) | A1 4 | aef |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F + T\cos30°\) | B1 | or \(45 \times 0.8\sin30°\) |
| \(45\sin30°\) must be involved in res. | B1 | \(T \times (1.2 + 0.8\cos30°)\) |
| resolving parallel to the slope | M1 | mom. about point of contact |
| \(F + T\cos30° = 45\sin30°\) aef | A1 | \(45\cdot0.8\sin30° = T(1.2+0.8\cos30°)\) |
| \(T = 9.51\) | A1 | |
| \(F = 14.3\) | A1 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T + F\cos30° = R\sin30°\) | B1 | res. horizontally |
| \(R\cos30° + F\sin30° = 45\) | B1 | res. vertically |
| \(\tan30° = \frac{T + F\cos30°}{45 - F\sin30°}\) | M1 | eliminating \(R\) |
## Question 8(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| com of hemisphere $0.3$ from $O$ | B1 | **or** $0.5$ from base |
| com of cylinder $h/2$ from $O$ | B1 | |
| $0.6\times45 = 40\times0.5 + (0.8+h/2)\times5$ **or** | M1 | **or** $40\times0.3 - 5\times h/2 = 45\times0.2$ |
| $45(h+0.2) = 5h/2 + 40(h+0.3)$ | A1 | **or** $5(0.2+h/2) = 40\times0.1$ |
| $27 = 20 + (0.8+h/2)\times5$ | M1 | solving |
| $h = 1.2$ | A1 **6** | **AG** |
## Question 8(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1.2\ T$ | B1 | |
| $0.8\ F$ | B1 | |
| $0.8F = 1.2T$ | M1 | |
| $F = 3T/2$ | A1 **4** | aef |
## Question 8(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F + T\cos30°$ | B1 | **or** $45 \times 0.8\sin30°$ |
| $45\sin30°$ must be involved in res. | B1 | $T \times (1.2 + 0.8\cos30°)$ |
| resolving parallel to the slope | M1 | mom. about point of contact |
| $F + T\cos30° = 45\sin30°$ aef | A1 | $45\cdot0.8\sin30° = T(1.2+0.8\cos30°)$ |
| $T = 9.51$ | A1 | |
| $F = 14.3$ | A1 **6** | |
## Question 8(iii) alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T + F\cos30° = R\sin30°$ | B1 | res. horizontally |
| $R\cos30° + F\sin30° = 45$ | B1 | res. vertically |
| $\tan30° = \frac{T + F\cos30°}{45 - F\sin30°}$ | M1 | eliminating $R$ |8
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9951c978-37e6-4d89-9fe3-c1e5e28b221e-4_451_481_274_833}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
An object consists of a uniform solid hemisphere of weight 40 N and a uniform solid cylinder of weight 5 N . The cylinder has height $h \mathrm {~m}$. The solids have the same base radius 0.8 m and are joined so that the hemisphere's plane face coincides with one of the cylinder's faces. The centre of the common face is the point $O$ (see Fig. 1). The centre of mass of the object lies inside the hemisphere and is at a distance of 0.2 m from $O$.\\
(i) Show that $h = 1.2$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9951c978-37e6-4d89-9fe3-c1e5e28b221e-4_620_1065_1297_541}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
One end of a light inextensible string is attached to a point on the circumference of the upper face of the cylinder. The string is horizontal and its other end is tied to a fixed point on a rough plane. The object rests in equilibrium on the plane with its axis of symmetry vertical. The plane makes an angle of $30 ^ { \circ }$ with the horizontal (see Fig. 2). The tension in the string is $T \mathrm {~N}$ and the frictional force acting on the object is $F \mathrm {~N}$.\\
(ii) By taking moments about $O$, express $F$ in terms of $T$.\\
(iii) Find another equation connecting $T$ and $F$. Hence calculate the tension and the frictional force.
\hfill \mbox{\textit{OCR M2 2007 Q8 [16]}}