| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with circular arc or semicircular arc components |
| Difficulty | Standard +0.3 Part (i) is a standard centre of mass formula application for a semicircular arc (likely given in formula booklet), requiring only substitution and verification. Part (ii) involves setting up equilibrium conditions with two different masses and using moments, which is routine M2 content but requires careful geometry and trigonometry across multiple steps. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
\begin{enumerate}[label=(\roman*)]
\item A uniform piece of wire, $ABC$, forms a semicircular arc of radius 6 cm. $O$ is the mid-point of $AC$ (see Fig. 1). Show that the distance from $O$ to the centre of mass of the wire is 3.82 cm, correct to 3 significant figures. [2]
\item Two semicircular pieces of wire, $ABC$ and $ADC$, are joined together at their ends to form a circular hoop of radius 6 cm. The mass of $ABC$ is 3 grams and the mass of $ADC$ is 5 grams. The hoop is freely suspended from $A$ (see Fig. 2). Calculate the angle which the diameter $AC$ makes with the vertical, giving your answer correct to the nearest degree. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2010 Q2 [7]}}