| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Resultant force on lamina |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring calculation of centre of mass position (part i, which is a 'show that' with the answer given) and then applying the equilibrium condition by comparing the position of the centre of mass relative to the pivot point. It involves standard M2 techniques with no novel insight required, but is slightly above average difficulty due to the two-part structure and need to interpret the physical situation correctly. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
(i) Show that the distance from the ball to the centre of mass of the toy is 10.7 cm , correct to 1 decimal place.\\
(ii)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6ae57fe9-3b6f-46c2-95b8-d48903ed796b-3_312_1051_1509_587}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The toy lies on horizontal ground in a position such that the ball is touching the ground (see Fig. 2). Determine whether the toy is lying in equilibrium or whether it will move to a position where the rod is vertical.
\hfill \mbox{\textit{OCR M2 2008 Q5 [8]}}