| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic potential energy calculations |
| Difficulty | Challenging +1.2 This is a multi-part mechanics question requiring energy methods and equilibrium analysis. While it involves elastic strings and potential energy (M4 topics), the mathematical steps are systematic: calculating gravitational PE of the rod's center of mass, elastic PE from extension, differentiating for equilibrium, and using second derivative test for stability. The algebra is moderate and the approach is standard for M4, making it somewhat above average difficulty but not requiring exceptional insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Potential energy (relative to \(\theta = 0\)) | M1, A1 | \(V = mga(\sin\theta) + \frac{1}{2} \times \frac{1}{3}mg(2a\sin\theta)^2 = mga[\sin^2\theta - \sin\theta]\) (shown) |
| Derivative for equilibrium | M1 | \(\frac{dV}{d\theta} = mga(2\sin\theta\cos\theta - \cos\theta) = mga\cos\theta(2\sin\theta - 1)\) |
| Equilibrium positions | A1 | For \(0 < \theta < \pi\): \(\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\) |
| Second derivative | M1 | \(\frac{d^2V}{d\theta^2} = mga(\sin\theta - 2\sin^2\theta + 2\cos^2\theta)\) |
| Stability at \(\theta = \frac{\pi}{6}\) | A1 | \(\frac{d^2V}{d\theta^2} = \frac{3}{2}mga > 0\) \(\therefore\) stable |
| Stability at \(\theta = \frac{\pi}{2}\) | A1 | \(\frac{d^2V}{d\theta^2} = -mga < 0\) \(\therefore\) unstable |
| Stability at \(\theta = \frac{5\pi}{6}\) | A1 | \(\frac{d^2V}{d\theta^2} = \frac{3}{2}mga > 0\) \(\therefore\) stable |
**Potential energy (relative to $\theta = 0$)** | M1, A1 | $V = mga(\sin\theta) + \frac{1}{2} \times \frac{1}{3}mg(2a\sin\theta)^2 = mga[\sin^2\theta - \sin\theta]$ (shown)
**Derivative for equilibrium** | M1 | $\frac{dV}{d\theta} = mga(2\sin\theta\cos\theta - \cos\theta) = mga\cos\theta(2\sin\theta - 1)$
**Equilibrium positions** | A1 | For $0 < \theta < \pi$: $\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$
**Second derivative** | M1 | $\frac{d^2V}{d\theta^2} = mga(\sin\theta - 2\sin^2\theta + 2\cos^2\theta)$
**Stability at $\theta = \frac{\pi}{6}$** | A1 | $\frac{d^2V}{d\theta^2} = \frac{3}{2}mga > 0$ $\therefore$ stable
**Stability at $\theta = \frac{\pi}{2}$** | A1 | $\frac{d^2V}{d\theta^2} = -mga < 0$ $\therefore$ unstable
**Stability at $\theta = \frac{5\pi}{6}$** | A1 | $\frac{d^2V}{d\theta^2} = \frac{3}{2}mga > 0$ $\therefore$ stable8\\
\includegraphics[max width=\textwidth, alt={}, center]{98647526-b52a-4316-9a09-48d756b8f599-3_493_748_1393_708}
The diagram shows a uniform rod $A B$, of mass $m$ and length $2 a$, free to rotate in a vertical plane about a fixed horizontal axis through $A$. A light elastic string has natural length $a$ and modulus of elasticity $\frac { 1 } { 2 } m g$. The string joins $B$ to a light ring $R$ which slides along a smooth horizontal wire fixed at a height $a$ above $A$ and in the same vertical plane as $A B$. The string $B R$ remains vertical. The angle between $A B$ and the horizontal is denoted by $\theta$, where $0 < \theta < \pi$.\\
(i) Taking the reference level for gravitational potential energy to be the horizontal through $A$, show that the total potential energy of the system is
$$m g a \left( \sin ^ { 2 } \theta - \sin \theta \right) .$$
(ii) Find the three values of $\theta$ for which the system is in equilibrium.\\
(iii) For each position of equilibrium, determine whether it is stable or unstable.
\hfill \mbox{\textit{OCR M4 2002 Q8 [12]}}