| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with elastic strings/springs |
| Difficulty | Challenging +1.8 This is a challenging M4 question requiring energy methods for a compound rigid body system with elastic strings. Part (i) demands careful geometric analysis to find positions and elastic extension, then combining gravitational PE (for two rods) with elastic PE. Part (ii) uses equilibrium condition dV/dθ=0 requiring differentiation of a complex expression. Part (iii) requires second derivative and stability analysis. The multi-component system, geometric complexity, and extended algebraic manipulation place this well above average difficulty, though it follows standard M4 energy-equilibrium methodology. |
| Spec | 1.07p Points of inflection: using second derivative6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(BD = 2a\cos\theta\) | B1 | Award if seen in EPE term |
| GPE for rod \(AB = (-)mg\left(a + \frac{1}{2}a\cos 2\theta\right)\) | B1 | \((-)mg\left(a + \frac{1}{2}a\sin(90 - 2\theta)\right)\) |
| GPE for rod \(BC = (-)mg\left(a + a\cos\theta - \frac{1}{2}a\sin 2\theta\right)\) | B1 | \((-)mg\left(a + a\sin(90 - \theta) - \frac{1}{2}a\cos(90 - 2\theta)\right)\) |
| \(EPE = \frac{\lambda mg(2a\cos\theta - a)^2}{2a}\) | M1 | Using \(\frac{\lambda x^2}{2a}\) with their BD |
| \(V = \frac{1}{2}mga(\sin 2\theta - 3\cos 2\theta) + \frac{1}{2}\lambda mga(2\cos\theta - 1)^2 - 2mga\) | A1 | AG Correctly shown |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{d\theta} = \frac{1}{2}mga(2\cos 2\theta + 6\sin 2\theta) +\) | M1 | Attempt at differentiation |
| \(\lambda mga(2\cos\theta - 1)(-2\sin\theta)\) | A1 | Correct derivative; \(... + \lambda mga(2\sin\theta - 2\sin 2\theta)\) |
| \(\frac{1}{2}(2(0) + 6) + \lambda\left(\frac{2\sqrt{2}}{2} - 1\right)\left(\frac{-2\sqrt{2}}{2}\right) = 0\) | M1 | Set \(\frac{dV}{d\theta} = 0\) and \(\theta = \frac{1}{4}\pi\) |
| \(\lambda = \frac{3}{2 - \sqrt{2}}\) | A1 | oe eg \(\frac{6 + 3\sqrt{2}}{2}\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2V}{d\theta^2} = \frac{1}{2}mga(-4\sin 2\theta + 12\cos 2\theta)\) | M1 | Attempt at second derivative |
| \(-2\lambda mga(2\cos 2\theta - \cos\theta)\) | A1 | oe; \(... - 2\lambda mga[(2\cos\theta - 1)\cos\theta - \sin^2\theta]\) |
| \(\frac{d^2V}{d\theta^2} = mga(1 + 3\sqrt{2}) > 0\), so equilibrium is stable | M1 | Set \(\theta = \frac{1}{4}\pi\) and using their \(\lambda\) (must be evidence of substitution) |
| A1 | Correct value of \(V''\) and \(> 0\); \(V'' = (5.24264...)mga\) | |
| [4] |
**(i)**
$BD = 2a\cos\theta$ | B1 | Award if seen in EPE term
GPE for rod $AB = (-)mg\left(a + \frac{1}{2}a\cos 2\theta\right)$ | B1 | $(-)mg\left(a + \frac{1}{2}a\sin(90 - 2\theta)\right)$
GPE for rod $BC = (-)mg\left(a + a\cos\theta - \frac{1}{2}a\sin 2\theta\right)$ | B1 | $(-)mg\left(a + a\sin(90 - \theta) - \frac{1}{2}a\cos(90 - 2\theta)\right)$
$EPE = \frac{\lambda mg(2a\cos\theta - a)^2}{2a}$ | M1 | Using $\frac{\lambda x^2}{2a}$ with their BD
$V = \frac{1}{2}mga(\sin 2\theta - 3\cos 2\theta) + \frac{1}{2}\lambda mga(2\cos\theta - 1)^2 - 2mga$ | A1 | AG Correctly shown
| [5] |
**(ii)**
$\frac{dV}{d\theta} = \frac{1}{2}mga(2\cos 2\theta + 6\sin 2\theta) +$ | M1 | Attempt at differentiation
$\lambda mga(2\cos\theta - 1)(-2\sin\theta)$ | A1 | Correct derivative; $... + \lambda mga(2\sin\theta - 2\sin 2\theta)$
$\frac{1}{2}(2(0) + 6) + \lambda\left(\frac{2\sqrt{2}}{2} - 1\right)\left(\frac{-2\sqrt{2}}{2}\right) = 0$ | M1 | Set $\frac{dV}{d\theta} = 0$ and $\theta = \frac{1}{4}\pi$
$\lambda = \frac{3}{2 - \sqrt{2}}$ | A1 | oe eg $\frac{6 + 3\sqrt{2}}{2}$
| [4] |
**(iii)**
$\frac{d^2V}{d\theta^2} = \frac{1}{2}mga(-4\sin 2\theta + 12\cos 2\theta)$ | M1 | Attempt at second derivative
$-2\lambda mga(2\cos 2\theta - \cos\theta)$ | A1 | oe; $... - 2\lambda mga[(2\cos\theta - 1)\cos\theta - \sin^2\theta]$
$\frac{d^2V}{d\theta^2} = mga(1 + 3\sqrt{2}) > 0$, so equilibrium is stable | M1 | Set $\theta = \frac{1}{4}\pi$ and using their $\lambda$ (must be evidence of substitution)
| A1 | Correct value of $V''$ and $> 0$; $V'' = (5.24264...)mga$
| [4] |\includegraphics{figure_3}
Two uniform rods $AB$ and $BC$, each of length $a$ and mass $m$, are rigidly joined together so that $AB$ is perpendicular to $BC$. The rod $AB$ is freely hinged to a fixed point at $A$. The rods can rotate in a vertical plane about a smooth fixed horizontal axis through $A$. One end of a light elastic string of natural length $a$ and modulus of elasticity $\lambda mg$ is attached to $B$. The other end of the string is attached to a fixed point $D$ vertically above $A$, where $AD = a$. The string $BD$ makes an angle $\theta$ radians with the downward vertical (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Taking $D$ as the reference level for gravitational potential energy, show that the total potential energy $V$ of the system is given by
$$V = \frac{1}{2}mga(\sin 2\theta - 3\cos 2\theta) + \frac{1}{2}\lambda mga(2\cos \theta - 1)^2 - 2mga.$$ [5]
\item Given that $\theta = \frac{1}{3}\pi$ is a position of equilibrium, find the exact value of $\lambda$. [4]
\item Find $\frac{d^2V}{d\theta^2}$ and hence determine whether the position of equilibrium at $\theta = \frac{1}{3}\pi$ is stable or unstable. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M4 2016 Q3 [13]}}