| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2008 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Force at pivot/axis |
| Difficulty | Challenging +1.3 This is a structured compound pendulum problem requiring moment of inertia calculation, application of energy conservation and rotational dynamics, and force analysis. While it involves multiple steps and M4-level content (small oscillations of rigid bodies), the question provides clear 'show that' targets and guides students through each stage systematically. The techniques are standard for Further Maths M4, though more demanding than typical A-level mechanics due to the rotational dynamics and simultaneous equations involved. |
| Spec | 3.04a Calculate moments: about a point6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I = \frac{1}{3}m\{a^2 + (\frac{3}{2}a)^2\} + m(\frac{1}{2}a)^2\) | M1, M1 | MI about perp axis through centre; using parallel axes rule |
| \(= \frac{13}{12}ma^2 + \frac{1}{4}ma^2 = \frac{4}{3}ma^2\) | A1 (ag) | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}(\frac{4}{3}ma^2)\omega^2 - \frac{1}{2}(\frac{4}{3}ma^2)\frac{9g}{10a} = mg(\frac{1}{2}a - \frac{1}{2}a\times\frac{3}{5})\) | M1, A1 | Equation involving KE and PE |
| \(\frac{2}{3}ma^2\omega^2 - \frac{2}{5}mga = \frac{1}{5}mga\) | ||
| \(\omega^2 = \frac{6g}{5a}\) | A1 (ag) | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mg\cos\theta - R = m(\frac{1}{2}a)\omega^2\) | M1 | Acceleration \(r\omega^2\) and three terms (one term must be \(R\)) |
| \(mg \times \frac{3}{5} - R = \frac{3}{5}mg\) | A1 | *SR* \(mg\cos\theta + R = m(\frac{1}{2}a)\omega^2 \Rightarrow R=0\) earns M1A0A1 |
| \(R = 0\) | A1 (ag) | |
| \(mg(\frac{1}{2}a\sin\theta) = I\alpha\) | M1A1 | Applying \(L = I\alpha\) |
| \(\alpha = \frac{3g}{10a}\) | A1 | |
| \(mg\sin\theta - S = m(\frac{1}{2}a)\alpha\) | M1A1 | Acceleration \(r\alpha\) and three terms (one term must be \(S\)); or \(S(\frac{1}{2}a) = I_G\alpha = \frac{13}{12}ma^2\alpha\) |
| \(S = \frac{4}{5}mg - \frac{3}{20}mg = \frac{13}{20}mg\) | A1 | Total: 9 |
# Question 6:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = \frac{1}{3}m\{a^2 + (\frac{3}{2}a)^2\} + m(\frac{1}{2}a)^2$ | M1, M1 | MI about perp axis through centre; using parallel axes rule |
| $= \frac{13}{12}ma^2 + \frac{1}{4}ma^2 = \frac{4}{3}ma^2$ | A1 (ag) | **Total: 3** |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}(\frac{4}{3}ma^2)\omega^2 - \frac{1}{2}(\frac{4}{3}ma^2)\frac{9g}{10a} = mg(\frac{1}{2}a - \frac{1}{2}a\times\frac{3}{5})$ | M1, A1 | Equation involving KE and PE |
| $\frac{2}{3}ma^2\omega^2 - \frac{2}{5}mga = \frac{1}{5}mga$ | | |
| $\omega^2 = \frac{6g}{5a}$ | A1 (ag) | **Total: 3** |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mg\cos\theta - R = m(\frac{1}{2}a)\omega^2$ | M1 | Acceleration $r\omega^2$ and three terms (one term must be $R$) |
| $mg \times \frac{3}{5} - R = \frac{3}{5}mg$ | A1 | *SR* $mg\cos\theta + R = m(\frac{1}{2}a)\omega^2 \Rightarrow R=0$ earns M1A0A1 |
| $R = 0$ | A1 (ag) | |
| $mg(\frac{1}{2}a\sin\theta) = I\alpha$ | M1A1 | Applying $L = I\alpha$ |
| $\alpha = \frac{3g}{10a}$ | A1 | |
| $mg\sin\theta - S = m(\frac{1}{2}a)\alpha$ | M1A1 | Acceleration $r\alpha$ and three terms (one term must be $S$); or $S(\frac{1}{2}a) = I_G\alpha = \frac{13}{12}ma^2\alpha$ |
| $S = \frac{4}{5}mg - \frac{3}{20}mg = \frac{13}{20}mg$ | A1 | **Total: 9** |
---(i) Show that the moment of inertia of the lamina about the axis through $X$ is $\frac { 4 } { 3 } m a ^ { 2 }$.\\
(ii) At an instant when $\cos \theta = \frac { 3 } { 5 }$, show that $\omega ^ { 2 } = \frac { 6 g } { 5 a }$.\\
(iii) At an instant when $\cos \theta = \frac { 3 } { 5 }$, show that $R = 0$, and given also that $\sin \theta = \frac { 4 } { 5 }$ find $S$ in terms of $m$ and $g$.
\hfill \mbox{\textit{OCR M4 2008 Q6 [15]}}