| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Small oscillations period |
| Difficulty | Challenging +1.2 This is a standard compound pendulum problem requiring calculation of moment of inertia using parallel axis theorem and application of the small oscillations formula. While it involves multiple steps (finding I about P, finding center of mass, applying T = 2π√(I/mgh)), these are routine techniques for M4 students with no novel insight required. The setup is slightly more complex than basic examples due to the added particle, placing it moderately above average difficulty. |
| Spec | 6.04a Centre of mass: gravitational effect6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(I = \frac{1}{2}(4m)(2a)^2 + (4m)a^2\) | M1 | Applying parallel axes rule |
| A1 | ||
| \(+ m(3a)^2\) | B1 | |
| \(= 21ma^2\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| From P, \(\bar{x} = \frac{(4m)a + m(3a)}{5m} = \frac{7a}{5}\) | M1 | |
| M1 | Correct formula \(2\pi\sqrt{\frac{I}{mgh}}\) seen | |
| Period is \(2\pi\sqrt{\frac{21ma^2}{5mg(\frac{7}{5}a)}}\) | A1 ft | *or* using \(L = I\ddot{\theta}\) and period \(2\pi/\omega\) |
| \(= 2\pi\sqrt{\frac{3a}{g}}\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-4mga\sin\theta - mg(3a)\sin\theta = (21ma^2)\ddot{\theta}\) | M1 | Using \(L = I\ddot{\theta}\) with three terms |
| M1 | Using period \(2\pi/\omega\) | |
| Period is \(2\pi\sqrt{\frac{21ma^2}{7mga}} = 2\pi\sqrt{\frac{3a}{g}}\) | A1 ft A1 |
## Question 3:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $I = \frac{1}{2}(4m)(2a)^2 + (4m)a^2$ | M1 | Applying parallel axes rule |
| | A1 | |
| $+ m(3a)^2$ | B1 | |
| $= 21ma^2$ | A1 | |
| | **[4]** | |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| From P, $\bar{x} = \frac{(4m)a + m(3a)}{5m} = \frac{7a}{5}$ | M1 | |
| | M1 | Correct formula $2\pi\sqrt{\frac{I}{mgh}}$ seen |
| Period is $2\pi\sqrt{\frac{21ma^2}{5mg(\frac{7}{5}a)}}$ | A1 ft | *or* using $L = I\ddot{\theta}$ and period $2\pi/\omega$ |
| $= 2\pi\sqrt{\frac{3a}{g}}$ | A1 | |
| | **[4]** | |
**Alternative for (ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $-4mga\sin\theta - mg(3a)\sin\theta = (21ma^2)\ddot{\theta}$ | M1 | Using $L = I\ddot{\theta}$ with three terms |
| | M1 | Using period $2\pi/\omega$ |
| Period is $2\pi\sqrt{\frac{21ma^2}{7mga}} = 2\pi\sqrt{\frac{3a}{g}}$ | A1 ft A1 | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{afecdb38-c372-480a-9d6d-fafe6a371dc2-2_664_623_904_760}
A uniform circular disc has mass $4 m$, radius $2 a$ and centre $O$. The points $A$ and $B$ are at opposite ends of a diameter of the disc, and the mid-point of $O A$ is $P$. A particle of mass $m$ is attached to the disc at $B$. The resulting compound pendulum is in a vertical plane and is free to rotate about a fixed horizontal axis passing through $P$ and perpendicular to the disc (see diagram). The pendulum makes small oscillations.\\
(i) Find the moment of inertia of the pendulum about the axis.\\
(ii) Find the approximate period of the small oscillations.
\hfill \mbox{\textit{OCR M4 2009 Q3 [8]}}