| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Challenging +1.2 This is a standard M4 compound pendulum question with routine calculations: part (i) is a straightforward parallel axis theorem application or direct integration (bookwork level), and part (ii) applies energy conservation with a constant couple—both are textbook exercises requiring methodical application of learned techniques rather than novel insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04a Centre of mass: gravitational effect |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I = \int_{-a}^{5a} \frac{m}{6a}x^2\,dx\) or \(\int_{-a}^{5a}\rho x^2\,dx\) | M1 | \((\delta m)x^2\) or \((\rho\,\delta x)x^2\) or integrating \(x^2\) |
| M1 | Using \(\delta m = \frac{m\,\delta x}{6a}\) or \(\rho = \frac{m}{6a}\) | |
| A1 | Correct integral expression for \(I\) | |
| \(= \left[\frac{m}{18a}x^3\right]_{-a}^{5a} = \frac{m}{18a}(125a^3 + a^3)\) or \(42\rho a^3\) | M1 | Evaluating definite integral (dep on integrating \(x^2\)) |
| \(= \frac{126ma^3}{18a} = 7ma^2\) | A1 ag [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| WD by couple is \(\frac{6mga}{\pi} \times 3\pi\) \((= 18mga)\) | M1, A1 | Using \(C\theta\) |
| Gain of PE is \(mg(4a)\) | B1 | |
| \(18mga = 4mga + \frac{1}{2}(7ma^2)\omega^2\) | M1, A1 ft | Equation involving WD, PE and \(\frac{1}{2}I\omega^2\) |
| Angular speed is \(\sqrt{\dfrac{4g}{a}}\) | A1 [6] |
# Question 5:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = \int_{-a}^{5a} \frac{m}{6a}x^2\,dx$ or $\int_{-a}^{5a}\rho x^2\,dx$ | M1 | $(\delta m)x^2$ or $(\rho\,\delta x)x^2$ or integrating $x^2$ |
| | M1 | Using $\delta m = \frac{m\,\delta x}{6a}$ or $\rho = \frac{m}{6a}$ |
| | A1 | Correct integral expression for $I$ |
| $= \left[\frac{m}{18a}x^3\right]_{-a}^{5a} = \frac{m}{18a}(125a^3 + a^3)$ or $42\rho a^3$ | M1 | Evaluating definite integral (dep on integrating $x^2$) |
| $= \frac{126ma^3}{18a} = 7ma^2$ | A1 ag [5] | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| WD by couple is $\frac{6mga}{\pi} \times 3\pi$ $(= 18mga)$ | M1, A1 | Using $C\theta$ |
| Gain of PE is $mg(4a)$ | B1 | |
| $18mga = 4mga + \frac{1}{2}(7ma^2)\omega^2$ | M1, A1 ft | Equation involving WD, PE and $\frac{1}{2}I\omega^2$ |
| Angular speed is $\sqrt{\dfrac{4g}{a}}$ | A1 [6] | |
---5 A uniform $\operatorname { rod } A B$ has mass $m$ and length $6 a$. The point $C$ on the rod is such that $A C = a$. The rod can rotate freely in a vertical plane about a fixed horizontal axis passing through $C$ and perpendicular to the rod.\\
(i) Show by integration that the moment of inertia of the rod about this axis is $7 m a ^ { 2 }$.
The rod starts at rest with $B$ vertically below $C$. A couple of constant moment $\frac { 6 m g a } { \pi }$ is then applied to the rod.\\
(ii) Find, in terms of $a$ and $g$, the angular speed of the rod when it has turned through one and a half revolutions.\\
\includegraphics[max width=\textwidth, alt={}, center]{ea62d6d9-ac2f-44e7-8bfb-ae9aeea7109b-3_721_621_872_762}
A light pulley of radius $a$ is free to rotate in a vertical plane about a fixed horizontal axis passing through its centre $O$. Two particles, $P$ of mass $5 m$ and $Q$ of mass $3 m$, are connected by a light inextensible string. The particle $P$ is attached to the circumference of the pulley, the string passes over the top of the pulley, and $Q$ hangs below the pulley on the opposite side to $P$. The section of string not in contact with the pulley is vertical. The fixed line $O X$ makes an angle $\alpha$ with the downward vertical, where $\cos \alpha = \frac { 4 } { 5 }$, and $O P$ makes an angle $\theta$ with $O X$ (see diagram).
You are given that the total potential energy of the system (using a suitable reference level) is $V$, where
$$V = m g a ( 3 \sin \theta - 4 \cos \theta - 3 \theta ) .$$
\hfill \mbox{\textit{OCR M4 2010 Q5 [11]}}