| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Solid of revolution MI |
| Difficulty | Challenging +1.8 This M4 question requires multiple advanced techniques: deriving moment of inertia via integration for a solid of revolution, applying parallel axis theorem, using compound pendulum period formula, then analyzing a rotating lamina with energy conservation and resolving forces. The multi-part structure, integration setup, and application of rotational dynamics principles place this well above average difficulty, though the individual steps follow standard M4 methods. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V = \int_a^{4a} \pi(ax)\,\mathrm{d}x\) | M1 | Omission of \(\pi\) is an accuracy error |
| \(= \left[\frac{1}{2}\pi a x^2\right]_a^{4a} = \frac{15}{2}\pi a^3\) | M1 | |
| Hence \(m = \frac{15}{2}\pi a^3\rho\) | M1 | |
| \(I = \sum \frac{1}{2}(\rho\pi y^2\delta x)y^2 = \int \frac{1}{2}\rho\pi y^4\,\mathrm{d}x\) | M1, A1 | For \(\int y^4\,\mathrm{d}x\) |
| \(= \int_a^{4a} \frac{1}{2}\rho\pi a^2 x^2\,\mathrm{d}x\) | A1 ft | Substitute for \(y^4\) and correct limits |
| \(= \left[\frac{1}{6}\rho\pi a^2 x^3\right]_a^{4a} = \frac{21}{2}\rho\pi a^5\) | A1 | |
| \(= \frac{7}{5}\left(\frac{15}{2}\pi a^3\rho\right)a^2 = \frac{7}{5}ma^2\) | A1 (ag) | Total: 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| MI about axis: \(I_A = \frac{7}{5}ma^2 + ma^2\) | M1 | Using parallel axes rule |
| \(= \frac{12}{5}ma^2\) | A1 | |
| Period is \(2\pi\sqrt{\frac{I}{mgh}}\) | M1 | |
| \(= 2\pi\sqrt{\frac{\frac{12}{5}ma^2}{mga}} = 2\pi\sqrt{\frac{12a}{5g}}\) | A1 ft | ft from any \(I\) with \(h = a\). Total: 4 |
# Question 5:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = \int_a^{4a} \pi(ax)\,\mathrm{d}x$ | M1 | Omission of $\pi$ is an accuracy error |
| $= \left[\frac{1}{2}\pi a x^2\right]_a^{4a} = \frac{15}{2}\pi a^3$ | M1 | |
| Hence $m = \frac{15}{2}\pi a^3\rho$ | M1 | |
| $I = \sum \frac{1}{2}(\rho\pi y^2\delta x)y^2 = \int \frac{1}{2}\rho\pi y^4\,\mathrm{d}x$ | M1, A1 | For $\int y^4\,\mathrm{d}x$ |
| $= \int_a^{4a} \frac{1}{2}\rho\pi a^2 x^2\,\mathrm{d}x$ | A1 ft | Substitute for $y^4$ and correct limits |
| $= \left[\frac{1}{6}\rho\pi a^2 x^3\right]_a^{4a} = \frac{21}{2}\rho\pi a^5$ | A1 | |
| $= \frac{7}{5}\left(\frac{15}{2}\pi a^3\rho\right)a^2 = \frac{7}{5}ma^2$ | A1 (ag) | **Total: 8** |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| MI about axis: $I_A = \frac{7}{5}ma^2 + ma^2$ | M1 | Using parallel axes rule |
| $= \frac{12}{5}ma^2$ | A1 | |
| Period is $2\pi\sqrt{\frac{I}{mgh}}$ | M1 | |
| $= 2\pi\sqrt{\frac{\frac{12}{5}ma^2}{mga}} = 2\pi\sqrt{\frac{12a}{5g}}$ | A1 ft | ft from any $I$ with $h = a$. **Total: 4** |
---5 The region bounded by the curve $y = \sqrt { a x }$ for $a \leqslant x \leqslant 4 a$ (where $a$ is a positive constant), the $x$-axis, and the lines $x = a$ and $x = 4 a$, is rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid of revolution of mass $m$.\\
(i) Show that the moment of inertia of this solid about the $x$-axis is $\frac { 7 } { 5 } m a ^ { 2 }$.
The solid is free to rotate about a fixed horizontal axis along the line $y = a$, and makes small oscillations as a compound pendulum.\\
(ii) Find, in terms of $a$ and $g$, the approximate period of these small oscillations.\\
\includegraphics[max width=\textwidth, alt={}, center]{a9e010ce-c3a8-4f95-a154-fd16ef3e5e5b-3_734_862_813_644}
A uniform rectangular lamina $A B C D$ has mass $m$ and sides $A B = 2 a$ and $B C = 3 a$. The mid-point of $A B$ is $P$ and the mid-point of $C D$ is $Q$. The lamina is rotating freely in a vertical plane about a fixed horizontal axis which is perpendicular to the lamina and passes through the point $X$ on $P Q$ where $P X = a$. Air resistance may be neglected. When $Q$ is vertically above $X$, the angular speed is $\sqrt { \frac { 9 g } { 10 a } }$. When $X Q$ makes an angle $\theta$ with the upward vertical, the angular speed is $\omega$, and the force acting on the lamina at $X$ has components $R$ parallel to $P Q$ and $S$ parallel to $B A$ (see diagram).\\
\hfill \mbox{\textit{OCR M4 2008 Q5 [12]}}