| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Challenging +1.8 This M4 mechanics question requires integration to find moment of inertia using parallel axis theorem, then applying compound pendulum theory. While conceptually demanding (requiring knowledge of perpendicular axis theorem for discs, parallel axis theorem, and compound pendulum period formula), the execution is methodical with a given starting formula. The multi-step nature and need to coordinate several advanced mechanics concepts places it well above average difficulty, though it follows a standard M4 pattern. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| MI of elemental disc about a diameter: \(\frac{1}{4}\left(\frac{M}{3a}\delta x\right)a^2\) | B1 | \(\frac{M}{3a}\) may be \(\rho\pi a^2\) throughout (condone use of \(\rho=1\)) |
| MI of elemental disc about \(AB\): \(\frac{1}{4}\left(\frac{M}{3a}\delta x\right)a^2 + \left(\frac{M}{3a}\delta x\right)x^2\) | M1 A1 | Using parallel axes rule (can award A1 for \(\frac{1}{4}ma^2 + mx^2\)) |
| \(I = \frac{M}{3a}\int_0^{3a}\left(\frac{1}{4}a^2 + x^2\right)dx\) | M1 A1 | Integrating MI of disc about \(AB\); Correct integral expression for \(I\) |
| \(= \frac{M}{3a}\left[\frac{1}{4}a^2x + \frac{1}{3}x^3\right]_0^{3a}\) | ||
| \(= \frac{M}{3a}\left(\frac{3}{4}a^3 + 9a^3\right) = M\left(\frac{1}{4}a^2 + 3a^2\right) = \frac{13}{4}Ma^2\) | M1 A1(ag) | Obtaining expression for \(I\) in terms of \(M\) and \(a\); Dependent on previous M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Period is \(2\pi\sqrt{\frac{I}{Mgh}}\) | M1 | or \(-Mgh\sin\theta = I\ddot{\theta}\) |
| \(= 2\pi\sqrt{\dfrac{\frac{13}{4}Ma^2}{Mg\frac{3}{2}a}}\) | A1 | |
| \(= 2\pi\sqrt{\dfrac{13a}{6g}}\) | A1 |
# Question 4:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| MI of elemental disc about a diameter: $\frac{1}{4}\left(\frac{M}{3a}\delta x\right)a^2$ | B1 | $\frac{M}{3a}$ may be $\rho\pi a^2$ throughout (condone use of $\rho=1$) |
| MI of elemental disc about $AB$: $\frac{1}{4}\left(\frac{M}{3a}\delta x\right)a^2 + \left(\frac{M}{3a}\delta x\right)x^2$ | M1 A1 | Using parallel axes rule (can award A1 for $\frac{1}{4}ma^2 + mx^2$) |
| $I = \frac{M}{3a}\int_0^{3a}\left(\frac{1}{4}a^2 + x^2\right)dx$ | M1 A1 | Integrating MI of disc about $AB$; Correct integral expression for $I$ |
| $= \frac{M}{3a}\left[\frac{1}{4}a^2x + \frac{1}{3}x^3\right]_0^{3a}$ | | |
| $= \frac{M}{3a}\left(\frac{3}{4}a^3 + 9a^3\right) = M\left(\frac{1}{4}a^2 + 3a^2\right) = \frac{13}{4}Ma^2$ | M1 A1(ag) | Obtaining expression for $I$ in terms of $M$ and $a$; Dependent on previous M1 |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Period is $2\pi\sqrt{\frac{I}{Mgh}}$ | M1 | or $-Mgh\sin\theta = I\ddot{\theta}$ |
| $= 2\pi\sqrt{\dfrac{\frac{13}{4}Ma^2}{Mg\frac{3}{2}a}}$ | A1 | |
| $= 2\pi\sqrt{\dfrac{13a}{6g}}$ | A1 | |
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\includegraphics[max width=\textwidth, alt={}, center]{181fad74-6e60-4435-a176-3edff5062c32-3_698_505_275_801}
A uniform solid cylinder has radius $a$, height $3 a$, and mass $M$. The line $A B$ is a diameter of one of the end faces of the cylinder (see diagram).\\
(i) Show by integration that the moment of inertia of the cylinder about $A B$ is $\frac { 13 } { 4 } M a ^ { 2 }$. (You may assume that the moment of inertia of a uniform disc of mass $m$ and radius $a$ about a diameter is $\frac { 1 } { 4 } m a ^ { 2 }$.)
The line $A B$ is now fixed in a horizontal position and the cylinder rotates freely about $A B$, making small oscillations as a compound pendulum.\\
(ii) Find the approximate period of these small oscillations, in terms of $a$ and $g$.
\hfill \mbox{\textit{OCR M4 2007 Q4 [15]}}