| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2016 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Angular speed and period |
| Difficulty | Challenging +1.2 This is a standard M4 compound pendulum problem requiring systematic application of parallel axis theorem, moment calculations, and SHM analysis. While it involves multiple components and algebraic manipulation, the techniques are routine for this module with no novel insights required. The optimization in part (iv) is straightforward calculus. More challenging than average due to length and algebraic complexity, but follows predictable M4 patterns. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.04b Find centre of mass: using symmetry6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_{particle} = x^2\) | B1 | |
| \(I_{rod} = \frac{1}{3}(3)\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2} - x\right)^2\) | M1 A1 | M1 for correct M of I about the centre of the rod or correct application of parallel axis theorem; \(I_{rod} = \frac{1}{4} + \frac{3}{4} - 3x + 3x^2\) |
| \(I_{disc} = \frac{1}{2}(6)\left(\frac{1}{3}\right)^2 + 6\left(\frac{1}{3} + 1 - x\right)^2\) | M1 A1 | M1 for correct M of I about the centre of the disc or correct application of parallel axis theorem; \(I_{disc} = \frac{1}{3} + \frac{32}{3} - 16x + 6x^2\) |
| \(I_{particle} + I_{rod} + I_{disc} = 10x^2 - 19x + 12\) | A1 | AG Correctly shown |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| C of M of pendulum (from A): | ||
| \(3\left(\frac{1}{2}\right) + 6\left(\frac{4}{3}\right) = 10x\) | M1 | C of M about any point on the pendulum |
| C of M from P: \(\frac{19}{20} - x\) | A1 | Or moment of all three weights separately (M1 A1 - lhs below correct (without \(g\sin\theta\))) |
| \((10x^2 - 19x + 12)\ddot{\theta} = -10g\left(\frac{19}{20} - x\right)\sin\theta\) | M1 | Applying \(C = I\ddot{\theta}\) with their C of M |
| \(\ddot{\theta} = -\frac{g}{2}\left(\frac{19 - 20x}{10x^2 - 19x + 12}\right)\sin\theta\) | A1 | Or with small angle approx. \(\ddot{\theta} = -\frac{g}{2}\left(\frac{19 - 20x}{10x^2 - 19x + 12}\right)\theta\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(E = xg\cos\theta - 3g\left(\frac{1}{2} - x\right)\cos\theta - 6g\left(\frac{4}{3} - x\right)\cos\theta + ...\) | M1 | \(E = T + V\) (4 terms) |
| \(...+ \frac{1}{2}(10x^2 - 19x + 12)\dot{\theta}^2\) | A1 | Cao |
| \((10x^2 - 19x + 12)\ddot{\theta} + \left(\frac{19}{2} - 10x\right)g\sin\theta = 0\) | M1 A1 | M1 for differentiating their energy equation |
| Answer | Marks | Guidance |
|---|---|---|
| For small \(\theta, \sin\theta \approx \theta\) | M1 | Apply small angle approximation and use of \(T = \frac{2\pi}{\omega}\) |
| \(\ddot{\theta} = -\frac{g}{2}\left(\frac{19 - 20x}{10x^2 - 19x + 12}\right)\theta\) | ||
| \(T = 2\pi\sqrt{\frac{20x^2 - 38x + 24}{g(19 - 20x)}}\) | A1 | AG Clearly shown – must state that the motion is (approx.) simple harmonic |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}\left(\frac{20x^2 - 38x + 24}{19 - 20x}\right) = 0\) | M1 | Clear attempt to differentiate \(\frac{20x^2 - 38x + 24}{19 - 20x}\) oe and putting this expression (or just numerator) equal to zero |
| \(200x^2 - 380x + 121 = 0\) | A1 | For a correct 3 term quadratic |
| \(x = 0.405\) (3 sf) | A1 | Only (not 1.4954...); \(x = 0.40456...\) or \(\frac{19 - \sqrt{119}}{20}\) |
| [3] |
**(i)**
$I_{particle} = x^2$ | B1 |
$I_{rod} = \frac{1}{3}(3)\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2} - x\right)^2$ | M1 A1 | M1 for correct M of I about the centre of the rod or correct application of parallel axis theorem; $I_{rod} = \frac{1}{4} + \frac{3}{4} - 3x + 3x^2$
$I_{disc} = \frac{1}{2}(6)\left(\frac{1}{3}\right)^2 + 6\left(\frac{1}{3} + 1 - x\right)^2$ | M1 A1 | M1 for correct M of I about the centre of the disc or correct application of parallel axis theorem; $I_{disc} = \frac{1}{3} + \frac{32}{3} - 16x + 6x^2$
$I_{particle} + I_{rod} + I_{disc} = 10x^2 - 19x + 12$ | A1 | AG Correctly shown
| [6] |
**(ii)**
C of M of pendulum (from A): |
$3\left(\frac{1}{2}\right) + 6\left(\frac{4}{3}\right) = 10x$ | M1 | C of M about any point on the pendulum
C of M from P: $\frac{19}{20} - x$ | A1 | Or moment of all three weights separately (M1 A1 - lhs below correct (without $g\sin\theta$))
$(10x^2 - 19x + 12)\ddot{\theta} = -10g\left(\frac{19}{20} - x\right)\sin\theta$ | M1 | Applying $C = I\ddot{\theta}$ with their C of M
$\ddot{\theta} = -\frac{g}{2}\left(\frac{19 - 20x}{10x^2 - 19x + 12}\right)\sin\theta$ | A1 | Or with small angle approx. $\ddot{\theta} = -\frac{g}{2}\left(\frac{19 - 20x}{10x^2 - 19x + 12}\right)\theta$
| [4] |
**OR**
$E = xg\cos\theta - 3g\left(\frac{1}{2} - x\right)\cos\theta - 6g\left(\frac{4}{3} - x\right)\cos\theta + ...$ | M1 | $E = T + V$ (4 terms)
$...+ \frac{1}{2}(10x^2 - 19x + 12)\dot{\theta}^2$ | A1 | Cao
$(10x^2 - 19x + 12)\ddot{\theta} + \left(\frac{19}{2} - 10x\right)g\sin\theta = 0$ | M1 A1 | M1 for differentiating their energy equation
**(iii)**
For small $\theta, \sin\theta \approx \theta$ | M1 | Apply small angle approximation and use of $T = \frac{2\pi}{\omega}$
$\ddot{\theta} = -\frac{g}{2}\left(\frac{19 - 20x}{10x^2 - 19x + 12}\right)\theta$ |
$T = 2\pi\sqrt{\frac{20x^2 - 38x + 24}{g(19 - 20x)}}$ | A1 | AG Clearly shown – must state that the motion is (approx.) simple harmonic
**(iv)**
$\frac{d}{dx}\left(\frac{20x^2 - 38x + 24}{19 - 20x}\right) = 0$ | M1 | Clear attempt to differentiate $\frac{20x^2 - 38x + 24}{19 - 20x}$ oe and putting this expression (or just numerator) equal to zero
$200x^2 - 380x + 121 = 0$ | A1 | For a correct 3 term quadratic
$x = 0.405$ (3 sf) | A1 | Only (not 1.4954...); $x = 0.40456...$ or $\frac{19 - \sqrt{119}}{20}$
| [3] |\includegraphics{figure_6}
A compound pendulum consists of a uniform rod $AB$ of length 1 m and mass 3 kg, a particle of mass 1 kg attached to the rod at $A$ and a circular disc of radius $\frac{1}{5}$ m, mass 6 kg and centre $C$. The end $B$ of the rod is rigidly attached to a point on the circumference of the disc in such a way that $ABC$ is a straight line. The pendulum is initially at rest with $B$ vertically below $A$ and it is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through the point $P$ on the rod where $AP = x$ m and $x < \frac{1}{3}$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the moment of inertia of the pendulum about the axis of rotation is $(10x^2 - 19x + 12)$ kg m$^2$. [6]
\end{enumerate}
The pendulum is making small oscillations about the equilibrium position, such that at time $t$ seconds the angular displacement that the pendulum makes with the downward vertical is $\theta$ radians.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the angular acceleration of the pendulum, in terms of $x$, $g$ and $\theta$. [4]
\item Show that the motion is approximately simple harmonic, and show that the approximate period of oscillations, in seconds, is given by $2\pi\sqrt{\frac{20x^2 - 38x + 24}{(19-20x)g}}$. [2]
\item Hence find the value of $x$ for which the approximate period of oscillations is least. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M4 2016 Q6 [15]}}