| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Composite body MI calculation |
| Difficulty | Standard +0.8 This is a multi-part compound pendulum problem requiring calculation of moment of inertia for a composite rigid body, derivation of equation of motion using torque methods, small angle approximation for SHM, and energy methods for angular motion. While systematic, it demands careful handling of multiple rods' contributions to inertia and center of mass, making it moderately challenging but still within standard M4 scope. |
| Spec | 3.02e Two-dimensional constant acceleration: with vectors3.02h Motion under gravity: vector form3.04a Calculate moments: about a point6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_{AD} = \frac{1}{3}(0.6)(0.75)^2\ (= 0.1125)\) | B1 | |
| \(I_{AB} = I_{CD} = 0.1125 + 0.6(0.75^2 + 0.75^2)\ (= 0.7875)\) | M1 | M0 for \(\frac{4}{3}(0.6)(0.75)^2 + (0.6)(0.75)^2\) |
| \(I_{BC} = 0.1125 + (0.6)(1.5)^2\ (= 1.4625)\) | M1 | |
| \(I = 0.1125 + 2\times0.7875 + 1.4625 = 3.15\) | A1 | AG |
| OR \(I = 4(0.1125 + 0.6\times0.75^2) + (2.4)(0.75)^2\) | M1 | For \(0.1125 + (0.6)(0.75)^2\) |
| \(= 1.8 + (2.4)(0.75)^2\) | M1 | For \(I_G + (2.4)(0.75)^2\) |
| \(= 3.15\) | A1 | AG |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For \(2.4\times9.8\times0.75\sin\theta\) | B1 | |
| \(-2.4\times9.8\times0.75\sin\theta = 3.15\dfrac{d^2\theta}{dt^2}\) | M1 | Equation of rotational motion |
| OR \(\frac{1}{2}I\omega^2 - mgh\cos\theta = K\) | M1 | Differentiating energy equation |
| \(I\omega\dot{\omega} + 2.4\times9.8\times0.75\sin\theta\,\dot{\theta} = 0\) | A1 | |
| \(\dfrac{d^2\theta}{dt^2} = -5.6\sin\theta\) | A1 | AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(\theta\) is small, \(\sin\theta \approx \theta\) | B1 | |
| \(\frac{d^2\theta}{dt^2} \approx -5.6\theta\), which is (approx.) SHM | B1 | |
| Period \(\left(\frac{2\pi}{\sqrt{5.6}}\right)\) is 2.66 s (3 sf) | B1 | Accept \(\pi\sqrt{\frac{5}{7}}\) etc \ |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| WD by couple is \(25\times1.2\ (=30)\) | B1 | |
| Change in PE is \(2.4\times9.8(0.75-0.75\cos1.2)\ (=11.25)\) | B1 | |
| \(\frac{1}{2}(3.15)\omega^2 = 30-11.25\) | M1 | Equation involving KE, WD and PE |
| Angular speed is \(3.45\ \text{rad s}^{-1}\) (3 sf) | A1 | |
| [4] |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_{AD} = \frac{1}{3}(0.6)(0.75)^2\ (= 0.1125)$ | B1 | |
| $I_{AB} = I_{CD} = 0.1125 + 0.6(0.75^2 + 0.75^2)\ (= 0.7875)$ | M1 | M0 for $\frac{4}{3}(0.6)(0.75)^2 + (0.6)(0.75)^2$ |
| $I_{BC} = 0.1125 + (0.6)(1.5)^2\ (= 1.4625)$ | M1 | |
| $I = 0.1125 + 2\times0.7875 + 1.4625 = 3.15$ | A1 | AG |
| **OR** $I = 4(0.1125 + 0.6\times0.75^2) + (2.4)(0.75)^2$ | M1 | For $0.1125 + (0.6)(0.75)^2$ |
| $= 1.8 + (2.4)(0.75)^2$ | M1 | For $I_G + (2.4)(0.75)^2$ |
| $= 3.15$ | A1 | AG |
| **[4]** | | |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For $2.4\times9.8\times0.75\sin\theta$ | B1 | |
| $-2.4\times9.8\times0.75\sin\theta = 3.15\dfrac{d^2\theta}{dt^2}$ | M1 | Equation of rotational motion |
| **OR** $\frac{1}{2}I\omega^2 - mgh\cos\theta = K$ | M1 | Differentiating energy equation |
| $I\omega\dot{\omega} + 2.4\times9.8\times0.75\sin\theta\,\dot{\theta} = 0$ | A1 | |
| $\dfrac{d^2\theta}{dt^2} = -5.6\sin\theta$ | A1 | AG |
| **[3]** | | |
## Question 5(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| When $\theta$ is small, $\sin\theta \approx \theta$ | B1 | |
| $\frac{d^2\theta}{dt^2} \approx -5.6\theta$, which is (approx.) SHM | B1 | |
| Period $\left(\frac{2\pi}{\sqrt{5.6}}\right)$ is 2.66 s (3 sf) | B1 | Accept $\pi\sqrt{\frac{5}{7}}$ etc \| Or $2\pi\sqrt{\frac{I}{mgh}} = 2\pi\sqrt{\frac{3.15}{2.4\times9.8\times0.75}}$ |
| **[3]** | | |
## Question 5(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| WD by couple is $25\times1.2\ (=30)$ | B1 | |
| Change in PE is $2.4\times9.8(0.75-0.75\cos1.2)\ (=11.25)$ | B1 | |
| $\frac{1}{2}(3.15)\omega^2 = 30-11.25$ | M1 | Equation involving KE, WD and PE |
| Angular speed is $3.45\ \text{rad s}^{-1}$ (3 sf) | A1 | |
| **[4]** | | |\begin{enumerate}[label=(\alph*)]
\item Find the magnitude and bearing of the velocity of $U$ relative to $P$.
\item Find the shortest distance between $P$ and $U$ in the subsequent motion.\\
(ii) Plane $Q$ is flying with constant velocity $160 \mathrm {~ms} ^ { - 1 }$ in the direction which brings it as close as possible to $U$.\\
(a) Find the bearing of the direction in which $Q$ is flying.\\
(b) Find the shortest distance between $Q$ and $U$ in the subsequent motion.\\
\includegraphics[max width=\textwidth, alt={}, center]{6e3d5f5e-7ffa-4111-903d-468fb4d20192-3_771_769_262_646}
A square frame $A B C D$ consists of four uniform rods $A B , B C , C D , D A$, rigidly joined at $A , B , C , D$. Each rod has mass 0.6 kg and length 1.5 m . The frame rotates freely in a vertical plane about a fixed horizontal axis passing through the mid-point $O$ of $A D$. At time $t$ seconds the angle between $A D$ and the horizontal, measured anticlockwise, is $\theta$ radians (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the moment of inertia of the frame about the axis through $O$ is $3.15 \mathrm {~kg} \mathrm {~m} ^ { 2 }$.
\item Show that $\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } = - 5.6 \sin \theta$.
\item Deduce that the frame can make small oscillations which are approximately simple harmonic, and find the period of these oscillations.
The frame is at rest with $A D$ horizontal. A couple of constant moment 25 Nm about the axis is then applied to the frame.
\item Find the angular speed of the frame when it has rotated through 1.2 radians.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR M4 2013 Q5 [14]}}