| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Energy methods for rotation |
| Difficulty | Challenging +1.8 This is a challenging M4 rotation problem requiring moment of inertia via parallel axis theorem, energy methods for angular velocity, and force resolution in a rotating frame. It demands multiple advanced mechanics techniques and careful coordinate work, placing it well above average difficulty but within reach of strong Further Maths students. |
| Spec | 6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I = \frac{1}{2}ma^2 + m\left(\frac{1}{2}a\right)^2\) | M1 | Using parallel axes rule |
| \(= \frac{3}{4}ma^2\) | A1 | |
| \(mg\left(\frac{1}{2}a\cos\theta\right) = I\alpha = \left(\frac{3}{4}ma^2\right)\alpha\) | M1 | Or differentiating the energy equation |
| \(\alpha = \dfrac{2g\cos\theta}{3a}\) | A1 (ag) | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}I\omega^2 = mg\left(\frac{1}{2}a\sin\theta\right)\) | M1 | Using \(\frac{1}{2}I\omega^2\) |
| A1 | ||
| \(\omega = \sqrt{\dfrac{4g\sin\theta}{3a}}\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R - mg\sin\theta = m\left(\frac{1}{2}a\right)\omega^2\) | B1 | For radial acc'n of \(C\) is \(\left(\frac{1}{2}a\right)\omega^2\) |
| M1 | \(\pm R \pm mg\sin\theta = mr\omega^2\) or \(kma\omega^2\) (with numerical \(k\)) | |
| \(R = \frac{5}{3}mg\sin\theta\) | A1 | |
| \(mg\cos\theta - S = m\left(\frac{1}{2}a\right)\alpha\) | B1 | For transverse acc'n of \(C\) is \(\left(\frac{1}{2}a\right)\alpha\); direction must be clear |
| M1 | as above | |
| \(S = \frac{2}{3}mg\cos\theta\) | A1 | Equations involving horizontal and vertical components can earn B1M1B1M1 |
| OR \(S\left(\frac{1}{2}a\right) = I_G\,\alpha\) | M1 | Must use \(I_G\) |
| \(S\left(\frac{1}{2}a\right) = \left(\frac{1}{2}ma^2\right)\alpha\) | A1 | |
| \(S = \frac{2}{3}mg\cos\theta\) | A1 | |
| Total: 6 |
# Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I = \frac{1}{2}ma^2 + m\left(\frac{1}{2}a\right)^2$ | M1 | Using parallel axes rule |
| $= \frac{3}{4}ma^2$ | A1 | |
| $mg\left(\frac{1}{2}a\cos\theta\right) = I\alpha = \left(\frac{3}{4}ma^2\right)\alpha$ | M1 | Or differentiating the energy equation |
| $\alpha = \dfrac{2g\cos\theta}{3a}$ | A1 (ag) | |
| **Total: 4** | | |
---
# Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}I\omega^2 = mg\left(\frac{1}{2}a\sin\theta\right)$ | M1 | Using $\frac{1}{2}I\omega^2$ |
| | A1 | |
| $\omega = \sqrt{\dfrac{4g\sin\theta}{3a}}$ | A1 | |
| **Total: 3** | | |
---
# Question 6(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R - mg\sin\theta = m\left(\frac{1}{2}a\right)\omega^2$ | B1 | For radial acc'n of $C$ is $\left(\frac{1}{2}a\right)\omega^2$ |
| | M1 | $\pm R \pm mg\sin\theta = mr\omega^2$ or $kma\omega^2$ (with numerical $k$) |
| $R = \frac{5}{3}mg\sin\theta$ | A1 | |
| $mg\cos\theta - S = m\left(\frac{1}{2}a\right)\alpha$ | B1 | For transverse acc'n of $C$ is $\left(\frac{1}{2}a\right)\alpha$; direction must be clear |
| | M1 | as above |
| $S = \frac{2}{3}mg\cos\theta$ | A1 | Equations involving horizontal and vertical components can earn B1M1B1M1 |
| **OR** $S\left(\frac{1}{2}a\right) = I_G\,\alpha$ | M1 | Must use $I_G$ |
| $S\left(\frac{1}{2}a\right) = \left(\frac{1}{2}ma^2\right)\alpha$ | A1 | |
| $S = \frac{2}{3}mg\cos\theta$ | A1 | |
| **Total: 6** | | |
---6 A uniform circular disc, of mass $m$ and radius $a$, has centre $C$. The disc can rotate freely in a vertical plane about a fixed horizontal axis through the point $A$ on the disc, where $C A = \frac { 1 } { 2 } a$. The disc is released from rest in the position with $C A$ horizontal. When the disc has rotated through an angle $\theta$,\\
(i) show that the angular acceleration of the disc is $\frac { 2 g \cos \theta } { 3 a }$,\\
(ii) find the angular speed of the disc,\\
(iii) find the components, parallel and perpendicular to $C A$, of the force acting on the disc at the axis.
\hfill \mbox{\textit{OCR M4 2005 Q6 [13]}}